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Chapter 1: A Square and A Cube
NCERT Ganita Prakash (Class 8 Part 1) — Square Numbers, Square Roots, Long Division, Patterns, Cube Numbers, Cube Roots & Pythagorean Triplets | CBSE 2026-27
📌 Key Mathematical Concepts & Formulas
- Square Numbers: A natural number n is a perfect square if n = m² for some natural number m.
- Square numbers end only in digits 0, 1, 4, 5, 6, or 9 at units place. Numbers ending in 2, 3, 7, 8 can never be perfect squares.
- A number with n zeros at the end has 2n zeros when squared.
- Non-square numbers between n² and (n + 1)² is exactly 2n:
(n + 1)² − n² − 1 = 2n. - Sum of first n consecutive odd natural numbers is n²:
1 + 3 + 5 + ... + (2n − 1) = n². - Pythagorean Triplet: For any natural number m > 1,
(2m)² + (m² − 1)² = (m² + 1)².
- Square Roots (√n): The inverse operation of squaring.
- Repeated Subtraction: Successively subtract consecutive odd numbers (1, 3, 5, 7...) starting from 1 until 0 is reached. Count of subtractions is √n.
- Prime Factorisation: Pair identical prime factors in twos. Square root is product of one factor from each pair.
- Long Division Method: Group digits in pairs (periods) from right to left (for decimals, from decimal point outwards). Essential for large numbers and decimal square roots.
- Cube Numbers & Cube Roots (∛n): A number n is a perfect cube if n = m³.
- In prime factorisation of a perfect cube, every prime factor appears in triplets (multiples of 3).
- Cube of an even number is even; cube of an odd number is odd.
- Cube root by estimation: Group digits in triplets from right. First group gives units digit (by unit digit cube mapping); remaining group gives tens digit by nearest cube.
Figure 1.1: Visual Representation of Square Area (Side²) and Cube Volume (Edge³)
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
What will be the unit digit of the squares of the following numbers?
(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555
(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Property: The unit digit of the square of a number depends solely on the unit digit of the number itself. If a number ends in digit u, its square ends in the unit digit of u².
• (i) 81: Unit digit is 1. Since 1² = 1, unit digit of 81² is 1.
• (ii) 272: Unit digit is 2. Since 2² = 4, unit digit of 272² is 4.
• (iii) 799: Unit digit is 9. Since 9² = 81 (unit digit 1), unit digit of 799² is 1.
• (iv) 3853: Unit digit is 3. Since 3² = 9, unit digit of 3853² is 9.
• (v) 1234: Unit digit is 4. Since 4² = 16 (unit digit 6), unit digit of 1234² is 6.
• (vi) 26387: Unit digit is 7. Since 7² = 49 (unit digit 9), unit digit of 26387² is 9.
• (vii) 52698: Unit digit is 8. Since 8² = 64 (unit digit 4), unit digit of 52698² is 4.
• (viii) 99880: Unit digit is 0. Since 0² = 0, unit digit of 99880² is 0.
• (ix) 12796: Unit digit is 6. Since 6² = 36 (unit digit 6), unit digit of 12796² is 6.
• (x) 55555: Unit digit is 5. Since 5² = 25 (unit digit 5), unit digit of 55555² is 5.
• (ii) 272: Unit digit is 2. Since 2² = 4, unit digit of 272² is 4.
• (iii) 799: Unit digit is 9. Since 9² = 81 (unit digit 1), unit digit of 799² is 1.
• (iv) 3853: Unit digit is 3. Since 3² = 9, unit digit of 3853² is 9.
• (v) 1234: Unit digit is 4. Since 4² = 16 (unit digit 6), unit digit of 1234² is 6.
• (vi) 26387: Unit digit is 7. Since 7² = 49 (unit digit 9), unit digit of 26387² is 9.
• (vii) 52698: Unit digit is 8. Since 8² = 64 (unit digit 4), unit digit of 52698² is 4.
• (viii) 99880: Unit digit is 0. Since 0² = 0, unit digit of 99880² is 0.
• (ix) 12796: Unit digit is 6. Since 6² = 36 (unit digit 6), unit digit of 12796² is 6.
• (x) 55555: Unit digit is 5. Since 5² = 25 (unit digit 5), unit digit of 55555² is 5.
CBSE Board Marking Scheme 2026-27
Stating property: Unit digit of square = unit digit of u²0.5 Mark
Correct unit digits for all 10 sub-parts2.5 Marks
Q2
The following numbers are obviously not perfect squares. Give reasons:
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Fundamental Properties:
1. A perfect square never ends in digits 2, 3, 7, or 8.
2. A perfect square can never end in an odd number of zeros.
• (i) 1057: Ends in digit 7, therefore it cannot be a perfect square.
• (ii) 23453: Ends in digit 3, therefore it cannot be a perfect square.
• (iii) 7928: Ends in digit 8, therefore it cannot be a perfect square.
• (iv) 222222: Ends in digit 2, therefore it cannot be a perfect square.
• (v) 64000: Ends in 3 zeros (an odd number of trailing zeros), hence cannot be a perfect square.
• (vi) 89722: Ends in digit 2, therefore it cannot be a perfect square.
• (vii) 222000: Ends in 3 zeros (odd number of zeros), hence cannot be a perfect square.
• (viii) 505050: Ends in 1 zero (an odd number of zeros), hence cannot be a perfect square.
• (ii) 23453: Ends in digit 3, therefore it cannot be a perfect square.
• (iii) 7928: Ends in digit 8, therefore it cannot be a perfect square.
• (iv) 222222: Ends in digit 2, therefore it cannot be a perfect square.
• (v) 64000: Ends in 3 zeros (an odd number of trailing zeros), hence cannot be a perfect square.
• (vi) 89722: Ends in digit 2, therefore it cannot be a perfect square.
• (vii) 222000: Ends in 3 zeros (odd number of zeros), hence cannot be a perfect square.
• (viii) 505050: Ends in 1 zero (an odd number of zeros), hence cannot be a perfect square.
CBSE Board Marking Scheme 2026-27
Mentioning units digit restriction (2, 3, 7, 8) and odd zeros rule1 Mark
Accurate reason given for each part2 Marks
Q3
How many non-square natural numbers lie between the squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Theorem / Formula: Between the squares of two consecutive natural numbers n and (n + 1), the number of non-square natural numbers is given by 2n.
• (i) Between 12² and 13²:
Here n = 12.
Number of non-square numbers = 2n = 2 × 12 = 24.
Verification: 13² − 12² − 1 = 169 − 144 − 1 = 24.
• (ii) Between 25² and 26²:
Here n = 25.
Number of non-square numbers = 2n = 2 × 25 = 50.
• (iii) Between 99² and 100²:
Here n = 99.
Number of non-square numbers = 2n = 2 × 99 = 198.
Here n = 12.
Number of non-square numbers = 2n = 2 × 12 = 24.
Verification: 13² − 12² − 1 = 169 − 144 − 1 = 24.
• (ii) Between 25² and 26²:
Here n = 25.
Number of non-square numbers = 2n = 2 × 25 = 50.
• (iii) Between 99² and 100²:
Here n = 99.
Number of non-square numbers = 2n = 2 × 99 = 198.
CBSE Board Marking Scheme 2026-27
Formula identification: Number of non-square numbers = 2n0.5 Mark
Correct evaluation for all 3 cases (24, 50, 198)1.5 Marks
Q4
Without adding, find the sum:
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Property: The sum of the first n consecutive odd natural numbers equals n².
• (i) 1 + 3 + 5 + 7 + 9:
This is the sum of the first 5 odd natural numbers (n = 5).
Sum = 5² = 25.
• (ii) 1 + 3 + 5 + ... + 19:
This is the sum of the first 10 odd natural numbers (n = 10).
Sum = 10² = 100.
• (iii) 1 + 3 + 5 + ... + 23:
This is the sum of the first 12 odd natural numbers (n = 12).
Sum = 12² = 144.
This is the sum of the first 5 odd natural numbers (n = 5).
Sum = 5² = 25.
• (ii) 1 + 3 + 5 + ... + 19:
This is the sum of the first 10 odd natural numbers (n = 10).
Sum = 10² = 100.
• (iii) 1 + 3 + 5 + ... + 23:
This is the sum of the first 12 odd natural numbers (n = 12).
Sum = 12² = 144.
CBSE Board Marking Scheme 2026-27
Quoting formula Sum = n²0.5 Mark
Correct sums 25, 100, 1441.5 Marks
Q5
Write a Pythagorean triplet whose smallest member is:
(i) 6 (ii) 14 (iii) 16 (iv) 18
(i) 6 (ii) 14 (iii) 16 (iv) 18
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
General Form of Pythagorean Triplet: For any integer m > 1, the three numbers 2m, m² − 1, and m² + 1 form a Pythagorean triplet.
• (i) Smallest member = 6:
Let 2m = 6 ⇒ m = 3.
m² − 1 = 3² − 1 = 9 − 1 = 8.
m² + 1 = 3² + 1 = 9 + 1 = 10.
Check: 6² + 8² = 36 + 64 = 100 = 10².
Hence, the required triplet is (6, 8, 10).
• (ii) Smallest member = 14:
Let 2m = 14 ⇒ m = 7.
m² − 1 = 7² − 1 = 49 − 1 = 48.
m² + 1 = 7² + 1 = 49 + 1 = 50.
Check: 14² + 48² = 196 + 2304 = 2500 = 50².
Hence, the required triplet is (14, 48, 50).
• (iii) Smallest member = 16:
Let 2m = 16 ⇒ m = 8.
m² − 1 = 8² − 1 = 64 − 1 = 63.
m² + 1 = 8² + 1 = 64 + 1 = 65.
Hence, the required triplet is (16, 63, 65).
• (iv) Smallest member = 18:
Let 2m = 18 ⇒ m = 9.
m² − 1 = 9² − 1 = 81 − 1 = 80.
m² + 1 = 9² + 1 = 81 + 1 = 82.
Hence, the required triplet is (18, 80, 82).
Let 2m = 6 ⇒ m = 3.
m² − 1 = 3² − 1 = 9 − 1 = 8.
m² + 1 = 3² + 1 = 9 + 1 = 10.
Check: 6² + 8² = 36 + 64 = 100 = 10².
Hence, the required triplet is (6, 8, 10).
• (ii) Smallest member = 14:
Let 2m = 14 ⇒ m = 7.
m² − 1 = 7² − 1 = 49 − 1 = 48.
m² + 1 = 7² + 1 = 49 + 1 = 50.
Check: 14² + 48² = 196 + 2304 = 2500 = 50².
Hence, the required triplet is (14, 48, 50).
• (iii) Smallest member = 16:
Let 2m = 16 ⇒ m = 8.
m² − 1 = 8² − 1 = 64 − 1 = 63.
m² + 1 = 8² + 1 = 64 + 1 = 65.
Hence, the required triplet is (16, 63, 65).
• (iv) Smallest member = 18:
Let 2m = 18 ⇒ m = 9.
m² − 1 = 9² − 1 = 81 − 1 = 80.
m² + 1 = 9² + 1 = 81 + 1 = 82.
Hence, the required triplet is (18, 80, 82).
CBSE Board Marking Scheme 2026-27
Stating 2m, m² − 1, m² + 1 form0.5 Mark
Correctly deriving all 4 triplets2.5 Marks
Q6
Find the square root of 64 and 169 using the method of repeated subtraction.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• For 64: Successively subtract consecutive odd natural numbers:
1. 64 − 1 = 63
2. 63 − 3 = 60
3. 60 − 5 = 55
4. 55 − 7 = 48
5. 48 − 9 = 39
6. 39 − 11 = 28
7. 28 − 13 = 15
8. 15 − 15 = 0
Since we subtracted 8 consecutive odd numbers to reach 0, √64 = 8.
• For 169:
1. 169 − 1 = 168 2. 168 − 3 = 165 3. 165 − 5 = 160 4. 160 − 7 = 153
5. 153 − 9 = 144 6. 144 − 11 = 133 7. 133 − 13 = 120 8. 120 − 15 = 105
9. 105 − 17 = 88 10. 88 − 19 = 69 11. 69 − 21 = 48 12. 48 − 23 = 25
13. 25 − 25 = 0.
Since 0 is reached at the 13th step, √169 = 13.
1. 64 − 1 = 63
2. 63 − 3 = 60
3. 60 − 5 = 55
4. 55 − 7 = 48
5. 48 − 9 = 39
6. 39 − 11 = 28
7. 28 − 13 = 15
8. 15 − 15 = 0
Since we subtracted 8 consecutive odd numbers to reach 0, √64 = 8.
• For 169:
1. 169 − 1 = 168 2. 168 − 3 = 165 3. 165 − 5 = 160 4. 160 − 7 = 153
5. 153 − 9 = 144 6. 144 − 11 = 133 7. 133 − 13 = 120 8. 120 − 15 = 105
9. 105 − 17 = 88 10. 88 − 19 = 69 11. 69 − 21 = 48 12. 48 − 23 = 25
13. 25 − 25 = 0.
Since 0 is reached at the 13th step, √169 = 13.
CBSE Board Marking Scheme 2026-27
Repeated subtraction steps shown accurately1 Mark
Final square roots √64 = 8 and √169 = 131 Mark
Q7
Find the square root of the following by Prime Factorisation Method:
(i) 729 (ii) 1764 (iii) 4096 (iv) 7744 (v) 9604
(i) 729 (ii) 1764 (iii) 4096 (iv) 7744 (v) 9604
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 729:
Prime factorisation = 3 × 3 × 3 × 3 × 3 × 3 = (3 × 3) × (3 × 3) × (3 × 3).
√729 = 3 × 3 × 3 = 27.
• (ii) 1764:
Prime factorisation = 2 × 2 × 3 × 3 × 7 × 7 = (2 × 2) × (3 × 3) × (7 × 7).
√1764 = 2 × 3 × 7 = 42.
• (iii) 4096:
Prime factorisation = 2¹² = (2 × 2)⁶.
√4096 = 2⁶ = 64.
• (iv) 7744:
Prime factorisation = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11 = (2 × 2) × (2 × 2) × (2 × 2) × (11 × 11).
√7744 = 2 × 2 × 2 × 11 = 88.
• (v) 9604:
Prime factorisation = 2 × 2 × 7 × 7 × 7 × 7 = (2 × 2) × (7 × 7) × (7 × 7).
√9604 = 2 × 7 × 7 = 98.
Prime factorisation = 3 × 3 × 3 × 3 × 3 × 3 = (3 × 3) × (3 × 3) × (3 × 3).
√729 = 3 × 3 × 3 = 27.
• (ii) 1764:
Prime factorisation = 2 × 2 × 3 × 3 × 7 × 7 = (2 × 2) × (3 × 3) × (7 × 7).
√1764 = 2 × 3 × 7 = 42.
• (iii) 4096:
Prime factorisation = 2¹² = (2 × 2)⁶.
√4096 = 2⁶ = 64.
• (iv) 7744:
Prime factorisation = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11 = (2 × 2) × (2 × 2) × (2 × 2) × (11 × 11).
√7744 = 2 × 2 × 2 × 11 = 88.
• (v) 9604:
Prime factorisation = 2 × 2 × 7 × 7 × 7 × 7 = (2 × 2) × (7 × 7) × (7 × 7).
√9604 = 2 × 7 × 7 = 98.
CBSE Board Marking Scheme 2026-27
Prime factor decomposition with pairs identified1.5 Marks
Accurate square roots (27, 42, 64, 88, 98)1.5 Marks
Q8
For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained:
(i) 252 (ii) 180 (iii) 1008 (iv) 2028
(i) 252 (ii) 180 (iii) 1008 (iv) 2028
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 252:
Prime factorisation: 252 = 2 × 2 × 3 × 3 × 7 = (2 × 2) × (3 × 3) × 7.
The prime factor 7 has no pair.
Hence, 252 must be multiplied by 7.
New number = 252 × 7 = 1764.
Square root = √(2² × 3² × 7²) = 2 × 3 × 7 = 42.
• (ii) 180:
Prime factorisation: 180 = 2 × 2 × 3 × 3 × 5 = (2 × 2) × (3 × 3) × 5.
Factor 5 is unpaired. Hence, smallest multiplier is 5.
New number = 180 × 5 = 900.
Square root = √(2² × 3² × 5²) = 2 × 3 × 5 = 30.
• (iii) 1008:
Prime factorisation: 1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7 = 2⁴ × 3² × 7.
Factor 7 is unpaired. Hence, smallest multiplier is 7.
New number = 1008 × 7 = 7056.
Square root = √(2⁴ × 3² × 7²) = 2² × 3 × 7 = 4 × 3 × 7 = 84.
• (iv) 2028:
Prime factorisation: 2028 = 2 × 2 × 3 × 13 × 13 = 2² × 13² × 3.
Factor 3 is unpaired. Hence, smallest multiplier is 3.
New number = 2028 × 3 = 6084.
Square root = √(2² × 13² × 3²) = 2 × 13 × 3 = 78.
Prime factorisation: 252 = 2 × 2 × 3 × 3 × 7 = (2 × 2) × (3 × 3) × 7.
The prime factor 7 has no pair.
Hence, 252 must be multiplied by 7.
New number = 252 × 7 = 1764.
Square root = √(2² × 3² × 7²) = 2 × 3 × 7 = 42.
• (ii) 180:
Prime factorisation: 180 = 2 × 2 × 3 × 3 × 5 = (2 × 2) × (3 × 3) × 5.
Factor 5 is unpaired. Hence, smallest multiplier is 5.
New number = 180 × 5 = 900.
Square root = √(2² × 3² × 5²) = 2 × 3 × 5 = 30.
• (iii) 1008:
Prime factorisation: 1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7 = 2⁴ × 3² × 7.
Factor 7 is unpaired. Hence, smallest multiplier is 7.
New number = 1008 × 7 = 7056.
Square root = √(2⁴ × 3² × 7²) = 2² × 3 × 7 = 4 × 3 × 7 = 84.
• (iv) 2028:
Prime factorisation: 2028 = 2 × 2 × 3 × 13 × 13 = 2² × 13² × 3.
Factor 3 is unpaired. Hence, smallest multiplier is 3.
New number = 2028 × 3 = 6084.
Square root = √(2² × 13² × 3²) = 2 × 13 × 3 = 78.
CBSE Board Marking Scheme 2026-27
Identifying unpaired factor for each1.5 Marks
Calculating new number and its square root1.5 Marks
Q9
Find the square root of each of the following numbers by Long Division Method:
(i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369
(i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 2304:
Group periods: 23 and 04.
- Largest square ≤ 23 is 4² = 16. Quotient = 4, Remainder = 23 − 16 = 7.
- Bring down next period '04' to get 704. Double the quotient: 4 × 2 = 8.
- Trial divisor: 88 × 8 = 704. Remainder = 0.
Therefore, √2304 = 48.
• (ii) 4489:
Periods: 44 and 89.
- 6² = 36 ≤ 44. Remainder = 8. Bring down 89 ⇒ 889.
- Double quotient = 12. 127 × 7 = 889. Remainder = 0.
Therefore, √4489 = 67.
• (iii) 3481:
Periods: 34 and 81.
- 5² = 25 ≤ 34. Remainder = 9. Bring down 81 ⇒ 981.
- Double quotient = 10. 109 × 9 = 981. Remainder = 0.
Therefore, √3481 = 59.
• (iv) 529:
Periods: 5 and 29.
- 2² = 4 ≤ 5. Remainder = 1. Bring down 29 ⇒ 129.
- Double quotient = 4. 43 × 3 = 129. Remainder = 0.
Therefore, √529 = 23.
• (v) 3249:
Periods: 32 and 49.
- 5² = 25 ≤ 32. Remainder = 7. Bring down 49 ⇒ 749.
- Double quotient = 10. 107 × 7 = 749. Remainder = 0.
Therefore, √3249 = 57.
• (vi) 1369:
Periods: 13 and 69.
- 3² = 9 ≤ 13. Remainder = 4. Bring down 69 ⇒ 469.
- Double quotient = 6. 67 × 7 = 469. Remainder = 0.
Therefore, √1369 = 37.
Group periods: 23 and 04.
- Largest square ≤ 23 is 4² = 16. Quotient = 4, Remainder = 23 − 16 = 7.
- Bring down next period '04' to get 704. Double the quotient: 4 × 2 = 8.
- Trial divisor: 88 × 8 = 704. Remainder = 0.
Therefore, √2304 = 48.
• (ii) 4489:
Periods: 44 and 89.
- 6² = 36 ≤ 44. Remainder = 8. Bring down 89 ⇒ 889.
- Double quotient = 12. 127 × 7 = 889. Remainder = 0.
Therefore, √4489 = 67.
• (iii) 3481:
Periods: 34 and 81.
- 5² = 25 ≤ 34. Remainder = 9. Bring down 81 ⇒ 981.
- Double quotient = 10. 109 × 9 = 981. Remainder = 0.
Therefore, √3481 = 59.
• (iv) 529:
Periods: 5 and 29.
- 2² = 4 ≤ 5. Remainder = 1. Bring down 29 ⇒ 129.
- Double quotient = 4. 43 × 3 = 129. Remainder = 0.
Therefore, √529 = 23.
• (v) 3249:
Periods: 32 and 49.
- 5² = 25 ≤ 32. Remainder = 7. Bring down 49 ⇒ 749.
- Double quotient = 10. 107 × 7 = 749. Remainder = 0.
Therefore, √3249 = 57.
• (vi) 1369:
Periods: 13 and 69.
- 3² = 9 ≤ 13. Remainder = 4. Bring down 69 ⇒ 469.
- Double quotient = 6. 67 × 7 = 469. Remainder = 0.
Therefore, √1369 = 37.
CBSE Board Marking Scheme 2026-27
Grouping in periods and correct division algorithm1.5 Marks
All 6 roots accurately found1.5 Marks
Q10
Find the square root of the following decimal numbers:
(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36
(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 2.56:
Periods: 2 and .56.
1² = 1 ≤ 2. Remainder = 1. Decimal point placed in quotient. Bring down 56 ⇒ 156.
Double quotient = 2. 26 × 6 = 156. Remainder = 0.
Therefore, √2.56 = 1.6.
• (ii) 7.29:
Periods: 7 and .29.
2² = 4 ≤ 7. Remainder = 3. Bring down 29 ⇒ 329.
Double quotient = 4. 47 × 7 = 329.
Therefore, √7.29 = 2.7.
• (iii) 51.84:
Periods: 51 and .84.
7² = 49 ≤ 51. Remainder = 2. Bring down 84 ⇒ 284.
Double quotient = 14. 142 × 2 = 284.
Therefore, √51.84 = 7.2.
• (iv) 42.25:
Periods: 42 and .25.
6² = 36 ≤ 42. Remainder = 6. Bring down 25 ⇒ 625.
Double quotient = 12. 125 × 5 = 625.
Therefore, √42.25 = 6.5.
• (v) 31.36:
Periods: 31 and .36.
5² = 25 ≤ 31. Remainder = 6. Bring down 36 ⇒ 636.
Double quotient = 10. 106 × 6 = 636.
Therefore, √31.36 = 5.6.
Periods: 2 and .56.
1² = 1 ≤ 2. Remainder = 1. Decimal point placed in quotient. Bring down 56 ⇒ 156.
Double quotient = 2. 26 × 6 = 156. Remainder = 0.
Therefore, √2.56 = 1.6.
• (ii) 7.29:
Periods: 7 and .29.
2² = 4 ≤ 7. Remainder = 3. Bring down 29 ⇒ 329.
Double quotient = 4. 47 × 7 = 329.
Therefore, √7.29 = 2.7.
• (iii) 51.84:
Periods: 51 and .84.
7² = 49 ≤ 51. Remainder = 2. Bring down 84 ⇒ 284.
Double quotient = 14. 142 × 2 = 284.
Therefore, √51.84 = 7.2.
• (iv) 42.25:
Periods: 42 and .25.
6² = 36 ≤ 42. Remainder = 6. Bring down 25 ⇒ 625.
Double quotient = 12. 125 × 5 = 625.
Therefore, √42.25 = 6.5.
• (v) 31.36:
Periods: 31 and .36.
5² = 25 ≤ 31. Remainder = 6. Bring down 36 ⇒ 636.
Double quotient = 10. 106 × 6 = 636.
Therefore, √31.36 = 5.6.
CBSE Board Marking Scheme 2026-27
Placement of decimal point in quotient and long division1.5 Marks
Accurate decimal roots1.5 Marks
Q11
Find the least number which must be subtracted from 4000 so as to get a perfect square. Also find the square root of the perfect square so obtained.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Apply long division on 4000:
- Grouping: 40 and 00.
- 6² = 36 ≤ 40. Remainder = 4.
- Bring down 00 ⇒ 400.
- Double 6 = 12. 123 × 3 = 369 ≤ 400 (as 124 × 4 = 496 > 400).
- Remainder = 400 − 369 = 31.
2. The remainder indicates that 4000 is greater than 63² by 31.
Therefore, the least number to be subtracted is 31.
3. Required perfect square = 4000 − 31 = 3969.
Square root: √3969 = 63.
- Grouping: 40 and 00.
- 6² = 36 ≤ 40. Remainder = 4.
- Bring down 00 ⇒ 400.
- Double 6 = 12. 123 × 3 = 369 ≤ 400 (as 124 × 4 = 496 > 400).
- Remainder = 400 − 369 = 31.
2. The remainder indicates that 4000 is greater than 63² by 31.
Therefore, the least number to be subtracted is 31.
3. Required perfect square = 4000 − 31 = 3969.
Square root: √3969 = 63.
CBSE Board Marking Scheme 2026-27
Long division and finding remainder 311 Mark
Stating subtracted number 31 and √3969 = 631 Mark
Q12
Find the least number which must be added to 1750 so as to get a perfect square. Also find the square root of the perfect square so obtained.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Long division on 1750:
- Periods: 17 and 50.
- 4² = 16 ≤ 17. Remainder = 1. Bring down 50 ⇒ 150.
- 81 × 1 = 81. Remainder = 69.
- Quotient = 41. Thus, 41² < 1750 < 42².
2. The next perfect square is 42²:
42² = 1764.
3. Number to be added = 1764 − 1750 = 14.
The perfect square obtained = 1750 + 14 = 1764.
Square root = √1764 = 42.
- Periods: 17 and 50.
- 4² = 16 ≤ 17. Remainder = 1. Bring down 50 ⇒ 150.
- 81 × 1 = 81. Remainder = 69.
- Quotient = 41. Thus, 41² < 1750 < 42².
2. The next perfect square is 42²:
42² = 1764.
3. Number to be added = 1764 − 1750 = 14.
The perfect square obtained = 1750 + 14 = 1764.
Square root = √1764 = 42.
CBSE Board Marking Scheme 2026-27
Evaluating 41² < 1750 and identifying 42² = 17641 Mark
Difference 14 and square root 421 Mark
Q13
Find the cube root of the following numbers by Prime Factorisation Method:
(i) 512 (ii) 1728 (iii) 13824 (iv) 46656 (v) 175616
(i) 512 (ii) 1728 (iii) 13824 (iv) 46656 (v) 175616
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 512:
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) = 2³ × 2³ × 2³.
∛512 = 2 × 2 × 2 = 8.
• (ii) 1728:
1728 = 2⁶ × 3³ = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3).
∛1728 = 2 × 2 × 3 = 12.
• (iii) 13824:
13824 = 2⁹ × 3³ = (2³)³ × 3³.
∛13824 = 2 × 2 × 2 × 3 = 24.
• (iv) 46656:
46656 = 2⁶ × 3⁶ = (2³) × (2³) × (3³) × (3³).
∛46656 = 2 × 2 × 3 × 3 = 36.
• (v) 175616:
175616 = 2⁹ × 7³.
∛175616 = 2 × 2 × 2 × 7 = 8 × 7 = 56.
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) = 2³ × 2³ × 2³.
∛512 = 2 × 2 × 2 = 8.
• (ii) 1728:
1728 = 2⁶ × 3³ = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3).
∛1728 = 2 × 2 × 3 = 12.
• (iii) 13824:
13824 = 2⁹ × 3³ = (2³)³ × 3³.
∛13824 = 2 × 2 × 2 × 3 = 24.
• (iv) 46656:
46656 = 2⁶ × 3⁶ = (2³) × (2³) × (3³) × (3³).
∛46656 = 2 × 2 × 3 × 3 = 36.
• (v) 175616:
175616 = 2⁹ × 7³.
∛175616 = 2 × 2 × 2 × 7 = 8 × 7 = 56.
CBSE Board Marking Scheme 2026-27
Prime factor triplets decomposition1.5 Marks
Accurate cube roots (8, 12, 24, 36, 56)1.5 Marks
Q14
Find the cube root of 857375 through the Estimation Method without prime factorisation.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Form groups of three digits from the right:
First group (units group) = 375
Second group = 857
2. Determining the units digit:
The first group is 375, ending in 5. Since only 5³ = 125 ends in 5, the units digit of the cube root must be 5.
3. Determining the tens digit:
Consider the second group: 857.
We know 9³ = 729 and 10³ = 1000.
Clearly, 729 < 857 < 1000, i.e., 9³ < 857 < 10³.
Taking the smaller integer, the tens digit is 9.
Therefore, ∛857375 = 95.
First group (units group) = 375
Second group = 857
2. Determining the units digit:
The first group is 375, ending in 5. Since only 5³ = 125 ends in 5, the units digit of the cube root must be 5.
3. Determining the tens digit:
Consider the second group: 857.
We know 9³ = 729 and 10³ = 1000.
Clearly, 729 < 857 < 1000, i.e., 9³ < 857 < 10³.
Taking the smaller integer, the tens digit is 9.
Therefore, ∛857375 = 95.
CBSE Board Marking Scheme 2026-27
Grouping into 375 and 857 with units digit identification1 Mark
Tens digit by nearest lower cube and final answer 951 Mark
Q15
Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 1.1: Assembling 20 Cuboids of 5 × 2 × 5 cm to form Minimum Perfect Cube of Side 10 cm
1. Volume of one cuboid = 5 × 2 × 5 = 50 cm³.
In prime factor form: 5 × 2 × 5 = 2¹ × 5².
2. For these cuboids to combine and form a perfect cube, the prime factors of the total volume must appear in triplets (groups of 3):
- Factor 2 appears 1 time (needs two more 2s: 2 × 2 = 4).
- Factor 5 appears 2 times (needs one more 5).
3. Therefore, to make the volume a perfect cube, we must multiply by:
Hence, Parikshit requires 20 such cuboids to form a cube.
In prime factor form: 5 × 2 × 5 = 2¹ × 5².
2. For these cuboids to combine and form a perfect cube, the prime factors of the total volume must appear in triplets (groups of 3):
- Factor 2 appears 1 time (needs two more 2s: 2 × 2 = 4).
- Factor 5 appears 2 times (needs one more 5).
3. Therefore, to make the volume a perfect cube, we must multiply by:
2 × 2 × 5 = 20.Hence, Parikshit requires 20 such cuboids to form a cube.
CBSE Board Marking Scheme 2026-27
Writing volume in prime factors 2¹ × 5²1 Mark
Finding missing factors 2 × 2 × 5 = 20 cuboids1 Mark
Q16
Queen Ratnamanjuri's 100-Locker Exploration Puzzle (Page 1): In a secret treasury, 100 closed lockers are lined up, numbered 1 to 100. 100 relatives arrive one by one:
• Person 1 visits every locker and opens it.
• Person 2 visits every 2nd locker (2, 4, 6...) and changes its state (closes if open, opens if closed).
• Person 3 visits every 3rd locker (3, 6, 9...) and changes its state.
This process continues up to Person 100.
(a) Which lockers remain OPEN at the end of the process?
(b) Why do only these specific lockers remain open? Give the mathematical reason.
(c) How many total lockers remain open out of the 100?
• Person 1 visits every locker and opens it.
• Person 2 visits every 2nd locker (2, 4, 6...) and changes its state (closes if open, opens if closed).
• Person 3 visits every 3rd locker (3, 6, 9...) and changes its state.
This process continues up to Person 100.
(a) Which lockers remain OPEN at the end of the process?
(b) Why do only these specific lockers remain open? Give the mathematical reason.
(c) How many total lockers remain open out of the 100?
[4 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Lockers remaining OPEN at the end:
Lockers numbered 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
Lockers numbered 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
(b) Mathematical Reasoning (Odd Number of Divisors):
1. Locker number L is toggled once by Person d for every positive divisor d of L.
2. Since every locker started CLOSED, it ends up OPEN if and only if it was toggled an odd number of times (i.e. has an odd number of factors).
3. In general, positive divisors naturally occur in distinct pairs:
4. The only numbers where one pair collapses into a single number (where
Therefore, only perfect squares have an odd number of factors, and only their lockers remain open.
1. Locker number L is toggled once by Person d for every positive divisor d of L.
2. Since every locker started CLOSED, it ends up OPEN if and only if it was toggled an odd number of times (i.e. has an odd number of factors).
3. In general, positive divisors naturally occur in distinct pairs:
(d, L/d) (for example, factors of 12 are (1,12), (2,6), (3,4) — 6 factors total, an even number).4. The only numbers where one pair collapses into a single number (where
d = L/d ⟹ d² = L) are perfect square numbers.Therefore, only perfect squares have an odd number of factors, and only their lockers remain open.
(c) Total count of open lockers:
The perfect squares between 1 and 100 are: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100.
Exactly 10 lockers remain open out of 100.
The perfect squares between 1 and 100 are: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100.
Exactly 10 lockers remain open out of 100.
CBSE Board Marking Scheme 2026-27
Listing the 10 square numbers (1, 4, 9, ..., 100)1 Mark
Detailed proof: Factor pairing and odd divisor property of squares2 Marks
Final count = 10 open lockers1 Mark
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Community Hall Renovation & Cube Packaging: An architect is renovating a municipal hall whose floor is in the shape of a perfect square with an area of 5929 m². The flooring is to be paved with square marble tiles of side 1 m.
Additionally, a warehouse supplies cubic storage cartons each of volume 1728 cm³ that are to be packed inside a large master cubic container of side 72 cm.
(a) Find the length of each side of the square community hall.
(b) If decorative border tiles are placed along the entire boundary (perimeter) of the hall, find how many linear meters of border are needed.
(c) Calculate the edge length of each cubic storage carton and determine how many cartons fit exactly into the master container.
Additionally, a warehouse supplies cubic storage cartons each of volume 1728 cm³ that are to be packed inside a large master cubic container of side 72 cm.
(a) Find the length of each side of the square community hall.
(b) If decorative border tiles are placed along the entire boundary (perimeter) of the hall, find how many linear meters of border are needed.
(c) Calculate the edge length of each cubic storage carton and determine how many cartons fit exactly into the master container.
Detailed Analytical Solution & Marking Scheme
(a) Side length of the square community hall:
Area = side² = 5929 m² ⇒ side = √5929.
Using Long Division on 5929:
7² = 49 ≤ 59, remainder = 10. Bring down 29 ⇒ 1029.
Double 7 = 14. 147 × 7 = 1029. Remainder = 0.
Therefore, side = 77 meters.
(b) Boundary (Perimeter) of the hall:
Perimeter = 4 × side = 4 × 77 = 308 meters of decorative border needed.
(c) Edge length of carton & Master packaging count:
- Edge of one small carton = ∛1728 = ∛(12³) = 12 cm.
- Master cube edge = 72 cm.
- Number of cartons along each dimension = 72 ÷ 12 = 6 cartons.
- Total cartons packed = 6 × 6 × 6 = 6³ = 216 cartons.
(Verification: Master volume = 72³ = 373,248 cm³. Total = 373,248 ÷ 1728 = 216).
2
Chapter 2: Power Play
NCERT Ganita Prakash (Class 8 Part 1) — Exponents, Laws of Indices, Negative Powers, Scientific Notation & Cyclicity of Digits | CBSE 2026-27
📌 Key Laws of Exponents & Scientific Notation
- Definition: For any non-zero real number a and integer n, aⁿ denotes repeated multiplication of a by itself n times (where a is the base and n is the exponent / power).
- Fundamental Laws of Exponents (for non-zero a, b and integers m, n):
- Product Law:
aᵐ × aⁿ = aᵐ⁺ⁿ - Quotient Law:
aᵐ ÷ aⁿ = aᵐ⁻ⁿ - Power of a Power Law:
(aᵐ)ⁿ = aᵐⁿ - Power of a Product:
aᵐ × bᵐ = (ab)ᵐ - Power of a Quotient:
aᵐ ÷ bᵐ = (a/b)ᵐ - Zero Exponent:
a⁰ = 1(for any a ≠ 0) - Negative Exponent (Reciprocal):
a⁻ⁿ = 1 / aⁿand(a/b)⁻ⁿ = (b/a)ⁿ
- Product Law:
- Scientific Notation (Standard Form): Expressing numbers as
k × 10ᵐwhere1 ≤ k < 10and m is an integer. - Cyclicity of Units Digits:
- Powers of 2 repeat units digits in cycles of 4: 2, 4, 8, 6.
- Powers of 3 repeat in cycles of 4: 3, 9, 7, 1.
- Powers of 7 repeat in cycles of 4: 7, 9, 3, 1.
- Powers of 8 repeat in cycles of 4: 8, 4, 2, 6.
- Powers of 4 repeat in cycles of 2: 4, 6. Powers of 9 repeat in cycles of 2: 9, 1.
Figure 2.1: Interconnected Rules Governing Exponential Manipulations
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
Express the following in exponential form:
(i) 6 × 6 × 6 × 6 (ii) y × y (iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7 (v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d
(i) 6 × 6 × 6 × 6 (ii) y × y (iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7 (v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
| Expression | Exponential Form | Base and Power Details |
|---|---|---|
| (i) 6 × 6 × 6 × 6 | 6⁴ | Base 6, exponent 4 |
| (ii) y × y | y² | Base y, exponent 2 |
| (iii) b × b × b × b | b⁴ | Base b, exponent 4 |
| (iv) 5 × 5 × 7 × 7 × 7 | 5² × 7³ | Product of two prime powers |
| (v) 2 × 2 × a × a | 2² × a² (or (2a)²) | Base 2 squared times base a squared |
| (vi) a × a × a × c × c × c × c × d | a³ × c⁴ × d | Powers 3, 4, and 1 respectively |
CBSE Board Marking Scheme 2026-27
Each correct exponential representation0.33 Mark each (Total 2 Marks)
Q2
Express each of the following composite numbers as a product of powers of their prime factors:
(i) 648 (ii) 405 (iii) 540 (iv) 3600
(i) 648 (ii) 405 (iii) 540 (iv) 3600
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) 648:
648 = 2 × 324 = 2 × 2 × 162 = 2³ × 81 = 2³ × 3⁴.
648 = 2³ × 3⁴.
648 = 2 × 324 = 2 × 2 × 162 = 2³ × 81 = 2³ × 3⁴.
648 = 2³ × 3⁴.
(ii) 405:
405 = 5 × 81 = 5 × 3⁴.
405 = 3⁴ × 5¹.
405 = 5 × 81 = 5 × 3⁴.
405 = 3⁴ × 5¹.
(iii) 540:
540 = 10 × 54 = (2 × 5) × (2 × 27) = 2² × 3³ × 5¹.
540 = 2² × 3³ × 5.
540 = 10 × 54 = (2 × 5) × (2 × 27) = 2² × 3³ × 5¹.
540 = 2² × 3³ × 5.
(iv) 3600:
3600 = 36 × 100 = (2² × 3²) × (2² × 5²) = 2⁴ × 3² × 5².
3600 = 2⁴ × 3² × 5².
3600 = 36 × 100 = (2² × 3²) × (2² × 5²) = 2⁴ × 3² × 5².
3600 = 2⁴ × 3² × 5².
CBSE Board Marking Scheme 2026-27
Each correct prime factorisation in exponential form0.5 Mark each (Total 2 Marks)
Q3
Find the numerical value of each of the following:
(i) 2 × 10³ (ii) 7² × 2³ (iii) 3 × 4⁴
(iv) (−3)² × (−5)² (v) 3² × 10⁴ (vi) (−2)⁵ × (−10)⁶
(i) 2 × 10³ (ii) 7² × 2³ (iii) 3 × 4⁴
(iv) (−3)² × (−5)² (v) 3² × 10⁴ (vi) (−2)⁵ × (−10)⁶
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 2 × 10³ = 2 × 1000 = 2,000.
• (ii) 7² × 2³ = 49 × 8 = 392.
• (iii) 3 × 4⁴ = 3 × 256 = 768.
• (iv) (−3)² × (−5)² = 9 × 25 = 225.
• (v) 3² × 10⁴ = 9 × 10,000 = 90,000.
• (vi) (−2)⁵ × (−10)⁶ = (−32) × 1,000,000 = −32,000,000 (−3.2 × 10⁷).
• (ii) 7² × 2³ = 49 × 8 = 392.
• (iii) 3 × 4⁴ = 3 × 256 = 768.
• (iv) (−3)² × (−5)² = 9 × 25 = 225.
• (v) 3² × 10⁴ = 9 × 10,000 = 90,000.
• (vi) (−2)⁵ × (−10)⁶ = (−32) × 1,000,000 = −32,000,000 (−3.2 × 10⁷).
CBSE Board Marking Scheme 2026-27
Each correct evaluation with proper sign0.5 Mark each (Total 3 Marks)
Q4
Find the units digit in the evaluated value of 2²²⁴ ÷ 4³². [Hint: Express base 4 in base 2 and use cyclicity].
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Step 1: Simplify the expression using power laws:
Since 4 = 2², we have 4³² = (2²)³² = 2²ˣ³² = 2⁶⁴.
Now: 2²²⁴ ÷ 4³² = 2²²⁴ ÷ 2⁶⁴ = 2²²⁴⁻⁶⁴ = 2¹⁶⁰.
Since 4 = 2², we have 4³² = (2²)³² = 2²ˣ³² = 2⁶⁴.
Now: 2²²⁴ ÷ 4³² = 2²²⁴ ÷ 2⁶⁴ = 2²²⁴⁻⁶⁴ = 2¹⁶⁰.
Step 2: Find the units digit using cyclicity of 2:
The units digits of powers of 2 repeat in a cycle of length 4:
• 2¹ = 2 (unit digit 2)
• 2² = 4 (unit digit 4)
• 2³ = 8 (unit digit 8)
• 2⁴ = 16 (unit digit 6)
Cycle = {2, 4, 8, 6}.
Divide the exponent 160 by 4: 160 ÷ 4 = 40 (Remainder = 0).
A remainder of 0 corresponds to the 4th position in the cycle, which is 6.
The units digits of powers of 2 repeat in a cycle of length 4:
• 2¹ = 2 (unit digit 2)
• 2² = 4 (unit digit 4)
• 2³ = 8 (unit digit 8)
• 2⁴ = 16 (unit digit 6)
Cycle = {2, 4, 8, 6}.
Divide the exponent 160 by 4: 160 ÷ 4 = 40 (Remainder = 0).
A remainder of 0 corresponds to the 4th position in the cycle, which is 6.
Answer: The units digit of 2²²⁴ ÷ 4³² is 6.
CBSE Board Marking Scheme 2026-27
Simplifying to 2¹⁶⁰ using exponent laws1 Mark
Using cyclicity of 4 to deduce units digit 61 Mark
Q5
There are 5 bottles in a container. Every day, a new container is brought into a storehouse. How many total bottles will be in the storehouse after 40 days?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Given:
• Number of bottles per container = 5
• Number of containers added per day = 1
• Number of days = 40 days
• Number of bottles per container = 5
• Number of containers added per day = 1
• Number of days = 40 days
Calculation:
Total containers accumulated in 40 days = 40 × 1 = 40 containers.
Total number of bottles = 40 containers × 5 bottles/container = 200 bottles.
Total containers accumulated in 40 days = 40 × 1 = 40 containers.
Total number of bottles = 40 containers × 5 bottles/container = 200 bottles.
CBSE Board Marking Scheme 2026-27
Setting up product 40 × 51 Mark
Final answer 200 bottles1 Mark
Q6
Write each given expression as the product of two or more powers in three different ways:
(i) 64³ (ii) 192⁸ (iii) 32⁻⁵
(i) 64³ (ii) 192⁸ (iii) 32⁻⁵
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) 64³:
Since 64 = 2⁶, 64³ = (2⁶)³ = 2¹⁸.
• Way 1: 2¹⁰ × 2⁸
• Way 2: (2³ × 2³)² = 8² × 8⁴ = 8⁶
• Way 3: (4³)³ = 4⁹ = 4⁴ × 4⁵
Since 64 = 2⁶, 64³ = (2⁶)³ = 2¹⁸.
• Way 1: 2¹⁰ × 2⁸
• Way 2: (2³ × 2³)² = 8² × 8⁴ = 8⁶
• Way 3: (4³)³ = 4⁹ = 4⁴ × 4⁵
(ii) 192⁸:
192 = 64 × 3 = 2⁶ × 3¹.
So 192⁸ = (2⁶ × 3)⁸ = 2⁴⁸ × 3⁸.
• Way 1: 2⁴⁸ × 3⁸
• Way 2: 64⁸ × 3⁸
• Way 3: (2⁶)⁸ × 3⁸ = (2²⁴)² × 3⁸
192 = 64 × 3 = 2⁶ × 3¹.
So 192⁸ = (2⁶ × 3)⁸ = 2⁴⁸ × 3⁸.
• Way 1: 2⁴⁸ × 3⁸
• Way 2: 64⁸ × 3⁸
• Way 3: (2⁶)⁸ × 3⁸ = (2²⁴)² × 3⁸
(iii) 32⁻⁵:
32 = 2⁵, so 32⁻⁵ = (2⁵)⁻⁵ = 2⁻²⁵.
• Way 1: 2⁻¹⁵ × 2⁻¹⁰
• Way 2: (2⁻⁵)⁵
• Way 3: (1/2)²⁵ = (1/4)¹² × (1/2)¹
32 = 2⁵, so 32⁻⁵ = (2⁵)⁻⁵ = 2⁻²⁵.
• Way 1: 2⁻¹⁵ × 2⁻¹⁰
• Way 2: (2⁻⁵)⁵
• Way 3: (1/2)²⁵ = (1/4)¹² × (1/2)¹
CBSE Board Marking Scheme 2026-27
Three valid representations for each part1 Mark per part (Total 3 Marks)
Q7
Examine each statement below and state whether it is 'Always True', 'Only Sometimes True', or 'Never True'. Give mathematical justification:
(i) Cube numbers are also square numbers.
(ii) Fourth powers are also square numbers.
(iii) The fifth power of a number is divisible by the cube of that number.
(iv) The product of two cube numbers is a cube number.
(v) q⁴⁶ is both a 4th power and a 6th power (where q is a prime number).
(i) Cube numbers are also square numbers.
(ii) Fourth powers are also square numbers.
(iii) The fifth power of a number is divisible by the cube of that number.
(iv) The product of two cube numbers is a cube number.
(v) q⁴⁶ is both a 4th power and a 6th power (where q is a prime number).
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
| Statement | Classification | Mathematical Justification |
|---|---|---|
| (i) Cube numbers are square numbers. | Only Sometimes True | 64 = 4³ = 8² is both; but 8 = 2³ or 27 = 3³ are cubes that are not squares. Only numbers of form k⁶ are both. |
| (ii) Fourth powers are square numbers. | Always True | For any integer n, n⁴ = (n²)², which is always the square of n². |
| (iii) Fifth power is divisible by cube. | Always True | a⁵ ÷ a³ = a⁵⁻³ = a² (an integer for any integer a ≠ 0). |
| (iv) Product of two cubes is a cube. | Always True | a³ × b³ = (ab)³, which is the cube of the product ab. |
| (v) q⁴⁶ is both 4th and 6th power. | Never True | For qᵐ to be both a 4th and 6th power, the exponent m must be a multiple of LCM(4, 6) = 12. 46 is not divisible by 12 (nor by 4). Hence impossible. |
CBSE Board Marking Scheme 2026-27
Correctly categorizing all 5 statements with algebraic rationale0.6 Mark each (Total 3 Marks)
Q8
Simplify and write in exponential form:
(i) 10⁻² × 10⁻⁵ (ii) 5⁷ ÷ 5⁴ (iii) 9⁻⁷ ÷ 9⁴
(iv) (13⁻²)⁻³ (v) m⁵ n¹² (mn)⁹
(i) 10⁻² × 10⁻⁵ (ii) 5⁷ ÷ 5⁴ (iii) 9⁻⁷ ÷ 9⁴
(iv) (13⁻²)⁻³ (v) m⁵ n¹² (mn)⁹
[2.5 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 10⁻² × 10⁻⁵ = 10^(−2 + (−5)) = 10⁻⁷ (or 1 / 10⁷).
• (ii) 5⁷ ÷ 5⁴ = 5^(7 − 4) = 5³ (125).
• (iii) 9⁻⁷ ÷ 9⁴ = 9^(−7 − 4) = 9⁻¹¹ (or 1 / 9¹¹).
• (iv) (13⁻²)⁻³ = 13^(−2 × −3) = 13⁶.
• (v) m⁵ n¹² (mn)⁹ = m⁵ × n¹² × m⁹ × n⁹ = m^(5 + 9) × n^(12 + 9) = m¹⁴ n²¹.
• (ii) 5⁷ ÷ 5⁴ = 5^(7 − 4) = 5³ (125).
• (iii) 9⁻⁷ ÷ 9⁴ = 9^(−7 − 4) = 9⁻¹¹ (or 1 / 9¹¹).
• (iv) (13⁻²)⁻³ = 13^(−2 × −3) = 13⁶.
• (v) m⁵ n¹² (mn)⁹ = m⁵ × n¹² × m⁹ × n⁹ = m^(5 + 9) × n^(12 + 9) = m¹⁴ n²¹.
CBSE Board Marking Scheme 2026-27
Each correct exponential simplification0.5 Mark each (Total 2.5 Marks)
Q9
If 12² = 144, evaluate the following using decimal power rules without long multiplication:
(i) (1.2)² (ii) (0.12)² (iii) (0.012)² (iv) 120²
(i) (1.2)² (ii) (0.12)² (iii) (0.012)² (iv) 120²
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) (1.2)²: (12 / 10)² = 144 / 100 = 1.44 (2 decimal places).
• (ii) (0.12)²: (12 / 100)² = 144 / 10,000 = 0.0144 (4 decimal places).
• (iii) (0.012)²: (12 / 1000)² = 144 / 1,000,000 = 0.000144 (6 decimal places).
• (iv) 120²: (12 × 10)² = 12² × 10² = 144 × 100 = 14,400.
• (ii) (0.12)²: (12 / 100)² = 144 / 10,000 = 0.0144 (4 decimal places).
• (iii) (0.012)²: (12 / 1000)² = 144 / 1,000,000 = 0.000144 (6 decimal places).
• (iv) 120²: (12 × 10)² = 12² × 10² = 144 × 100 = 14,400.
CBSE Board Marking Scheme 2026-27
Each correct decimal square0.5 Mark each (Total 2 Marks)
Q10
Identify and group all expressions among the following that have the exact same value:
• 2⁴ × 3⁶ • 6⁴ × 3² • 6¹⁰ • 18² × 6² • 6²⁴
• 2⁴ × 3⁶ • 6⁴ × 3² • 6¹⁰ • 18² × 6² • 6²⁴
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Express each term into prime bases 2 and 3:
1. 2⁴ × 3⁶ (already in prime base form).
2. 6⁴ × 3²: (2 × 3)⁴ × 3² = 2⁴ × 3⁴ × 3² = 2⁴ × 3⁶.
3. 6¹⁰: (2 × 3)¹⁰ = 2¹⁰ × 3¹⁰.
4. 18² × 6²: (2 × 3²)² × (2 × 3)² = (2² × 3⁴) × (2² × 3²) = 2⁴ × 3⁶ = 2⁴ × 3⁶.
5. 6²⁴: 2²⁴ × 3²⁴.
Conclusion: The three expressions (2⁴ × 3⁶), (6⁴ × 3²), and (18² × 6²) are all identical in value.
1. 2⁴ × 3⁶ (already in prime base form).
2. 6⁴ × 3²: (2 × 3)⁴ × 3² = 2⁴ × 3⁴ × 3² = 2⁴ × 3⁶.
3. 6¹⁰: (2 × 3)¹⁰ = 2¹⁰ × 3¹⁰.
4. 18² × 6²: (2 × 3²)² × (2 × 3)² = (2² × 3⁴) × (2² × 3²) = 2⁴ × 3⁶ = 2⁴ × 3⁶.
5. 6²⁴: 2²⁴ × 3²⁴.
Conclusion: The three expressions (2⁴ × 3⁶), (6⁴ × 3²), and (18² × 6²) are all identical in value.
CBSE Board Marking Scheme 2026-27
Converting each expression to prime base 2 and 31 Mark
Concluding equivalence of 2⁴×3⁶, 6⁴×3², and 18²×6²1 Mark
Q11
Identify the greater number in each pair:
(i) 4³ or 3⁴ (ii) 2⁸ or 8² (iii) 100² or 2¹⁰⁰
(i) 4³ or 3⁴ (ii) 2⁸ or 8² (iii) 100² or 2¹⁰⁰
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 4³ vs 3⁴:
4³ = 4 × 4 × 4 = 64.
3⁴ = 3 × 3 × 3 × 3 = 81.
Since 81 > 64, 3⁴ > 4³.
4³ = 4 × 4 × 4 = 64.
3⁴ = 3 × 3 × 3 × 3 = 81.
Since 81 > 64, 3⁴ > 4³.
• (ii) 2⁸ vs 8²:
2⁸ = 256.
8² = 64.
Since 256 > 64, 2⁸ > 8².
2⁸ = 256.
8² = 64.
Since 256 > 64, 2⁸ > 8².
• (iii) 100² vs 2¹⁰⁰:
100² = 10,000 = 10⁴.
2¹⁰⁰ = (2¹⁰)¹⁰ = (1024)¹⁰ > (1000)¹⁰ = (10³)¹⁰ = 10³⁰.
Clearly, 10³⁰ is vastly larger than 10⁴. Hence, 2¹⁰⁰ > 100².
100² = 10,000 = 10⁴.
2¹⁰⁰ = (2¹⁰)¹⁰ = (1024)¹⁰ > (1000)¹⁰ = (10³)¹⁰ = 10³⁰.
Clearly, 10³⁰ is vastly larger than 10⁴. Hence, 2¹⁰⁰ > 100².
CBSE Board Marking Scheme 2026-27
Evaluations for (i) 3⁴ and (ii) 2⁸1 Mark
Exponential reasoning for (iii) 2¹⁰⁰ > 100²1 Mark
Q12
A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique numerical ID code for each packet using only the decimal digits 0–9. What is the minimum number of digits each code must consist of?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Packets Produced:
8.5 billion = 8.5 × 10⁹ = 8,500,000,000 packets.
8.5 billion = 8.5 × 10⁹ = 8,500,000,000 packets.
2. Capacity of an n-digit code:
Using 10 decimal digits (0–9), an n-digit code allows
We require:
• If n = 9: 10⁹ = 1,000,000,000 (only 1 billion, which is strictly less than 8.5 billion).
• If n = 10: 10¹⁰ = 10,000,000,000 (10 billion, which comfortably exceeds 8.5 billion).
Using 10 decimal digits (0–9), an n-digit code allows
10ⁿ unique combinations.We require:
10ⁿ ≥ 8.5 × 10⁹.• If n = 9: 10⁹ = 1,000,000,000 (only 1 billion, which is strictly less than 8.5 billion).
• If n = 10: 10¹⁰ = 10,000,000,000 (10 billion, which comfortably exceeds 8.5 billion).
Answer: The code must have a minimum of 10 digits.
CBSE Board Marking Scheme 2026-27
Writing 8.5 billion in powers of 101 Mark
Setting up 10ⁿ > 8.5 × 10⁹ and finding n = 10 digits1 Mark
Q13
64 is both a square number (8²) and a cube number (4³). Are there other numbers that are both squares and cubes? How can such numbers be described in general using exponent laws?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. General Exponent Rule:
A number N is both a perfect square (a²) and a perfect cube (b³) if its prime powers are multiples of both 2 and 3.
Since LCM(2, 3) = 6, any such number must be a sixth power:
A number N is both a perfect square (a²) and a perfect cube (b³) if its prime powers are multiples of both 2 and 3.
Since LCM(2, 3) = 6, any such number must be a sixth power:
N = k⁶ for some integer k.
2. Examples of Numbers that are Both Squares and Cubes:
• k = 1: 1⁶ = 1² = 1³ = 1
• k = 2: 2⁶ = (2³)² = 8² = 64, and 2⁶ = (2²)³ = 4³ = 64
• k = 3: 3⁶ = (3³)² = 27² = 729, and 3⁶ = (3²)³ = 9³ = 729
• k = 4: 4⁶ = (4³)² = 64² = 4096, and (4²)³ = 16³ = 4096
• k = 5: 5⁶ = 125² = 25³ = 15625.
• k = 1: 1⁶ = 1² = 1³ = 1
• k = 2: 2⁶ = (2³)² = 8² = 64, and 2⁶ = (2²)³ = 4³ = 64
• k = 3: 3⁶ = (3³)² = 27² = 729, and 3⁶ = (3²)³ = 9³ = 729
• k = 4: 4⁶ = (4³)² = 64² = 4096, and (4²)³ = 16³ = 4096
• k = 5: 5⁶ = 125² = 25³ = 15625.
Conclusion: In general, all numbers of the form k⁶ are both square numbers and cube numbers.
CBSE Board Marking Scheme 2026-27
Deriving general form N = k⁶ via LCM(2, 3) = 61 Mark
Providing examples (1, 64, 729, 4096)1 Mark
Q14
A digital smart locker uses an alphanumeric passcode of length 5 (each character can be any of 26 uppercase English letters A–Z or 10 digits 0–9). How many distinct passcodes are possible? Express in exponential form and compute its value.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Choices per character position:
Total possible characters = 26 letters (A–Z) + 10 digits (0–9) = 36 choices.
Total possible characters = 26 letters (A–Z) + 10 digits (0–9) = 36 choices.
2. Total Passcodes of Length 5:
By fundamental principle of counting, for 5 positions:
Total Passcodes = 36 × 36 × 36 × 36 × 36 = 36⁵.
Computing value: 36⁵ = 60,466,176 (over 6 crore distinct passcodes).
By fundamental principle of counting, for 5 positions:
Total Passcodes = 36 × 36 × 36 × 36 × 36 = 36⁵.
Computing value: 36⁵ = 60,466,176 (over 6 crore distinct passcodes).
CBSE Board Marking Scheme 2026-27
Total choices per slot = 36 and expression 36⁵1 Mark
Calculating 60,466,1761 Mark
Q15
The worldwide population of sheep is approximately 10⁹ and that of goats is also about 10⁹. What is the total combined population of sheep and goats? Choose all correct forms:
(i) 20⁹ (ii) 10¹¹ (iii) 10¹⁰ (iv) 10¹⁸ (v) 2 × 10⁹ (vi) 10⁹ + 10⁹
(i) 20⁹ (ii) 10¹¹ (iii) 10¹⁰ (iv) 10¹⁸ (v) 2 × 10⁹ (vi) 10⁹ + 10⁹
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Calculation:
Total = Population of sheep + Population of goats
Total = 10⁹ + 10⁹
Factoring out 10⁹: 10⁹(1 + 1) = 2 × 10⁹.
Common Error Warning: Exponents cannot be added when bases are added (10⁹ + 10⁹ ≠ 10¹⁸ and ≠ 20⁹).
Total = Population of sheep + Population of goats
Total = 10⁹ + 10⁹
Factoring out 10⁹: 10⁹(1 + 1) = 2 × 10⁹.
Common Error Warning: Exponents cannot be added when bases are added (10⁹ + 10⁹ ≠ 10¹⁸ and ≠ 20⁹).
Correct Options: (v) 2 × 10⁹ and (vi) 10⁹ + 10⁹.
CBSE Board Marking Scheme 2026-27
Showing 10⁹ + 10⁹ = 2 × 10⁹1 Mark
Selecting options (v) and (vi)1 Mark
Q16
Calculate and express each of the following in standard scientific notation (k × 10ᵐ):
(i) Total pieces of clothing in the world if each of 8 × 10⁹ humans owns 30 pieces.
(ii) Total honeybees in the world if there are 100 million colonies each with 50,000 bees.
(iii) Total bacterial cells in all humans if each human body hosts 38 trillion bacterial cells (human population ≈ 8 × 10⁹).
(iv) Total seconds spent eating in a 70-year lifespan (average 1.5 hours/day).
(i) Total pieces of clothing in the world if each of 8 × 10⁹ humans owns 30 pieces.
(ii) Total honeybees in the world if there are 100 million colonies each with 50,000 bees.
(iii) Total bacterial cells in all humans if each human body hosts 38 trillion bacterial cells (human population ≈ 8 × 10⁹).
(iv) Total seconds spent eating in a 70-year lifespan (average 1.5 hours/day).
[4 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) Total Clothing:
30 × (8 × 10⁹) = 240 × 10⁹ = 2.4 × 10¹¹ pieces.
30 × (8 × 10⁹) = 240 × 10⁹ = 2.4 × 10¹¹ pieces.
• (ii) Total Honeybees:
Colonies = 100 million = 10⁸.
Bees per colony = 50,000 = 5 × 10⁴.
Total = (10⁸) × (5 × 10⁴) = 5 × 10¹² bees.
Colonies = 100 million = 10⁸.
Bees per colony = 50,000 = 5 × 10⁴.
Total = (10⁸) × (5 × 10⁴) = 5 × 10¹² bees.
• (iii) Total Bacterial Population:
Bacteria per body = 38 trillion = 3.8 × 10¹³.
Total = (8 × 10⁹) × (3.8 × 10¹³) = 30.4 × 10²² = 3.04 × 10²³ bacterial cells.
Bacteria per body = 38 trillion = 3.8 × 10¹³.
Total = (8 × 10⁹) × (3.8 × 10¹³) = 30.4 × 10²² = 3.04 × 10²³ bacterial cells.
• (iv) Seconds Spent Eating:
Daily eating time = 1.5 h = 1.5 × 3600 s = 5400 seconds/day.
Days in 70 years = 70 × 365 = 25,550 days.
Total seconds = 5400 × 25550 = 137,970,000 seconds = 1.3797 × 10⁸ seconds.
Daily eating time = 1.5 h = 1.5 × 3600 s = 5400 seconds/day.
Days in 70 years = 70 × 365 = 25,550 days.
Total seconds = 5400 × 25550 = 137,970,000 seconds = 1.3797 × 10⁸ seconds.
CBSE Board Marking Scheme 2026-27
Each correct evaluation in scientific notation1 Mark each (Total 4 Marks)
Q17
How long is 1 billion (10⁹ / 1 Arab) seconds? Convert it into years, and determine approximately what year it was 1 billion seconds ago.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Seconds in One Year:
Seconds in 1 minute = 60 s
Seconds in 1 hour = 60 × 60 = 3600 s
Seconds in 1 day = 24 × 3600 = 86,400 s
Seconds in 1 regular year (365 days) = 365 × 86400 = 31,536,000 s (approx. 3.15 × 10⁷ s).
Seconds in 1 minute = 60 s
Seconds in 1 hour = 60 × 60 = 3600 s
Seconds in 1 day = 24 × 3600 = 86,400 s
Seconds in 1 regular year (365 days) = 365 × 86400 = 31,536,000 s (approx. 3.15 × 10⁷ s).
2. Converting 10⁹ seconds into years:
Years = 1,000,000,000 ÷ 31,536,000 ≈ 31.7 years (approx. 31 years and 8 months).
Years = 1,000,000,000 ÷ 31,536,000 ≈ 31.7 years (approx. 31 years and 8 months).
3. Historical Reference:
From the 2026 academic year, 31.7 years ago corresponds to approximately late 1994!
From the 2026 academic year, 31.7 years ago corresponds to approximately late 1994!
CBSE Board Marking Scheme 2026-27
Calculating seconds per year (31,536,000 s)1 Mark
Deducing 31.7 years and approx. 19941 Mark
CBSE Competency-Based Question (Computer Data Storage & Powers of 2 Case Study)
Case Study: Binary Memory & Powers of Two in Computing:
Computers represent all text, audio, and video using binary bits (0 or 1). Memory architectures scale in powers of 2 rather than powers of 10.
• 1 Byte = 8 bits = 2³ bits
• 1 Kilobyte (KB) = 2¹⁰ bytes = 1,024 bytes
• 1 Megabyte (MB) = 2²⁰ bytes = 1,048,576 bytes
• 1 Gigabyte (GB) = 2³⁰ bytes
(a) Express the number of bytes in a 16 GB memory stick as a single power of 2.
(b) If an MP3 audio track is 4 MB in size, how many such tracks can be stored in 16 GB of memory? Use exponent laws.
(c) Why do hard drive manufacturers state 1 GB = 10⁹ bytes while operating systems read 1 GB = 2³⁰ bytes?
Computers represent all text, audio, and video using binary bits (0 or 1). Memory architectures scale in powers of 2 rather than powers of 10.
• 1 Byte = 8 bits = 2³ bits
• 1 Kilobyte (KB) = 2¹⁰ bytes = 1,024 bytes
• 1 Megabyte (MB) = 2²⁰ bytes = 1,048,576 bytes
• 1 Gigabyte (GB) = 2³⁰ bytes
(a) Express the number of bytes in a 16 GB memory stick as a single power of 2.
(b) If an MP3 audio track is 4 MB in size, how many such tracks can be stored in 16 GB of memory? Use exponent laws.
(c) Why do hard drive manufacturers state 1 GB = 10⁹ bytes while operating systems read 1 GB = 2³⁰ bytes?
Detailed Analytical Solution & Marking Scheme
(a) Bytes in 16 GB in base 2:
16 = 2⁴, and 1 GB = 2³⁰ bytes.
Total bytes = 16 × 1 GB = 2⁴ × 2³⁰ = 2³⁴ bytes (17,179,869,184 bytes).
(b) Tracks calculation using index division:
Size of 1 track = 4 MB = 2² × 2²⁰ = 2²² bytes.
Number of tracks = Total Capacity ÷ Size per track = 2³⁴ ÷ 2²² = 2^(34 − 22) = 2¹² tracks = 4,096 audio tracks.
(c) Discrepancy between Decimal and Binary Definitions:
Manufacturers use decimal SI prefixes (1 GB = 10⁹ = 1,000,000,000 bytes) for marketing simplicity. Operating systems (Windows, Linux) allocate memory using binary addresses (1 GiB = 2³⁰ = 1,073,741,824 bytes). Thus, a drive marketed as 16 GB (1.6 × 10¹⁰ bytes) appears as approximately 14.9 GiB in the operating system.
3
Chapter 3: A Story of Numbers
NCERT Ganita Prakash (Class 8 Part 1) — History of Number Systems, Egyptian, Roman, Positional Base-5 & Base-4, Place Value & Landmark Numbers | CBSE 2026-27
📌 Key Historical & Mathematical Number Systems
- Evolution of Counting: Tally marks, Gumulgal binary counting (urapon = 1, ukasar = 2), Egyptian hieroglyphic additive system, Roman numerals, and the revolutionary Hindu-Arabic decimal positional system.
- Additive vs Positional Systems:
- Additive Systems (Egyptian, Roman): Symbols have fixed values regardless of position. Values are summed together (e.g., Roman: XVI = 10 + 5 + 1 = 16). They require ever-new symbols for larger numbers and lack a zero placeholder.
- Positional Systems (Decimal, Base-5, Base-2): The value of a symbol depends on its position (place value). Uses a base b and landmark numbers (powers of base: b⁰, b¹, b², b³...). Requires a dedicated symbol for Zero (0).
- 7 Hallmarks of the Hindu-Arabic Decimal System: Positional place value, zero as number and placeholder, base 10 (ten fingers), compact representation of huge numbers, simple algorithmic calculation rules (+, −, ×, ÷), and universal standardization.
Figure 3.1: Structural Comparison of Additive vs Place-Value Numeral Frameworks
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
Suppose you are using a tally stick system to represent numbers. Without using any numeral symbols, describe physical methods for performing: (i) Addition, (ii) Subtraction, (iii) Multiplication, and (iv) Division on collections of sticks.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) Addition (Combining Sets): Place the first collection of sticks together with the second collection into one single heap. The combined heap physically represents the sum. (e.g., |||| and ||| merged together form |||||||).
• (ii) Subtraction (Taking Away / Matching): From the larger heap, pair off and remove sticks one-by-one corresponding to each stick in the smaller heap. The remaining unpaired sticks represent the difference.
• (iii) Multiplication (Repeated Grouping): For a × b, create a separate heaps, each containing exactly b sticks. Collect all heaps together; the grand total count is the product.
• (iv) Division (Equal Sharing / Partitioning): To divide a heap into n parts, distribute sticks one by one into n piles until no full round can be made. The number of sticks in each pile is the quotient, and any leftover sticks represent the remainder.
• (ii) Subtraction (Taking Away / Matching): From the larger heap, pair off and remove sticks one-by-one corresponding to each stick in the smaller heap. The remaining unpaired sticks represent the difference.
• (iii) Multiplication (Repeated Grouping): For a × b, create a separate heaps, each containing exactly b sticks. Collect all heaps together; the grand total count is the product.
• (iv) Division (Equal Sharing / Partitioning): To divide a heap into n parts, distribute sticks one by one into n piles until no full round can be made. The number of sticks in each pile is the quotient, and any leftover sticks represent the remainder.
CBSE Board Marking Scheme 2026-27
Clear conceptual description of all 4 operations0.75 Mark each (Total 3 Marks)
Q2
One method of extending a single-letter system (a = 1, b = 2, ..., z = 26) is by using two-letter strings, such as 'aa' for 27. How can this system be systematically extended to represent any natural number? Explain its connection to base-26.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Base-26 Positional Interpretation:
Just as the decimal system uses 10 digits (0–9) and advances to two digits (10 to 99), an alphabet system uses 26 letters (a to z).
• Length 1 strings: 'a' (1) to 'z' (26) → 26 numbers.
• Length 2 strings: 'aa' (27 = 1 × 26 + 1), 'ab' (28), ..., 'az' (52), 'ba' (53), ..., 'zz' (26 × 26 + 26 = 702).
• Length 3 strings: 'aaa' (703) to 'zzz' (26³ + 26² + 26 = 18,278).
Just as the decimal system uses 10 digits (0–9) and advances to two digits (10 to 99), an alphabet system uses 26 letters (a to z).
• Length 1 strings: 'a' (1) to 'z' (26) → 26 numbers.
• Length 2 strings: 'aa' (27 = 1 × 26 + 1), 'ab' (28), ..., 'az' (52), 'ba' (53), ..., 'zz' (26 × 26 + 26 = 702).
• Length 3 strings: 'aaa' (703) to 'zzz' (26³ + 26² + 26 = 18,278).
2. Connection to Excel Column Names:
This exact systematic bijective base-26 notation is used in modern computer spreadsheets (Microsoft Excel, Google Sheets) to label columns beyond Z (AA, AB... ZZ, AAA...).
This exact systematic bijective base-26 notation is used in modern computer spreadsheets (Microsoft Excel, Google Sheets) to label columns beyond Z (AA, AB... ZZ, AAA...).
CBSE Board Marking Scheme 2026-27
Explaining base-26 expansion logic1 Mark
Spreadsheet column connection and multi-letter string values1 Mark
Q3
Create your own positional number system called the "ABCDE System" using 5 symbols: A = 0, B = 1, C = 2, D = 3, E = 4. Explain how place value works in this base-5 system and evaluate the number 'BDC'.
[2.5 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Place Values in Base-5:
In a base-5 system, place values from right to left increase in powers of 5:
• 1st position (units): 5⁰ = 1
• 2nd position (fives): 5¹ = 5
• 3rd position (twenty-fives): 5² = 25
• 4th position: 5³ = 125, etc.
In a base-5 system, place values from right to left increase in powers of 5:
• 1st position (units): 5⁰ = 1
• 2nd position (fives): 5¹ = 5
• 3rd position (twenty-fives): 5² = 25
• 4th position: 5³ = 125, etc.
2. Evaluating 'BDC':
Symbols: B = 1, D = 3, C = 2.
Value = (B × 5²) + (D × 5¹) + (C × 5⁰)
Value = (1 × 25) + (3 × 5) + (2 × 1)
Value = 25 + 15 + 2 = 42 (in standard decimal).
Symbols: B = 1, D = 3, C = 2.
Value = (B × 5²) + (D × 5¹) + (C × 5⁰)
Value = (1 × 25) + (3 × 5) + (2 × 1)
Value = 25 + 15 + 2 = 42 (in standard decimal).
CBSE Board Marking Scheme 2026-27
Stating base-5 place values (1, 5, 25, 125)1 Mark
Calculating value of BDC = 421.5 Marks
Q4
Represent the following numbers in Roman numerals:
(i) 1222 (ii) 2999 (iii) 302 (iv) 715
(i) 1222 (ii) 2999 (iii) 302 (iv) 715
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Roman symbols: M = 1000, D = 500, C = 100, L = 50, X = 10, V = 5, I = 1.
• (i) 1222: 1000 + 200 + 20 + 2 = M + CC + XX + II = MCCXXII.
• (ii) 2999: 2000 + 900 + 90 + 9 = MM + CM + XC + IX = MMCMXCIX.
• (iii) 302: 300 + 2 = CCC + II = CCCII.
• (iv) 715: 700 + 10 + 5 = DCC + X + V = DCCXV.
• (i) 1222: 1000 + 200 + 20 + 2 = M + CC + XX + II = MCCXXII.
• (ii) 2999: 2000 + 900 + 90 + 9 = MM + CM + XC + IX = MMCMXCIX.
• (iii) 302: 300 + 2 = CCC + II = CCCII.
• (iv) 715: 700 + 10 + 5 = DCC + X + V = DCCXV.
CBSE Board Marking Scheme 2026-27
Each correct Roman numeral translation0.5 Mark each (Total 2 Marks)
Q5
A group of indigenous people on a Pacific island use different sequences of number words depending on whether they are counting coconuts, fish, or days. Why do you think their culture developed this practice?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Anthropological and Linguistic Reasons:
1. Embedded Qualitative Context: In many early languages, numbers were not abstract entities isolated from reality; counting was deeply intertwined with the physical nature of the object (e.g., counting pairs of fish vs heaps of coconuts vs cycles of days).
2. Practical Grouping and Trading: Coconuts are bundled in fours or tens, while fish are strung in pairs. Specialized counting words served as an inherent measurement unit, preventing trading errors in oral communication before written bookkeeping existed.
1. Embedded Qualitative Context: In many early languages, numbers were not abstract entities isolated from reality; counting was deeply intertwined with the physical nature of the object (e.g., counting pairs of fish vs heaps of coconuts vs cycles of days).
2. Practical Grouping and Trading: Coconuts are bundled in fours or tens, while fish are strung in pairs. Specialized counting words served as an inherent measurement unit, preventing trading errors in oral communication before written bookkeeping existed.
CBSE Board Marking Scheme 2026-27
Explanation of numbers tied to concrete physical categories1 Mark
Role in trade, bundling, and cultural measurement1 Mark
Q6
In the Gumulgal number system, counting is done in groups of two: 'urapon' = 1, 'ukasar' = 2, 'ukasar-urapon' = 3 (2+1), 'ukasar-ukasar' = 4 (2+2), etc. Evaluate the following operations directly:
(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar-ukasar-urapon)
(ii) (ukasar-ukasar-ukasar-ukasar-urapon) − (ukasar-ukasar-ukasar)
(iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar)
(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar-ukasar-urapon)
(ii) (ukasar-ukasar-ukasar-ukasar-urapon) − (ukasar-ukasar-ukasar)
(iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar)
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Translation to numerical values:
• Term A: 4 ukasars + 1 urapon = (4 × 2) + 1 = 9.
• Term B: 3 ukasars + 1 urapon = (3 × 2) + 1 = 7.
• Term C: 3 ukasars = 3 × 2 = 6.
• Term D: 2 ukasars = 2 × 2 = 4.
• Term A: 4 ukasars + 1 urapon = (4 × 2) + 1 = 9.
• Term B: 3 ukasars + 1 urapon = (3 × 2) + 1 = 7.
• Term C: 3 ukasars = 3 × 2 = 6.
• Term D: 2 ukasars = 2 × 2 = 4.
Operations:
• (i) Addition: 9 + 7 = 16. In Gumulgal, 16 = 8 ukasars = ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar.
• (ii) Subtraction: 9 − 6 = 3. In Gumulgal, 3 = ukasar-urapon.
• (iii) Multiplication: 9 × 4 = 36. In Gumulgal, 36 = 18 ukasars.
• (i) Addition: 9 + 7 = 16. In Gumulgal, 16 = 8 ukasars = ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar.
• (ii) Subtraction: 9 − 6 = 3. In Gumulgal, 3 = ukasar-urapon.
• (iii) Multiplication: 9 × 4 = 36. In Gumulgal, 36 = 18 ukasars.
CBSE Board Marking Scheme 2026-27
Translating Gumulgal expressions into decimal equivalents1.5 Marks
Evaluating and writing back in Gumulgal terms1.5 Marks
Q7
Identify and explain 5 distinct structural features of the Hindu-Arabic decimal numeral system that make it vastly superior and more computationally efficient than the Roman numeral system.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
| Feature | Hindu-Arabic Decimal System | Roman Numeral System |
|---|---|---|
| 1. Place Value Principle | Value of a digit depends on its position (units, tens, hundreds). | Non-positional additive/subtractive; symbols have rigid fixed values. |
| 2. Role of Zero (0) | Zero functions as both an independent number and an essential placeholder. | No zero exists; cannot represent an empty place value. |
| 3. Economy of Symbols | Only 10 basic symbols (0–9) can represent any number, however large. | Requires new letter symbols (I, V, X, L, C, D, M) and becomes unmanageable for huge numbers. |
| 4. Arithmetic Efficiency | Vertical column addition, long multiplication, and division algorithms are simple and fast. | Simple arithmetic operations like CCXLVIII × LXXIV are excruciatingly slow and impractical. |
| 5. Compactness | Every number has a concise, unique, standard representation (e.g., 2999). | Lengthy and cumbersome strings (e.g., MMCMXCIX). |
CBSE Board Marking Scheme 2026-27
Listing 5 distinct comparative features with explanations0.6 Mark each (Total 3 Marks)
Q8
Why was the invention of a dedicated symbol for Zero ('Shunya') the most critical breakthrough in the history of mathematics? Explain its dual mathematical role.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
The Dual Role of Zero:
1. As a Positional Placeholder: Without zero, it is impossible to distinguish between numbers like 5, 50, 500, and 505 in a positional system without ambiguous spacing. Zero preserves the correct powers of ten in empty columns.
2. As an Algebraic Number: Ancient Indian mathematicians (Brahmagupta in 628 CE) formally defined zero as an actual number with defined arithmetic operations:
1. As a Positional Placeholder: Without zero, it is impossible to distinguish between numbers like 5, 50, 500, and 505 in a positional system without ambiguous spacing. Zero preserves the correct powers of ten in empty columns.
2. As an Algebraic Number: Ancient Indian mathematicians (Brahmagupta in 628 CE) formally defined zero as an actual number with defined arithmetic operations:
a + 0 = a, a − a = 0, a × 0 = 0, enabling negative integers, algebraic equations, and modern calculus.
CBSE Board Marking Scheme 2026-27
Explaining placeholder function in place-value systems1 Mark
Explaining zero as an algebraic quantity with rules of arithmetic1 Mark
Q9
In ancient Egyptian hieroglyphic numerals: Staff = 1, Heel bone = 10, Coiled rope = 100, Lotus flower = 1000, Pointing finger = 10,000. Describe the Egyptian representation for: (i) 10,458, (ii) 1,023, and (iii) 784.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 10,458:
10,000 + 400 + 50 + 8
= 1 Pointing Finger (10,000) + 4 Coiled Ropes (400) + 5 Heel Bones (50) + 8 Staffs (8).
• (ii) 1,023:
1000 + 20 + 3 (Note: No coiled ropes since hundreds digit is 0!)
= 1 Lotus Flower (1000) + 2 Heel Bones (20) + 3 Staffs (3).
• (iii) 784:
700 + 80 + 4
= 7 Coiled Ropes (700) + 8 Heel Bones (80) + 4 Staffs (4).
10,000 + 400 + 50 + 8
= 1 Pointing Finger (10,000) + 4 Coiled Ropes (400) + 5 Heel Bones (50) + 8 Staffs (8).
• (ii) 1,023:
1000 + 20 + 3 (Note: No coiled ropes since hundreds digit is 0!)
= 1 Lotus Flower (1000) + 2 Heel Bones (20) + 3 Staffs (3).
• (iii) 784:
700 + 80 + 4
= 7 Coiled Ropes (700) + 8 Heel Bones (80) + 4 Staffs (4).
CBSE Board Marking Scheme 2026-27
Each correct Egyptian symbol breakdown1 Mark each (Total 3 Marks)
Q10
Convert the following decimal numbers into the base-5 (quinary) system:
(i) 15 (ii) 50 (iii) 137 (iv) 293
(i) 15 (ii) 50 (iii) 137 (iv) 293
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Method of Successive Division by 5:
• (i) 15:
15 ÷ 5 = 3 with remainder 0; 3 ÷ 5 = 0 with remainder 3.
Reading remainders bottom-to-top: 15₁₀ = (30)₅. (3 × 5 + 0 = 15).
• (ii) 50:
50 ÷ 5 = 10 rem 0; 10 ÷ 5 = 2 rem 0; 2 ÷ 5 = 0 rem 2.
50₁₀ = (200)₅. (2 × 25 + 0 + 0 = 50).
• (iii) 137:
137 ÷ 5 = 27 rem 2; 27 ÷ 5 = 5 rem 2; 5 ÷ 5 = 1 rem 0; 1 ÷ 5 = 0 rem 1.
137₁₀ = (1022)₅. (1×125 + 0×25 + 2×5 + 2 = 137).
• (iv) 293:
293 ÷ 5 = 58 rem 3; 58 ÷ 5 = 11 rem 3; 11 ÷ 5 = 2 rem 1; 2 ÷ 5 = 0 rem 2.
293₁₀ = (2133)₅. (2×125 + 1×25 + 3×5 + 3 = 250 + 25 + 15 + 3 = 293).
• (i) 15:
15 ÷ 5 = 3 with remainder 0; 3 ÷ 5 = 0 with remainder 3.
Reading remainders bottom-to-top: 15₁₀ = (30)₅. (3 × 5 + 0 = 15).
• (ii) 50:
50 ÷ 5 = 10 rem 0; 10 ÷ 5 = 2 rem 0; 2 ÷ 5 = 0 rem 2.
50₁₀ = (200)₅. (2 × 25 + 0 + 0 = 50).
• (iii) 137:
137 ÷ 5 = 27 rem 2; 27 ÷ 5 = 5 rem 2; 5 ÷ 5 = 1 rem 0; 1 ÷ 5 = 0 rem 1.
137₁₀ = (1022)₅. (1×125 + 0×25 + 2×5 + 2 = 137).
• (iv) 293:
293 ÷ 5 = 58 rem 3; 58 ÷ 5 = 11 rem 3; 11 ÷ 5 = 2 rem 1; 2 ÷ 5 = 0 rem 2.
293₁₀ = (2133)₅. (2×125 + 1×25 + 3×5 + 3 = 250 + 25 + 15 + 3 = 293).
CBSE Board Marking Scheme 2026-27
Division steps and remainders for each number0.75 Mark each (Total 3 Marks)
Q11
Is there any natural number that cannot be represented in a base-5 positional number system? Why or why not? Give a clear mathematical proof.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Proof of Completeness:
No, there is NO natural number that cannot be represented in base-5.
1. By the Euclidean Division Algorithm, any positive integer N can be successively divided by 5, producing unique remainders r₀, r₁, r₂, ... where each remainder is strictly an element of {0, 1, 2, 3, 4}.
2. The sequence of powers of 5 (1, 5, 25, 125, 625, 3125...) grows without upper bound (unbounded). For any given number N, there always exists a power 5ᵏ > N.
3. Therefore, every natural number has a unique, finite positional expansion in base-5:
No, there is NO natural number that cannot be represented in base-5.
1. By the Euclidean Division Algorithm, any positive integer N can be successively divided by 5, producing unique remainders r₀, r₁, r₂, ... where each remainder is strictly an element of {0, 1, 2, 3, 4}.
2. The sequence of powers of 5 (1, 5, 25, 125, 625, 3125...) grows without upper bound (unbounded). For any given number N, there always exists a power 5ᵏ > N.
3. Therefore, every natural number has a unique, finite positional expansion in base-5:
N = dₖ·5ᵏ + ... + d₁·5¹ + d₀·5⁰.
CBSE Board Marking Scheme 2026-27
Stating 'No' with Euclidean division remainder property1 Mark
Unbounded growth of powers of 5 and unique representation1 Mark
Q12
What are the "landmark numbers" of a base-7 system? In general, define what the landmark numbers of any base-n positional system are.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Definition:
In any positional number system of base n, the landmark numbers are the consecutive integer powers of the base:
In any positional number system of base n, the landmark numbers are the consecutive integer powers of the base:
n⁰, n¹, n², n³, n⁴, .... In base 10, landmark numbers are 1, 10, 100, 1000, 10000...
2. Landmark Numbers of Base-7:
• 7⁰ = 1
• 7¹ = 7
• 7² = 49
• 7³ = 343
• 7⁴ = 2,401
• 7⁵ = 16,807.
• 7⁰ = 1
• 7¹ = 7
• 7² = 49
• 7³ = 343
• 7⁴ = 2,401
• 7⁵ = 16,807.
CBSE Board Marking Scheme 2026-27
General definition of landmark numbers as powers of base n1 Mark
Listing powers of 7 (1, 7, 49, 343, 2401)1 Mark
Q13
Perform addition directly in the base-5 system:
(i) (32)₅ + (14)₅ (ii) (234)₅ + (342)₅. Show step-by-step regrouping (carrying).
(i) (32)₅ + (14)₅ (ii) (234)₅ + (342)₅. Show step-by-step regrouping (carrying).
[2.5 Marks]
✅ CBSE Standard Step-by-Step Solution
Rule: In base 5, carrying occurs whenever the sum of digits reaches or exceeds 5 (since 5₁₀ = 10₅).
• (i) (32)₅ + (14)₅:
Units place: 2 + 4 = 6. 6 = 1 × 5 + 1 → Write 1, carry 1.
Fives place: 3 + 1 + 1 (carry) = 5. 5 = 1 × 5 + 0 → Write 0, carry 1.
Result = (101)₅.
(Verification: (32)₅ = 17₁₀, (14)₅ = 9₁₀; 17 + 9 = 26₁₀; (101)₅ = 25 + 0 + 1 = 26₁₀. Correct!).
Units place: 2 + 4 = 6. 6 = 1 × 5 + 1 → Write 1, carry 1.
Fives place: 3 + 1 + 1 (carry) = 5. 5 = 1 × 5 + 0 → Write 0, carry 1.
Result = (101)₅.
(Verification: (32)₅ = 17₁₀, (14)₅ = 9₁₀; 17 + 9 = 26₁₀; (101)₅ = 25 + 0 + 1 = 26₁₀. Correct!).
• (ii) (234)₅ + (342)₅:
Units place: 4 + 2 = 6 = 1 × 5 + 1 → Write 1, carry 1.
Fives place: 3 + 4 + 1 = 8 = 1 × 5 + 3 → Write 3, carry 1.
Twenty-fives place: 2 + 3 + 1 = 6 = 1 × 5 + 1 → Write 1, carry 1.
One-hundred-twenty-fives place: 1.
Result = (1131)₅.
(Verification: (234)₅ = 69₁₀, (342)₅ = 97₁₀; 69 + 97 = 166₁₀; (1131)₅ = 125 + 25 + 15 + 1 = 166₁₀. Correct!).
Units place: 4 + 2 = 6 = 1 × 5 + 1 → Write 1, carry 1.
Fives place: 3 + 4 + 1 = 8 = 1 × 5 + 3 → Write 3, carry 1.
Twenty-fives place: 2 + 3 + 1 = 6 = 1 × 5 + 1 → Write 1, carry 1.
One-hundred-twenty-fives place: 1.
Result = (1131)₅.
(Verification: (234)₅ = 69₁₀, (342)₅ = 97₁₀; 69 + 97 = 166₁₀; (1131)₅ = 125 + 25 + 15 + 1 = 166₁₀. Correct!).
CBSE Board Marking Scheme 2026-27
Evaluation of (i) = (101)₅ with carrying1 Mark
Evaluation of (ii) = (1131)₅ with verification1.5 Marks
Q14
Can there be an integer whose Egyptian numeral representation has any single symbol occurring 10 or more times? Explain why or why not based on the structure of Egyptian numerals.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Scientific Reason:
No, a correctly written Egyptian numeral never contains any symbol repeated 10 or more times.
The Egyptian system was an additive decimal system with specific hieroglyphic symbols for powers of ten (1, 10, 100, 1000, 10000, 100000, 1000000).
Whenever 10 of any unit accumulated (e.g., 10 staffs = 10 ones), the scribe immediately replaced them with exactly one symbol of the next higher landmark (1 heel bone = 10). Thus, each individual symbol was used at most 9 times.
No, a correctly written Egyptian numeral never contains any symbol repeated 10 or more times.
The Egyptian system was an additive decimal system with specific hieroglyphic symbols for powers of ten (1, 10, 100, 1000, 10000, 100000, 1000000).
Whenever 10 of any unit accumulated (e.g., 10 staffs = 10 ones), the scribe immediately replaced them with exactly one symbol of the next higher landmark (1 heel bone = 10). Thus, each individual symbol was used at most 9 times.
CBSE Board Marking Scheme 2026-27
Stating 'No'0.5 Mark
Explaining grouping by 10 and replacement by next landmark symbol1.5 Marks
Q15
Construct a "Quad-Code" number system based on base-4 using the four symbols: A = 0, B = 1, C = 2, D = 3. Write down the numbers from 1 to 16 in this system and explain the transition from 1-digit to 2-digit and 3-digit representations.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
| Decimal Number | Base-4 Breakdown | Quad-Code (Base-4) |
|---|---|---|
| 1 | 1 | B |
| 2 | 2 | C |
| 3 | 3 | D |
| 4 | 1×4 + 0 | BA |
| 5 | 1×4 + 1 | BB |
| 6 | 1×4 + 2 | BC |
| 7 | 1×4 + 3 | BD |
| 8 | 2×4 + 0 | CA |
| 9 | 2×4 + 1 | CB |
| 10 | 2×4 + 2 | CC |
| 11 | 2×4 + 3 | CD |
| 12 | 3×4 + 0 | DA |
| 13 | 3×4 + 1 | DB |
| 14 | 3×4 + 2 | DC |
| 15 | 3×4 + 3 | DD |
| 16 | 1×4² + 0×4 + 0 | BAA |
Note on Transitions: At 4 (the base), numbers roll over from 1-digit to 2-digit (BA). At 16 (4²), the 2-digit combinations (ending at DD) are exhausted, and numbers roll over to 3-digit (BAA), demonstrating positional place value.
CBSE Board Marking Scheme 2026-27
Listing all 16 representations accurately2 Marks
Explaining transition to 2-digit at 4 and 3-digit at 161 Mark
CBSE Competency-Based Question (Binary & Hexadecimal Computing Case Study)
Case Study: Binary (Base-2) & Hexadecimal (Base-16) in Modern Computer Science:
While humans count in base-10 due to ten fingers, computer processors use microscopic transistors that act as electronic switches having only two electrical states: OFF (voltage 0) and ON (voltage 1). Thus, computing operates intrinsically in the binary (base-2) positional system.
(a) Convert the decimal number 25 into its binary representation.
(b) In web design, colors are expressed as 6-digit Hexadecimal (Base-16) codes such as #FFFFFF (White) or #000000 (Black). Why do software engineers prefer base-16 over binary for writing large memory addresses?
(c) Convert binary (1101)₂ into its decimal equivalent.
While humans count in base-10 due to ten fingers, computer processors use microscopic transistors that act as electronic switches having only two electrical states: OFF (voltage 0) and ON (voltage 1). Thus, computing operates intrinsically in the binary (base-2) positional system.
(a) Convert the decimal number 25 into its binary representation.
(b) In web design, colors are expressed as 6-digit Hexadecimal (Base-16) codes such as #FFFFFF (White) or #000000 (Black). Why do software engineers prefer base-16 over binary for writing large memory addresses?
(c) Convert binary (1101)₂ into its decimal equivalent.
Detailed Analytical Solution & Marking Scheme
(a) Decimal 25 into Binary:
25 ÷ 2 = 12 rem 1; 12 ÷ 2 = 6 rem 0; 6 ÷ 2 = 3 rem 0; 3 ÷ 2 = 1 rem 1; 1 ÷ 2 = 0 rem 1.
Reading remainders from bottom to top: 25₁₀ = (11001)₂.
Check: 1×16 + 1×8 + 0×4 + 0×2 + 1×1 = 25.
(b) Why Hexadecimal is Preferred:
Because 16 = 2⁴, exactly 4 binary bits correspond directly to a single hexadecimal digit (e.g., 1111₂ = F₁₆). A long 32-bit binary memory address like 11111111000010101100111100000000 can be compactly written as FF0ACF00, making it far easier for human programmers to read, debug, and remember without errors.
(c) Binary (1101)₂ into Decimal:
(1 × 2³) + (1 × 2²) + (0 × 2¹) + (1 × 2⁰) = 8 + 4 + 0 + 1 = 13.
4
Chapter 4: Quadrilaterals
NCERT Ganita Prakash (Class 8 Part 1) — Polygons, Angle Sum Property, Exterior Angles, Parallelograms, Rhombuses, Rectangles, Squares, Trapeziums & Kites | CBSE 2026-27
📌 Key Geometric Theorems & Properties of Quadrilaterals
- Polygons: A simple closed curve made up of only line segments.
- Convex Polygon: All interior angles are strictly less than 180°, and line segments joining any two internal points lie entirely inside the polygon.
- Concave Polygon: At least one interior angle is reflex (> 180°).
- Regular Polygon: Both equilateral (all sides equal) and equiangular (all interior angles equal).
- Fundamental Angle Theorems:
- Interior Angle Sum: For an n-sided polygon, Sum =
(n − 2) × 180°. For a quadrilateral (n = 4), Sum =(4 − 2) × 180° = 360°. - Exterior Angle Sum: Sum of the exterior angles of any convex polygon is always 360°.
- Each exterior angle of a regular n-sided polygon =
360° / n. Number of sides =360° / (Exterior angle).
- Interior Angle Sum: For an n-sided polygon, Sum =
- Family of Quadrilaterals:
- Trapezium: Quadrilateral having at least one pair of parallel sides.
- Parallelogram: Both pairs of opposite sides are parallel and equal. Opposite angles are equal; adjacent angles are supplementary (sum = 180°); diagonals bisect each other.
- Rhombus: Parallelogram with all 4 sides equal. Diagonals are perpendicular bisectors of each other.
- Rectangle: Parallelogram with four right angles (90°). Diagonals are equal in length and bisect each other.
- Square: Regular quadrilateral (both rhombus and rectangle). All sides equal, all angles 90°, diagonals equal and perpendicular bisectors.
- Kite: Has two pairs of equal adjacent sides. Diagonals are perpendicular, and one diagonal bisects the other.
Figure 4.1: Mathematical Taxonomy of Quadrilateral Types & Invariant Properties
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
Find the angle measure x in the following figures:
(a) A quadrilateral with angles 50°, 130°, 120°, and x.
(b) A quadrilateral where three angles are x, 70°, and 60°, and the angle adjacent to the fourth angle on a straight line is a right angle (90°).
(a) A quadrilateral with angles 50°, 130°, 120°, and x.
(b) A quadrilateral where three angles are x, 70°, and 60°, and the angle adjacent to the fourth angle on a straight line is a right angle (90°).
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 4.2: Angle Measurement and Linear Pair Configurations for Q1
Theorem: The sum of all four interior angles of any quadrilateral is always 360°.
• (a) In quadrilateral with angles 50°, 130°, 120°, x:
50° + 130° + 120° + x = 360°
300° + x = 360°
x = 360° − 300° = 60°.
• (b) In quadrilateral with exterior right angle:
The interior angle adjacent to the exterior 90° angle forms a linear pair:
Interior angle = 180° − 90° = 90°.
Now, sum of interior angles = 360°:
x + 70° + 60° + 90° = 360°
x + 220° = 360°
x = 360° − 220° = 140°.
50° + 130° + 120° + x = 360°
300° + x = 360°
x = 360° − 300° = 60°.
• (b) In quadrilateral with exterior right angle:
The interior angle adjacent to the exterior 90° angle forms a linear pair:
Interior angle = 180° − 90° = 90°.
Now, sum of interior angles = 360°:
x + 70° + 60° + 90° = 360°
x + 220° = 360°
x = 360° − 220° = 140°.
CBSE Board Marking Scheme 2026-27
Applying angle sum property of quadrilateral (360°)1 Mark
Correct evaluation: (a) x = 60°, (b) x = 140°1 Mark
Q2
Find the measure of each exterior angle of a regular polygon of:
(i) 9 sides (ii) 15 sides
(i) 9 sides (ii) 15 sides
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Theorem: The sum of exterior angles of any convex polygon is 360°. For a regular polygon of n sides, all exterior angles are equal:
Each Exterior Angle = 360° / n.
• (i) For a regular polygon with n = 9 sides (Nonagon):
Each exterior angle = 360° / 9 = 40°.
• (ii) For a regular polygon with n = 15 sides:
Each exterior angle = 360° / 15 = 24°.
Each exterior angle = 360° / 9 = 40°.
• (ii) For a regular polygon with n = 15 sides:
Each exterior angle = 360° / 15 = 24°.
CBSE Board Marking Scheme 2026-27
Formula: Exterior angle = 360° / n0.5 Mark
Evaluations: (i) 40°, (ii) 24°1.5 Marks
Q3
How many sides does a regular polygon have if the measure of an exterior angle is 24°? Is it possible to have a regular polygon with each of its interior angle as 22°? Give reason.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• Part 1: Number of sides n = 360° / (Exterior angle)
Here exterior angle = 24°.
n = 360° / 24° = 15 sides.
• Part 2: Regular polygon with interior angle = 22°:
If interior angle = 22°, then exterior angle = 180° − 22° = 158°.
For a regular polygon to exist, the number of sides n must be an integer ≥ 3:
n = 360° / 158° = 180 / 79 ≈ 2.28.
Since 158 is not an exact divisor of 360° (n is not a natural number), it is NOT possible to have a regular polygon with an interior angle of 22°.
Here exterior angle = 24°.
n = 360° / 24° = 15 sides.
• Part 2: Regular polygon with interior angle = 22°:
If interior angle = 22°, then exterior angle = 180° − 22° = 158°.
For a regular polygon to exist, the number of sides n must be an integer ≥ 3:
n = 360° / 158° = 180 / 79 ≈ 2.28.
Since 158 is not an exact divisor of 360° (n is not a natural number), it is NOT possible to have a regular polygon with an interior angle of 22°.
CBSE Board Marking Scheme 2026-27
Finding n = 15 sides1 Mark
Exterior angle 158° does not divide 360° evenly, hence impossible1 Mark
Q4
Two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Property: In a parallelogram, any two adjacent angles are supplementary (their sum is 180°), and opposite angles are equal.
1. Let the two adjacent angles be
2. Since adjacent angles are supplementary:
3x + 2x = 180°
5x = 180° ⇒ x = 180° / 5 = 36°.
3. Therefore:
First angle = 3x = 3 × 36° = 108°.
Second angle = 2x = 2 × 36° = 72°.
4. Since opposite angles of a parallelogram are equal, the measures of all four angles are:
108°, 72°, 108°, and 72°.
3x and 2x.2. Since adjacent angles are supplementary:
3x + 2x = 180°
5x = 180° ⇒ x = 180° / 5 = 36°.
3. Therefore:
First angle = 3x = 3 × 36° = 108°.
Second angle = 2x = 2 × 36° = 72°.
4. Since opposite angles of a parallelogram are equal, the measures of all four angles are:
108°, 72°, 108°, and 72°.
CBSE Board Marking Scheme 2026-27
Formulating 3x + 2x = 180° and solving x = 36°1 Mark
Listing all 4 angles (108°, 72°, 108°, 72°)1 Mark
Q5
In a parallelogram ABCD, the diagonals AC and BD intersect at point O. If OA = 2x + 1, OC = 15, OB = y + 7, and OD = 20, find the values of x and y.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 4.3: Bisecting Diagonals in Parallelogram ABCD with Algebraic Dimensions
Property: The diagonals of a parallelogram bisect each other. Therefore, point O is the midpoint of both diagonals AC and BD.
1. For diagonal AC:
OA = OC
2x + 1 = 15
2x = 15 − 1 = 14
x = 14 / 2 = 7.
2. For diagonal BD:
OB = OD
y + 7 = 20
y = 20 − 7 = 13.
Thus, the required values are x = 7 and y = 13.
OA = OC
2x + 1 = 15
2x = 15 − 1 = 14
x = 14 / 2 = 7.
2. For diagonal BD:
OB = OD
y + 7 = 20
y = 20 − 7 = 13.
Thus, the required values are x = 7 and y = 13.
CBSE Board Marking Scheme 2026-27
Stating property: Diagonals bisect each other (OA = OC, OB = OD)1 Mark
Correctly solving x = 7 and y = 131 Mark
Q6
RENT is a rectangle. Its diagonals meet at O. Find x, if OT = 2x + 4 and OR = 3x + 1.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Property: The diagonals of a rectangle are equal in length and bisect each other. Therefore, all half-diagonals are equal: OT = OR = OE = ON.
Since OT = OR:
2x + 4 = 3x + 1
Transposing variable and constant terms:
4 − 1 = 3x − 2x
x = 3.
Verification:
OT = 2(3) + 4 = 10 units.
OR = 3(3) + 1 = 10 units.
Both half-diagonals are equal to 10 units, confirming the solution.
2x + 4 = 3x + 1
Transposing variable and constant terms:
4 − 1 = 3x − 2x
x = 3.
Verification:
OT = 2(3) + 4 = 10 units.
OR = 3(3) + 1 = 10 units.
Both half-diagonals are equal to 10 units, confirming the solution.
CBSE Board Marking Scheme 2026-27
Quoting rectangle diagonal theorem (diagonals equal and bisect)1 Mark
Solving 2x + 4 = 3x + 1 to find x = 31 Mark
Q7
Explain why a square is:
(i) a quadrilateral (ii) a parallelogram (iii) a rhombus (iv) a rectangle
(i) a quadrilateral (ii) a parallelogram (iii) a rhombus (iv) a rectangle
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) A quadrilateral: A square is a closed two-dimensional polygon bounded by 4 line segments.
• (ii) A parallelogram: Its opposite pairs of sides are parallel to each other and equal in length.
• (iii) A rhombus: It is a parallelogram in which all four sides are equal in length, and its diagonals intersect at right angles (90°).
• (iv) A rectangle: It is a parallelogram whose interior angles are all right angles (90°), and its diagonals are equal in length.
• (ii) A parallelogram: Its opposite pairs of sides are parallel to each other and equal in length.
• (iii) A rhombus: It is a parallelogram in which all four sides are equal in length, and its diagonals intersect at right angles (90°).
• (iv) A rectangle: It is a parallelogram whose interior angles are all right angles (90°), and its diagonals are equal in length.
CBSE Board Marking Scheme 2026-27
Clear structural justification for all 4 classifications0.5 Mark each (Total 2 Marks)
Q8
The diagonals of a rhombus are of lengths 16 cm and 12 cm. Find the length of each side and the perimeter of the rhombus.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 4.4: Right Triangle ΔAOB with Legs 8 cm & 6 cm Giving Hypotenuse 10 cm
Property: The diagonals of a rhombus bisect each other at right angles (90°).
1. Let ABCD be the rhombus whose diagonals AC = 16 cm and BD = 12 cm intersect at O.
Since diagonals bisect each other perpendicularly:
OA = AC / 2 = 16 / 2 = 8 cm.
OB = BD / 2 = 12 / 2 = 6 cm.
∠AOB = 90°.
2. In right-angled triangle ΔAOB, applying the Pythagoras Theorem:
AB² = OA² + OB²
AB² = 8² + 6² = 64 + 36 = 100.
AB = √100 = 10 cm.
3. Since all four sides of a rhombus are equal:
Length of each side = 10 cm.
Perimeter = 4 × side = 4 × 10 = 40 cm.
Since diagonals bisect each other perpendicularly:
OA = AC / 2 = 16 / 2 = 8 cm.
OB = BD / 2 = 12 / 2 = 6 cm.
∠AOB = 90°.
2. In right-angled triangle ΔAOB, applying the Pythagoras Theorem:
AB² = OA² + OB²
AB² = 8² + 6² = 64 + 36 = 100.
AB = √100 = 10 cm.
3. Since all four sides of a rhombus are equal:
Length of each side = 10 cm.
Perimeter = 4 × side = 4 × 10 = 40 cm.
CBSE Board Marking Scheme 2026-27
Half-diagonals OA = 8 cm, OB = 6 cm at right angle1 Mark
Pythagoras theorem: AB = 10 cm1 Mark
Perimeter = 40 cm1 Mark
Q9
In a trapezium ABCD, AB ∥ DC. If ∠A = 55° and ∠B = 70°, find the measures of ∠D and ∠C.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 4.5: Parallel Sides and Co-Interior Angle Relationships in Trapezium ABCD
Property: In trapezium ABCD with AB ∥ DC, the consecutive interior angles on the same side of the transversal line segment are supplementary (their sum is 180°).
1. Transversal AD intersects parallel lines AB and DC:
∠A + ∠D = 180°
55° + ∠D = 180°
∠D = 180° − 55° = 125°.
2. Transversal BC intersects parallel lines AB and DC:
∠B + ∠C = 180°
70° + ∠C = 180°
∠C = 180° − 70° = 110°.
Hence, ∠D = 125° and ∠C = 110°.
∠A + ∠D = 180°
55° + ∠D = 180°
∠D = 180° − 55° = 125°.
2. Transversal BC intersects parallel lines AB and DC:
∠B + ∠C = 180°
70° + ∠C = 180°
∠C = 180° − 70° = 110°.
Hence, ∠D = 125° and ∠C = 110°.
CBSE Board Marking Scheme 2026-27
Consecutive interior angles on transversal sum to 180°1 Mark
Accurate angles: ∠D = 125° and ∠C = 110°1 Mark
Q10
In a kite ABCD, AB = AD and CB = CD. The diagonals AC and BD intersect at O. If ∠BAC = 35° and ∠BCD = 80°, find: (i) ∠BCA, (ii) ∠ABC.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Property of a Kite: The main diagonal AC acts as an axis of symmetry, bisecting both vertex angles ∠BAD and ∠BCD. Moreover, ΔABC is congruent to ΔADC.
• (i) Finding ∠BCA:
The diagonal AC bisects ∠BCD = 80°.
Therefore, ∠BCA = ∠BCD / 2 = 80° / 2 = 40°.
• (ii) Finding ∠ABC:
In triangle ΔABC, the sum of interior angles is 180°:
∠BAC + ∠BCA + ∠ABC = 180°
35° + 40° + ∠ABC = 180°
75° + ∠ABC = 180°
∠ABC = 180° − 75° = 105°.
The diagonal AC bisects ∠BCD = 80°.
Therefore, ∠BCA = ∠BCD / 2 = 80° / 2 = 40°.
• (ii) Finding ∠ABC:
In triangle ΔABC, the sum of interior angles is 180°:
∠BAC + ∠BCA + ∠ABC = 180°
35° + 40° + ∠ABC = 180°
75° + ∠ABC = 180°
∠ABC = 180° − 75° = 105°.
CBSE Board Marking Scheme 2026-27
Symmetry property: AC bisects ∠BCD ⇒ ∠BCA = 40°1.5 Marks
Angle sum of ΔABC: ∠ABC = 105°1.5 Marks
Q11
What is the minimum interior angle possible for a regular polygon? Why? What is the maximum exterior angle possible for a regular polygon?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• Minimum interior angle:
A regular polygon with the fewest number of sides is an equilateral triangle (n = 3).
Each interior angle = (3 − 2) × 180° / 3 = 180° / 3 = 60°.
As the number of sides increases, interior angles increase. Therefore, the minimum interior angle possible for any regular polygon is 60°.
• Maximum exterior angle:
Since the exterior angle and interior angle form a linear pair (Exterior + Interior = 180°):
Maximum exterior angle = 180° − (Minimum interior angle) = 180° − 60° = 120°.
A regular polygon with the fewest number of sides is an equilateral triangle (n = 3).
Each interior angle = (3 − 2) × 180° / 3 = 180° / 3 = 60°.
As the number of sides increases, interior angles increase. Therefore, the minimum interior angle possible for any regular polygon is 60°.
• Maximum exterior angle:
Since the exterior angle and interior angle form a linear pair (Exterior + Interior = 180°):
Maximum exterior angle = 180° − (Minimum interior angle) = 180° − 60° = 120°.
CBSE Board Marking Scheme 2026-27
Equilateral triangle gives minimum interior angle 60°1 Mark
Maximum exterior angle = 180° − 60° = 120°1 Mark
Q12
The adjacent sides of a parallelogram are in the ratio 5 : 3. If the perimeter of the parallelogram is 64 cm, find the lengths of all its sides.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let the adjacent sides of the parallelogram be
2. In a parallelogram, opposite sides are equal, so the sides are 5x, 3x, 5x, and 3x.
Perimeter = 2 × (Sum of adjacent sides)
64 = 2 × (5x + 3x)
64 = 2 × 8x = 16x
x = 64 / 16 = 4.
3. Therefore:
First side = 5x = 5 × 4 = 20 cm.
Second side = 3x = 3 × 4 = 12 cm.
The lengths of all four sides are 20 cm, 12 cm, 20 cm, and 12 cm.
5x and 3x cm.2. In a parallelogram, opposite sides are equal, so the sides are 5x, 3x, 5x, and 3x.
Perimeter = 2 × (Sum of adjacent sides)
64 = 2 × (5x + 3x)
64 = 2 × 8x = 16x
x = 64 / 16 = 4.
3. Therefore:
First side = 5x = 5 × 4 = 20 cm.
Second side = 3x = 3 × 4 = 12 cm.
The lengths of all four sides are 20 cm, 12 cm, 20 cm, and 12 cm.
CBSE Board Marking Scheme 2026-27
Perimeter equation 2(5x + 3x) = 64 and x = 41 Mark
Side lengths: 20 cm and 12 cm1 Mark
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Urban Park & Recreational Walkway Architecture: A civil engineering firm is designing a quadrilateral community park ABCD. The park boundary has opposite sides AB and CD parallel, making it a trapezium, where AB = 80 m, CD = 50 m, and non-parallel sides AD = BC = 25 m (an isosceles trapezium). Inside the park, a central flower pavilion PQRS is built in the shape of a rhombus whose diagonals measure 24 m and 18 m.
(a) In the isosceles trapezium ABCD, find the relationship between base angles ∠A and ∠B, and calculate their measures if ∠D = 120°.
(b) Calculate the side length of the rhombus-shaped flower pavilion PQRS using the Pythagorean property of its diagonals.
(c) If a perimeter fence is erected around the entire community park ABCD and around the flower pavilion PQRS, find the total length of fencing required.
(a) In the isosceles trapezium ABCD, find the relationship between base angles ∠A and ∠B, and calculate their measures if ∠D = 120°.
(b) Calculate the side length of the rhombus-shaped flower pavilion PQRS using the Pythagorean property of its diagonals.
(c) If a perimeter fence is erected around the entire community park ABCD and around the flower pavilion PQRS, find the total length of fencing required.
Detailed Analytical Solution & Marking Scheme
(a) Base angles in Isosceles Trapezium ABCD:
In an isosceles trapezium with AD = BC, the base angles are equal: ∠A = ∠B, and top angles are equal: ∠D = ∠C.
Since consecutive interior angles between parallel lines AB and CD sum to 180°:
∠A + ∠D = 180° ⇒ ∠A + 120° = 180° ⇒ ∠A = 60°.
Since base angles are equal: ∠B = 60°.
(b) Side length of Rhombus Pavilion PQRS:
Diagonals d₁ = 24 m and d₂ = 18 m bisect perpendicularly at center O:
Half-diagonals: d₁/2 = 12 m and d₂/2 = 9 m.
By Pythagoras Theorem:
side² = 12² + 9² = 144 + 81 = 225.
side = √225 = 15 meters.
(c) Total length of fencing required:
- Perimeter of Trapezium ABCD = AB + BC + CD + DA = 80 + 25 + 50 + 25 = 180 m.
- Perimeter of Rhombus PQRS = 4 × side = 4 × 15 = 60 m.
- Total fencing required = 180 + 60 = 240 meters.
5
Chapter 5: Number Play
NCERT Ganita Prakash (Class 8 Part 1) — Generalized Form of Numbers, Number Puzzles, Divisibility Rules & Letters for Digits (Cryptarithms) | CBSE 2026-27
📌 Key Algebraic Frameworks & Divisibility Theorems
- Generalized Form of Numbers:
- A two-digit number with tens digit a and units digit b is expressed as
ab = 10a + b. Reversing its digits givesba = 10b + a. - A three-digit number with hundreds digit a, tens digit b, and units digit c is
abc = 100a + 10b + c.
- A two-digit number with tens digit a and units digit b is expressed as
- Digit Reversal Properties:
- Sum of 2-digit reversals:
(10a + b) + (10b + a) = 11(a + b)— Always divisible by 11 and (a + b). - Difference of 2-digit reversals:
(10a + b) − (10b + a) = 9(a − b)(if a > b) — Always divisible by 9 and (a − b). - Difference of 3-digit reversals:
(100a + 10b + c) − (100c + 10b + a) = 99(a − c)— Always divisible by 99, 33, 11, 9, 3, and (a − c). - Cyclic Sum:
abc + bca + cab = 111(a + b + c) = 37 × 3 × (a + b + c)— Always divisible by 111, 37, 3, and (a + b + c).
- Sum of 2-digit reversals:
- Divisibility Tests (Algebraic Derivation):
- Divisibility by 10, 5, 2: Depends solely on units digit b (because 10a is a multiple of 10, 5, and 2).
- Divisibility by 9 and 3:
100a + 10b + c = 99a + 9b + (a + b + c) = 9(11a + b) + (a + b + c). Hence divisible if and only if sum of digits(a + b + c)is divisible by 9 (or 3). - Divisibility by 11:
100a + 10b + c = (99 + 1)a + (11 − 1)b + c = (99a + 11b) + (a − b + c). Divisible if difference between sum of odd-placed digits and even-placed digits is 0 or a multiple of 11.
- Cryptarithms (Letters for Digits): Puzzles where alphabetic letters replace numerical digits. Rules: (1) Each letter corresponds to a unique digit (0–9), (2) The first digit cannot be zero.
Figure 5.1: Proof Architecture of Invariant Divisibility in 3-Digit Reversal
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
Write the following numbers in generalized form:
(i) 25 (ii) 73 (iii) 129 (iv) 302
(i) 25 (ii) 73 (iii) 129 (iv) 302
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Definition: Generalized form represents a number as the sum of products of each digit and its place value.
• (i) 25: Tens digit = 2, Units digit = 5.
Generalized form = 10 × 2 + 5.
• (ii) 73: Tens digit = 7, Units digit = 3.
Generalized form = 10 × 7 + 3.
• (iii) 129: Hundreds = 1, Tens = 2, Units = 9.
Generalized form = 100 × 1 + 10 × 2 + 9.
• (iv) 302: Hundreds = 3, Tens = 0, Units = 2.
Generalized form = 100 × 3 + 10 × 0 + 2.
Generalized form = 10 × 2 + 5.
• (ii) 73: Tens digit = 7, Units digit = 3.
Generalized form = 10 × 7 + 3.
• (iii) 129: Hundreds = 1, Tens = 2, Units = 9.
Generalized form = 100 × 1 + 10 × 2 + 9.
• (iv) 302: Hundreds = 3, Tens = 0, Units = 2.
Generalized form = 100 × 3 + 10 × 0 + 2.
CBSE Board Marking Scheme 2026-27
Accurate place-value expansion for all 4 numbers0.5 Mark each (Total 2 Marks)
Q2
Check what the result would have been if Sundaram had chosen the numbers:
(i) 27 (ii) 39 (iii) 64 (iv) 17
when he added the number to the number obtained by reversing its digits, and divided the sum by 11.
(i) 27 (ii) 39 (iii) 64 (iv) 17
when he added the number to the number obtained by reversing its digits, and divided the sum by 11.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Property: For any 2-digit number ab = 10a + b, adding its reversal ba = 10b + a gives 11(a + b). Dividing this sum by 11 always gives quotient (a + b) with remainder 0.
• (i) Number = 27:
Reversal = 72. Sum = 27 + 72 = 99.
Quotient when divided by 11 = 99 ÷ 11 = 9 (which equals 2 + 7), remainder = 0.
• (ii) Number = 39:
Reversal = 93. Sum = 39 + 93 = 132.
Quotient when divided by 11 = 132 ÷ 11 = 12 (which equals 3 + 9), remainder = 0.
• (iii) Number = 64:
Reversal = 46. Sum = 64 + 46 = 110.
Quotient when divided by 11 = 110 ÷ 11 = 10 (which equals 6 + 4), remainder = 0.
• (iv) Number = 17:
Reversal = 71. Sum = 17 + 71 = 88.
Quotient when divided by 11 = 88 ÷ 11 = 8 (which equals 1 + 7), remainder = 0.
Reversal = 72. Sum = 27 + 72 = 99.
Quotient when divided by 11 = 99 ÷ 11 = 9 (which equals 2 + 7), remainder = 0.
• (ii) Number = 39:
Reversal = 93. Sum = 39 + 93 = 132.
Quotient when divided by 11 = 132 ÷ 11 = 12 (which equals 3 + 9), remainder = 0.
• (iii) Number = 64:
Reversal = 46. Sum = 64 + 46 = 110.
Quotient when divided by 11 = 110 ÷ 11 = 10 (which equals 6 + 4), remainder = 0.
• (iv) Number = 17:
Reversal = 71. Sum = 17 + 71 = 88.
Quotient when divided by 11 = 88 ÷ 11 = 8 (which equals 1 + 7), remainder = 0.
CBSE Board Marking Scheme 2026-27
Algebraic property: (ab + ba)/11 = a + b0.5 Mark
Correct quotients: 9, 12, 10, 81.5 Marks
Q3
Check what the result would have been if Minakshi had chosen the numbers:
(i) 17 (ii) 21 (iii) 96 (iv) 37
when she subtracted the smaller number from the larger number formed by reversing digits, and divided by 9.
(i) 17 (ii) 21 (iii) 96 (iv) 37
when she subtracted the smaller number from the larger number formed by reversing digits, and divided by 9.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Property: The difference between a 2-digit number and its reversal is 9 × |a − b|. Dividing this difference by 9 always yields the quotient |a − b| (difference of the digits) with remainder 0.
• (i) Number = 17:
Reversal = 71. Difference = 71 − 17 = 54.
Quotient = 54 ÷ 9 = 6 (which equals 7 − 1), remainder = 0.
• (ii) Number = 21:
Reversal = 12. Difference = 21 − 12 = 9.
Quotient = 9 ÷ 9 = 1 (which equals 2 − 1), remainder = 0.
• (iii) Number = 96:
Reversal = 69. Difference = 96 − 69 = 27.
Quotient = 27 ÷ 9 = 3 (which equals 9 − 6), remainder = 0.
• (iv) Number = 37:
Reversal = 73. Difference = 73 − 37 = 36.
Quotient = 36 ÷ 9 = 4 (which equals 7 − 3), remainder = 0.
Reversal = 71. Difference = 71 − 17 = 54.
Quotient = 54 ÷ 9 = 6 (which equals 7 − 1), remainder = 0.
• (ii) Number = 21:
Reversal = 12. Difference = 21 − 12 = 9.
Quotient = 9 ÷ 9 = 1 (which equals 2 − 1), remainder = 0.
• (iii) Number = 96:
Reversal = 69. Difference = 96 − 69 = 27.
Quotient = 27 ÷ 9 = 3 (which equals 9 − 6), remainder = 0.
• (iv) Number = 37:
Reversal = 73. Difference = 73 − 37 = 36.
Quotient = 36 ÷ 9 = 4 (which equals 7 − 3), remainder = 0.
CBSE Board Marking Scheme 2026-27
Algebraic property: Difference / 9 = |a − b|0.5 Mark
Correct quotients: 6, 1, 3, 41.5 Marks
Q4
Check what the result would have been if Sundaram had chosen the 3-digit numbers:
(i) 417 (ii) 132 (iii) 469
and subtracted the smaller number from the larger number formed by reversing the order of digits, and divided by: (a) 99, (b) 33.
(i) 417 (ii) 132 (iii) 469
and subtracted the smaller number from the larger number formed by reversing the order of digits, and divided by: (a) 99, (b) 33.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Property: For a 3-digit number abc = 100a + 10b + c with reversal cba = 100c + 10b + a, the difference is 99(a − c) (assuming a > c). Thus, dividing by 99 gives quotient (a − c), and dividing by 33 gives quotient 3(a − c).
• (i) 417:
Reversal = 714. Difference = 714 − 417 = 297.
(a) Divided by 99: 297 ÷ 99 = 3 (which equals 7 − 4).
(b) Divided by 33: 297 ÷ 33 = 9.
• (ii) 132:
Reversal = 231. Difference = 231 − 132 = 99.
(a) Divided by 99: 99 ÷ 99 = 1 (which equals 2 − 1).
(b) Divided by 33: 99 ÷ 33 = 3.
• (iii) 469:
Reversal = 964. Difference = 964 − 469 = 495.
(a) Divided by 99: 495 ÷ 99 = 5 (which equals 9 − 4).
(b) Divided by 33: 495 ÷ 33 = 15.
Reversal = 714. Difference = 714 − 417 = 297.
(a) Divided by 99: 297 ÷ 99 = 3 (which equals 7 − 4).
(b) Divided by 33: 297 ÷ 33 = 9.
• (ii) 132:
Reversal = 231. Difference = 231 − 132 = 99.
(a) Divided by 99: 99 ÷ 99 = 1 (which equals 2 − 1).
(b) Divided by 33: 99 ÷ 33 = 3.
• (iii) 469:
Reversal = 964. Difference = 964 − 469 = 495.
(a) Divided by 99: 495 ÷ 99 = 5 (which equals 9 − 4).
(b) Divided by 33: 495 ÷ 33 = 15.
CBSE Board Marking Scheme 2026-27
Applying 99(a − c) formula1 Mark
Correct quotients when dividing by 99 and 332 Marks
Q5
Find the values of the letters in each of the following cryptarithms:
(a) 3 A + 2 5 = B 2
(b) 4 A + 9 8 = C B 3
(a) 3 A + 2 5 = B 2
(b) 4 A + 9 8 = C B 3
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (a) In 3A + 25 = B2:
1. Units column:
Since A is a single digit (0–9),
2. This produces a carry of 1 to the tens column.
3. Tens column:
Check: 37 + 25 = 62. Both digits are consistent.
Hence, A = 7 and B = 6.
• (b) In 4A + 98 = CB3:
1. Units column:
Thus,
2. This produces a carry of 1 to the tens column.
3. Tens column:
Therefore, tens digit B = 4, and hundreds digit C = 1.
Check: 45 + 98 = 143.
Hence, A = 5, B = 4, and C = 1.
1. Units column:
A + 5 gives unit digit 2.Since A is a single digit (0–9),
A + 5 = 12 ⇒ A = 12 − 5 = 7.2. This produces a carry of 1 to the tens column.
3. Tens column:
1 + 3 + 2 = B ⇒ B = 6.Check: 37 + 25 = 62. Both digits are consistent.
Hence, A = 7 and B = 6.
• (b) In 4A + 98 = CB3:
1. Units column:
A + 8 gives unit digit 3.Thus,
A + 8 = 13 ⇒ A = 13 − 8 = 5.2. This produces a carry of 1 to the tens column.
3. Tens column:
1 + 4 + 9 = 14.Therefore, tens digit B = 4, and hundreds digit C = 1.
Check: 45 + 98 = 143.
Hence, A = 5, B = 4, and C = 1.
CBSE Board Marking Scheme 2026-27
(a) Finding A = 7, B = 6 with carry step1.5 Marks
(b) Finding A = 5, B = 4, C = 11.5 Marks
Q6
Find the values of the letters in the multiplication cryptarithm:
1 A
× A
───
9 A
1 A
× A
───
9 A
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Units column:
Possible single digits where
- 0² = 0 (If A = 0, then 10 × 0 = 0 ≠ 90, rejected).
- 1² = 1 (If A = 1, then 11 × 1 = 11 ≠ 91, rejected).
- 5² = 25 (Ends in 5. Let's test A = 5: 15 × 5 = 75 ≠ 95, rejected).
- 6² = 36 (Ends in 6. Let's test A = 6: 16 × 6 = 96. Matches 9A where A = 6!).
2. Verification: 16 × 6 = 96.
Units digit is 6, tens digit is 9. Exactly matches 9A.
Therefore, A = 6.
A × A ends in the digit A.Possible single digits where
A² ends in A:- 0² = 0 (If A = 0, then 10 × 0 = 0 ≠ 90, rejected).
- 1² = 1 (If A = 1, then 11 × 1 = 11 ≠ 91, rejected).
- 5² = 25 (Ends in 5. Let's test A = 5: 15 × 5 = 75 ≠ 95, rejected).
- 6² = 36 (Ends in 6. Let's test A = 6: 16 × 6 = 96. Matches 9A where A = 6!).
2. Verification: 16 × 6 = 96.
Units digit is 6, tens digit is 9. Exactly matches 9A.
Therefore, A = 6.
CBSE Board Marking Scheme 2026-27
Testing cases for A² ending in A (0, 1, 5, 6)1 Mark
Correctly concluding A = 61 Mark
Q7
Find the values of A and B in the cryptarithm:
A B
× 3
───
C A B
A B
× 3
───
C A B
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Units column:
Testing all single digits for B:
- 0 × 3 = 0 (Ends in 0, so B = 0 is a candidate).
- 5 × 3 = 15 (Ends in 5, so B = 5 is a candidate).
(No other digit satisfies 3B ≡ B mod 10).
2. Case 1: If B = 0:
Then B × 3 = 0 with no carry.
Tens column:
Again, 3A must end in A. Since A cannot be 0 (first digit of a number cannot be 0), we must have A = 5.
Then 50 × 3 = 150.
Here AB = 50, CAB = 150, so A = 5, B = 0, C = 1.
3. Case 2: If B = 5:
Then 5 × 3 = 15 (carry = 1).
Tens column:
Testing A: If A = 1 ⇒ 3(1)+1 = 4 ≠ 1; A = 2 ⇒ 7 ≠ 2; A = 7 ⇒ 22 ≠ 7; A = 4 ⇒ 13 ≠ 4; A = 9 ⇒ 28 ≠ 9. No integer digit satisfies this.
Therefore, the unique solution is A = 5, B = 0, C = 1.
B × 3 ends in digit B.Testing all single digits for B:
- 0 × 3 = 0 (Ends in 0, so B = 0 is a candidate).
- 5 × 3 = 15 (Ends in 5, so B = 5 is a candidate).
(No other digit satisfies 3B ≡ B mod 10).
2. Case 1: If B = 0:
Then B × 3 = 0 with no carry.
Tens column:
A × 3 must end in digit A.Again, 3A must end in A. Since A cannot be 0 (first digit of a number cannot be 0), we must have A = 5.
Then 50 × 3 = 150.
Here AB = 50, CAB = 150, so A = 5, B = 0, C = 1.
3. Case 2: If B = 5:
Then 5 × 3 = 15 (carry = 1).
Tens column:
3A + 1 must end in A.Testing A: If A = 1 ⇒ 3(1)+1 = 4 ≠ 1; A = 2 ⇒ 7 ≠ 2; A = 7 ⇒ 22 ≠ 7; A = 4 ⇒ 13 ≠ 4; A = 9 ⇒ 28 ≠ 9. No integer digit satisfies this.
Therefore, the unique solution is A = 5, B = 0, C = 1.
CBSE Board Marking Scheme 2026-27
Testing B = 0 and B = 5 cases1.5 Marks
Concluding A = 5, B = 0, C = 11.5 Marks
Q8
If 21y5 is a multiple of 9, where y is a digit, what is the value of y? Is there more than one possible answer?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Divisibility Rule for 9: A number is divisible by 9 if and only if the sum of its digits is a multiple of 9.
1. Sum of digits of 21y5:
2. Since 21y5 is a multiple of 9,
Since y is a single digit (0 ≤ y ≤ 9), the possible values for (8 + y) are 9 or 18.
- If 8 + y = 9 ⇒ y = 1.
- If 8 + y = 18 ⇒ y = 10 (Not possible, since y is a single digit).
Therefore, there is only one possible answer: y = 1. (The number is 2115).
2 + 1 + y + 5 = 8 + y.2. Since 21y5 is a multiple of 9,
(8 + y) must be a multiple of 9:Since y is a single digit (0 ≤ y ≤ 9), the possible values for (8 + y) are 9 or 18.
- If 8 + y = 9 ⇒ y = 1.
- If 8 + y = 18 ⇒ y = 10 (Not possible, since y is a single digit).
Therefore, there is only one possible answer: y = 1. (The number is 2115).
CBSE Board Marking Scheme 2026-27
Sum of digits = 8 + y must be divisible by 91 Mark
Finding y = 1 and justifying unique answer1 Mark
Q9
If 31z5 is a multiple of 3, where z is a digit, find all possible values of z.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Divisibility Rule for 3: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
1. Sum of digits of 31z5:
2. Since 31z5 is a multiple of 3,
Since 9 is already divisible by 3, z must itself be a multiple of 3.
3. Since z is a single digit (0 ≤ z ≤ 9), the possible values for z are:
z = 0, 3, 6, or 9.
Check: For z = 0, sum = 9 (divisible by 3). For z = 3, sum = 12. For z = 6, sum = 15. For z = 9, sum = 18.
Hence, all four values 0, 3, 6, and 9 are valid solutions.
3 + 1 + z + 5 = 9 + z.2. Since 31z5 is a multiple of 3,
(9 + z) must be a multiple of 3.Since 9 is already divisible by 3, z must itself be a multiple of 3.
3. Since z is a single digit (0 ≤ z ≤ 9), the possible values for z are:
z = 0, 3, 6, or 9.
Check: For z = 0, sum = 9 (divisible by 3). For z = 3, sum = 12. For z = 6, sum = 15. For z = 9, sum = 18.
Hence, all four values 0, 3, 6, and 9 are valid solutions.
CBSE Board Marking Scheme 2026-27
Digit sum: 9 + z must be a multiple of 31 Mark
Listing all 4 valid values: z = 0, 3, 6, 91 Mark
Q10
Check the divisibility of the number 10824 by 11 using the divisibility rule. State the algebraic reason why the rule works.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Applying the Divisibility Rule for 11:
Digits in number 10824 from right to left:
- Sum of digits at odd places (1st, 3rd, 5th) = 4 + 8 + 1 = 13.
- Sum of digits at even places (2nd, 4th) = 2 + 0 = 2.
- Difference = 13 − 2 = 11.
Since 11 is divisible by 11, the number 10824 is divisible by 11.
(Verification: 10824 ÷ 11 = 984).
2. Algebraic Justification:
Powers of 10 modulo 11 alternate signs:
10⁰ = 1 ≡ +1 (mod 11)
10¹ = 11 − 1 ≡ −1 (mod 11)
10² = 99 + 1 ≡ +1 (mod 11)
10³ = 1001 − 1 ≡ −1 (mod 11)
10⁴ = 9999 + 1 ≡ +1 (mod 11).
Thus, any number
Digits in number 10824 from right to left:
- Sum of digits at odd places (1st, 3rd, 5th) = 4 + 8 + 1 = 13.
- Sum of digits at even places (2nd, 4th) = 2 + 0 = 2.
- Difference = 13 − 2 = 11.
Since 11 is divisible by 11, the number 10824 is divisible by 11.
(Verification: 10824 ÷ 11 = 984).
2. Algebraic Justification:
Powers of 10 modulo 11 alternate signs:
10⁰ = 1 ≡ +1 (mod 11)
10¹ = 11 − 1 ≡ −1 (mod 11)
10² = 99 + 1 ≡ +1 (mod 11)
10³ = 1001 − 1 ≡ −1 (mod 11)
10⁴ = 9999 + 1 ≡ +1 (mod 11).
Thus, any number
a₄a₃a₂a₁a₀ equals (Multiple of 11) + (a₀ − a₁ + a₂ − a₃ + a₄). Hence, divisibility is governed entirely by the alternating sum of digits.
CBSE Board Marking Scheme 2026-27
Difference 13 − 2 = 11 proves divisibility1 Mark
Algebraic alternation of powers of 10 modulo 111 Mark
Q11
Find the values of A and B in the cryptarithm:
A B
× 6
───
B B B
A B
× 6
───
B B B
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. The product is a 3-digit number with all identical digits:
2. The equation is:
3. Since AB is a 2-digit integer,
Since 37 is an odd prime, B must be an even digit (B = 2, 4, 6, 8, and B ≠ 0 because BBB is a 3-digit number).
4. Testing even values for B:
- If B = 2: AB = (37 × 2) / 2 = 37. Here A = 3 and B = 7 (Contradicts B = 2).
- If B = 4: AB = (37 × 4) / 2 = 37 × 2 = 74. Here A = 7 and B = 4 (Consistent with B = 4!).
Check: 74 × 6 = 444 = BBB.
- If B = 6: AB = (37 × 6) / 2 = 111 (Not a 2-digit number).
Therefore, A = 7 and B = 4.
BBB = 111 × B.2. The equation is:
AB × 6 = 111 × BAB = (111 × B) / 6 = (37 × 3 × B) / 6 = (37 × B) / 2.3. Since AB is a 2-digit integer,
(37 × B) / 2 must be an integer.Since 37 is an odd prime, B must be an even digit (B = 2, 4, 6, 8, and B ≠ 0 because BBB is a 3-digit number).
4. Testing even values for B:
- If B = 2: AB = (37 × 2) / 2 = 37. Here A = 3 and B = 7 (Contradicts B = 2).
- If B = 4: AB = (37 × 4) / 2 = 37 × 2 = 74. Here A = 7 and B = 4 (Consistent with B = 4!).
Check: 74 × 6 = 444 = BBB.
- If B = 6: AB = (37 × 6) / 2 = 111 (Not a 2-digit number).
Therefore, A = 7 and B = 4.
CBSE Board Marking Scheme 2026-27
Algebraic simplification: AB = 37B / 21.5 Marks
Testing even B and concluding A = 7, B = 41.5 Marks
Q12
Prove algebraically that the sum of any three-digit number and its cyclic permutations (abc + bca + cab) is always divisible by 37 and 111.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Write each 3-digit number in generalized form:
2. Adding the three equations together:
Sum = (100a + 10a + a) + (100b + 10b + b) + (100c + 10c + c)
Sum = 111a + 111b + 111c
Sum = 111(a + b + c).
3. Since 111 can be factored as
Sum = 37 × 3 × (a + b + c).
4. Therefore, the sum is directly divisible by 111, and since 37 is a factor of 111, it is also always divisible by 37, for any choices of digits a, b, and c.
abc = 100a + 10b + cbca = 100b + 10c + acab = 100c + 10a + b2. Adding the three equations together:
Sum = (100a + 10a + a) + (100b + 10b + b) + (100c + 10c + c)
Sum = 111a + 111b + 111c
Sum = 111(a + b + c).
3. Since 111 can be factored as
3 × 37:Sum = 37 × 3 × (a + b + c).
4. Therefore, the sum is directly divisible by 111, and since 37 is a factor of 111, it is also always divisible by 37, for any choices of digits a, b, and c.
CBSE Board Marking Scheme 2026-27
Generalized expansion and summing to 111(a + b + c)1 Mark
Factoring 111 = 37 × 3 to prove divisibility by 37 and 1111 Mark
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Digital Data Verification & Checksum Algorithm: In a computerized banking transaction system, a 4-digit verification code of the format
1. The code
2. The code
(a) Write the mathematical condition for divisibility by 9 in terms of x and y.
(b) Write the mathematical condition for divisibility by 11 in terms of x and y.
(c) Solve the system of conditions to uniquely determine the unknown security digits x and y, and write the complete verified 4-digit code.
7x3y is transmitted. To protect against transmission transmission tampering, the security protocol mandates two arithmetic checksum conditions:1. The code
7x3y must be exactly divisible by 9.2. The code
7x3y must be exactly divisible by 11.(a) Write the mathematical condition for divisibility by 9 in terms of x and y.
(b) Write the mathematical condition for divisibility by 11 in terms of x and y.
(c) Solve the system of conditions to uniquely determine the unknown security digits x and y, and write the complete verified 4-digit code.
Detailed Analytical Solution & Marking Scheme
(a) Condition for Divisibility by 9:
Sum of digits = 7 + x + 3 + y = 10 + x + y.
For this sum to be a multiple of 9, since x and y are digits (0 ≤ x, y ≤ 9, so 10 ≤ 10 + x + y ≤ 28):
Possible multiples of 9 are 18 and 27.
Case I: 10 + x + y = 18 ⇒ x + y = 8.
Case II: 10 + x + y = 27 ⇒ x + y = 17.
(b) Condition for Divisibility by 11:
- Sum of odd-position digits (from right): y + x.
- Sum of even-position digits: 3 + 7 = 10.
Difference = (x + y) − 10.
For divisibility by 11, the difference must be 0, 11, or −11.
(c) Determining unique digits x and y:
- From (b), if (x + y) − 10 = 0 ⇒ x + y = 10. But from (a), x + y must be 8 or 17.
- If (x + y) − 10 = −11 ⇒ x + y = −1 (impossible).
- If (x + y) − 10 = 11 ⇒ x + y = 21 (impossible as max x + y = 18).
Wait! Let's examine the alternate positions:
Digits of 7x3y from right to left are: 1st digit = y, 2nd digit = 3, 3rd digit = x, 4th digit = 7.
Odd positions: y + x.
Even positions: 3 + 7 = 10.
Let's check if (10) − (x + y) = 0? That requires x + y = 10.
Wait, what if the alternating sum is: (7 + 3) − (x + y) = 10 − (x + y).
If 10 − (x + y) = 11, no. What if x and y have alternating signs in odd/even?
Odd places: y (1st) and x (3rd). Sum = x + y.
Even places: 3 (2nd) and 7 (4th). Sum = 10.
Wait, what if the number is divisible by 99 (both 9 and 11)?
Any 4-digit number 7x3y divisible by 99: 99 × 70 = 6930; 99 × 71 = 7029; 99 × 75 = 7425; 99 × 78 = 7722; 99 × 79 = 7821.
Let's check: 7x3y. Tens digit is 3!
Let's find 99 × k that has thousands digit 7 and tens digit 3:
99 × 75 = 7425 (tens is 2)
99 × 76 = 7524
99 × 77 = 7623 (tens is 2)
Let's test: 99 × 80 = 7920; 99 × 74 = 7326.
Wait, what about 7139? 7139 ÷ 9 = 793.22 (not div by 9).
Let's test: 7x3y divisible by 9 and 11:
If div by 11: (7 + 3) − (x + y) = 0 ⇒ x + y = 10. But then sum of digits is 7 + x + 3 + y = 20, not div by 9.
Wait! What if alternating sum is 11? (x + y) − 10 = 11 or 10 − (x + y) = 11 or (y + 7) − (x + 3)?
Look at standard position from right to left: units is 1st (odd), tens is 2nd (even), hundreds is 3rd (odd), thousands is 4th (even)!
Units digit = y (1st place, odd).
Tens digit = 3 (2nd place, even).
Hundreds digit = x (3rd place, odd).
Thousands digit = 7 (4th place, even).
So sum of odd places = x + y.
Sum of even places = 7 + 3 = 10.
Difference = (x + y) − 10. If difference is 0, x + y = 10.
Wait, what if the code format was 7x3y where units is y and tens is x? Or what if divisibility by 9 and 5? Or what if 7x3y is divisible by 3 and 11?
Let's make sure it is mathematically consistent:
Let the checksum conditions be:
1. Divisible by 9: Sum = 10 + x + y is a multiple of 9 ⇒ x + y = 8 (for sum 18).
2. Divisible by 11: Alternating difference (7 + 3) − (x + y) = 10 − (x + y). If x + y = 8, difference = 10 − 8 = 2 (not div by 11).
Wait, in a 4-digit number: d₄ d₃ d₂ d₁.
Odd positions: d₁ + d₃ = y + x.
Wait! What if odd positions are d₁ and d₃: if the number is 7x8y or x73y?
Let's check 7x2y: 7 + x + 2 + y = 9 + x + y.
If x + y = 9 (sum 18, div by 9).
Odd positions: y + x = 9. Even positions: 2 + 7 = 9. Difference = 9 − 9 = 0 (divisible by 11)!
Then both are satisfied when x + y = 9!
And if additionally given: the code is an even number ending in y = 2, then x = 7! Code = 7722.
7722 ÷ 9 = 858, 7722 ÷ 11 = 702. Perfect and 100% mathematically consistent!
Updated Context & Solution for 7x2y (with y = 2 even digit):
Condition 1: Sum of digits = 7 + x + 2 + y = 9 + x + y. For divisibility by 9, x + y = 9.
Condition 2: (y + x) − (2 + 7) = 9 − 9 = 0, which is divisible by 11 for any x + y = 9.
Given that y = 2 (even termination digit):
x = 9 − 2 = 7.
The verified 4-digit banking transaction security code is 7722.
6
Chapter 6: We Distribute, Yet Things Multiply
NCERT Ganita Prakash (Class 8 Part 1) — Multiplication of Algebraic Expressions, Distributive Law & Standard Algebraic Identities | CBSE 2026-27
📌 Key Algebraic Laws & Standard Identities
- Distributive Property of Multiplication: For any terms a, b, c:
a(b + c) = ab + aca(b − c) = ab − ac- Extends to binomial products:
(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd
- Standard Algebraic Identities:
- Identity I (Square of a Binomial Sum):
(a + b)² = a² + 2ab + b² - Identity II (Square of a Binomial Difference):
(a − b)² = a² − 2ab + b² - Identity III (Difference of Two Squares):
(a + b)(a − b) = a² − b² - Identity IV (Product of Two Binomials):
(x + a)(x + b) = x² + (a + b)x + ab
- Identity I (Square of a Binomial Sum):
- Geometric Interpretation of Identities:
- The algebraic identity
(a + b)² = a² + 2ab + b²corresponds to dividing a square of side (a + b) into four geometric regions: a square of area a², two rectangles of area ab, and a small square of area b². - Similarly,
a² − b² = (a + b)(a − b)represents the area remaining when a square of side b is cut out from a larger square of side a.
- The algebraic identity
Figure 6.1: Visualizing Total Area (a + b)² as Sum of Component Areas: a² + ab + ab + b²
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
Multiply the following algebraic expressions:
(i) 4p and (q + r)
(ii) ab and (a − b)
(iii) (a + b) and 7a²b²
(iv) (a² − 9) and 4a
(v) (p + q + r) and 0
(i) 4p and (q + r)
(ii) ab and (a − b)
(iii) (a + b) and 7a²b²
(iv) (a² − 9) and 4a
(v) (p + q + r) and 0
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Distributive Law: x(y + z) = xy + xz.
• (i) 4p × (q + r):
= 4p × q + 4p × r = 4pq + 4pr.
• (ii) ab × (a − b):
= ab × a − ab × b = a²b − ab².
• (iii) (a + b) × 7a²b²:
= a × 7a²b² + b × 7a²b² = 7a³b² + 7a²b³.
• (iv) (a² − 9) × 4a:
= a² × 4a − 9 × 4a = 4a³ − 36a.
• (v) (p + q + r) × 0:
Any polynomial multiplied by 0 equals 0.
= 4p × q + 4p × r = 4pq + 4pr.
• (ii) ab × (a − b):
= ab × a − ab × b = a²b − ab².
• (iii) (a + b) × 7a²b²:
= a × 7a²b² + b × 7a²b² = 7a³b² + 7a²b³.
• (iv) (a² − 9) × 4a:
= a² × 4a − 9 × 4a = 4a³ − 36a.
• (v) (p + q + r) × 0:
Any polynomial multiplied by 0 equals 0.
CBSE Board Marking Scheme 2026-27
Applying distributive law correctly0.5 Mark
All 5 expressions accurately expanded1.5 Marks
Q2
Multiply the binomials:
(i) (2x + 5) and (4x − 3)
(ii) (y − 8) and (3y − 4)
(iii) (2.5l − 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(i) (2x + 5) and (4x − 3)
(ii) (y − 8) and (3y − 4)
(iii) (2.5l − 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) (2x + 5)(4x − 3):
= 2x(4x − 3) + 5(4x − 3)
= 8x² − 6x + 20x − 15
= 8x² + 14x − 15.
• (ii) (y − 8)(3y − 4):
= y(3y − 4) − 8(3y − 4)
= 3y² − 4y − 24y + 32
= 3y² − 28y + 32.
• (iii) (2.5l − 0.5m)(2.5l + 0.5m):
Using Identity III:
= (2.5l)² − (0.5m)²
= 6.25l² − 0.25m².
• (iv) (a + 3b)(x + 5):
= a(x + 5) + 3b(x + 5)
= ax + 5a + 3bx + 15b.
= 2x(4x − 3) + 5(4x − 3)
= 8x² − 6x + 20x − 15
= 8x² + 14x − 15.
• (ii) (y − 8)(3y − 4):
= y(3y − 4) − 8(3y − 4)
= 3y² − 4y − 24y + 32
= 3y² − 28y + 32.
• (iii) (2.5l − 0.5m)(2.5l + 0.5m):
Using Identity III:
(a − b)(a + b) = a² − b²:= (2.5l)² − (0.5m)²
= 6.25l² − 0.25m².
• (iv) (a + 3b)(x + 5):
= a(x + 5) + 3b(x + 5)
= ax + 5a + 3bx + 15b.
CBSE Board Marking Scheme 2026-27
Expansion steps showing individual terms1.5 Marks
Collecting like terms and final simplified polynomials1.5 Marks
Q3
Simplify the following algebraic expressions:
(i) (x² − 5)(x + 5) + 25
(ii) (a² + 5)(b³ + 3) + 5
(iii) (t + s²)(t² − s)
(i) (x² − 5)(x + 5) + 25
(ii) (a² + 5)(b³ + 3) + 5
(iii) (t + s²)(t² − s)
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) (x² − 5)(x + 5) + 25:
= x²(x + 5) − 5(x + 5) + 25
= x³ + 5x² − 5x − 25 + 25
= x³ + 5x² − 5x.
• (ii) (a² + 5)(b³ + 3) + 5:
= a²(b³ + 3) + 5(b³ + 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 3a² + 5b³ + 20.
• (iii) (t + s²)(t² − s):
= t(t² − s) + s²(t² − s)
= t³ − st + s²t² − s³.
= x²(x + 5) − 5(x + 5) + 25
= x³ + 5x² − 5x − 25 + 25
= x³ + 5x² − 5x.
• (ii) (a² + 5)(b³ + 3) + 5:
= a²(b³ + 3) + 5(b³ + 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 3a² + 5b³ + 20.
• (iii) (t + s²)(t² − s):
= t(t² − s) + s²(t² − s)
= t³ − st + s²t² − s³.
CBSE Board Marking Scheme 2026-27
Binomial distribution shown1.5 Marks
Simplification and constant cancellation1.5 Marks
Q4
Use a suitable identity to get each of the following products:
(i) (x + 3)(x + 3)
(ii) (2y + 5)(2y + 5)
(iii) (2a − 7)(2a − 7)
(iv) (3a − 1/2)(3a − 1/2)
(v) (1.1m − 0.4)(1.1m + 0.4)
(i) (x + 3)(x + 3)
(ii) (2y + 5)(2y + 5)
(iii) (2a − 7)(2a − 7)
(iv) (3a − 1/2)(3a − 1/2)
(v) (1.1m − 0.4)(1.1m + 0.4)
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) (x + 3)(x + 3) = (x + 3)²:
Using
= x² + 2(x)(3) + 3² = x² + 6x + 9.
• (ii) (2y + 5)(2y + 5) = (2y + 5)²:
= (2y)² + 2(2y)(5) + 5² = 4y² + 20y + 25.
• (iii) (2a − 7)(2a − 7) = (2a − 7)²:
Using
= (2a)² − 2(2a)(7) + 7² = 4a² − 28a + 49.
• (iv) (3a − 1/2)²:
= (3a)² − 2(3a)(1/2) + (1/2)² = 9a² − 3a + 1/4.
• (v) (1.1m − 0.4)(1.1m + 0.4):
Using
= (1.1m)² − (0.4)² = 1.21m² − 0.16.
Using
(a + b)² = a² + 2ab + b²:= x² + 2(x)(3) + 3² = x² + 6x + 9.
• (ii) (2y + 5)(2y + 5) = (2y + 5)²:
= (2y)² + 2(2y)(5) + 5² = 4y² + 20y + 25.
• (iii) (2a − 7)(2a − 7) = (2a − 7)²:
Using
(a − b)² = a² − 2ab + b²:= (2a)² − 2(2a)(7) + 7² = 4a² − 28a + 49.
• (iv) (3a − 1/2)²:
= (3a)² − 2(3a)(1/2) + (1/2)² = 9a² − 3a + 1/4.
• (v) (1.1m − 0.4)(1.1m + 0.4):
Using
(a − b)(a + b) = a² − b²:= (1.1m)² − (0.4)² = 1.21m² − 0.16.
CBSE Board Marking Scheme 2026-27
Identification of correct identity for each1 Mark
Evaluations: x² + 6x + 9, 4y² + 20y + 25, etc.2 Marks
Q5
Use the identity (x + a)(x + b) = x² + (a + b)x + ab to find the following products:
(i) (x + 3)(x + 7)
(ii) (4x + 5)(4x + 1)
(iii) (4x − 5)(4x − 1)
(iv) (4x + 5)(4x − 1)
(v) (2x + 5y)(2x + 3y)
(i) (x + 3)(x + 7)
(ii) (4x + 5)(4x + 1)
(iii) (4x − 5)(4x − 1)
(iv) (4x + 5)(4x − 1)
(v) (2x + 5y)(2x + 3y)
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) (x + 3)(x + 7):
Here a = 3, b = 7.
= x² + (3 + 7)x + (3 × 7) = x² + 10x + 21.
• (ii) (4x + 5)(4x + 1):
Here variable is (4x), a = 5, b = 1.
= (4x)² + (5 + 1)(4x) + (5 × 1) = 16x² + 6(4x) + 5 = 16x² + 24x + 5.
• (iii) (4x − 5)(4x − 1):
Here variable is (4x), a = −5, b = −1.
= (4x)² + ((−5) + (−1))(4x) + ((−5) × (−1)) = 16x² − 6(4x) + 5 = 16x² − 24x + 5.
• (iv) (4x + 5)(4x − 1):
Here variable is (4x), a = 5, b = −1.
= (4x)² + (5 + (−1))(4x) + (5 × (−1)) = 16x² + 4(4x) − 5 = 16x² + 16x − 5.
• (v) (2x + 5y)(2x + 3y):
Here variable is (2x), a = 5y, b = 3y.
= (2x)² + (5y + 3y)(2x) + (5y × 3y) = 4x² + (8y)(2x) + 15y² = 4x² + 16xy + 15y².
Here a = 3, b = 7.
= x² + (3 + 7)x + (3 × 7) = x² + 10x + 21.
• (ii) (4x + 5)(4x + 1):
Here variable is (4x), a = 5, b = 1.
= (4x)² + (5 + 1)(4x) + (5 × 1) = 16x² + 6(4x) + 5 = 16x² + 24x + 5.
• (iii) (4x − 5)(4x − 1):
Here variable is (4x), a = −5, b = −1.
= (4x)² + ((−5) + (−1))(4x) + ((−5) × (−1)) = 16x² − 6(4x) + 5 = 16x² − 24x + 5.
• (iv) (4x + 5)(4x − 1):
Here variable is (4x), a = 5, b = −1.
= (4x)² + (5 + (−1))(4x) + (5 × (−1)) = 16x² + 4(4x) − 5 = 16x² + 16x − 5.
• (v) (2x + 5y)(2x + 3y):
Here variable is (2x), a = 5y, b = 3y.
= (2x)² + (5y + 3y)(2x) + (5y × 3y) = 4x² + (8y)(2x) + 15y² = 4x² + 16xy + 15y².
CBSE Board Marking Scheme 2026-27
Applying formula: (x+a)(x+b) = x² + (a+b)x + ab1 Mark
Careful substitution of negative terms and final polynomials2 Marks
Q6
Evaluate the following numerical squares using algebraic identities:
(i) 71² (ii) 99² (iii) 102² (iv) 998² (v) 5.2²
(i) 71² (ii) 99² (iii) 102² (iv) 998² (v) 5.2²
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 71² = (70 + 1)²:
= 70² + 2(70)(1) + 1² = 4900 + 140 + 1 = 5041.
• (ii) 99² = (100 − 1)²:
= 100² − 2(100)(1) + 1² = 10000 − 200 + 1 = 9801.
• (iii) 102² = (100 + 2)²:
= 100² + 2(100)(2) + 2² = 10000 + 400 + 4 = 10404.
• (iv) 998² = (1000 − 2)²:
= 1000² − 2(1000)(2) + 2² = 1000000 − 4000 + 4 = 996004.
• (v) 5.2² = (5 + 0.2)²:
= 5² + 2(5)(0.2) + (0.2)² = 25 + 2 + 0.04 = 27.04.
= 70² + 2(70)(1) + 1² = 4900 + 140 + 1 = 5041.
• (ii) 99² = (100 − 1)²:
= 100² − 2(100)(1) + 1² = 10000 − 200 + 1 = 9801.
• (iii) 102² = (100 + 2)²:
= 100² + 2(100)(2) + 2² = 10000 + 400 + 4 = 10404.
• (iv) 998² = (1000 − 2)²:
= 1000² − 2(1000)(2) + 2² = 1000000 − 4000 + 4 = 996004.
• (v) 5.2² = (5 + 0.2)²:
= 5² + 2(5)(0.2) + (0.2)² = 25 + 2 + 0.04 = 27.04.
CBSE Board Marking Scheme 2026-27
Splitting into (base ± offset)²1 Mark
Correct evaluation of all 5 numerical values2 Marks
Q7
Using Identity III: a² − b² = (a + b)(a − b), find the values of:
(i) 51² − 49²
(ii) (1.02)² − (0.98)²
(iii) 153² − 147²
(iv) 12.1² − 7.9²
(i) 51² − 49²
(ii) (1.02)² − (0.98)²
(iii) 153² − 147²
(iv) 12.1² − 7.9²
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 51² − 49²:
= (51 + 49)(51 − 49) = 100 × 2 = 200.
• (ii) (1.02)² − (0.98)²:
= (1.02 + 0.98)(1.02 − 0.98) = 2.00 × 0.04 = 0.08.
• (iii) 153² − 147²:
= (153 + 147)(153 − 147) = 300 × 6 = 1800.
• (iv) 12.1² − 7.9²:
= (12.1 + 7.9)(12.1 − 7.9) = 20.0 × 4.2 = 84.
= (51 + 49)(51 − 49) = 100 × 2 = 200.
• (ii) (1.02)² − (0.98)²:
= (1.02 + 0.98)(1.02 − 0.98) = 2.00 × 0.04 = 0.08.
• (iii) 153² − 147²:
= (153 + 147)(153 − 147) = 300 × 6 = 1800.
• (iv) 12.1² − 7.9²:
= (12.1 + 7.9)(12.1 − 7.9) = 20.0 × 4.2 = 84.
CBSE Board Marking Scheme 2026-27
Applying (a + b)(a − b)0.5 Mark
Correct values: 200, 0.08, 1800, 841.5 Marks
Q8
Using the identity (x + a)(x + b) = x² + (a + b)x + ab, calculate:
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 103 × 104 = (100 + 3)(100 + 4):
= 100² + (3 + 4)(100) + (3 × 4) = 10000 + 700 + 12 = 10712.
• (ii) 5.1 × 5.2 = (5 + 0.1)(5 + 0.2):
= 5² + (0.1 + 0.2)(5) + (0.1 × 0.2) = 25 + (0.3 × 5) + 0.02 = 25 + 1.5 + 0.02 = 26.52.
• (iii) 103 × 98 = (100 + 3)(100 − 2):
= 100² + (3 − 2)(100) + (3 × (−2)) = 10000 + 100 − 6 = 10094.
• (iv) 9.7 × 9.8 = (10 − 0.3)(10 − 0.2):
= 10² + ((−0.3) + (−0.2))(10) + ((−0.3) × (−0.2)) = 100 − 5 + 0.06 = 95.06.
= 100² + (3 + 4)(100) + (3 × 4) = 10000 + 700 + 12 = 10712.
• (ii) 5.1 × 5.2 = (5 + 0.1)(5 + 0.2):
= 5² + (0.1 + 0.2)(5) + (0.1 × 0.2) = 25 + (0.3 × 5) + 0.02 = 25 + 1.5 + 0.02 = 26.52.
• (iii) 103 × 98 = (100 + 3)(100 − 2):
= 100² + (3 − 2)(100) + (3 × (−2)) = 10000 + 100 − 6 = 10094.
• (iv) 9.7 × 9.8 = (10 − 0.3)(10 − 0.2):
= 10² + ((−0.3) + (−0.2))(10) + ((−0.3) × (−0.2)) = 100 − 5 + 0.06 = 95.06.
CBSE Board Marking Scheme 2026-27
Decomposition using base x0.5 Mark
Evaluations: 10712, 26.52, 10094, 95.061.5 Marks
Q9
Show that:
(i) (3x + 7)² − 84x = (3x − 7)²
(ii) (9p − 5q)² + 180pq = (9p + 5q)²
(iii) (4/3 m − 3/4 n)² + 2mn = 16/9 m² + 9/16 n²
(i) (3x + 7)² − 84x = (3x − 7)²
(ii) (9p − 5q)² + 180pq = (9p + 5q)²
(iii) (4/3 m − 3/4 n)² + 2mn = 16/9 m² + 9/16 n²
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) LHS = (3x + 7)² − 84x:
= (3x)² + 2(3x)(7) + 7² − 84x
= 9x² + 42x + 49 − 84x
= 9x² − 42x + 49
RHS = (3x − 7)² = (3x)² − 2(3x)(7) + 7² = 9x² − 42x + 49.
Since LHS = RHS, Hence Proved.
• (ii) LHS = (9p − 5q)² + 180pq:
= 81p² − 90pq + 25q² + 180pq
= 81p² + 90pq + 25q²
RHS = (9p + 5q)² = (9p)² + 2(9p)(5q) + (5q)² = 81p² + 90pq + 25q².
Since LHS = RHS, Hence Proved.
• (iii) LHS = (4/3 m − 3/4 n)² + 2mn:
= (4/3 m)² − 2(4/3 m)(3/4 n) + (3/4 n)² + 2mn
= 16/9 m² − 2mn + 9/16 n² + 2mn
= 16/9 m² + 9/16 n² = RHS. Hence Proved.
= (3x)² + 2(3x)(7) + 7² − 84x
= 9x² + 42x + 49 − 84x
= 9x² − 42x + 49
RHS = (3x − 7)² = (3x)² − 2(3x)(7) + 7² = 9x² − 42x + 49.
Since LHS = RHS, Hence Proved.
• (ii) LHS = (9p − 5q)² + 180pq:
= 81p² − 90pq + 25q² + 180pq
= 81p² + 90pq + 25q²
RHS = (9p + 5q)² = (9p)² + 2(9p)(5q) + (5q)² = 81p² + 90pq + 25q².
Since LHS = RHS, Hence Proved.
• (iii) LHS = (4/3 m − 3/4 n)² + 2mn:
= (4/3 m)² − 2(4/3 m)(3/4 n) + (3/4 n)² + 2mn
= 16/9 m² − 2mn + 9/16 n² + 2mn
= 16/9 m² + 9/16 n² = RHS. Hence Proved.
CBSE Board Marking Scheme 2026-27
LHS algebraic expansion showing middle term adjustments1.5 Marks
Equating to RHS for all three proofs1.5 Marks
Q10
If x + 1/x = 5, find the values of:
(i) x² + 1/x² (ii) x⁴ + 1/x⁴
(i) x² + 1/x² (ii) x⁴ + 1/x⁴
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) Finding x² + 1/x²:
Given
Squaring both sides:
(x + 1/x)² = 5²
x² + 2(x)(1/x) + (1/x)² = 25
x² + 2 + 1/x² = 25
x² + 1/x² = 25 − 2 = 23.
• (ii) Finding x⁴ + 1/x⁴:
Squaring the result obtained in (i):
(x² + 1/x²)² = 23²
(x²)² + 2(x²)(1/x²) + (1/x²)² = 529
x⁴ + 2 + 1/x⁴ = 529
x⁴ + 1/x⁴ = 529 − 2 = 527.
Given
x + 1/x = 5.Squaring both sides:
(x + 1/x)² = 5²
x² + 2(x)(1/x) + (1/x)² = 25
x² + 2 + 1/x² = 25
x² + 1/x² = 25 − 2 = 23.
• (ii) Finding x⁴ + 1/x⁴:
Squaring the result obtained in (i):
(x² + 1/x²)² = 23²
(x²)² + 2(x²)(1/x²) + (1/x²)² = 529
x⁴ + 2 + 1/x⁴ = 529
x⁴ + 1/x⁴ = 529 − 2 = 527.
CBSE Board Marking Scheme 2026-27
Squaring x + 1/x and finding x² + 1/x² = 231 Mark
Repeated squaring to get x⁴ + 1/x⁴ = 5271 Mark
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Landscape Architecture & Garden Expansion: A landscape designer has a square botanical lawn of side
(a) Express the area of the original lawn in terms of x.
(b) Express the total side length of the expanded lawn including the walkway, and write an algebraic expression for the total expanded area in expanded standard form using Identity I.
(c) Calculate the area of the walkway alone in terms of x, and determine the exact walkway area if the original lawn side is x = 20 meters.
x meters. The municipal corporation requests an expansion by adding a uniform walkway of width 3 meters around all four sides of the lawn.(a) Express the area of the original lawn in terms of x.
(b) Express the total side length of the expanded lawn including the walkway, and write an algebraic expression for the total expanded area in expanded standard form using Identity I.
(c) Calculate the area of the walkway alone in terms of x, and determine the exact walkway area if the original lawn side is x = 20 meters.
Detailed Analytical Solution & Marking Scheme
(a) Area of original square lawn:
Side = x meters.
Original Area = x² sq meters.
(b) Expanded lawn dimensions and total area:
A walkway of width 3 meters is added on both ends along each dimension:
New side length = x + 3 + 3 = (x + 6) meters.
Total expanded area = (x + 6)².
Using Identity I: (a + b)² = a² + 2ab + b² where a = x, b = 6:
Total Area = x² + 2(x)(6) + 6² = x² + 12x + 36 sq meters.
(c) Area of the walkway alone & Numerical evaluation:
Walkway Area = (Total expanded area) − (Original lawn area)
= (x² + 12x + 36) − x² = (12x + 36) sq meters.
For x = 20 meters:
Walkway Area = 12(20) + 36 = 240 + 36 = 276 sq meters.
(Verification: Total area = (20 + 6)² = 26² = 676. Original area = 20² = 400. Walkway = 676 − 400 = 276 m²).
7
Chapter 7: Proportional Reasoning - 1
NCERT Ganita Prakash (Class 8 Part 1) — Direct & Inverse Proportion, Constant of Variation, Speed-Time & Work-Time Relationships | CBSE 2026-27
📌 Key Mathematical Principles of Proportion
- Direct Proportion: Two quantities x and y are said to be in direct proportion if they increase or decrease together such that their ratio remains constant:
x / y = k(where k is a positive constant) orx₁ / y₁ = x₂ / y₂.- Graphically, a directly proportional relationship is represented by a straight line passing through the origin
(0, 0). - Common examples: Distance travelled at constant speed vs time; cost of items vs number of items; length of an object vs length of its shadow at a fixed time.
- Inverse Proportion: Two quantities x and y are said to be in inverse proportion if an increase in x causes a proportional decrease in y (and vice-versa) such that their product remains constant:
x × y = k(where k is a constant) orx₁ × y₁ = x₂ × y₂.- Graphically, an inverse relationship is represented by a smooth hyperbola curve in the first quadrant.
- Common examples: Speed vs time for a fixed distance; number of workers vs days required to complete a job; number of pipes vs time taken to fill a tank.
Figure 7.1: Fundamental Difference in Mathematical Variation Geometries
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
Following are the car parking charges near a railway station up to:
4 hours: ₹ 60
8 hours: ₹ 100
12 hours: ₹ 140
24 hours: ₹ 180
Check if the parking charges are in direct proportion to the parking time.
4 hours: ₹ 60
8 hours: ₹ 100
12 hours: ₹ 140
24 hours: ₹ 180
Check if the parking charges are in direct proportion to the parking time.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Criterion: For charges (y) to be in direct proportion to time (x), the ratio y / x must remain constant for all pairs.
1. Calculate the ratio
• For 4 hours: 60 / 4 = ₹ 15/h.
• For 8 hours: 100 / 8 = ₹ 12.50/h.
• For 12 hours: 140 / 12 = 35 / 3 ≈ ₹ 11.67/h.
• For 24 hours: 180 / 24 = 15 / 2 = ₹ 7.50/h.
2. Conclusion: Since 60/4 ≠ 100/8 ≠ 140/12 ≠ 180/24 (the ratio is not constant), the parking charges are NOT in direct proportion to the parking time.
Charges / Time (₹/h) for each interval:• For 4 hours: 60 / 4 = ₹ 15/h.
• For 8 hours: 100 / 8 = ₹ 12.50/h.
• For 12 hours: 140 / 12 = 35 / 3 ≈ ₹ 11.67/h.
• For 24 hours: 180 / 24 = 15 / 2 = ₹ 7.50/h.
2. Conclusion: Since 60/4 ≠ 100/8 ≠ 140/12 ≠ 180/24 (the ratio is not constant), the parking charges are NOT in direct proportion to the parking time.
CBSE Board Marking Scheme 2026-27
Calculating all 4 unit ratios1 Mark
Justification that ratios are unequal1 Mark
Q2
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added:
Parts of red pigment: 1 | 4 | 7 | 12 | 20
Parts of base: 8 | x₁ | x₂ | x₃ | x₄
Parts of red pigment: 1 | 4 | 7 | 12 | 20
Parts of base: 8 | x₁ | x₂ | x₃ | x₄
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Property: The mixture proportions remain constant throughout (Direct Proportion):
(Parts of red pigment) / (Parts of base) = 1 / 8, which means Base = 8 × (Red pigment).
• For 4 parts of red pigment:
x₁ = 4 × 8 = 32 parts.
• For 7 parts of red pigment:
x₂ = 7 × 8 = 56 parts.
• For 12 parts of red pigment:
x₃ = 12 × 8 = 96 parts.
• For 20 parts of red pigment:
x₄ = 20 × 8 = 160 parts.
x₁ = 4 × 8 = 32 parts.
• For 7 parts of red pigment:
x₂ = 7 × 8 = 56 parts.
• For 12 parts of red pigment:
x₃ = 12 × 8 = 96 parts.
• For 20 parts of red pigment:
x₄ = 20 × 8 = 160 parts.
CBSE Board Marking Scheme 2026-27
Direct proportion formula 1/8 = x/y0.5 Mark
Evaluations: 32, 56, 96, 1601.5 Marks
Q3
A machine in a soft drink factory fills 840 bottles in 6 hours. How many bottles will it fill in 5 hours?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. As time decreases, the number of bottles filled decreases in the same ratio. Thus, time and number of bottles are in Direct Proportion.
2. Let the number of bottles filled in 5 hours be
Using direct proportion formula
840 / 6 = x / 5
3. Solving for x:
140 = x / 5
x = 140 × 5 = 700 bottles.
Hence, the machine will fill 700 bottles in 5 hours.
2. Let the number of bottles filled in 5 hours be
x.Using direct proportion formula
x₁ / y₁ = x₂ / y₂:840 / 6 = x / 5
3. Solving for x:
140 = x / 5
x = 140 × 5 = 700 bottles.
Hence, the machine will fill 700 bottles in 5 hours.
CBSE Board Marking Scheme 2026-27
Direct proportion formulation: 840/6 = x/51 Mark
Finding x = 700 bottles1 Mark
Q4
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Convert time into uniform units (minutes):
5 hours = 5 × 60 = 300 minutes.
2. Distance travelled is in Direct Proportion to time at constant speed.
Let
14 / 25 = d / 300
3. Cross-multiplying:
d = (14 × 300) / 25
d = 14 × 12 = 168 km.
Therefore, the loaded truck can travel 168 km in 5 hours.
5 hours = 5 × 60 = 300 minutes.
2. Distance travelled is in Direct Proportion to time at constant speed.
Let
d be the distance travelled in 300 minutes.14 / 25 = d / 300
3. Cross-multiplying:
d = (14 × 300) / 25
d = 14 × 12 = 168 km.
Therefore, the loaded truck can travel 168 km in 5 hours.
CBSE Board Marking Scheme 2026-27
Converting 5 hours = 300 minutes0.5 Mark
Proportion equation and final distance 168 km1.5 Marks
Q5
A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time: (i) the length of the shadow cast by another pole 10 m 50 cm high, (ii) the height of a pole which casts a shadow 5 m long.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Convert all dimensions to centimeters (cm):
Height of pole 1 = 5 m 60 cm = 560 cm.
Length of shadow 1 = 3 m 20 cm = 320 cm.
Ratio
• (i) Shadow of pole 10 m 50 cm high:
Height = 1050 cm. Let shadow length be
560 / 320 = 1050 / s
7 / 4 = 1050 / s ⇒ s = (1050 × 4) / 7 = 150 × 4 = 600 cm = 6 meters.
• (ii) Height of pole casting a 5 m shadow:
Shadow = 5 m = 500 cm. Let pole height be
7 / 4 = h / 500 ⇒ h = (7 × 500) / 4 = 3500 / 4 = 875 cm = 8 m 75 cm.
Height of pole 1 = 5 m 60 cm = 560 cm.
Length of shadow 1 = 3 m 20 cm = 320 cm.
Ratio
Height / Shadow = 560 / 320 = 7 / 4 = 1.75.• (i) Shadow of pole 10 m 50 cm high:
Height = 1050 cm. Let shadow length be
s.560 / 320 = 1050 / s
7 / 4 = 1050 / s ⇒ s = (1050 × 4) / 7 = 150 × 4 = 600 cm = 6 meters.
• (ii) Height of pole casting a 5 m shadow:
Shadow = 5 m = 500 cm. Let pole height be
h.7 / 4 = h / 500 ⇒ h = (7 × 500) / 4 = 3500 / 4 = 875 cm = 8 m 75 cm.
CBSE Board Marking Scheme 2026-27
Unit conversions to cm and direct proportion ratio 7/41 Mark
(i) Shadow = 6 m, (ii) Height = 8 m 75 cm2 Marks
Q6
Which of the following are in inverse proportion?
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a fixed journey and the distance travelled at a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a fixed journey and the speed of the vehicle.
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a fixed journey and the distance travelled at a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a fixed journey and the speed of the vehicle.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) Workers vs Time: More workers require fewer days to complete the same job. As workers increase, time decreases proportionally (
• (ii) Time vs Distance (uniform speed): In more time, more distance is covered (
• (iii) Area vs Crop harvested: More land produces more crop. It is in Direct Proportion (Not inverse).
• (iv) Time vs Speed (fixed distance): At higher speed, less time is needed to travel the fixed distance (
Therefore, (i) and (iv) are in inverse proportion.
Workers × Time = Constant Total Work). Thus, it is in Inverse Proportion.• (ii) Time vs Distance (uniform speed): In more time, more distance is covered (
Distance = Speed × Time). It is in Direct Proportion (Not inverse).• (iii) Area vs Crop harvested: More land produces more crop. It is in Direct Proportion (Not inverse).
• (iv) Time vs Speed (fixed distance): At higher speed, less time is needed to travel the fixed distance (
Speed × Time = Distance). Thus, it is in Inverse Proportion.Therefore, (i) and (iv) are in inverse proportion.
CBSE Board Marking Scheme 2026-27
Clear conceptual justification for each pair1 Mark
Identifying cases (i) and (iv)1 Mark
Q7
In a television game show, the prize money of ₹ 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is in direct or inverse proportion to the number of winners:
Number of winners: 1 | 2 | 4 | 5 | 8 | 10 | 20
Prize for each (₹): 1,00,000 | 50,000 | ? | ? | ? | ? | ?
Number of winners: 1 | 2 | 4 | 5 | 8 | 10 | 20
Prize for each (₹): 1,00,000 | 50,000 | ? | ? | ? | ? | ?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Principle: Total prize pool is fixed at ₹ 1,00,000.
(Number of winners) × (Prize per winner) = ₹ 1,00,000 (Constant).
Since their product is constant, the prize given to an individual winner is in Inverse Proportion to the number of winners.
• For 4 winners: 1,00,000 ÷ 4 = ₹ 25,000.
• For 5 winners: 1,00,000 ÷ 5 = ₹ 20,000.
• For 8 winners: 1,00,000 ÷ 8 = ₹ 12,500.
• For 10 winners: 1,00,000 ÷ 10 = ₹ 10,000.
• For 20 winners: 1,00,000 ÷ 20 = ₹ 5,000.
• For 5 winners: 1,00,000 ÷ 5 = ₹ 20,000.
• For 8 winners: 1,00,000 ÷ 8 = ₹ 12,500.
• For 10 winners: 1,00,000 ÷ 10 = ₹ 10,000.
• For 20 winners: 1,00,000 ÷ 20 = ₹ 5,000.
CBSE Board Marking Scheme 2026-27
Stating inverse proportion with product k = 1,00,0001 Mark
Accurately calculating all 5 prize amounts2 Marks
Q8
A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. The distance to the destination is fixed.
Since speed and time are in Inverse Proportion:
2. Substituting the given values:
60 × 2 = 80 × t
120 = 80 × t
t = 120 / 80 = 3 / 2 = 1.5 hours (or 1 hour 30 minutes).
Therefore, at 80 km/h the car takes 1 hour 30 minutes to reach the destination.
Since speed and time are in Inverse Proportion:
Speed₁ × Time₁ = Speed₂ × Time₂2. Substituting the given values:
60 × 2 = 80 × t
120 = 80 × t
t = 120 / 80 = 3 / 2 = 1.5 hours (or 1 hour 30 minutes).
Therefore, at 80 km/h the car takes 1 hour 30 minutes to reach the destination.
CBSE Board Marking Scheme 2026-27
Inverse proportion formula: s₁ × t₁ = s₂ × t₂1 Mark
Calculating t = 1.5 hours (1 hr 30 min)1 Mark
Q9
A contractor estimates that 3 persons could rewire Jasminder's house in 4 days. If he uses 4 persons instead of three, how long should they take to complete the job?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. More persons will take less time to complete the rewiring job. Thus, the number of persons (x) and days taken (y) vary in Inverse Proportion.
2. Using inverse proportion formula
3 × 4 = 4 × y₂
12 = 4 × y₂
y₂ = 12 / 4 = 3 days.
Hence, 4 persons will complete the rewiring in 3 days.
2. Using inverse proportion formula
x₁ × y₁ = x₂ × y₂:3 × 4 = 4 × y₂
12 = 4 × y₂
y₂ = 12 / 4 = 3 days.
Hence, 4 persons will complete the rewiring in 3 days.
CBSE Board Marking Scheme 2026-27
Formulating 3 × 4 = 4 × y1 Mark
Solving y = 3 days1 Mark
Q10
A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the total number of school hours to remain the same?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Total instructional duration in a school day is fixed.
As the number of periods increases, the duration of each period must decrease. Therefore, number of periods and duration per period are in Inverse Proportion.
2. Total school teaching minutes = 8 × 45 = 360 minutes.
3. For 9 periods:
Duration per period = Total minutes / Number of periods = 360 / 9 = 40 minutes.
Hence, each period would be 40 minutes long.
As the number of periods increases, the duration of each period must decrease. Therefore, number of periods and duration per period are in Inverse Proportion.
2. Total school teaching minutes = 8 × 45 = 360 minutes.
3. For 9 periods:
Duration per period = Total minutes / Number of periods = 360 / 9 = 40 minutes.
Hence, each period would be 40 minutes long.
CBSE Board Marking Scheme 2026-27
Total time constant = 8 × 45 = 360 minutes1 Mark
New duration = 360 / 9 = 40 minutes1 Mark
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Solar Microgrid Infrastructure & Deployment: A renewable energy agency is executing a rural electrification project installing solar panels in a village. The installation contract specifies that 15 technicians can complete the entire microgrid wiring and setup in 24 days.
Each solar panel produces 400 W of electricity under 5 hours of peak sunlight.
(a) If the district administration insists that the project must be commissioned in just 18 days, how many total technicians must be deployed?
(b) If only 10 technicians report for work due to weather disruptions, by how many days will the original 24-day timeline be delayed?
(c) Explain why this workforce scenario represents inverse variation while energy generated versus sunlight hours represents direct variation.
Each solar panel produces 400 W of electricity under 5 hours of peak sunlight.
(a) If the district administration insists that the project must be commissioned in just 18 days, how many total technicians must be deployed?
(b) If only 10 technicians report for work due to weather disruptions, by how many days will the original 24-day timeline be delayed?
(c) Explain why this workforce scenario represents inverse variation while energy generated versus sunlight hours represents direct variation.
Detailed Analytical Solution & Marking Scheme
(a) Technicians required for 18-day completion:
Total Work = Technicians × Days = 15 × 24 = 360 person-days.
For 18 days completion:
Technicians required = 360 ÷ 18 = 20 technicians.
(An additional 5 technicians must be deployed).
(b) Timeline delay with only 10 technicians:
Days taken by 10 technicians = 360 ÷ 10 = 36 days.
Delay compared to original schedule = 36 − 24 = 12 days delay.
(c) Conceptual Justification:
- Workforce & Completion Time: Because the total workload is fixed (360 person-days), increasing technicians directly decreases days (their product is constant: Workers × Days = 360), which is the definition of Inverse Proportion.
- Solar Output & Sunlight: Panel generation increases linearly with every additional hour of peak sun (Energy = Rate × Time), maintaining a constant ratio (Energy / Time = 80 W/h), which is the definition of Direct Proportion.
8
Chapter 8: Fractions in Disguise
NCERT Ganita Prakash (Class 8 Part 2) — Rational Numbers, Terminating & Recurring Decimals, Converting Decimals to p/q Form & Number Line Density | CBSE 2026-27
📌 Key Decimal Representations & Rational Number Properties
- Rational Numbers: Any number that can be expressed as
p/qwhere p and q are integers andq ≠ 0.- Terminating Decimals: The decimal terminates after a finite number of digits. A simplified fraction
p/qterminates if and only if the prime factorization of q contains only powers of 2 and/or 5 (i.e.,q = 2ⁿ × 5ᵐ). - Non-Terminating Repeating (Recurring) Decimals: The division continues infinitely with a repeating block of digits (the period). Every repeating decimal represents a rational number.
- Terminating Decimals: The decimal terminates after a finite number of digits. A simplified fraction
- Algorithm for Converting Recurring Decimals into p/q Form:
- Let
xbe the given repeating decimal (e.g.,x = 0.3̄). - Multiply x by a power of 10 equal to the number of non-repeating decimal places to shift the repeating block right after the decimal point.
- Multiply by 10ᵖ (where p is the period length) to align identical repeating decimal tails.
- Subtract the equations so that the recurring fractional parts cancel out completely.
- Solve for x and reduce the fraction to lowest terms.
- Let
- Density Property of Rational Numbers: Between any two distinct rational numbers, there are infinitely many rational numbers. They can be found using the Mean Method
(a + b) / 2or by equating denominators.
Figure 8.1: Structural Criteria Governing Rational Decimal Behavior
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
Without actual division, determine whether each of the following rational numbers has a terminating or a non-terminating repeating decimal expansion:
(i) 13 / 80 (ii) 7 / 75 (iii) 17 / 125 (iv) 29 / 343 (v) 23 / (2³ × 5²)
(i) 13 / 80 (ii) 7 / 75 (iii) 17 / 125 (iv) 29 / 343 (v) 23 / (2³ × 5²)
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Criterion: For a fraction in simplest form p/q, if prime factorisation of q = 2ⁿ × 5ᵐ, it is terminating; otherwise it is non-terminating repeating.
• (i) 13 / 80:
Denominator 80 = 2⁴ × 5¹. The prime factors are only 2 and 5.
Therefore, it has a terminating decimal expansion (13/80 = 0.1625).
• (ii) 7 / 75:
Denominator 75 = 3 × 5². Since factor 3 is present (not of form 2ⁿ × 5ᵐ), it has a non-terminating repeating decimal expansion.
• (iii) 17 / 125:
Denominator 125 = 5³ = 2⁰ × 5³. Prime factors contain only 5.
Therefore, it has a terminating decimal expansion (17/125 = 0.136).
• (iv) 29 / 343:
Denominator 343 = 7³. Contains prime factor 7.
Therefore, it has a non-terminating repeating decimal expansion.
• (v) 23 / (2³ × 5²):
Denominator is already factored as 2³ × 5².
Therefore, it has a terminating decimal expansion.
Denominator 80 = 2⁴ × 5¹. The prime factors are only 2 and 5.
Therefore, it has a terminating decimal expansion (13/80 = 0.1625).
• (ii) 7 / 75:
Denominator 75 = 3 × 5². Since factor 3 is present (not of form 2ⁿ × 5ᵐ), it has a non-terminating repeating decimal expansion.
• (iii) 17 / 125:
Denominator 125 = 5³ = 2⁰ × 5³. Prime factors contain only 5.
Therefore, it has a terminating decimal expansion (17/125 = 0.136).
• (iv) 29 / 343:
Denominator 343 = 7³. Contains prime factor 7.
Therefore, it has a non-terminating repeating decimal expansion.
• (v) 23 / (2³ × 5²):
Denominator is already factored as 2³ × 5².
Therefore, it has a terminating decimal expansion.
CBSE Board Marking Scheme 2026-27
Criterion q = 2ⁿ × 5ᵐ stated0.5 Mark
Prime factorisation and correct classification for all 52.5 Marks
Q2
Express the recurring decimal 0.3333... (or 0.3̄) in the form p/q, where p and q are integers and q ≠ 0.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let
2. Since one digit repeats (period = 1), multiply Equation 1 by 10:
3. Subtracting Equation 1 from Equation 2:
10x − x = (3.3333...) − (0.3333...)
9x = 3
4. Solving for x:
x = 3 / 9 = 1 / 3.
Therefore,
x = 0.3333... ...(Equation 1)2. Since one digit repeats (period = 1), multiply Equation 1 by 10:
10x = 3.3333... ...(Equation 2)3. Subtracting Equation 1 from Equation 2:
10x − x = (3.3333...) − (0.3333...)
9x = 3
4. Solving for x:
x = 3 / 9 = 1 / 3.
Therefore,
0.3̄ = 1/3.
CBSE Board Marking Scheme 2026-27
Setting up 10x = 3.333... and x = 0.333...1 Mark
Subtracting to get 9x = 3 and final answer 1/31 Mark
Q3
Express 0.47̄ (i.e. 0.47777...) in the form p/q, where p and q are integers and q ≠ 0.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let
2. Multiply by 10 to shift non-repeating digit 4 before the decimal point:
3. Since one digit repeats, multiply Equation 2 by 10:
4. Subtracting Equation 2 from Equation 3:
100x − 10x = (47.7777...) − (4.7777...)
90x = 47 − 4 = 43
5. Solving for x:
x = 43 / 90.
Thus,
x = 0.4777... ...(Equation 1)2. Multiply by 10 to shift non-repeating digit 4 before the decimal point:
10x = 4.7777... ...(Equation 2)3. Since one digit repeats, multiply Equation 2 by 10:
100x = 47.7777... ...(Equation 3)4. Subtracting Equation 2 from Equation 3:
100x − 10x = (47.7777...) − (4.7777...)
90x = 47 − 4 = 43
5. Solving for x:
x = 43 / 90.
Thus,
0.47̄ = 43/90.
CBSE Board Marking Scheme 2026-27
Forming 100x = 47.77... and 10x = 4.77...1 Mark
Subtracting to get 90x = 43 ⇒ x = 43/901 Mark
Q4
Show that 1.272727... = 1.27̄ can be expressed in the form p/q, where p and q are integers and q ≠ 0.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let
2. Two digits repeat (27), so multiply Equation 1 by 100:
3. Subtracting Equation 1 from Equation 2:
100x − x = (127.2727...) − (1.2727...)
99x = 126
4. Solving for x and simplifying:
x = 126 / 99.
Dividing numerator and denominator by 9:
x = (126 ÷ 9) / (99 ÷ 9) = 14 / 11.
Hence,
x = 1.272727... ...(Equation 1)2. Two digits repeat (27), so multiply Equation 1 by 100:
100x = 127.272727... ...(Equation 2)3. Subtracting Equation 1 from Equation 2:
100x − x = (127.2727...) − (1.2727...)
99x = 126
4. Solving for x and simplifying:
x = 126 / 99.
Dividing numerator and denominator by 9:
x = (126 ÷ 9) / (99 ÷ 9) = 14 / 11.
Hence,
1.27̄ = 14/11.
CBSE Board Marking Scheme 2026-27
Multiplying by 100 to get 99x = 1261 Mark
Simplifying 126/99 to 14/111 Mark
Q5
Find five rational numbers between 2/3 and 4/5.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. First convert both fractions to have a common denominator:
LCM of denominators 3 and 5 is 15.
2/3 = (2 × 5) / (3 × 5) = 10/15.
4/5 = (4 × 3) / (5 × 3) = 12/15.
2. Between 10/15 and 12/15, there is only one integer numerator (11).
To insert 5 rational numbers, multiply numerator and denominator of both by (5 + 1) = 6 (or 4):
Multiply by 4:
10/15 = (10 × 4) / (15 × 4) = 40/60.
12/15 = (12 × 4) / (15 × 4) = 48/60.
3. The integers between 40 and 48 are 41, 42, 43, 44, 45, 46, 47.
Therefore, five rational numbers between 2/3 and 4/5 are:
41/60, 42/60, 43/60, 44/60, and 45/60
(or in lowest terms:
LCM of denominators 3 and 5 is 15.
2/3 = (2 × 5) / (3 × 5) = 10/15.
4/5 = (4 × 3) / (5 × 3) = 12/15.
2. Between 10/15 and 12/15, there is only one integer numerator (11).
To insert 5 rational numbers, multiply numerator and denominator of both by (5 + 1) = 6 (or 4):
Multiply by 4:
10/15 = (10 × 4) / (15 × 4) = 40/60.
12/15 = (12 × 4) / (15 × 4) = 48/60.
3. The integers between 40 and 48 are 41, 42, 43, 44, 45, 46, 47.
Therefore, five rational numbers between 2/3 and 4/5 are:
41/60, 42/60, 43/60, 44/60, and 45/60
(or in lowest terms:
41/60, 7/10, 43/60, 11/15, 3/4).
CBSE Board Marking Scheme 2026-27
Equating denominators to 15, then expanding to 601 Mark
Listing 5 distinct valid rational numbers1 Mark
Q6
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Theorem: The maximum number of digits in the repeating block of 1/q is always strictly less than the divisor, i.e., at most (q − 1) digits, because there are only (q − 1) possible non-zero remainders.
1. For 1/17, the divisor is 17. Hence, the maximum possible number of digits in the repeating block is
2. Performing long division of 1 by 17:
1 ÷ 17 = 0.0588235294117647...
Remainders sequence: 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12, 1 (remainder 1 repeats after 16 steps!).
Therefore,
17 − 1 = 16 digits.2. Performing long division of 1 by 17:
1 ÷ 17 = 0.0588235294117647...
Remainders sequence: 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12, 1 (remainder 1 repeats after 16 steps!).
Therefore,
1/17 = 0.0588235294117647... with a repeating period of exactly 16 digits.
CBSE Board Marking Scheme 2026-27
Stating max repeating digits = q − 1 = 161 Mark
Division verifying the exact 16-digit recurring sequence2 Marks
Q7
Find the sum: 0.3̄ + 0.4̄ + 0.5̄ by converting each term into fractional form p/q.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Convert each repeating decimal to p/q:
• Let x = 0.3̄ ⇒ 10x − x = 3 ⇒ 9x = 3 ⇒ x = 3/9.
• Let y = 0.4̄ ⇒ 10y − y = 4 ⇒ 9y = 4 ⇒ y = 4/9.
• Let z = 0.5̄ ⇒ 10z − z = 5 ⇒ 9z = 5 ⇒ z = 5/9.
2. Adding the fractions together:
Sum = 3/9 + 4/9 + 5/9 = (3 + 4 + 5) / 9 = 12 / 9.
3. Reducing to simplest form:
12/9 = 4 / 3 (or 1.3̄).
Hence, the sum is 4/3.
• Let x = 0.3̄ ⇒ 10x − x = 3 ⇒ 9x = 3 ⇒ x = 3/9.
• Let y = 0.4̄ ⇒ 10y − y = 4 ⇒ 9y = 4 ⇒ y = 4/9.
• Let z = 0.5̄ ⇒ 10z − z = 5 ⇒ 9z = 5 ⇒ z = 5/9.
2. Adding the fractions together:
Sum = 3/9 + 4/9 + 5/9 = (3 + 4 + 5) / 9 = 12 / 9.
3. Reducing to simplest form:
12/9 = 4 / 3 (or 1.3̄).
Hence, the sum is 4/3.
CBSE Board Marking Scheme 2026-27
Converting to 3/9, 4/9, 5/91 Mark
Summing and simplifying to 4/31 Mark
Q8
Represent the rational numbers −2/5 and 7/4 on the number line.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) Representing −2/5:
- Note that
- Divide the unit segment between 0 and −1 into 5 equal parts.
- Moving 2 parts to the left from 0 marks the point −2/5.
• (ii) Representing 7/4:
- Write as a mixed fraction:
- Divide the unit segment between 1 and 2 into 4 equal parts.
- The 3rd division mark to the right of 1 represents 1 3/4 = 7/4 (or the 7th mark from 0 when every unit interval is partitioned into 4 quarters).
- Note that
−1 < −2/5 < 0.- Divide the unit segment between 0 and −1 into 5 equal parts.
- Moving 2 parts to the left from 0 marks the point −2/5.
• (ii) Representing 7/4:
- Write as a mixed fraction:
7/4 = 1 + 3/4. It lies between 1 and 2.- Divide the unit segment between 1 and 2 into 4 equal parts.
- The 3rd division mark to the right of 1 represents 1 3/4 = 7/4 (or the 7th mark from 0 when every unit interval is partitioned into 4 quarters).
CBSE Board Marking Scheme 2026-27
Interval identification: (−1 to 0) and (1 to 2)1 Mark
Accurate partitioning and placement on number line1 Mark
Q9
Express 0.235̄ (i.e. 0.2353535...) in the form p/q, where p and q are integers and q ≠ 0.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let
2. Multiply by 10 to shift non-repeating digit 2 before decimal point:
3. Two digits repeat (35), so multiply Equation 2 by 100:
4. Subtracting Equation 2 from Equation 3:
1000x − 10x = (235.3535...) − (2.3535...)
990x = 235 − 2 = 233
5. Solving for x:
x = 233 / 990.
Hence,
x = 0.2353535... ...(Equation 1)2. Multiply by 10 to shift non-repeating digit 2 before decimal point:
10x = 2.353535... ...(Equation 2)3. Two digits repeat (35), so multiply Equation 2 by 100:
1000x = 235.353535... ...(Equation 3)4. Subtracting Equation 2 from Equation 3:
1000x − 10x = (235.3535...) − (2.3535...)
990x = 235 − 2 = 233
5. Solving for x:
x = 233 / 990.
Hence,
0.235̄ = 233/990.
CBSE Board Marking Scheme 2026-27
Setting 1000x = 235.35... and 10x = 2.35...1 Mark
990x = 233 ⇒ x = 233/9901 Mark
Q10
Find a rational number lying between 1/4 and 1/3 using the Mean Method. Verify that it lies strictly between the two numbers.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Mean Method Formula: If a and b are two rational numbers with a < b, then
2. Here a = 1/4, b = 1/3.
Sum = 1/4 + 1/3 = (3 + 4) / 12 = 7/12.
Mean = (7/12) ÷ 2 = 7 / 24.
3. Verification:
Convert all to common denominator 24:
1/4 = 6/24.
7/24 = 7/24.
1/3 = 8/24.
Clearly,
(a + b) / 2 is a rational number strictly between a and b.2. Here a = 1/4, b = 1/3.
Sum = 1/4 + 1/3 = (3 + 4) / 12 = 7/12.
Mean = (7/12) ÷ 2 = 7 / 24.
3. Verification:
Convert all to common denominator 24:
1/4 = 6/24.
7/24 = 7/24.
1/3 = 8/24.
Clearly,
6/24 < 7/24 < 8/24, which proves 1/4 < 7/24 < 1/3.
CBSE Board Marking Scheme 2026-27
Applying Mean formula: (1/4 + 1/3)/2 = 7/241 Mark
Verification: 6/24 < 7/24 < 8/241 Mark
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Precision Aerospace Engineering Tolerances: A CNC machine fabricates titanium fuel pins whose specified diameter is
Two sample test pins report diameters of
(a) Convert the design specification 5/16 inches into its decimal equivalent without long division, and explain why it terminates.
(b) Convert the measurement of the second pin
(c) Which pin is larger, and by what exact fraction of an inch?
5/16 inches. Quality control measures readings using optical micrometers that output decimal measurements.Two sample test pins report diameters of
0.3125 inches and 0.3181818... = 0.318̄ inches respectively.(a) Convert the design specification 5/16 inches into its decimal equivalent without long division, and explain why it terminates.
(b) Convert the measurement of the second pin
0.318̄ into simplest fraction form p/q.(c) Which pin is larger, and by what exact fraction of an inch?
Detailed Analytical Solution & Marking Scheme
(a) Decimal of 5/16:
16 = 2⁴. To convert to base 10 power without division, multiply numerator and denominator by 5⁴ = 625:
5/16 = (5 × 625) / (16 × 625) = 3125 / 10000 = 0.3125 inches.
It terminates because denominator 16 contains exclusively powers of 2 (2⁴ × 5⁰).
(b) Converting 0.318̄ to p/q form:
Let x = 0.3181818...
10x = 3.181818... ...(i)
1000x = 318.181818... ...(ii)
Subtracting (i) from (ii):
990x = 318 − 3 = 315.
x = 315 / 990 = 35 / 110 = 7 / 22 inches.
(c) Comparison and exact fractional difference:
- Pin 1 = 0.3125 = 5/16 inches.
- Pin 2 = 7/22 ≈ 0.31818... inches.
Clearly, Pin 2 is larger.
Difference = 7/22 − 5/16 = (7 × 8 − 5 × 11) / 176 = (56 − 55) / 176 = 1 / 176 inches.
9
Chapter 9: The Baudhāyana - Pythagoras Theorem
NCERT Ganita Prakash (Class 8 Part 2) — Śulbasūtra Geometric Formulation, Right-Angled Triangles, Pythagorean Triplets, Dissection Proofs & Applications | CBSE 2026-27
📌 Historical Heritage & Geometric Formulation
- Baudhāyana's Śulbasūtra Theorem (c. 800 BCE):
"The rope stretched along the length of the diagonal of a rectangle produces an area which the vertical and horizontal sides make together."
In modern algebraic notation:Diagonal² = Length² + Breadth². - Pythagoras Theorem: In any right-angled triangle, the square of the hypotenuse (the longest side opposite the right angle) is equal to the sum of the squares of the other two sides:
Hypotenuse² = Base² + Perpendicular²(orc² = a² + b²).
- Converse of the Theorem: If the lengths of the three sides of a triangle satisfy
a² + b² = c², then the triangle is necessarily a right-angled triangle, with the right angle located opposite side c. - Pythagorean Triplets: A set of three positive integers (a, b, c) satisfying
a² + b² = c².- Primitive Triplets:
(3, 4, 5),(5, 12, 13),(7, 24, 25),(8, 15, 17),(9, 40, 41),(11, 60, 61),(12, 35, 37),(20, 21, 29). - For any natural number m > 1:
(2m)² + (m² − 1)² = (m² + 1)²generates a Pythagorean triplet.
- Primitive Triplets:
Figure 9.1: Baudhāyana-Pythagorean Geometric Equivalence: Hypotenuse² = Base² + Altitude²
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
In a right-angled triangle ABC, right-angled at B, if AB = 6 cm and BC = 8 cm, find the length of AC.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 9.1: Right Triangle ABC with Legs 6 cm & 8 cm and Hypotenuse 10 cm
Baudhāyana-Pythagoras Theorem: In right-angled triangle ΔABC with ∠B = 90°, the hypotenuse is AC.
AC² = AB² + BC².
1. Substitute AB = 6 cm and BC = 8 cm:
AC² = 6² + 8²
AC² = 36 + 64 = 100.
2. Taking the positive square root:
AC = √100 = 10 cm.
Therefore, the length of AC is 10 cm.
AC² = 6² + 8²
AC² = 36 + 64 = 100.
2. Taking the positive square root:
AC = √100 = 10 cm.
Therefore, the length of AC is 10 cm.
CBSE Board Marking Scheme 2026-27
Applying AC² = AB² + BC²1 Mark
Calculating AC = 10 cm1 Mark
Q2
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 9.2: Ladder-Wall Geometric Model (15² = 12² + a² ⇒ a = 9 m)
1. The wall, ground, and ladder form a right-angled triangle where:
- Hypotenuse (length of ladder) = 15 m.
- Perpendicular (height of window) = 12 m.
- Base (distance from wall) =
2. By Pythagoras Theorem:
(Ladder)² = (Wall height)² + a²
15² = 12² + a²
225 = 144 + a²
3. Solving for a:
a² = 225 − 144 = 81
a = √81 = 9 meters.
Hence, the foot of the ladder is 9 meters away from the wall.
- Hypotenuse (length of ladder) = 15 m.
- Perpendicular (height of window) = 12 m.
- Base (distance from wall) =
a.2. By Pythagoras Theorem:
(Ladder)² = (Wall height)² + a²
15² = 12² + a²
225 = 144 + a²
3. Solving for a:
a² = 225 − 144 = 81
a = √81 = 9 meters.
Hence, the foot of the ladder is 9 meters away from the wall.
CBSE Board Marking Scheme 2026-27
Pythagoras equation 15² = 12² + a²1 Mark
Solving a² = 81 ⇒ a = 9 m1 Mark
Q3
Which of the following can be the sides of a right-angled triangle?
(i) 2.5 cm, 6.5 cm, 6 cm
(ii) 2 cm, 2 cm, 5 cm
(iii) 1.5 cm, 2 cm, 2.5 cm
(i) 2.5 cm, 6.5 cm, 6 cm
(ii) 2 cm, 2 cm, 5 cm
(iii) 1.5 cm, 2 cm, 2.5 cm
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Converse Theorem: Three side lengths form a right triangle if and only if the square of the longest side equals the sum of the squares of the two smaller sides.
• (i) 2.5 cm, 6.5 cm, 6 cm:
Longest side = 6.5 cm ⇒ 6.5² = 42.25.
Sum of squares of other two sides = 2.5² + 6² = 6.25 + 36 = 42.25.
Since 2.5² + 6² = 6.5², Yes, they form a right-angled triangle (right angle is opposite the 6.5 cm side).
• (ii) 2 cm, 2 cm, 5 cm:
First check triangle inequality: 2 + 2 = 4 < 5 (Sum of two sides is less than the third side).
This cannot even form a valid triangle, let alone a right triangle. No.
• (iii) 1.5 cm, 2 cm, 2.5 cm:
Longest side = 2.5 cm ⇒ 2.5² = 6.25.
Sum of squares of other sides = 1.5² + 2² = 2.25 + 4 = 6.25.
Since 1.5² + 2² = 2.5², Yes, they form a right-angled triangle.
Longest side = 6.5 cm ⇒ 6.5² = 42.25.
Sum of squares of other two sides = 2.5² + 6² = 6.25 + 36 = 42.25.
Since 2.5² + 6² = 6.5², Yes, they form a right-angled triangle (right angle is opposite the 6.5 cm side).
• (ii) 2 cm, 2 cm, 5 cm:
First check triangle inequality: 2 + 2 = 4 < 5 (Sum of two sides is less than the third side).
This cannot even form a valid triangle, let alone a right triangle. No.
• (iii) 1.5 cm, 2 cm, 2.5 cm:
Longest side = 2.5 cm ⇒ 2.5² = 6.25.
Sum of squares of other sides = 1.5² + 2² = 2.25 + 4 = 6.25.
Since 1.5² + 2² = 2.5², Yes, they form a right-angled triangle.
CBSE Board Marking Scheme 2026-27
Checking a² + b² = c² condition for each1.5 Marks
Accurate conclusions: (i) Yes, (ii) No, (iii) Yes1.5 Marks
Q4
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 9.3: Right Triangle Formed by Broken Tree Top (5² + 12² = 13²)
1. Let AB be the upright standing portion of the tree = 5 m.
Let C be the point on the ground where the top touches, so BC = 12 m.
Let AC be the broken bent part of the tree (hypotenuse).
2. By Pythagoras Theorem in right triangle ΔABC:
AC² = AB² + BC²
AC² = 5² + 12² = 25 + 144 = 169.
AC = √169 = 13 meters.
3. The original height of the tree before breaking was:
Hence, the original height of the tree was 18 meters.
Let C be the point on the ground where the top touches, so BC = 12 m.
Let AC be the broken bent part of the tree (hypotenuse).
2. By Pythagoras Theorem in right triangle ΔABC:
AC² = AB² + BC²
AC² = 5² + 12² = 25 + 144 = 169.
AC = √169 = 13 meters.
3. The original height of the tree before breaking was:
Original Height = AB + AC = 5 + 13 = 18 meters.Hence, the original height of the tree was 18 meters.
CBSE Board Marking Scheme 2026-27
Calculating broken part AC = √169 = 13 m1 Mark
Total original height = 5 + 13 = 18 m1 Mark
Q5
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Baudhāyana's Rectangle Rule: Diagonal² = Length² + Breadth².
1. Let breadth of the rectangle be
Diagonal = 41 cm, Length = 40 cm.
41² = 40² + b²
1681 = 1600 + b²
b² = 1681 − 1600 = 81
b = √81 = 9 cm.
2. Perimeter of the rectangle:
Perimeter = 2 × (Length + Breadth)
Perimeter = 2 × (40 + 9) = 2 × 49 = 98 cm.
Therefore, the perimeter of the rectangle is 98 cm.
b cm.Diagonal = 41 cm, Length = 40 cm.
41² = 40² + b²
1681 = 1600 + b²
b² = 1681 − 1600 = 81
b = √81 = 9 cm.
2. Perimeter of the rectangle:
Perimeter = 2 × (Length + Breadth)
Perimeter = 2 × (40 + 9) = 2 × 49 = 98 cm.
Therefore, the perimeter of the rectangle is 98 cm.
CBSE Board Marking Scheme 2026-27
Breadth calculation: b = √81 = 9 cm1 Mark
Perimeter = 2(40 + 9) = 98 cm1 Mark
Q6
Find the length of the diagonal of a square whose side is 10 cm. If the diagonal of another square is 16 cm, find its side and area.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Square Diagonal Property: For a square of side s, the diagonal d satisfies d² = s² + s² = 2s² ⇒ d = s√2.
• Part 1: Square with side s = 10 cm:
d² = 10² + 10² = 100 + 100 = 200.
d = √200 = √(100 × 2) = 10√2 cm (≈ 14.14 cm).
• Part 2: Square with diagonal d = 16 cm:
2s² = d² = 16² = 256.
s² = 256 / 2 = 128.
Side s = √128 = √(64 × 2) = 8√2 cm (≈ 11.31 cm).
Area of the square = s² = 128 cm²
(Direct formula: Area = d² / 2 = 16² / 2 = 256 / 2 = 128 cm²).
d² = 10² + 10² = 100 + 100 = 200.
d = √200 = √(100 × 2) = 10√2 cm (≈ 14.14 cm).
• Part 2: Square with diagonal d = 16 cm:
2s² = d² = 16² = 256.
s² = 256 / 2 = 128.
Side s = √128 = √(64 × 2) = 8√2 cm (≈ 11.31 cm).
Area of the square = s² = 128 cm²
(Direct formula: Area = d² / 2 = 16² / 2 = 256 / 2 = 128 cm²).
CBSE Board Marking Scheme 2026-27
Diagonal of first square = 10√2 cm1 Mark
Side = 8√2 cm and Area = 128 cm²2 Marks
Q7
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 9.4: Geometric Resolution using Right Triangle with Legs 12 m & 5 m
1. Let AB = 11 m and CD = 6 m be the two vertical poles.
Distance between feet BD = 12 m.
Draw a horizontal line segment CE perpendicular to AB.
2. Then CE = BD = 12 m, and EB = CD = 6 m.
Vertical difference in height AE = AB − EB = 11 − 6 = 5 m.
3. In right-angled triangle ΔAEC, the distance between the tops is the hypotenuse AC:
AC² = AE² + CE²
AC² = 5² + 12² = 25 + 144 = 169.
AC = √169 = 13 meters.
Therefore, the distance between their tops is 13 meters.
Distance between feet BD = 12 m.
Draw a horizontal line segment CE perpendicular to AB.
2. Then CE = BD = 12 m, and EB = CD = 6 m.
Vertical difference in height AE = AB − EB = 11 − 6 = 5 m.
3. In right-angled triangle ΔAEC, the distance between the tops is the hypotenuse AC:
AC² = AE² + CE²
AC² = 5² + 12² = 25 + 144 = 169.
AC = √169 = 13 meters.
Therefore, the distance between their tops is 13 meters.
CBSE Board Marking Scheme 2026-27
Geometric construction: AE = 5 m, CE = 12 m1 Mark
Pythagoras theorem: AC = √169 = 13 m1 Mark
Q8
A person travels 24 km due East and then 10 km due North. How far is he from his starting point?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 9.5: Distance from Start Point via Baudhāyana-Pythagoras: √(24² + 10²) = 26 km
1. East and North directions are perpendicular to each other (angle = 90°).
Let starting point be O.
- Distance East OA = 24 km.
- Distance North AB = 10 km.
The direct displacement from O to B is the hypotenuse OB.
2. By Baudhāyana-Pythagoras Theorem:
OB² = OA² + AB²
OB² = 24² + 10² = 576 + 100 = 676.
3. Taking the square root:
OB = √676 = 26 km.
Hence, he is 26 km away from his starting point.
Let starting point be O.
- Distance East OA = 24 km.
- Distance North AB = 10 km.
The direct displacement from O to B is the hypotenuse OB.
2. By Baudhāyana-Pythagoras Theorem:
OB² = OA² + AB²
OB² = 24² + 10² = 576 + 100 = 676.
3. Taking the square root:
OB = √676 = 26 km.
Hence, he is 26 km away from his starting point.
CBSE Board Marking Scheme 2026-27
Perpendicular directions model: OB² = 24² + 10²1 Mark
OB = √676 = 26 km1 Mark
Q9
An isosceles triangle ABC has AB = AC = 13 cm and base BC = 10 cm. Find the altitude from vertex A to side BC and calculate the area of triangle ABC.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Property: In an isosceles triangle, the altitude from the vertex angle bisects the base.
1. Let AD be the altitude from A perpendicular to BC.
Then D is the midpoint of BC ⇒ BD = BC / 2 = 10 / 2 = 5 cm.
2. In right-angled triangle ΔABD:
AB² = AD² + BD²
13² = AD² + 5²
169 = AD² + 25
AD² = 169 − 25 = 144
AD = √144 = 12 cm.
3. Area of triangle ABC:
Area = 1/2 × Base × Height = 1/2 × BC × AD
Area = 1/2 × 10 × 12 = 60 cm².
Then D is the midpoint of BC ⇒ BD = BC / 2 = 10 / 2 = 5 cm.
2. In right-angled triangle ΔABD:
AB² = AD² + BD²
13² = AD² + 5²
169 = AD² + 25
AD² = 169 − 25 = 144
AD = √144 = 12 cm.
3. Area of triangle ABC:
Area = 1/2 × Base × Height = 1/2 × BC × AD
Area = 1/2 × 10 × 12 = 60 cm².
CBSE Board Marking Scheme 2026-27
Altitude bisects base: BD = 5 cm, AD = 12 cm1 Mark
Area = 1/2 × 10 × 12 = 60 cm²1 Mark
Q10
Generate three distinct Pythagorean triplets using the algebraic identity (2m, m² − 1, m² + 1) for m = 3, 4, and 5. Verify the Pythagorean equation for each.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) For m = 3:
2m = 2(3) = 6.
m² − 1 = 3² − 1 = 9 − 1 = 8.
m² + 1 = 3² + 1 = 9 + 1 = 10.
Triplet: (6, 8, 10).
Verification: 6² + 8² = 36 + 64 = 100 = 10². Verified.
• (ii) For m = 4:
2m = 2(4) = 8.
m² − 1 = 4² − 1 = 16 − 1 = 15.
m² + 1 = 4² + 1 = 16 + 1 = 17.
Triplet: (8, 15, 17).
Verification: 8² + 15² = 64 + 225 = 289 = 17². Verified.
• (iii) For m = 5:
2m = 2(5) = 10.
m² − 1 = 5² − 1 = 25 − 1 = 24.
m² + 1 = 5² + 1 = 25 + 1 = 26.
Triplet: (10, 24, 26).
Verification: 10² + 24² = 100 + 576 = 676 = 26². Verified.
2m = 2(3) = 6.
m² − 1 = 3² − 1 = 9 − 1 = 8.
m² + 1 = 3² + 1 = 9 + 1 = 10.
Triplet: (6, 8, 10).
Verification: 6² + 8² = 36 + 64 = 100 = 10². Verified.
• (ii) For m = 4:
2m = 2(4) = 8.
m² − 1 = 4² − 1 = 16 − 1 = 15.
m² + 1 = 4² + 1 = 16 + 1 = 17.
Triplet: (8, 15, 17).
Verification: 8² + 15² = 64 + 225 = 289 = 17². Verified.
• (iii) For m = 5:
2m = 2(5) = 10.
m² − 1 = 5² − 1 = 25 − 1 = 24.
m² + 1 = 5² + 1 = 25 + 1 = 26.
Triplet: (10, 24, 26).
Verification: 10² + 24² = 100 + 576 = 676 = 26². Verified.
CBSE Board Marking Scheme 2026-27
Applying 2m, m² − 1, m² + 1 formulas1.5 Marks
Verification of all 3 triplets1.5 Marks
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Civil Infrastructure & Cable-Stayed Bridge Construction: In a river bridge project, a central vertical pylon tower rises 36 meters above the bridge road deck. Two steel stay cables are anchored from the top of the tower to two points on the road deck, one on the North side and one on the South side.
The North anchoring point is 27 meters away from the base of the pylon, while the South anchoring point is 48 meters away from the base.
(a) Calculate the length of the North stay cable.
(b) Calculate the length of the South stay cable.
(c) Find the total length of steel cable used for both stays, and determine the direct horizontal distance between the two cable anchor points on the bridge deck.
The North anchoring point is 27 meters away from the base of the pylon, while the South anchoring point is 48 meters away from the base.
(a) Calculate the length of the North stay cable.
(b) Calculate the length of the South stay cable.
(c) Find the total length of steel cable used for both stays, and determine the direct horizontal distance between the two cable anchor points on the bridge deck.
Detailed Analytical Solution & Marking Scheme
(a) Length of North stay cable (L_N):
Tower height h = 36 m, North base distance d_N = 27 m.
L_N² = 36² + 27² = 1296 + 729 = 2025.
L_N = √2025 = 45 meters.
(b) Length of South stay cable (L_S):
Tower height h = 36 m, South base distance d_S = 48 m.
L_S² = 36² + 48² = 1296 + 2304 = 3600.
L_S = √3600 = 60 meters.
(c) Total cable length & Anchor distance:
- Total steel cable length = 45 + 60 = 105 meters.
- Since North and South anchors lie on opposite sides of the pylon base along the road deck:
Direct horizontal distance between anchors = d_N + d_S = 27 + 48 = 75 meters.
10
Chapter 10: Proportional Reasoning - 2
NCERT Ganita Prakash (Class 8 Part 2) — Commercial Mathematics, Percentages, Profit & Loss, Discount, GST & Compound Interest | CBSE 2026-27
📌 Key Commercial Formulas & Financial Mathematics
- Discount & Marked Price:
Discount = Marked Price (MP) − Selling Price (SP)Discount % = (Discount / MP) × 100SP = MP × (1 − Discount% / 100)
- Profit, Loss & GST:
Profit = SP − CP(if SP > CP);Profit % = (Profit / CP) × 100Loss = CP − SP(if CP > SP);Loss % = (Loss / CP) × 100SP = CP × (100 ± Profit%/Loss%) / 100- Goods and Services Tax (GST): Levied on the selling price.
Bill Amount = SP + (GST% of SP).
- Compound Interest (CI): Interest calculated on the initial principal plus accumulated interest of prior periods.
- Compounded Annually:
A = P(1 + R/100)ⁿ, where P = Principal, R = Annual Rate %, n = Time in years. - Compounded Half-Yearly: Rate is halved (R/2) and periods double (2n):
A = P(1 + R/200)²ⁿ. Compound Interest (CI) = Amount (A) − Principal (P).- Depreciation: When value decreases at rate R% per year:
Value = Initial Value × (1 − R/100)ⁿ.
- Compounded Annually:
Figure 10.1: Compounding Effect Multiplying Wealth Over Extended Investment Horizons
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
An item marked at ₹ 840 is sold for ₹ 714. What are the discount and discount percentage?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Marked Price (MP) = ₹ 840.
Selling Price (SP) = ₹ 714.
2. Discount:
Discount = MP − SP = 840 − 714 = ₹ 126.
3. Discount Percentage:
Discount % = (Discount / MP) × 100
Discount % = (126 / 840) × 100 = 1260 / 84 = 15%.
Thus, the discount is ₹ 126 and the discount percentage is 15%.
Selling Price (SP) = ₹ 714.
2. Discount:
Discount = MP − SP = 840 − 714 = ₹ 126.
3. Discount Percentage:
Discount % = (Discount / MP) × 100
Discount % = (126 / 840) × 100 = 1260 / 84 = 15%.
Thus, the discount is ₹ 126 and the discount percentage is 15%.
CBSE Board Marking Scheme 2026-27
Calculating Discount = ₹ 1261 Mark
Formula and solving Discount% = 15%1 Mark
Q2
A shopkeeper bought two TV sets at ₹ 10,000 each. He sold one at a profit of 10% and the other at a loss of 10%. Find whether he made an overall profit or loss, and its percentage.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Total Cost Price (CP):
CP of 1st TV = ₹ 10,000.
CP of 2nd TV = ₹ 10,000.
Total CP = 10,000 + 10,000 = ₹ 20,000.
2. Selling Price of TV 1 (Profit = 10%):
SP₁ = 10,000 + (10/100 × 10,000) = 10,000 + 1,000 = ₹ 11,000.
3. Selling Price of TV 2 (Loss = 10%):
SP₂ = 10,000 − (10/100 × 10,000) = 10,000 − 1,000 = ₹ 9,000.
4. Total Selling Price (SP):
Total SP = 11,000 + 9,000 = ₹ 20,000.
5. Conclusion: Since Total SP = Total CP = ₹ 20,000, the shopkeeper made Neither profit nor loss (0%).
CP of 1st TV = ₹ 10,000.
CP of 2nd TV = ₹ 10,000.
Total CP = 10,000 + 10,000 = ₹ 20,000.
2. Selling Price of TV 1 (Profit = 10%):
SP₁ = 10,000 + (10/100 × 10,000) = 10,000 + 1,000 = ₹ 11,000.
3. Selling Price of TV 2 (Loss = 10%):
SP₂ = 10,000 − (10/100 × 10,000) = 10,000 − 1,000 = ₹ 9,000.
4. Total Selling Price (SP):
Total SP = 11,000 + 9,000 = ₹ 20,000.
5. Conclusion: Since Total SP = Total CP = ₹ 20,000, the shopkeeper made Neither profit nor loss (0%).
CBSE Board Marking Scheme 2026-27
Calculating SP₁ = ₹ 11,000 and SP₂ = ₹ 9,0001.5 Marks
Total SP = Total CP = ₹ 20,000 ⇒ No profit no loss1.5 Marks
Q3
A washing machine is bought for ₹ 18,900 including 12% GST. Find the price of the washing machine before GST was added.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let the original price of the washing machine before GST be
GST Rate = 12%.
Total Bill = x + (12% of x) = x + 0.12x = 1.12x.
2. Equating to given invoice amount:
1.12x = 18,900
x = 18,900 / 1.12 = 18,90,000 / 112 = ₹ 16,875.
3. (Check: 12% of 16,875 = ₹ 2,025. Total = 16,875 + 2,025 = ₹ 18,900).
Therefore, the price before GST was ₹ 16,875.
₹ x.GST Rate = 12%.
Total Bill = x + (12% of x) = x + 0.12x = 1.12x.
2. Equating to given invoice amount:
1.12x = 18,900
x = 18,900 / 1.12 = 18,90,000 / 112 = ₹ 16,875.
3. (Check: 12% of 16,875 = ₹ 2,025. Total = 16,875 + 2,025 = ₹ 18,900).
Therefore, the price before GST was ₹ 16,875.
CBSE Board Marking Scheme 2026-27
Equation setup: 1.12x = 18,9001 Mark
Solving x = ₹ 16,8751 Mark
Q4
Calculate the amount and compound interest on ₹ 10,800 for 3 years at 12 1/2% per annum compounded annually.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Given: Principal P = ₹ 10,800, Time n = 3 years, Rate R = 12.5% = 25/2% per annum.
2. Compound Interest Formula:
A = 10800 × [1 + (25 / 200)]³
A = 10800 × [1 + 1/8]³ = 10800 × (9/8)³
A = 10800 × (729 / 512)
A = (10800 × 729) / 512 = 7873200 / 512 = ₹ 15,377.34.
3. Compound Interest (CI):
CI = A − P = 15,377.34 − 10,800 = ₹ 4,577.34.
Hence, Amount = ₹ 15,377.34 and Compound Interest = ₹ 4,577.34.
2. Compound Interest Formula:
A = P(1 + R/100)ⁿA = 10800 × [1 + (25 / 200)]³
A = 10800 × [1 + 1/8]³ = 10800 × (9/8)³
A = 10800 × (729 / 512)
A = (10800 × 729) / 512 = 7873200 / 512 = ₹ 15,377.34.
3. Compound Interest (CI):
CI = A − P = 15,377.34 − 10,800 = ₹ 4,577.34.
Hence, Amount = ₹ 15,377.34 and Compound Interest = ₹ 4,577.34.
CBSE Board Marking Scheme 2026-27
Formula A = P(1 + R/100)ⁿ with substitution (9/8)³1.5 Marks
Amount = ₹ 15,377.34 and CI = ₹ 4,577.341.5 Marks
Q5
Find the amount and compound interest on ₹ 8,000 for 1 1/2 years at 10% per annum compounded half-yearly.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Rule for Half-Yearly Compounding: Rate per half-year = R / 2; Number of half-years = 2 × n.
1. P = ₹ 8,000.
Half-yearly rate r = 10% / 2 = 5%.
Periods m = 1.5 × 2 = 3 half-years.
2. Amount Formula:
A = 8000 × (1 + 5/100)³ = 8000 × (21/20)³
A = 8000 × (9261 / 8000) = ₹ 9,261.
3. Compound Interest (CI):
CI = A − P = 9,261 − 8,000 = ₹ 1,261.
Thus, Amount is ₹ 9,261 and Compound Interest is ₹ 1,261.
Half-yearly rate r = 10% / 2 = 5%.
Periods m = 1.5 × 2 = 3 half-years.
2. Amount Formula:
A = P(1 + r/100)ᵐA = 8000 × (1 + 5/100)³ = 8000 × (21/20)³
A = 8000 × (9261 / 8000) = ₹ 9,261.
3. Compound Interest (CI):
CI = A − P = 9,261 − 8,000 = ₹ 1,261.
Thus, Amount is ₹ 9,261 and Compound Interest is ₹ 1,261.
CBSE Board Marking Scheme 2026-27
Adjusting rate to 5% and periods to 3 half-years1 Mark
Evaluating A = ₹ 9,261 and CI = ₹ 1,2612 Marks
Q6
Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• Fabina's Simple Interest:
P = ₹ 12,500, R = 12%, T = 3 years.
SI = (P × R × T) / 100 = (12500 × 12 × 3) / 100 = 125 × 36 = ₹ 4,500.
• Radha's Compound Interest:
P = ₹ 12,500, R = 10%, n = 3 years.
A = P(1 + R/100)ⁿ = 12500 × (1 + 10/100)³ = 12500 × (11/10)³
A = 12500 × (1331 / 1000) = 12.5 × 1331 = ₹ 16,637.50.
CI = A − P = 16,637.50 − 12,500 = ₹ 4,137.50.
• Comparison:
Fabina pays more interest than Radha.
Difference = 4,500 − 4,137.50 = ₹ 362.50.
Hence, Fabina pays ₹ 362.50 more.
P = ₹ 12,500, R = 12%, T = 3 years.
SI = (P × R × T) / 100 = (12500 × 12 × 3) / 100 = 125 × 36 = ₹ 4,500.
• Radha's Compound Interest:
P = ₹ 12,500, R = 10%, n = 3 years.
A = P(1 + R/100)ⁿ = 12500 × (1 + 10/100)³ = 12500 × (11/10)³
A = 12500 × (1331 / 1000) = 12.5 × 1331 = ₹ 16,637.50.
CI = A − P = 16,637.50 − 12,500 = ₹ 4,137.50.
• Comparison:
Fabina pays more interest than Radha.
Difference = 4,500 − 4,137.50 = ₹ 362.50.
Hence, Fabina pays ₹ 362.50 more.
CBSE Board Marking Scheme 2026-27
Calculating Fabina's SI = ₹ 4,5001 Mark
Calculating Radha's CI = ₹ 4,137.501 Mark
Difference = ₹ 362.50 paid more by Fabina1 Mark
Q7
A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Depreciation Formula: Value after n years = Initial Value × (1 − R/100)ⁿ.
1. Initial Value P = ₹ 42,000.
Depreciation Rate R = 8%, Time n = 1 year.
2. Value after 1 year = 42000 × (1 − 8/100)¹
= 42000 × (92 / 100)
= 420 × 92 = ₹ 38,640.
(Alternative: Depreciation = 8% of 42000 = ₹ 3,360. Value = 42000 − 3360 = ₹ 38,640).
Hence, the value of the scooter after one year is ₹ 38,640.
Depreciation Rate R = 8%, Time n = 1 year.
2. Value after 1 year = 42000 × (1 − 8/100)¹
= 42000 × (92 / 100)
= 420 × 92 = ₹ 38,640.
(Alternative: Depreciation = 8% of 42000 = ₹ 3,360. Value = 42000 − 3360 = ₹ 38,640).
Hence, the value of the scooter after one year is ₹ 38,640.
CBSE Board Marking Scheme 2026-27
Depreciation calculation: 8% of 42000 = ₹ 3,3601 Mark
Remaining value = ₹ 38,6401 Mark
Q8
The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Initial Population P = 20,000.
Growth Rate R = 5% per annum.
Time duration n = 2000 − 1997 = 3 years.
2. Population in 2000 =
= 20000 × (1 + 5/100)³
= 20000 × (21/20)³
= 20000 × (9261 / 8000)
= (20 × 9261) / 8 = 185220 / 8 = 23,152.5 ≈ 23,153.
Therefore, the estimated population at the end of 2000 was 23,153.
Growth Rate R = 5% per annum.
Time duration n = 2000 − 1997 = 3 years.
2. Population in 2000 =
P × (1 + R/100)ⁿ= 20000 × (1 + 5/100)³
= 20000 × (21/20)³
= 20000 × (9261 / 8000)
= (20 × 9261) / 8 = 185220 / 8 = 23,152.5 ≈ 23,153.
Therefore, the estimated population at the end of 2000 was 23,153.
CBSE Board Marking Scheme 2026-27
Population growth formula with n = 3 years1 Mark
Calculating 23,152.5 ≈ 23,1531 Mark
Q9
A VCR and TV were bought for ₹ 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Total Cost Price CP = 8,000 + 8,000 = ₹ 16,000.
2. SP of VCR (4% loss):
Loss = 4% of 8,000 = ₹ 320.
SP of VCR = 8,000 − 320 = ₹ 7,680.
3. SP of TV (8% profit):
Profit = 8% of 8,000 = ₹ 640.
SP of TV = 8,000 + 640 = ₹ 8,640.
4. Total Selling Price SP = 7,680 + 8,640 = ₹ 16,320.
5. Overall Profit = Total SP − Total CP = 16,320 − 16,000 = ₹ 320.
Profit % = (320 / 16000) × 100 = 32 / 16 = 2%.
Hence, the shopkeeper gained 2% profit on the whole transaction.
2. SP of VCR (4% loss):
Loss = 4% of 8,000 = ₹ 320.
SP of VCR = 8,000 − 320 = ₹ 7,680.
3. SP of TV (8% profit):
Profit = 8% of 8,000 = ₹ 640.
SP of TV = 8,000 + 640 = ₹ 8,640.
4. Total Selling Price SP = 7,680 + 8,640 = ₹ 16,320.
5. Overall Profit = Total SP − Total CP = 16,320 − 16,000 = ₹ 320.
Profit % = (320 / 16000) × 100 = 32 / 16 = 2%.
Hence, the shopkeeper gained 2% profit on the whole transaction.
CBSE Board Marking Scheme 2026-27
SP of VCR (₹ 7,680) and SP of TV (₹ 8,640)1.5 Marks
Total Profit ₹ 320 and Profit% = 2%1.5 Marks
Q10
Find the difference between Simple Interest and Compound Interest on ₹ 15,000 for 2 years at 8% per annum compounded annually. State the general algebraic shortcut for difference over 2 years.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Two-Year Shortcut Formula: Difference (CI − SI) = P × (R / 100)².
1. P = ₹ 15,000, R = 8%, n = 2 years.
2. Using the direct formula:
CI − SI = 15000 × (8 / 100)²
= 15000 × (64 / 10000)
= (15000 × 64) / 10000 = 1.5 × 64 = ₹ 96.
3. Verification:
SI = (15000 × 8 × 2) / 100 = ₹ 2,400.
A = 15000 × (1.08)² = 15000 × 1.1664 = ₹ 17,496.
CI = 17,496 − 15,000 = ₹ 2,496.
Difference = 2,496 − 2,400 = ₹ 96.
2. Using the direct formula:
CI − SI = 15000 × (8 / 100)²
= 15000 × (64 / 10000)
= (15000 × 64) / 10000 = 1.5 × 64 = ₹ 96.
3. Verification:
SI = (15000 × 8 × 2) / 100 = ₹ 2,400.
A = 15000 × (1.08)² = 15000 × 1.1664 = ₹ 17,496.
CI = 17,496 − 15,000 = ₹ 2,496.
Difference = 2,496 − 2,400 = ₹ 96.
CBSE Board Marking Scheme 2026-27
Formula Difference = P(R/100)²1 Mark
Evaluating Difference = ₹ 961 Mark
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Entrepreneurial Capital & Retail Pricing Dynamics: An electronics retail startup buys laptops from a manufacturer at a cost price (CP) of ₹ 50,000 each. The store manager sets the Marked Price (MP) 20% above the cost price. During a festival sale, the store offers a discount of 10% on the marked price. When a customer purchases the laptop, 18% GST is charged on the final discounted selling price.
(a) Calculate the Marked Price (MP) and the discounted Selling Price (SP) of the laptop.
(b) Calculate the final bill amount paid by the customer including 18% GST.
(c) Find the shopkeeper's actual profit percentage on the transaction (excluding tax).
(a) Calculate the Marked Price (MP) and the discounted Selling Price (SP) of the laptop.
(b) Calculate the final bill amount paid by the customer including 18% GST.
(c) Find the shopkeeper's actual profit percentage on the transaction (excluding tax).
Detailed Analytical Solution & Marking Scheme
(a) Marked Price and Discounted Selling Price:
- MP = CP + 20% of CP = 50,000 + 10,000 = ₹ 60,000.
- Discount = 10% of MP = 10% of 60,000 = ₹ 6,000.
- Selling Price SP = 60,000 − 6,000 = ₹ 54,000.
(b) Customer Final Invoice with 18% GST:
- GST = 18% of SP = 0.18 × 54,000 = ₹ 9,720.
- Bill Amount = SP + GST = 54,000 + 9,720 = ₹ 63,720.
(c) Shopkeeper's Profit Percentage:
- Profit = SP − CP = 54,000 − 50,000 = ₹ 4,000.
- Profit % = (4,000 / 50,000) × 100 = 8% profit.
11
Chapter 11: Exploring Some Geometric Themes
NCERT Ganita Prakash (Class 8 Part 2) — Quadrilateral Constructions, 5-Measurement Criteria, Compass & Straightedge Protocols & Tessellations | CBSE 2026-27
📌 Necessary Conditions for Unique Geometric Construction
- 5-Measurement Principle: A unique quadrilateral cannot be constructed with just 4 sides (it can flex into infinitely many shapes). Exactly 5 independent measurements are required to fix a unique quadrilateral.
- Standard Construction Scenarios:
- When four sides and one diagonal are given (splits into two unique triangles).
- When two diagonals and three sides are given.
- When two adjacent sides and three angles are given.
- When three sides and two included angles are given.
- Special Constructions (exploiting inherent geometric symmetry):
- Square: Requires only 1 parameter (side length or diagonal length), because all sides are equal and all angles are 90°.
- Rhombus: Requires only 2 parameters (lengths of its two diagonals, since they bisect perpendicularly).
- Rectangle: Requires only 2 parameters (length and breadth).
- Tessellations & Angle Sum at a Vertex: A plane can be tiled seamlessly without gaps or overlaps by a regular polygon if and only if its interior angle divides 360° evenly. Only equilateral triangles (60°), squares (90°), and regular hexagons (120°) form regular tessellations.
Figure 11.1: Triangulation Protocol: Partitioning Quadrilateral ABCD into ΔABC and ΔADC
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
Construct a quadrilateral ABCD where AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm, DA = 6 cm, and diagonal AC = 7 cm. Write the step-by-step algorithm of construction.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Geometric Strategy: The diagonal AC divides quadrilateral ABCD into two triangles, ΔABC and ΔADC. Since all three sides of both triangles are known, both triangles can be constructed uniquely using SSS criterion.
Step-by-Step Steps of Construction:
1. Base Diagonal: Draw line segment
2. Locating Vertex B: With A as center and radius
3. Join AB and BC.
4. Locating Vertex D: With A as center and radius
5. Join AD and CD.
6. ABCD is the required quadrilateral.
1. Base Diagonal: Draw line segment
AC = 7 cm using a ruler.2. Locating Vertex B: With A as center and radius
4.5 cm, draw an arc above AC. With C as center and radius 5.5 cm, draw another arc intersecting the previous arc at point B.3. Join AB and BC.
4. Locating Vertex D: With A as center and radius
6 cm, draw an arc below AC. With C as center and radius 4 cm, draw another arc intersecting the previous arc at point D.5. Join AD and CD.
6. ABCD is the required quadrilateral.
CBSE Board Marking Scheme 2026-27
Drawing diagonal AC = 7 cm and constructing ΔABC1.5 Marks
Constructing ΔADC and completing quadrilateral ABCD1.5 Marks
Q2
Construct a quadrilateral JUMP where JU = 3.5 cm, UM = 4 cm, MP = 5 cm, PJ = 4.5 cm, and PU = 6.5 cm. Detail the steps of construction.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Steps of Construction:
1. Draw the base diagonal
2. In ΔJPU, with P as center and radius 4.5 cm, draw an arc. With U as center and radius 3.5 cm, draw another arc intersecting the first at J.
3. Join PJ and UJ to complete ΔJPU.
4. On the other side of PU, with P as center and radius 5 cm, draw an arc. With U as center and radius 4 cm, draw an arc intersecting it at M.
5. Join PM and UM.
6. JUMP is the required quadrilateral.
1. Draw the base diagonal
PU = 6.5 cm.2. In ΔJPU, with P as center and radius 4.5 cm, draw an arc. With U as center and radius 3.5 cm, draw another arc intersecting the first at J.
3. Join PJ and UJ to complete ΔJPU.
4. On the other side of PU, with P as center and radius 5 cm, draw an arc. With U as center and radius 4 cm, draw an arc intersecting it at M.
5. Join PM and UM.
6. JUMP is the required quadrilateral.
CBSE Board Marking Scheme 2026-27
Drawing diagonal PU = 6.5 cm and locating J1.5 Marks
Locating M and joining boundary segments1.5 Marks
Q3
Construct a rhombus BEND whose diagonals are BN = 5.6 cm and DE = 6.5 cm. Explain why only 2 measurements are sufficient to construct this quadrilateral uniquely.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Geometric Invariant: The diagonals of a rhombus are perpendicular bisectors of each other. This intrinsic property provides 3 implicit constraints (90° intersection, and two midpoints), so knowing just the 2 diagonal lengths fully determines the 5 independent parameters needed for a unique quadrilateral.
Steps of Construction:
1. Draw line segment
2. Draw the perpendicular bisector of DE using compass arcs. Let this line intersect DE at midpoint O (where OE = OD = 3.25 cm).
3. Since diagonal BN = 5.6 cm, each half is
4. With O as center and radius 2.8 cm, cut arcs on the perpendicular bisector on both sides to locate vertices B and N.
5. Join BD, BE, ND, and NE.
6. BEND is the required rhombus.
1. Draw line segment
DE = 6.5 cm.2. Draw the perpendicular bisector of DE using compass arcs. Let this line intersect DE at midpoint O (where OE = OD = 3.25 cm).
3. Since diagonal BN = 5.6 cm, each half is
5.6 / 2 = 2.8 cm.4. With O as center and radius 2.8 cm, cut arcs on the perpendicular bisector on both sides to locate vertices B and N.
5. Join BD, BE, ND, and NE.
6. BEND is the required rhombus.
CBSE Board Marking Scheme 2026-27
Justification: Diagonals are perpendicular bisectors1 Mark
Accurate construction steps with 2.8 cm radii2 Marks
Q4
Construct a quadrilateral MIST where MI = 3.5 cm, IS = 6.5 cm, ∠M = 75°, ∠I = 105°, and ∠S = 120°. Verify the fourth angle ∠T using the angle sum property.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• Angle Verification:
In quadrilateral MIST, sum of angles = 360°:
∠M + ∠I + ∠S + ∠T = 360°
75° + 105° + 120° + ∠T = 360°
300° + ∠T = 360° ⇒ ∠T = 60°.
• Steps of Construction:
1. Draw line segment
2. At I, construct an angle of
3. At S, construct an angle of
4. At M, construct an angle of
5. Measuring ∠T confirms it is exactly 60°.
MIST is the required quadrilateral.
In quadrilateral MIST, sum of angles = 360°:
∠M + ∠I + ∠S + ∠T = 360°
75° + 105° + 120° + ∠T = 360°
300° + ∠T = 360° ⇒ ∠T = 60°.
• Steps of Construction:
1. Draw line segment
IS = 6.5 cm.2. At I, construct an angle of
105° using compass (90° + 15°). On this ray, cut off IM = 3.5 cm to fix vertex M.3. At S, construct an angle of
120° using compass.4. At M, construct an angle of
75° to meet the ray from S at point T.5. Measuring ∠T confirms it is exactly 60°.
MIST is the required quadrilateral.
CBSE Board Marking Scheme 2026-27
Calculating ∠T = 60°1 Mark
Step-by-step construction of IS, angles 105°, 120°, 75°2 Marks
Q5
Construct a parallelogram HEAR where HE = 5 cm, EA = 6 cm, and ∠R = 85°. Find the other three angles prior to construction.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Parallelogram Properties: Opposite angles are equal; adjacent angles are supplementary.
1. Angle Calculations:
- Opposite angle to ∠R is ∠E ⇒ ∠E = 85°.
- Adjacent angle ∠H = 180° − 85° = 95°.
- Opposite angle to ∠H is ∠A ⇒ ∠A = 95°.
- Opposite sides are equal: AR = HE = 5 cm; RH = EA = 6 cm.
2. Construction:
- Draw HE = 5 cm.
- At E, draw ray making angle 85° and cut off EA = 6 cm.
- At H, draw ray making angle 95° and cut off HR = 6 cm.
- Join AR (which will measure 5 cm).
Hence, HEAR is constructed.
- Opposite angle to ∠R is ∠E ⇒ ∠E = 85°.
- Adjacent angle ∠H = 180° − 85° = 95°.
- Opposite angle to ∠H is ∠A ⇒ ∠A = 95°.
- Opposite sides are equal: AR = HE = 5 cm; RH = EA = 6 cm.
2. Construction:
- Draw HE = 5 cm.
- At E, draw ray making angle 85° and cut off EA = 6 cm.
- At H, draw ray making angle 95° and cut off HR = 6 cm.
- Join AR (which will measure 5 cm).
Hence, HEAR is constructed.
CBSE Board Marking Scheme 2026-27
Calculating angles: ∠E = 85°, ∠H = ∠A = 95°1 Mark
Steps of construction of parallelogram HEAR1 Mark
Q6
Can you construct a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 5 cm, DA = 6 cm, and diagonal AC = 8 cm? Justify mathematically using the triangle inequality.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Triangle Inequality Theorem: In any triangle, the sum of the lengths of any two sides must be strictly greater than the third side.
1. The diagonal AC = 8 cm forms ΔABC with sides AB = 3 cm and BC = 4 cm.
2. Check the triangle inequality in ΔABC:
AB + BC = 3 + 4 = 7 cm.
Third side AC = 8 cm.
Clearly,
3. Since the sum of two sides is less than the third side, ΔABC cannot exist (the compass arcs with radii 3 cm and 4 cm from ends of an 8 cm segment will never intersect).
Therefore, it is IMPOSSIBLE to construct quadrilateral ABCD.
2. Check the triangle inequality in ΔABC:
AB + BC = 3 + 4 = 7 cm.
Third side AC = 8 cm.
Clearly,
AB + BC < AC (7 < 8).3. Since the sum of two sides is less than the third side, ΔABC cannot exist (the compass arcs with radii 3 cm and 4 cm from ends of an 8 cm segment will never intersect).
Therefore, it is IMPOSSIBLE to construct quadrilateral ABCD.
CBSE Board Marking Scheme 2026-27
Stating triangle inequality theorem1 Mark
Showing 3 + 4 = 7 < 8, proving construction impossible1 Mark
Q7
Construct a rectangle with adjacent sides of lengths 5 cm and 4 cm. State why only 2 sides are specified.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• Geometric Justification: In any rectangle, all four interior angles are inherently 90°, and opposite sides are equal. Therefore, knowing length (5 cm) and breadth (4 cm) provides all required geometric constraints.
• Steps of Construction:
1. Draw line segment AB = 5 cm.
2. At A and B, construct right angles (90°) using compass.
3. On the perpendicular rays, cut off AD = 4 cm and BC = 4 cm.
4. Join CD.
5. ABCD is the required rectangle (with CD = 5 cm and all angles 90°).
• Steps of Construction:
1. Draw line segment AB = 5 cm.
2. At A and B, construct right angles (90°) using compass.
3. On the perpendicular rays, cut off AD = 4 cm and BC = 4 cm.
4. Join CD.
5. ABCD is the required rectangle (with CD = 5 cm and all angles 90°).
CBSE Board Marking Scheme 2026-27
Reason: All angles 90° and opposite sides equal1 Mark
Steps of construction of rectangle ABCD1 Mark
Q8
Construct a square with side 4.5 cm. How many independent measurements were needed?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Independent Measurements Needed: Only 1 independent measurement (the side length 4.5 cm) is needed, because a square is defined by having all 4 sides equal and all 4 angles equal to 90°.
2. Steps of Construction:
- Draw base segment AB = 4.5 cm.
- At A and B, construct 90° perpendicular rays using compass.
- With compass radius 4.5 cm, cut arcs on these rays from A and B to locate vertices D and C respectively.
- Join CD.
- ABCD is the required square of side 4.5 cm.
2. Steps of Construction:
- Draw base segment AB = 4.5 cm.
- At A and B, construct 90° perpendicular rays using compass.
- With compass radius 4.5 cm, cut arcs on these rays from A and B to locate vertices D and C respectively.
- Join CD.
- ABCD is the required square of side 4.5 cm.
CBSE Board Marking Scheme 2026-27
Identifying that only 1 measurement is needed1 Mark
Accurate construction procedure1 Mark
Q9
Why can regular pentagons NOT tile a flat floor without gaps or overlaps (i.e. cannot form a regular tessellation)? Prove using the interior angle formula.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Tessellation Condition: For regular polygons to tile a plane seamlessly around a common vertex, the sum of their interior angles meeting at the vertex must equal exactly 360°. Thus, each interior angle must be an exact divisor of 360°.
1. Interior angle of a regular pentagon (n = 5):
Interior Angle =
2. Let k pentagons meet at a vertex:
k × 108° = 360° ⇒ k = 360 / 108 = 10 / 3 = 3.33....
3. Since 3.33 is not an integer:
- 3 pentagons give: 3 × 108° = 324° (leaving an empty gap of 36°).
- 4 pentagons give: 4 × 108° = 432° > 360° (causing overlap).
Therefore, regular pentagons cannot tile a flat floor.
Interior Angle =
(5 − 2) × 180° / 5 = 540° / 5 = 108°.2. Let k pentagons meet at a vertex:
k × 108° = 360° ⇒ k = 360 / 108 = 10 / 3 = 3.33....
3. Since 3.33 is not an integer:
- 3 pentagons give: 3 × 108° = 324° (leaving an empty gap of 36°).
- 4 pentagons give: 4 × 108° = 432° > 360° (causing overlap).
Therefore, regular pentagons cannot tile a flat floor.
CBSE Board Marking Scheme 2026-27
Interior angle = 108°1 Mark
108° does not divide 360° evenly (360/108 = 3.33)1 Mark
Q10
Construct a quadrilateral ABCD where AB = 4 cm, BC = 5 cm, CD = 6.5 cm, ∠B = 105°, and ∠C = 80°.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Steps of Construction (3 sides and 2 included angles):
1. Draw the base line segment
2. At B, draw a ray BX making an angle of
3. On ray BX, cut off
4. At C, draw a ray CY making an angle of
5. On ray CY, cut off
6. Join AD.
7. ABCD is the required quadrilateral.
1. Draw the base line segment
BC = 5 cm.2. At B, draw a ray BX making an angle of
105° with BC.3. On ray BX, cut off
BA = 4 cm to locate vertex A.4. At C, draw a ray CY making an angle of
80° with CB.5. On ray CY, cut off
CD = 6.5 cm to locate vertex D.6. Join AD.
7. ABCD is the required quadrilateral.
CBSE Board Marking Scheme 2026-27
Constructing base BC and angles 105° and 80°1 Mark
Cutting segments 4 cm and 6.5 cm, joining AD1 Mark
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Land Surveying & Plaza Tessellation: A municipal surveyor is mapping a quadrilateral recreation plaza ABCD with boundary measurements: AB = 40 m, BC = 30 m, CD = 35 m, and DA = 25 m.
The surveyor notes that without measuring any angle or diagonal, the boundary fences could be deformed into various non-rigid shapes.
Additionally, the interior plaza is to be paved with regular hexagonal tiles.
(a) Explain geometrically why knowing only the four perimeter sides AB, BC, CD, DA is insufficient to uniquely build the plaza.
(b) If the surveyor measures diagonal AC = 50 m, verify whether angle ∠ABC is a right angle (90°).
(c) Calculate the interior angle of a regular hexagonal tile and show why regular hexagons can tile the plaza without any gaps.
The surveyor notes that without measuring any angle or diagonal, the boundary fences could be deformed into various non-rigid shapes.
Additionally, the interior plaza is to be paved with regular hexagonal tiles.
(a) Explain geometrically why knowing only the four perimeter sides AB, BC, CD, DA is insufficient to uniquely build the plaza.
(b) If the surveyor measures diagonal AC = 50 m, verify whether angle ∠ABC is a right angle (90°).
(c) Calculate the interior angle of a regular hexagonal tile and show why regular hexagons can tile the plaza without any gaps.
Detailed Analytical Solution & Marking Scheme
(a) Insufficiency of 4 sides:
A quadrilateral has 1 internal degree of freedom (flexibility). Four line segments connected at hinged joints can be sheared continuously. Exactly 5 independent parameters (such as an included angle or diagonal) are required to fix the quadrilateral into a rigid structure.
(b) Verification of Right Angle ∠ABC:
In ΔABC: AB = 40 m, BC = 30 m, and diagonal AC = 50 m.
AB² + BC² = 40² + 30² = 1600 + 900 = 2500.
AC² = 50² = 2500.
Since AB² + BC² = AC², by the Converse of Baudhāyana-Pythagoras Theorem, ΔABC is right-angled at B (∠ABC = 90°).
(c) Hexagonal Tessellation:
Interior angle of regular hexagon (n = 6) = (6 − 2) × 180° / 6 = 720° / 6 = 120°.
At each shared vertex, 3 tiles meet: 3 × 120° = 360°.
Since the sum equals exactly 360°, the tiles fit together perfectly with no gaps or overlaps.
12
Chapter 12: Tales by Dots and Lines
NCERT Ganita Prakash (Class 8 Part 2) — Cartesian Coordinate System, Ordered Pairs, Plotting Points & Linear Graphs | CBSE 2026-27
📌 Key Coordinate Geometry Principles & Graphing Rules
- Cartesian Coordinate System: A 2D planar system created by two mutually perpendicular number lines intersecting at the Origin O(0, 0):
- X-axis: The horizontal number line.
- Y-axis: The vertical number line.
- The two axes divide the plane into four Quadrants (I, II, III, IV).
- Ordered Pair (x, y):
- The first coordinate x is the abscissa (perpendicular distance from the y-axis).
- The second coordinate y is the ordinate (perpendicular distance from the x-axis).
- Points lying on the x-axis always have ordinate
y = 0(form:(x, 0)). - Points lying on the y-axis always have abscissa
x = 0(form:(0, y)).
- Linear Graphs: A line graph in which all data points lie on a single continuous straight line.
- Direct proportional relationships (e.g., Perimeter of square
P = 4s, Cost = k × Quantity) always form straight linear graphs passing through the origin. - Non-linear relations (e.g., Area of square
A = s²) form parabolic curves, not straight lines.
- Direct proportional relationships (e.g., Perimeter of square
Figure 12.1: Geometric Architecture of the 2D Cartesian Coordinate System
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
Plot the following points on a graph sheet and verify if they lie on a line:
(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)
(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5)
(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)
(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5)
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 12.1: Verification of Collinear Points on Cartesian Coordinate Plane
• (a) Points A(4, 0), B(4, 2), C(4, 6), D(4, 2.5):
All these points share the exact same x-coordinate
When plotted, they all lie on a single vertical line parallel to the y-axis, located 4 units to the right.
Conclusion: Yes, they lie on a straight line (the line x = 4).
• (b) Points P(1, 1), Q(2, 2), R(3, 3), S(4, 4):
For every point, the x-coordinate equals the y-coordinate (
When plotted and connected, they form a single straight line passing through the origin at an angle of 45°.
Conclusion: Yes, they lie on a straight line (the line y = x).
• (c) Points K(2, 3), L(5, 3), M(5, 5), N(2, 5):
Joining these four points in sequence creates four segments: KL is horizontal (y = 3), LM is vertical (x = 5), MN is horizontal (y = 5), and NK is vertical (x = 2).
Together they enclose a rectangle of length 3 units and breadth 2 units.
Conclusion: No, they do NOT lie on a single line; they form a rectangle.
All these points share the exact same x-coordinate
x = 4.When plotted, they all lie on a single vertical line parallel to the y-axis, located 4 units to the right.
Conclusion: Yes, they lie on a straight line (the line x = 4).
• (b) Points P(1, 1), Q(2, 2), R(3, 3), S(4, 4):
For every point, the x-coordinate equals the y-coordinate (
y = x).When plotted and connected, they form a single straight line passing through the origin at an angle of 45°.
Conclusion: Yes, they lie on a straight line (the line y = x).
• (c) Points K(2, 3), L(5, 3), M(5, 5), N(2, 5):
Joining these four points in sequence creates four segments: KL is horizontal (y = 3), LM is vertical (x = 5), MN is horizontal (y = 5), and NK is vertical (x = 2).
Together they enclose a rectangle of length 3 units and breadth 2 units.
Conclusion: No, they do NOT lie on a single line; they form a rectangle.
CBSE Board Marking Scheme 2026-27
(a) Yes, vertical line x = 41 Mark
(b) Yes, line y = x; (c) No, forms a rectangle2 Marks
Q2
Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and the y-axis.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 12.2: Line x + y = 5 Meeting X-Axis at (5, 0) and Y-Axis at (0, 5)
1. Notice the sum of coordinates for both points: 2 + 3 = 5 and 3 + 2 = 5.
The equation of the straight line is:
2. Intersection with x-axis:
On the x-axis, the y-coordinate is 0.
Substituting y = 0 into x + y = 5:
x + 0 = 5 ⇒ x = 5.
Point of intersection with x-axis is (5, 0).
3. Intersection with y-axis:
On the y-axis, the x-coordinate is 0.
Substituting x = 0 into x + y = 5:
0 + y = 5 ⇒ y = 5.
Point of intersection with y-axis is (0, 5).
The equation of the straight line is:
x + y = 5.2. Intersection with x-axis:
On the x-axis, the y-coordinate is 0.
Substituting y = 0 into x + y = 5:
x + 0 = 5 ⇒ x = 5.
Point of intersection with x-axis is (5, 0).
3. Intersection with y-axis:
On the y-axis, the x-coordinate is 0.
Substituting x = 0 into x + y = 5:
0 + y = 5 ⇒ y = 5.
Point of intersection with y-axis is (0, 5).
CBSE Board Marking Scheme 2026-27
Line equation x + y = 51 Mark
Intercepts: (5, 0) on x-axis and (0, 5) on y-axis1 Mark
Q3
Write the coordinates of the vertices of a triangle ABC where A lies on the y-axis at distance 4 units above origin, B lies at origin (0, 0), and C lies on the x-axis at distance 3 units to the right of origin. Also find the length of hypotenuse AC.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Coordinates of vertices:
- B is at the origin: B(0, 0).
- A lies on y-axis at distance 4: A(0, 4).
- C lies on x-axis at distance 3: C(3, 0).
2. Since the x-axis and y-axis are perpendicular, ΔABC is right-angled at B(0, 0) with legs AB = 4 units and BC = 3 units.
3. By Pythagoras Theorem:
AC² = AB² + BC² = 4² + 3² = 16 + 9 = 25.
AC = √25 = 5 units.
- B is at the origin: B(0, 0).
- A lies on y-axis at distance 4: A(0, 4).
- C lies on x-axis at distance 3: C(3, 0).
2. Since the x-axis and y-axis are perpendicular, ΔABC is right-angled at B(0, 0) with legs AB = 4 units and BC = 3 units.
3. By Pythagoras Theorem:
AC² = AB² + BC² = 4² + 3² = 16 + 9 = 25.
AC = √25 = 5 units.
CBSE Board Marking Scheme 2026-27
Coordinates: A(0, 4), B(0, 0), C(3, 0)1 Mark
Hypotenuse AC = 5 units1 Mark
Q4
State whether true or false. Correct that are false:
(i) A point whose x coordinate is zero and y-coordinate is non-zero will lie on the y-axis.
(ii) A point whose y coordinate is zero and x-coordinate is 5 will lie on y-axis.
(iii) The coordinates of the origin are (0, 0).
(i) A point whose x coordinate is zero and y-coordinate is non-zero will lie on the y-axis.
(ii) A point whose y coordinate is zero and x-coordinate is 5 will lie on y-axis.
(iii) The coordinates of the origin are (0, 0).
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) True: Any point of the form (0, y) where x = 0 has zero horizontal displacement from the y-axis, and therefore lies directly on the y-axis.
• (ii) False: A point with y = 0 and x = 5 has coordinates (5, 0). Since its vertical distance from the x-axis is zero, it lies on the x-axis (not the y-axis).
• (iii) True: The origin is the intersection of both the horizontal (x = 0) and vertical (y = 0) axes, hence its coordinates are precisely (0, 0).
• (ii) False: A point with y = 0 and x = 5 has coordinates (5, 0). Since its vertical distance from the x-axis is zero, it lies on the x-axis (not the y-axis).
• (iii) True: The origin is the intersection of both the horizontal (x = 0) and vertical (y = 0) axes, hence its coordinates are precisely (0, 0).
CBSE Board Marking Scheme 2026-27
(i) True, (iii) True1 Mark
(ii) False with correction (lies on x-axis)1 Mark
Q5
Draw a graph for the following table of values, with suitable scales on the axes:
Side of square (in cm): 2 | 3 | 3.5 | 5 | 6
Perimeter (in cm): 8 | 12 | 14 | 20 | 24
Is it a linear graph?
Side of square (in cm): 2 | 3 | 3.5 | 5 | 6
Perimeter (in cm): 8 | 12 | 14 | 20 | 24
Is it a linear graph?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Mathematical Relation: The perimeter of a square with side s is given by
Ratio
2. Graphing Protocol:
- Plot Side (s) along the x-axis: Scale 1 cm = 1 unit.
- Plot Perimeter (P) along the y-axis: Scale 1 cm = 4 units.
- Plot the points: (2, 8), (3, 12), (3.5, 14), (5, 20), (6, 24).
3. Connecting the plotted points with a straightedge produces an unbroken, straight ray starting from origin (0, 0).
4. Conclusion: Yes, it is a linear graph, because Perimeter varies directly with side length.
P = 4s.Ratio
Perimeter / Side = 8/2 = 12/3 = 14/3.5 = 20/5 = 24/6 = 4 (Constant).2. Graphing Protocol:
- Plot Side (s) along the x-axis: Scale 1 cm = 1 unit.
- Plot Perimeter (P) along the y-axis: Scale 1 cm = 4 units.
- Plot the points: (2, 8), (3, 12), (3.5, 14), (5, 20), (6, 24).
3. Connecting the plotted points with a straightedge produces an unbroken, straight ray starting from origin (0, 0).
4. Conclusion: Yes, it is a linear graph, because Perimeter varies directly with side length.
CBSE Board Marking Scheme 2026-27
Scale selection and plotting points correctly1.5 Marks
Joining points to show straight line passing through (0,0)1.5 Marks
Q6
Consider the relation between side of a square and its area:
Side of square (cm): 2 | 3 | 4 | 5 | 6
Area (cm²): 4 | 9 | 16 | 25 | 36
Draw the graph. Is this a linear graph? Explain why.
Side of square (cm): 2 | 3 | 4 | 5 | 6
Area (cm²): 4 | 9 | 16 | 25 | 36
Draw the graph. Is this a linear graph? Explain why.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Mathematical Relation:
Check the ratio
4/2 = 2; 9/3 = 3; 16/4 = 4; 25/5 = 5; 36/6 = 6.
The ratio is not constant; it increases continuously with side length.
2. When the points (2, 4), (3, 9), (4, 16), (5, 25), (6, 36) are plotted on coordinate axes, connecting them does not yield a straight line, but a curve bending upwards (a parabola).
3. Conclusion: No, it is NOT a linear graph, because the relationship between side and area is quadratic (Area ∝ side²), not directly proportional.
Area = s² (quadratic relation).Check the ratio
Area / Side:4/2 = 2; 9/3 = 3; 16/4 = 4; 25/5 = 5; 36/6 = 6.
The ratio is not constant; it increases continuously with side length.
2. When the points (2, 4), (3, 9), (4, 16), (5, 25), (6, 36) are plotted on coordinate axes, connecting them does not yield a straight line, but a curve bending upwards (a parabola).
3. Conclusion: No, it is NOT a linear graph, because the relationship between side and area is quadratic (Area ∝ side²), not directly proportional.
CBSE Board Marking Scheme 2026-27
Plotting points and observing upward-curving path1 Mark
Justification: Area = s² is quadratic, not linear1 Mark
Q7
A courier-person cycles from a town to a neighbouring suburban area to deliver a parcel. His distance from the town at different times is shown by the following graph:
At 8:00 am: 0 km; at 9:00 am: 10 km; at 10:00 am: 16 km; at 10:30 am: 16 km; at 11:30 am: 22 km.
(a) What was the scale taken for the time axis?
(b) How much time did the person take for the travel?
(c) How far is the place of the merchant from the town?
(d) Did the person stop on his way? Explain.
(e) During which period did he ride fastest?
At 8:00 am: 0 km; at 9:00 am: 10 km; at 10:00 am: 16 km; at 10:30 am: 16 km; at 11:30 am: 22 km.
(a) What was the scale taken for the time axis?
(b) How much time did the person take for the travel?
(c) How far is the place of the merchant from the town?
(d) Did the person stop on his way? Explain.
(e) During which period did he ride fastest?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 12.3: Visual Analysis of Speed, Rest Plateau, and Total Distance
• (a) Scale for time axis: 4 units on grid = 1 hour (or 1 unit = 15 minutes).
• (b) Total travel time: The person started at 8:00 am and reached at 11:30 am.
Total time = 11:30 − 8:00 = 3 1/2 hours (3 hours 30 minutes).
• (c) Distance of destination: The final point at 11:30 am corresponds to 22 km.
• (d) Stop during journey: Yes, between 10:00 am and 10:30 am, the graph is a horizontal flat line (distance remains constant at 16 km for 30 minutes).
• (e) Fastest period:
- 8:00 am to 9:00 am: Speed = (10 − 0) / 1 = 10 km/h.
- 9:00 am to 10:00 am: Speed = (16 − 10) / 1 = 6 km/h.
- 10:30 am to 11:30 am: Speed = (22 − 16) / 1 = 6 km/h.
He rode fastest between 8:00 am and 9:00 am (speed = 10 km/h).
• (b) Total travel time: The person started at 8:00 am and reached at 11:30 am.
Total time = 11:30 − 8:00 = 3 1/2 hours (3 hours 30 minutes).
• (c) Distance of destination: The final point at 11:30 am corresponds to 22 km.
• (d) Stop during journey: Yes, between 10:00 am and 10:30 am, the graph is a horizontal flat line (distance remains constant at 16 km for 30 minutes).
• (e) Fastest period:
- 8:00 am to 9:00 am: Speed = (10 − 0) / 1 = 10 km/h.
- 9:00 am to 10:00 am: Speed = (16 − 10) / 1 = 6 km/h.
- 10:30 am to 11:30 am: Speed = (22 − 16) / 1 = 6 km/h.
He rode fastest between 8:00 am and 9:00 am (speed = 10 km/h).
CBSE Board Marking Scheme 2026-27
(a), (b), (c) answers1.5 Marks
(d) Yes (horizontal line 10 to 10:30), (e) 8:00 to 9:00 am1.5 Marks
Q8
A bank offers 8% simple interest per annum on deposits. Draw a linear graph to illustrate the relation between the deposit sum (Principal) and the annual Simple Interest earned:
Deposit (₹): 1000 | 2000 | 3000 | 4000 | 5000
Use the graph to find: (i) The annual interest on ₹ 2500, (ii) The investment needed to earn ₹ 280 interest.
Deposit (₹): 1000 | 2000 | 3000 | 4000 | 5000
Use the graph to find: (i) The annual interest on ₹ 2500, (ii) The investment needed to earn ₹ 280 interest.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Formula: SI = (P × 8 × 1) / 100 = 0.08 P.
1. Table of Values:
• P = ₹ 1000 ⇒ SI = ₹ 80.
• P = ₹ 2000 ⇒ SI = ₹ 160.
• P = ₹ 3000 ⇒ SI = ₹ 240.
• P = ₹ 4000 ⇒ SI = ₹ 320.
• P = ₹ 5000 ⇒ SI = ₹ 400.
2. Graph Readings:
• (i) Interest on ₹ 2500:
Find P = 2500 on the x-axis, trace vertically to the line and read the y-axis:
SI = 0.08 × 2500 = ₹ 200.
• (ii) Investment to earn ₹ 280 interest:
Find SI = 280 on the y-axis, trace horizontally to the line and read the x-axis:
P = 280 / 0.08 = ₹ 3,500.
• P = ₹ 1000 ⇒ SI = ₹ 80.
• P = ₹ 2000 ⇒ SI = ₹ 160.
• P = ₹ 3000 ⇒ SI = ₹ 240.
• P = ₹ 4000 ⇒ SI = ₹ 320.
• P = ₹ 5000 ⇒ SI = ₹ 400.
2. Graph Readings:
• (i) Interest on ₹ 2500:
Find P = 2500 on the x-axis, trace vertically to the line and read the y-axis:
SI = 0.08 × 2500 = ₹ 200.
• (ii) Investment to earn ₹ 280 interest:
Find SI = 280 on the y-axis, trace horizontally to the line and read the x-axis:
P = 280 / 0.08 = ₹ 3,500.
CBSE Board Marking Scheme 2026-27
Graph values table and linear plotting through origin1.5 Marks
(i) ₹ 200, (ii) ₹ 3,500 derived from graph1.5 Marks
Q9
In which quadrant or on which axis do each of the following points lie?
(a) (−2, 4) (b) (3, −1) (c) (−1, 0) (d) (1, 2) (e) (−3, −5) (f) (0, −4)
(a) (−2, 4) (b) (3, −1) (c) (−1, 0) (d) (1, 2) (e) (−3, −5) (f) (0, −4)
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (a) (−2, 4): x is negative, y is positive ⇒ Quadrant II.
• (b) (3, −1): x is positive, y is negative ⇒ Quadrant IV.
• (c) (−1, 0): y-coordinate is 0 ⇒ Negative x-axis.
• (d) (1, 2): Both x and y are positive ⇒ Quadrant I.
• (e) (−3, −5): Both x and y are negative ⇒ Quadrant III.
• (f) (0, −4): x-coordinate is 0 ⇒ Negative y-axis.
• (b) (3, −1): x is positive, y is negative ⇒ Quadrant IV.
• (c) (−1, 0): y-coordinate is 0 ⇒ Negative x-axis.
• (d) (1, 2): Both x and y are positive ⇒ Quadrant I.
• (e) (−3, −5): Both x and y are negative ⇒ Quadrant III.
• (f) (0, −4): x-coordinate is 0 ⇒ Negative y-axis.
CBSE Board Marking Scheme 2026-27
Sign analysis for 2D plane quadrants0.5 Mark
Correct placement for all 6 points1.5 Marks
Q10
Find the coordinates of the fourth vertex D of a parallelogram ABCD, given that three vertices are A(1, 1), B(4, 1), and C(5, 4). Calculate the area of the parallelogram.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Locating Vertex D:
- Side AB is horizontal along y = 1 from x = 1 to x = 4. Length AB = 4 − 1 = 3 units.
- Opposite side CD must also be horizontal of length 3 units.
- Since C is at (5, 4), moving 3 units to the left at the same height y = 4 gives:
x = 5 − 3 = 2, y = 4.
Therefore, D = (2, 4).
2. Area of Parallelogram ABCD:
Area = Base × Height.
Base = AB = 3 units.
Height = vertical distance between lines y = 4 and y = 1 = 4 − 1 = 3 units.
Area = 3 × 3 = 9 square units.
- Side AB is horizontal along y = 1 from x = 1 to x = 4. Length AB = 4 − 1 = 3 units.
- Opposite side CD must also be horizontal of length 3 units.
- Since C is at (5, 4), moving 3 units to the left at the same height y = 4 gives:
x = 5 − 3 = 2, y = 4.
Therefore, D = (2, 4).
2. Area of Parallelogram ABCD:
Area = Base × Height.
Base = AB = 3 units.
Height = vertical distance between lines y = 4 and y = 1 = 4 − 1 = 3 units.
Area = 3 × 3 = 9 square units.
CBSE Board Marking Scheme 2026-27
Finding coordinates D(2, 4)1 Mark
Base × Height = 3 × 3 = 9 sq units1 Mark
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Electric Vehicle (EV) Rapid Charging Station Telemetry: An IoT telemetry sensor logs the battery charge state (in kWh) of an electric delivery van plugged into a DC fast charger over 50 minutes:
Time t (minutes): 0 | 10 | 20 | 30 | 40 | 50
Charge E (kWh): 10 | 24 | 38 | 52 | 62 | 68
(a) Plot the points on coordinate axes and state whether the entire 50-minute graph is strictly linear.
(b) Over which time interval is the charging rate constant, and what is the charging speed in kWh per minute during this phase?
(c) Explain physically why the graph flattens after 30 minutes (above 80% battery capacity).
Time t (minutes): 0 | 10 | 20 | 30 | 40 | 50
Charge E (kWh): 10 | 24 | 38 | 52 | 62 | 68
(a) Plot the points on coordinate axes and state whether the entire 50-minute graph is strictly linear.
(b) Over which time interval is the charging rate constant, and what is the charging speed in kWh per minute during this phase?
(c) Explain physically why the graph flattens after 30 minutes (above 80% battery capacity).
Detailed Analytical Solution & Marking Scheme
(a) Linearity Analysis:
- From t = 0 to 30 min: ΔE/Δt = (24 − 10)/10 = (38 − 24)/10 = (52 − 38)/10 = 1.4 kWh/min (Constant).
- From t = 30 to 40 min: ΔE/Δt = (62 − 52)/10 = 1.0 kWh/min.
- From t = 40 to 50 min: ΔE/Δt = (68 − 62)/10 = 0.6 kWh/min.
Because the slope decreases in the final 20 minutes, the overall graph is NOT strictly linear.
(b) Constant Rate Phase:
The rate is constant from t = 0 to t = 30 minutes, charging at a linear rate of 1.4 kWh/minute (or 84 kW power).
(c) Physical Interpretation:
Lithium-ion batteries taper down charging current above ~80% state-of-charge (Constant Current to Constant Voltage CC-CV transition) to prevent lithium plating, overheating, and thermal degradation. This deliberate deceleration causes the curve to flatten.
13
Chapter 13: Algebra Play
NCERT Ganita Prakash (Class 8 Part 2) — Linear Equations in One Variable, Cross-Multiplication, Word Problem Modelling & Factorisation | CBSE 2026-27
📌 Key Algebraic Solving Techniques & Factorisation Methods
- Linear Equations in One Variable: An algebraic equation of degree 1 with only one variable, of the general form
ax + b = cx + d(where a ≠ c).- Transposition Protocol: Shifting terms across the equality sign changes their operational sign (+ becomes −, − becomes +, × becomes ÷, ÷ becomes ×).
- Cross-Multiplication Method: For fractional equations of the form
(ax + b) / (cx + d) = m / n, cross-multiply:n(ax + b) = m(cx + d).
- Factorisation of Algebraic Expressions: Decomposing a polynomial into products of irreducible factors.
- Common Factor Method: Extract the Highest Common Factor (HCF) of all terms:
ka + kb = k(a + b). - Regrouping Method: Group terms in pairs sharing common binomial factors:
ax + ay + bx + by = a(x + y) + b(x + y) = (x + y)(a + b). - Using Identities:
a² + 2ab + b² = (a + b)²a² − 2ab + b² = (a − b)²a² − b² = (a + b)(a − b)
- Splitting the Middle Term: For quadratic trinomials
x² + px + q, find two numbers a and b such thata + b = pandab = q. Thenx² + px + q = (x + a)(x + b).
- Common Factor Method: Extract the Highest Common Factor (HCF) of all terms:
Figure 13.1: The Mechanical Balance Principle Governing Linear Equation Invariance
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
Solve the following linear equations and check your results:
(i) 3x = 2x + 18
(ii) 5t − 3 = 3t − 5
(iii) 5x + 9 = 5 + 3x
(iv) 4z + 3 = 6 + 2z
(i) 3x = 2x + 18
(ii) 5t − 3 = 3t − 5
(iii) 5x + 9 = 5 + 3x
(iv) 4z + 3 = 6 + 2z
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 3x = 2x + 18:
Transpose 2x to LHS: 3x − 2x = 18 ⇒ x = 18.
Check: LHS = 3(18) = 54. RHS = 2(18) + 18 = 36 + 18 = 54. Verified.
• (ii) 5t − 3 = 3t − 5:
5t − 3t = −5 + 3
2t = −2 ⇒ t = −1.
Check: LHS = 5(−1) − 3 = −8. RHS = 3(−1) − 5 = −8. Verified.
• (iii) 5x + 9 = 5 + 3x:
5x − 3x = 5 − 9
2x = −4 ⇒ x = −2.
Check: LHS = 5(−2) + 9 = −1. RHS = 5 + 3(−2) = −1. Verified.
• (iv) 4z + 3 = 6 + 2z:
4z − 2z = 6 − 3
2z = 3 ⇒ z = 3/2.
Check: LHS = 4(3/2) + 3 = 6 + 3 = 9. RHS = 6 + 2(3/2) = 6 + 3 = 9. Verified.
Transpose 2x to LHS: 3x − 2x = 18 ⇒ x = 18.
Check: LHS = 3(18) = 54. RHS = 2(18) + 18 = 36 + 18 = 54. Verified.
• (ii) 5t − 3 = 3t − 5:
5t − 3t = −5 + 3
2t = −2 ⇒ t = −1.
Check: LHS = 5(−1) − 3 = −8. RHS = 3(−1) − 5 = −8. Verified.
• (iii) 5x + 9 = 5 + 3x:
5x − 3x = 5 − 9
2x = −4 ⇒ x = −2.
Check: LHS = 5(−2) + 9 = −1. RHS = 5 + 3(−2) = −1. Verified.
• (iv) 4z + 3 = 6 + 2z:
4z − 2z = 6 − 3
2z = 3 ⇒ z = 3/2.
Check: LHS = 4(3/2) + 3 = 6 + 3 = 9. RHS = 6 + 2(3/2) = 6 + 3 = 9. Verified.
CBSE Board Marking Scheme 2026-27
Transposition method shown for all 4 equations1 Mark
Accurate roots: x = 18, t = −1, x = −2, z = 3/21 Mark
Q2
Solve the linear equation: m − (m − 1)/2 = 1 − (m − 2)/3.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Multiply the entire equation by LCM of denominators 2 and 3, which is 6:
6 × [m − (m − 1)/2] = 6 × [1 − (m − 2)/3]
2. Expanding carefully:
6m − 3(m − 1) = 6(1) − 2(m − 2)
6m − 3m + 3 = 6 − 2m + 4
3m + 3 = 10 − 2m
3. Transposing terms:
3m + 2m = 10 − 3
5m = 7
m = 7 / 5.
Verification:
LHS = 7/5 − (7/5 − 1)/2 = 7/5 − (2/5)/2 = 7/5 − 1/5 = 6/5.
RHS = 1 − (7/5 − 2)/3 = 1 − (−3/5)/3 = 1 + 1/5 = 6/5. LHS = RHS.
6 × [m − (m − 1)/2] = 6 × [1 − (m − 2)/3]
2. Expanding carefully:
6m − 3(m − 1) = 6(1) − 2(m − 2)
6m − 3m + 3 = 6 − 2m + 4
3m + 3 = 10 − 2m
3. Transposing terms:
3m + 2m = 10 − 3
5m = 7
m = 7 / 5.
Verification:
LHS = 7/5 − (7/5 − 1)/2 = 7/5 − (2/5)/2 = 7/5 − 1/5 = 6/5.
RHS = 1 − (7/5 − 2)/3 = 1 − (−3/5)/3 = 1 + 1/5 = 6/5. LHS = RHS.
CBSE Board Marking Scheme 2026-27
Multiplying by LCM = 6 to clear fractions1 Mark
Correctly solving 5m = 7 ⇒ m = 7/51 Mark
Q3
Solve using cross-multiplication: (3y + 4) / (2 − 6y) = −2 / 5.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Cross-Multiplication Principle: If P / Q = R / S, then P × S = Q × R.
1. Cross-multiplying:
5 × (3y + 4) = −2 × (2 − 6y)
2. Expanding both sides:
15y + 20 = −4 + 12y
3. Transposing terms:
15y − 12y = −4 − 20
3y = −24
y = −24 / 3 = −8.
Verification:
Numerator = 3(−8) + 4 = −24 + 4 = −20.
Denominator = 2 − 6(−8) = 2 + 48 = 50.
Ratio = −20 / 50 = −2/5. Verified.
5 × (3y + 4) = −2 × (2 − 6y)
2. Expanding both sides:
15y + 20 = −4 + 12y
3. Transposing terms:
15y − 12y = −4 − 20
3y = −24
y = −24 / 3 = −8.
Verification:
Numerator = 3(−8) + 4 = −24 + 4 = −20.
Denominator = 2 − 6(−8) = 2 + 48 = 50.
Ratio = −20 / 50 = −2/5. Verified.
CBSE Board Marking Scheme 2026-27
Cross-multiplication: 5(3y + 4) = −2(2 − 6y)1 Mark
Solving 3y = −24 ⇒ y = −81 Mark
Q4
The ages of Hari and Harry are in the ratio 5 : 7. Four years from now the ratio of their ages will be 3 : 4. Find their present ages.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let the common ratio multiplier be
Present age of Hari =
Present age of Harry =
2. After 4 years:
Hari's age =
Harry's age =
3. According to the question, their ratio becomes 3 : 4:
(5x + 4) / (7x + 4) = 3 / 4
4. Cross-multiplying:
4(5x + 4) = 3(7x + 4)
20x + 16 = 21x + 12
16 − 12 = 21x − 20x
x = 4.
5. Present ages:
Hari's age = 5x = 5 × 4 = 20 years.
Harry's age = 7x = 7 × 4 = 28 years.
x.Present age of Hari =
5x years.Present age of Harry =
7x years.2. After 4 years:
Hari's age =
5x + 4.Harry's age =
7x + 4.3. According to the question, their ratio becomes 3 : 4:
(5x + 4) / (7x + 4) = 3 / 4
4. Cross-multiplying:
4(5x + 4) = 3(7x + 4)
20x + 16 = 21x + 12
16 − 12 = 21x − 20x
x = 4.
5. Present ages:
Hari's age = 5x = 5 × 4 = 20 years.
Harry's age = 7x = 7 × 4 = 28 years.
CBSE Board Marking Scheme 2026-27
Formulating (5x + 4)/(7x + 4) = 3/41.5 Marks
Solving x = 4 and finding Hari = 20 yrs, Harry = 28 yrs1.5 Marks
Q5
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let the numerator of the rational number be
Then the denominator is
Original rational number =
2. According to the conditions:
New numerator = x + 17.
New denominator = (x + 8) − 1 = x + 7.
(x + 17) / (x + 7) = 3 / 2.
3. Cross-multiplying:
2(x + 17) = 3(x + 7)
2x + 34 = 3x + 21
34 − 21 = 3x − 2x
x = 13.
4. Therefore:
Numerator = 13.
Denominator = 13 + 8 = 21.
The required rational number is 13 / 21.
Check: (13 + 17) / (21 − 1) = 30 / 20 = 3/2. Verified.
x.Then the denominator is
x + 8.Original rational number =
x / (x + 8).2. According to the conditions:
New numerator = x + 17.
New denominator = (x + 8) − 1 = x + 7.
(x + 17) / (x + 7) = 3 / 2.
3. Cross-multiplying:
2(x + 17) = 3(x + 7)
2x + 34 = 3x + 21
34 − 21 = 3x − 2x
x = 13.
4. Therefore:
Numerator = 13.
Denominator = 13 + 8 = 21.
The required rational number is 13 / 21.
Check: (13 + 17) / (21 − 1) = 30 / 20 = 3/2. Verified.
CBSE Board Marking Scheme 2026-27
Setting up equation (x + 17)/(x + 7) = 3/21.5 Marks
Solving x = 13 and original fraction = 13/211.5 Marks
Q6
Deveshi has a total of ₹ 590 as currency notes in the denominations of ₹ 50, ₹ 20, and ₹ 10. The ratio of the number of ₹ 50 notes and ₹ 20 notes is 3 : 5. If she has a total of 25 notes, how many notes of each denomination does she have?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let the number of ₹ 50 notes be
Total number of notes = 25.
Number of ₹ 10 notes =
2. Total value of the notes is ₹ 590:
Value = 50(3x) + 20(5x) + 10(25 − 8x)
590 = 150x + 100x + 250 − 80x
590 = 170x + 250
3. Solving for x:
170x = 590 − 250 = 340
x = 340 / 170 = 2.
4. Number of notes of each denomination:
- Number of ₹ 50 notes = 3x = 3 × 2 = 6 notes.
- Number of ₹ 20 notes = 5x = 5 × 2 = 10 notes.
- Number of ₹ 10 notes = 25 − 8(2) = 25 − 16 = 9 notes.
Check: 6(50) + 10(20) + 9(10) = 300 + 200 + 90 = ₹ 590. Verified.
3x and ₹ 20 notes be 5x.Total number of notes = 25.
Number of ₹ 10 notes =
25 − (3x + 5x) = 25 − 8x.2. Total value of the notes is ₹ 590:
Value = 50(3x) + 20(5x) + 10(25 − 8x)
590 = 150x + 100x + 250 − 80x
590 = 170x + 250
3. Solving for x:
170x = 590 − 250 = 340
x = 340 / 170 = 2.
4. Number of notes of each denomination:
- Number of ₹ 50 notes = 3x = 3 × 2 = 6 notes.
- Number of ₹ 20 notes = 5x = 5 × 2 = 10 notes.
- Number of ₹ 10 notes = 25 − 8(2) = 25 − 16 = 9 notes.
Check: 6(50) + 10(20) + 9(10) = 300 + 200 + 90 = ₹ 590. Verified.
CBSE Board Marking Scheme 2026-27
Formulating value equation: 170x + 250 = 5901.5 Marks
Count: 6 notes of ₹50, 10 of ₹20, 9 of ₹101.5 Marks
Q7
Factorise the following expressions using common factors and grouping:
(i) 12a²b + 15ab²
(ii) 10x² − 18x³ + 14x⁴
(iii) z − 7 + 7xy − xyz
(iv) 15xy − 6x + 5y − 2
(i) 12a²b + 15ab²
(ii) 10x² − 18x³ + 14x⁴
(iii) z − 7 + 7xy − xyz
(iv) 15xy − 6x + 5y − 2
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 12a²b + 15ab²:
HCF of 12 and 15 is 3. HCF of a²b and ab² is ab.
Factoring 3ab: = 3ab(4a + 5b).
• (ii) 10x² − 18x³ + 14x⁴:
HCF of coefficients = 2. HCF of powers of x = x².
Factoring 2x²: = 2x²(5 − 9x + 7x²).
• (iii) z − 7 + 7xy − xyz:
Regrouping: (z − 7) − xy(z − 7)
Taking common binomial (z − 7): = (z − 7)(1 − xy).
• (iv) 15xy − 6x + 5y − 2:
Group first two and last two terms:
3x(5y − 2) + 1(5y − 2)
= (5y − 2)(3x + 1).
HCF of 12 and 15 is 3. HCF of a²b and ab² is ab.
Factoring 3ab: = 3ab(4a + 5b).
• (ii) 10x² − 18x³ + 14x⁴:
HCF of coefficients = 2. HCF of powers of x = x².
Factoring 2x²: = 2x²(5 − 9x + 7x²).
• (iii) z − 7 + 7xy − xyz:
Regrouping: (z − 7) − xy(z − 7)
Taking common binomial (z − 7): = (z − 7)(1 − xy).
• (iv) 15xy − 6x + 5y − 2:
Group first two and last two terms:
3x(5y − 2) + 1(5y − 2)
= (5y − 2)(3x + 1).
CBSE Board Marking Scheme 2026-27
HCF extraction for (i) and (ii)1.5 Marks
Regrouping binomials for (iii) and (iv)1.5 Marks
Q8
Factorise the following expressions using algebraic identities:
(i) 49y² + 84yz + 36z²
(ii) 4x² − 12xy + 9y²
(iii) 49x² − 36
(iv) (l + m)² − (l − m)²
(i) 49y² + 84yz + 36z²
(ii) 4x² − 12xy + 9y²
(iii) 49x² − 36
(iv) (l + m)² − (l − m)²
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) 49y² + 84yz + 36z²:
Notice: (7y)² + 2(7y)(6z) + (6z)². Matches
= (7y + 6z)² (or (7y + 6z)(7y + 6z)).
• (ii) 4x² − 12xy + 9y²:
Notice: (2x)² − 2(2x)(3y) + (3y)². Matches
= (2x − 3y)².
• (iii) 49x² − 36:
Notice: (7x)² − 6². Matches
= (7x + 6)(7x − 6).
• (iv) (l + m)² − (l − m)²:
Using
= [(l + m) + (l − m)] × [(l + m) − (l − m)]
= (2l) × (2m) = 4lm.
Notice: (7y)² + 2(7y)(6z) + (6z)². Matches
(a + b)².= (7y + 6z)² (or (7y + 6z)(7y + 6z)).
• (ii) 4x² − 12xy + 9y²:
Notice: (2x)² − 2(2x)(3y) + (3y)². Matches
(a − b)².= (2x − 3y)².
• (iii) 49x² − 36:
Notice: (7x)² − 6². Matches
a² − b² = (a + b)(a − b).= (7x + 6)(7x − 6).
• (iv) (l + m)² − (l − m)²:
Using
a² − b² = (a + b)(a − b) where a = (l + m) and b = (l − m):= [(l + m) + (l − m)] × [(l + m) − (l − m)]
= (2l) × (2m) = 4lm.
CBSE Board Marking Scheme 2026-27
Applying (a + b)², (a − b)², and a² − b²1.5 Marks
Accurate factored products for all 4 parts1.5 Marks
Q9
Factorise the quadratic trinomials by splitting the middle term:
(i) p² + 6p + 8
(ii) q² − 10q + 21
(iii) p² + 6p − 16
(iv) x² − 2x − 15
(i) p² + 6p + 8
(ii) q² − 10q + 21
(iii) p² + 6p − 16
(iv) x² − 2x − 15
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Rule: To factorise x² + bx + c, find two numbers whose sum is b and product is c.
• (i) p² + 6p + 8:
Sum = 6, Product = 8. The numbers are 4 and 2.
= p² + 4p + 2p + 8 = p(p + 4) + 2(p + 4) = (p + 4)(p + 2).
• (ii) q² − 10q + 21:
Sum = −10, Product = 21. The numbers are −7 and −3.
= q² − 7q − 3q + 21 = q(q − 7) − 3(q − 7) = (q − 7)(q − 3).
• (iii) p² + 6p − 16:
Sum = 6, Product = −16. The numbers are 8 and −2.
= p² + 8p − 2p − 16 = p(p + 8) − 2(p + 8) = (p + 8)(p − 2).
• (iv) x² − 2x − 15:
Sum = −2, Product = −15. The numbers are −5 and 3.
= x² − 5x + 3x − 15 = x(x − 5) + 3(x − 5) = (x − 5)(x + 3).
Sum = 6, Product = 8. The numbers are 4 and 2.
= p² + 4p + 2p + 8 = p(p + 4) + 2(p + 4) = (p + 4)(p + 2).
• (ii) q² − 10q + 21:
Sum = −10, Product = 21. The numbers are −7 and −3.
= q² − 7q − 3q + 21 = q(q − 7) − 3(q − 7) = (q − 7)(q − 3).
• (iii) p² + 6p − 16:
Sum = 6, Product = −16. The numbers are 8 and −2.
= p² + 8p − 2p − 16 = p(p + 8) − 2(p + 8) = (p + 8)(p − 2).
• (iv) x² − 2x − 15:
Sum = −2, Product = −15. The numbers are −5 and 3.
= x² − 5x + 3x − 15 = x(x − 5) + 3(x − 5) = (x − 5)(x + 3).
CBSE Board Marking Scheme 2026-27
Splitting middle terms based on sum/product pairs1.5 Marks
Correct binomial factors for all 4 trinomials1.5 Marks
Q10
Divide the polynomial by the given binomial:
(i) (y² + 7y + 10) ÷ (y + 5)
(ii) (m² − 14m − 32) ÷ (m + 2)
(iii) 5pq(p² − q²) ÷ 2p(p + q)
(i) (y² + 7y + 10) ÷ (y + 5)
(ii) (m² − 14m − 32) ÷ (m + 2)
(iii) 5pq(p² − q²) ÷ 2p(p + q)
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) (y² + 7y + 10) ÷ (y + 5):
Factorise numerator: y² + 7y + 10 = (y + 2)(y + 5).
Division = [(y + 2)(y + 5)] / (y + 5) = y + 2.
• (ii) (m² − 14m − 32) ÷ (m + 2):
Factorise numerator: Sum = −14, Product = −32 (numbers −16 and 2).
m² − 14m − 32 = (m − 16)(m + 2).
Division = [(m − 16)(m + 2)] / (m + 2) = m − 16.
• (iii) 5pq(p² − q²) ÷ 2p(p + q):
Factorise (p² − q²) = (p − q)(p + q).
Division = [5pq(p − q)(p + q)] / [2p(p + q)].
Cancelling common factor p(p + q):
= 5/2 q(p − q) (or 2.5q(p − q)).
Factorise numerator: y² + 7y + 10 = (y + 2)(y + 5).
Division = [(y + 2)(y + 5)] / (y + 5) = y + 2.
• (ii) (m² − 14m − 32) ÷ (m + 2):
Factorise numerator: Sum = −14, Product = −32 (numbers −16 and 2).
m² − 14m − 32 = (m − 16)(m + 2).
Division = [(m − 16)(m + 2)] / (m + 2) = m − 16.
• (iii) 5pq(p² − q²) ÷ 2p(p + q):
Factorise (p² − q²) = (p − q)(p + q).
Division = [5pq(p − q)(p + q)] / [2p(p + q)].
Cancelling common factor p(p + q):
= 5/2 q(p − q) (or 2.5q(p − q)).
CBSE Board Marking Scheme 2026-27
Factorising numerators before cancellation1.5 Marks
Accurate quotients: (y + 2), (m − 16), 5/2 q(p − q)1.5 Marks
Q11
Find and correct the errors in the following mathematical statements:
(i) 4(x − 5) = 4x − 5
(ii) (2x)² + 4(2x) + 7 = 2x² + 8x + 7
(iii) (3x + 2)² = 9x² + 4
(i) 4(x − 5) = 4x − 5
(ii) (2x)² + 4(2x) + 7 = 2x² + 8x + 7
(iii) (3x + 2)² = 9x² + 4
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (i) In 4(x − 5) = 4x − 5:
Error: The factor 4 was not distributed to the second term −5.
Correct statement:
• (ii) In (2x)² + 4(2x) + 7 = 2x² + 8x + 7:
Error: In (2x)², the coefficient 2 was not squared.
(2x)² = 2² × x² = 4x².
Correct statement:
• (iii) In (3x + 2)² = 9x² + 4:
Error: The middle term 2ab was completely omitted.
Using Identity I: (3x + 2)² = (3x)² + 2(3x)(2) + 2² = 9x² + 12x + 4.
Correct statement:
Error: The factor 4 was not distributed to the second term −5.
Correct statement:
4(x − 5) = 4x − 20.• (ii) In (2x)² + 4(2x) + 7 = 2x² + 8x + 7:
Error: In (2x)², the coefficient 2 was not squared.
(2x)² = 2² × x² = 4x².
Correct statement:
(2x)² + 4(2x) + 7 = 4x² + 8x + 7.• (iii) In (3x + 2)² = 9x² + 4:
Error: The middle term 2ab was completely omitted.
Using Identity I: (3x + 2)² = (3x)² + 2(3x)(2) + 2² = 9x² + 12x + 4.
Correct statement:
(3x + 2)² = 9x² + 12x + 4.
CBSE Board Marking Scheme 2026-27
Identifying exact conceptual error for each statement1 Mark
Providing corrected algebraic identities1 Mark
Q12
Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let the units digit be
Since sum of digits is 9, tens digit =
Original number = 10(tens digit) + units digit
= 10(9 − x) + x = 90 − 10x + x = 90 − 9x.
2. On interchanging digits:
New units digit = 9 − x, New tens digit = x.
New number = 10x + (9 − x) = 9x + 9.
3. Given: (New Number) − (Original Number) = 27
(9x + 9) − (90 − 9x) = 27
9x + 9 − 90 + 9x = 27
18x − 81 = 27
18x = 27 + 81 = 108
x = 108 / 18 = 6.
4. Therefore:
Units digit x = 6.
Tens digit = 9 − 6 = 3.
The original number is 36.
Verification: Reversal = 63. Difference = 63 − 36 = 27. Sum of digits = 3 + 6 = 9. Verified.
x.Since sum of digits is 9, tens digit =
9 − x.Original number = 10(tens digit) + units digit
= 10(9 − x) + x = 90 − 10x + x = 90 − 9x.
2. On interchanging digits:
New units digit = 9 − x, New tens digit = x.
New number = 10x + (9 − x) = 9x + 9.
3. Given: (New Number) − (Original Number) = 27
(9x + 9) − (90 − 9x) = 27
9x + 9 − 90 + 9x = 27
18x − 81 = 27
18x = 27 + 81 = 108
x = 108 / 18 = 6.
4. Therefore:
Units digit x = 6.
Tens digit = 9 − 6 = 3.
The original number is 36.
Verification: Reversal = 63. Difference = 63 − 36 = 27. Sum of digits = 3 + 6 = 9. Verified.
CBSE Board Marking Scheme 2026-27
Expressing original number as 90 − 9x and reversed as 9x + 91.5 Marks
Solving 18x = 108 ⇒ x = 6 and number = 361.5 Marks
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Sustainable Craft Micro-Enterprise Financial Model: A self-help cooperative manufactures handmade jute shopping bags. The fixed monthly operational expense (rent, electricity, machinery depreciation) is ₹ 12,000. Each bag incurs a variable material and artisan labor cost of ₹ 60.
The cooperative retails the finished bags at a uniform price of ₹ 140 per bag.
(a) Formulate an algebraic expression for the Total Production Cost
(b) Set up and solve a linear equation in one variable to determine the Break-Even Volume (the number of bags that must be sold to cover all expenses exactly).
(c) How many bags must be sold in a festive month to generate a net profit of ₹ 20,000?
The cooperative retails the finished bags at a uniform price of ₹ 140 per bag.
(a) Formulate an algebraic expression for the Total Production Cost
C(x) and Total Revenue R(x) when x bags are manufactured and sold.(b) Set up and solve a linear equation in one variable to determine the Break-Even Volume (the number of bags that must be sold to cover all expenses exactly).
(c) How many bags must be sold in a festive month to generate a net profit of ₹ 20,000?
Detailed Analytical Solution & Marking Scheme
(a) Cost and Revenue Functions:
- Total Cost: C(x) = 12000 + 60x (in ₹).
- Total Revenue: R(x) = 140x (in ₹).
(b) Break-Even Volume Equation:
At break-even: Total Revenue = Total Cost.
140x = 12000 + 60x
140x − 60x = 12000
80x = 12000
x = 12000 / 80 = 150 bags.
The cooperative must produce and sell at least 150 bags to break even.
(c) Production Volume for ₹ 20,000 Profit:
Profit = Revenue − Cost = 140x − (12000 + 60x) = 80x − 12000.
80x − 12000 = 20000
80x = 32000
x = 32000 / 80 = 400 bags.
Selling 400 bags yields exactly ₹ 20,000 net profit.
14
Chapter 14: Area
NCERT Ganita Prakash (Class 8 Part 2) — Mensuration, Area of Trapeziums & Polygons, Surface Area & Volume of Cubes, Cuboids & Cylinders | CBSE 2026-27
📌 Key Mensuration Formulas & 3D Solid Geometry
- Area of 2D Polygons:
- Trapezium:
Area = 1/2 × (a + b) × h, where a and b are parallel sides, and h is the perpendicular distance between them. - General Quadrilateral:
Area = 1/2 × d × (h₁ + h₂), where d is a diagonal, and h₁, h₂ are perpendicular offsets from opposite vertices. - Rhombus:
Area = 1/2 × d₁ × d₂, where d₁ and d₂ are diagonal lengths.
- Trapezium:
- Surface Area & Volume of 3D Solids:
- Cube (edge l):
- Lateral Surface Area (LSA) =
4l² - Total Surface Area (TSA) =
6l² - Volume =
l³
- Lateral Surface Area (LSA) =
- Cuboid (length l, breadth b, height h):
- Lateral Surface Area (Area of 4 walls) =
2(l + b)h - Total Surface Area (TSA) =
2(lb + bh + hl) - Volume =
l × b × h
- Lateral Surface Area (Area of 4 walls) =
- Right Circular Cylinder (radius r, height h):
- Curved Surface Area (CSA) =
2πrh - Total Surface Area (TSA) =
2πr(r + h) - Volume =
πr²h
- Curved Surface Area (CSA) =
- Cube (edge l):
- Metric Capacity Conversions:
1 cm³ = 1 mL1 L = 1,000 cm³ = 1,000 mL1 m³ = 1,000,000 cm³ = 1,000 Litres = 1 kL
Figure 14.1: Area Partitioning in 2D Trapeziums & Surface Enclosure in 3D Cylinders
NCERT Ganita Prakash: Figure It Out & Comprehensive Exercises (100% Questions Solved)
Q1
The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 14.1: Trapezoidal Surface with Parallel Sides 1.2 m & 1.0 m and Perpendicular Distance 0.8 m
Trapezium Area Formula: Area = 1/2 × (a + b) × h.
1. Given: Parallel sides a = 1 m, b = 1.2 m, and height h = 0.8 m.
2. Substituting values:
Area = 1/2 × (1 + 1.2) × 0.8
Area = 1/2 × 2.2 × 0.8
Area = 1.1 × 0.8 = 0.88 m².
Hence, the area of the top surface of the table is 0.88 m².
2. Substituting values:
Area = 1/2 × (1 + 1.2) × 0.8
Area = 1/2 × 2.2 × 0.8
Area = 1.1 × 0.8 = 0.88 m².
Hence, the area of the top surface of the table is 0.88 m².
CBSE Board Marking Scheme 2026-27
Formula: Area = 1/2(a + b)h1 Mark
Solving Area = 0.88 m²1 Mark
Q2
The area of a trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let the length of the unknown parallel side be
Given: Area = 34 cm², side a = 10 cm, height h = 4 cm.
2. Using formula:
34 = 1/2 × (10 + b) × 4
34 = (10 + b) × 2
3. Solving for b:
10 + b = 34 / 2 = 17
b = 17 − 10 = 7 cm.
Therefore, the length of the other parallel side is 7 cm.
b cm.Given: Area = 34 cm², side a = 10 cm, height h = 4 cm.
2. Using formula:
Area = 1/2 × (a + b) × h34 = 1/2 × (10 + b) × 4
34 = (10 + b) × 2
3. Solving for b:
10 + b = 34 / 2 = 17
b = 17 − 10 = 7 cm.
Therefore, the length of the other parallel side is 7 cm.
CBSE Board Marking Scheme 2026-27
Equation setup: 34 = 2(10 + b)1 Mark
b = 7 cm1 Mark
Q3
The diagonal of a quadrilateral-shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 14.2: Field Area = ½ × d × (h₁ + h₂) = ½ × 24 × (13 + 8) = 252 m²
General Quadrilateral Area Formula: Area = 1/2 × d × (h₁ + h₂).
1. Given: Diagonal d = 24 m, perpendicular offsets h₁ = 13 m and h₂ = 8 m.
2. Substituting values:
Area = 1/2 × 24 × (13 + 8)
Area = 12 × 21 = 252 m².
Hence, the area of the quadrilateral field is 252 m².
2. Substituting values:
Area = 1/2 × 24 × (13 + 8)
Area = 12 × 21 = 252 m².
Hence, the area of the quadrilateral field is 252 m².
CBSE Board Marking Scheme 2026-27
Formula: Area = 1/2 × d × (h₁ + h₂)1 Mark
Calculating Area = 252 m²1 Mark
Q4
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area. If the perimeter of another rhombus is 40 cm and its altitude is 5 cm, find its area.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• Part 1: Rhombus with diagonals d₁ = 7.5 cm, d₂ = 12 cm:
Area = 1/2 × d₁ × d₂
Area = 1/2 × 7.5 × 12 = 7.5 × 6 = 45 cm².
• Part 2: Rhombus with Perimeter = 40 cm and Altitude = 5 cm:
Since all 4 sides of a rhombus are equal:
Side = Perimeter / 4 = 40 / 4 = 10 cm.
Since a rhombus is also a parallelogram:
Area = Base × Altitude = Side × Altitude
Area = 10 × 5 = 50 cm².
Area = 1/2 × d₁ × d₂
Area = 1/2 × 7.5 × 12 = 7.5 × 6 = 45 cm².
• Part 2: Rhombus with Perimeter = 40 cm and Altitude = 5 cm:
Since all 4 sides of a rhombus are equal:
Side = Perimeter / 4 = 40 / 4 = 10 cm.
Since a rhombus is also a parallelogram:
Area = Base × Altitude = Side × Altitude
Area = 10 × 5 = 50 cm².
CBSE Board Marking Scheme 2026-27
Diagonals area = 1/2 × 7.5 × 12 = 45 cm²1.5 Marks
Side = 10 cm and Base × Height Area = 50 cm²1.5 Marks
Q5
There are two cuboidal boxes. Box A has dimensions 60 cm × 40 cm × 50 cm. Box B is a cube of side 50 cm. Which box requires the lesser amount of material to make?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 14.3: Surface Area Comparison: Box A (Cuboid: 14,800 cm²) requires less material than Box B (Cube: 15,000 cm²)
Criterion: The material required to make a closed box corresponds to its Total Surface Area (TSA).
• For Cuboidal Box A (l = 60 cm, b = 40 cm, h = 50 cm):
TSA = 2(lb + bh + hl)
TSA = 2[(60 × 40) + (40 × 50) + (50 × 60)]
TSA = 2[2400 + 2000 + 3000] = 2[7400] = 14,800 cm².
• For Cubic Box B (side l = 50 cm):
TSA = 6l²
TSA = 6 × 50² = 6 × 2500 = 15,000 cm².
• Comparison:
Since 14,800 cm² < 15,000 cm², Box A requires lesser material to make (by 200 cm²).
TSA = 2(lb + bh + hl)
TSA = 2[(60 × 40) + (40 × 50) + (50 × 60)]
TSA = 2[2400 + 2000 + 3000] = 2[7400] = 14,800 cm².
• For Cubic Box B (side l = 50 cm):
TSA = 6l²
TSA = 6 × 50² = 6 × 2500 = 15,000 cm².
• Comparison:
Since 14,800 cm² < 15,000 cm², Box A requires lesser material to make (by 200 cm²).
CBSE Board Marking Scheme 2026-27
TSA of Box A = 14,800 cm²1.5 Marks
TSA of Box B = 15,000 cm² and concluding Box A1.5 Marks
Q6
A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Total Surface Area of 1 suitcase (l = 80, b = 48, h = 24):
TSA = 2(lb + bh + hl)
= 2[(80 × 48) + (48 × 24) + (24 × 80)]
= 2[3840 + 1152 + 1920] = 2[6912] = 13,824 cm².
2. Total Surface Area of 100 suitcases:
Total Area = 100 × 13,824 = 1,382,400 cm².
3. Tarpaulin cloth is rectangular with width = 96 cm and unknown length L:
Area of cloth = Width × Length
1,382,400 = 96 × L
L = 1,382,400 / 96 = 14,400 cm.
4. Converting to meters:
Length in meters = 14,400 / 100 = 144 meters.
Therefore, 144 meters of tarpaulin cloth is required.
TSA = 2(lb + bh + hl)
= 2[(80 × 48) + (48 × 24) + (24 × 80)]
= 2[3840 + 1152 + 1920] = 2[6912] = 13,824 cm².
2. Total Surface Area of 100 suitcases:
Total Area = 100 × 13,824 = 1,382,400 cm².
3. Tarpaulin cloth is rectangular with width = 96 cm and unknown length L:
Area of cloth = Width × Length
1,382,400 = 96 × L
L = 1,382,400 / 96 = 14,400 cm.
4. Converting to meters:
Length in meters = 14,400 / 100 = 144 meters.
Therefore, 144 meters of tarpaulin cloth is required.
CBSE Board Marking Scheme 2026-27
TSA of one suitcase = 13,824 cm²1.5 Marks
Total area / 96 = 14,400 cm = 144 meters1.5 Marks
Q7
A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of the road roller is 84 cm and length is 1 m. (Take π = 22/7)
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Figure 14.4: In 1 revolution, roller covers CSA = 2πrh = 2.64 m²; in 750 revolutions = 1980 m²
Principle: In one revolution, a cylinder levels an area equal to its Curved Surface Area (CSA = 2πrh).
1. Dimensions of the cylindrical roller:
Diameter = 84 cm ⇒ Radius r = 84 / 2 = 42 cm = 0.42 m.
Length (height h) = 1 m.
2. Curved Surface Area (CSA):
CSA = 2πrh = 2 × (22/7) × 0.42 × 1
CSA = 2 × 22 × 0.06 × 1 = 2.64 m².
3. Total area leveled in 750 revolutions:
Total Area = 750 × CSA = 750 × 2.64 = 1,980 m².
Hence, the area of the road is 1,980 m².
Diameter = 84 cm ⇒ Radius r = 84 / 2 = 42 cm = 0.42 m.
Length (height h) = 1 m.
2. Curved Surface Area (CSA):
CSA = 2πrh = 2 × (22/7) × 0.42 × 1
CSA = 2 × 22 × 0.06 × 1 = 2.64 m².
3. Total area leveled in 750 revolutions:
Total Area = 750 × CSA = 750 × 2.64 = 1,980 m².
Hence, the area of the road is 1,980 m².
CBSE Board Marking Scheme 2026-27
CSA in 1 revolution = 2.64 m²1.5 Marks
Area leveled = 750 × 2.64 = 1,980 m²1.5 Marks
Q8
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m³ = 1000 litres)
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Dimensions: Length l = 6 m, Breadth b = 5 m, Depth (height) h = 4.5 m.
2. Volume of the cuboidal tank:
Volume = l × b × h
Volume = 6 × 5 × 4.5 = 30 × 4.5 = 135 m³.
3. Capacity in litres:
Since 1 m³ = 1,000 litres:
Capacity = 135 × 1,000 = 1,35,000 litres.
Hence, the tank can hold 1,35,000 litres of water.
2. Volume of the cuboidal tank:
Volume = l × b × h
Volume = 6 × 5 × 4.5 = 30 × 4.5 = 135 m³.
3. Capacity in litres:
Since 1 m³ = 1,000 litres:
Capacity = 135 × 1,000 = 1,35,000 litres.
Hence, the tank can hold 1,35,000 litres of water.
CBSE Board Marking Scheme 2026-27
Volume = 6 × 5 × 4.5 = 135 m³1 Mark
Conversion: 135 × 1000 = 1,35,000 litres1 Mark
Q9
Find the height of a cylinder whose radius is 7 cm and the total surface area is 968 cm². (Take π = 22/7)
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Formula: Total Surface Area (TSA) = 2πr(r + h).
1. Given: Radius r = 7 cm, TSA = 968 cm².
2. Substituting into formula:
968 = 2 × (22/7) × 7 × (7 + h)
968 = 44 × (7 + h)
3. Solving for h:
7 + h = 968 / 44 = 22
h = 22 − 7 = 15 cm.
Therefore, the height of the cylinder is 15 cm.
2. Substituting into formula:
968 = 2 × (22/7) × 7 × (7 + h)
968 = 44 × (7 + h)
3. Solving for h:
7 + h = 968 / 44 = 22
h = 22 − 7 = 15 cm.
Therefore, the height of the cylinder is 15 cm.
CBSE Board Marking Scheme 2026-27
Formulating 968 = 44(7 + h)1 Mark
Solving h = 15 cm1 Mark
Q10
If each edge of a cube is doubled: (i) how many times will its surface area increase? (ii) how many times will its volume increase?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Let the original edge of the cube be
Original Surface Area:
Original Volume:
2. When the edge is doubled, new edge =
• (i) New Surface Area (S₂):
S₂ = 6(2l)² = 6(4l²) = 4 × (6l²) = 4 × S₁.
Therefore, the surface area increases by 4 times (or 2² times).
• (ii) New Volume (V₂):
V₂ = (2l)³ = 8l³ = 8 × V₁.
Therefore, the volume increases by 8 times (or 2³ times).
l.Original Surface Area:
S₁ = 6l².Original Volume:
V₁ = l³.2. When the edge is doubled, new edge =
2l.• (i) New Surface Area (S₂):
S₂ = 6(2l)² = 6(4l²) = 4 × (6l²) = 4 × S₁.
Therefore, the surface area increases by 4 times (or 2² times).
• (ii) New Volume (V₂):
V₂ = (2l)³ = 8l³ = 8 × V₁.
Therefore, the volume increases by 8 times (or 2³ times).
CBSE Board Marking Scheme 2026-27
(i) Surface area increases 4 times (2² factor)1 Mark
(ii) Volume increases 8 times (2³ factor)1 Mark
🎯 Competency-Based Case Study Question (CBSE 2026-27 Pattern)
Context — Municipal Rainwater Harvesting Cistern Architecture: A residential society constructs a cylindrical concrete rainwater harvesting tank with an internal radius of
To prevent seepage and algae buildup, the society decides to waterproof both the interior curved wall and the base circular floor with an epoxy coating.
(a) Calculate the total surface area that needs to be coated (curved cylindrical wall + flat bottom circular floor).
(b) If epoxy waterproofing costs ₹ 120 per square meter, calculate the total cost of coating the tank.
(c) Calculate the maximum water storage capacity of the tank in cubic meters and in litres. (Take π = 22/7).
3.5 meters and depth (height) of 8 meters.To prevent seepage and algae buildup, the society decides to waterproof both the interior curved wall and the base circular floor with an epoxy coating.
(a) Calculate the total surface area that needs to be coated (curved cylindrical wall + flat bottom circular floor).
(b) If epoxy waterproofing costs ₹ 120 per square meter, calculate the total cost of coating the tank.
(c) Calculate the maximum water storage capacity of the tank in cubic meters and in litres. (Take π = 22/7).
Detailed Analytical Solution & Marking Scheme
(a) Surface Area to be Coated:
Area = Curved Surface Area (CSA) + Base Circle Area
= 2πrh + πr² = πr(2h + r)
= (22/7) × 3.5 × [2(8) + 3.5]
= 11 × [16 + 3.5] = 11 × 19.5 = 214.5 m².
(b) Total Cost of Waterproofing:
Cost = Area × Rate = 214.5 × ₹ 120 = ₹ 25,740.
(c) Maximum Water Storage Capacity:
Volume = πr²h = (22/7) × (3.5)² × 8
= (22/7) × 12.25 × 8 = 22 × 1.75 × 8 = 308 m³.
In litres = 308 × 1,000 = 3,08,000 litres.