1
Chapter 1: Patterns in Mathematics
NCERT Ganita Prakash (Class 6) — Number Sequences, Triangular & Square Numbers, Matchstick Patterns, Palindromes & Rule Formulation | CBSE 2026-27
📌 Key Mathematical Concepts & Formulas
- Patterns: A sequence of numbers, geometric shapes, or designs that follow a definite, repeatable rule.
- Triangular Numbers: Numbers that can be arranged in the shape of an equilateral triangle: 1, 3, 6, 10, 15, 21, 28, 36, ... The n-th triangular number is given by $T_n = \frac{n(n + 1)}{2}$.
- Square Numbers: Numbers formed by multiplying an integer by itself: $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, ...$ ($S_n = n^2$). Sum of two consecutive triangular numbers gives a square number: $T_{n-1} + T_n = n^2$.
- Sum of First n Odd Numbers: The sum of the first $n$ odd consecutive numbers equals $n^2$: $1 + 3 + 5 + \dots + (2n - 1) = n^2$.
- Matchstick Patterns & Algebraic Rules: If each additional unit requires a fixed number of sticks $d$, the rule for $n$ shapes is given by $d \cdot n + c$. For example, a row of $n$ squares requires $3n + 1$ matchsticks; a row of $n$ triangles requires $2n + 1$ matchsticks.
Figure 1.1: Dot Array Models for Triangular Numbers Sequence
NCERT Exercise 1.1: Number Patterns & Triangular Numbers
Q1
Observe the number sequence: 1, 3, 6, 10, 15, ...
(i) Write the next three numbers of this sequence.
(ii) What is the special mathematical name given to these numbers?
(iii) Show that the sum of the 3ʳᵈ and 4ᵗʰ numbers forms a perfect square.
(i) Write the next three numbers of this sequence.
(ii) What is the special mathematical name given to these numbers?
(iii) Show that the sum of the 3ʳᵈ and 4ᵗʰ numbers forms a perfect square.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Pattern Rule & Next Three Numbers:
Let us find the difference between consecutive terms:
3 - 1 = 2, 6 - 3 = 3, 10 - 6 = 4, 15 - 10 = 5.
Each step adds the next natural number (+2, +3, +4, +5).
Therefore:
- Next term: 15 + 6 = 21
- Next term: 21 + 7 = 28
- Next term: 28 + 8 = 36
The next three numbers are 21, 28, and 36.
Let us find the difference between consecutive terms:
3 - 1 = 2, 6 - 3 = 3, 10 - 6 = 4, 15 - 10 = 5.
Each step adds the next natural number (+2, +3, +4, +5).
Therefore:
- Next term: 15 + 6 = 21
- Next term: 21 + 7 = 28
- Next term: 28 + 8 = 36
The next three numbers are 21, 28, and 36.
(ii) Mathematical Name:
These numbers are called Triangular Numbers because dots representing them can be arranged into an equilateral triangle.
These numbers are called Triangular Numbers because dots representing them can be arranged into an equilateral triangle.
(iii) Sum of 3ʳᵈ and 4ᵗʰ Triangular Numbers:
3ʳᵈ triangular number = 6; 4ᵗʰ triangular number = 10.
Sum = 6 + 10 = 16 = 4².
Since 16 = 4 × 4, the sum is indeed a perfect square.
3ʳᵈ triangular number = 6; 4ᵗʰ triangular number = 10.
Sum = 6 + 10 = 16 = 4².
Since 16 = 4 × 4, the sum is indeed a perfect square.
CBSE Board Marking Scheme 2026-27
Part (i): Identifying pattern and writing 21, 28, 361 Mark
Part (ii): Stating "Triangular Numbers"1 Mark
Part (iii): Computing 6 + 10 = 16 and verifying 4²1 Mark
Q2
Find the 10ᵗʰ triangular number using the formula $T_n = \frac{n(n+1)}{2}$. Verify your answer by summing the first 10 natural numbers.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Step 1: Formula Calculation:
Given $n = 10$,
$T_{10} = \frac{10 \times (10 + 1)}{2} = \frac{10 \times 11}{2} = 5 \times 11 = \mathbf{55}$.
Given $n = 10$,
$T_{10} = \frac{10 \times (10 + 1)}{2} = \frac{10 \times 11}{2} = 5 \times 11 = \mathbf{55}$.
Step 2: Verification by Addition:
$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = (1 + 10) + (2 + 9) + (3 + 8) + (4 + 7) + (5 + 6)$
$= 11 \times 5 = \mathbf{55}$.
Both methods yield 55. Hence verified.
$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = (1 + 10) + (2 + 9) + (3 + 8) + (4 + 7) + (5 + 6)$
$= 11 \times 5 = \mathbf{55}$.
Both methods yield 55. Hence verified.
CBSE Board Marking Scheme 2026-27
Correct substitution in formula yielding 551 Mark
Verification by grouping or consecutive addition1 Mark
Figure 1.2: Matchstick Pattern of Connected Squares
NCERT Exercise 1.2: Matchstick Patterns & Algebraic Generalization
Q3
A student makes a row of connected squares using matchsticks as shown in Figure 1.2.
(i) Find the number of matchsticks required for 1 square, 2 squares, 3 squares, and 4 squares.
(ii) Write the general rule for the number of matchsticks required to make n squares.
(iii) How many matchsticks are needed to make a row of 25 connected squares?
(i) Find the number of matchsticks required for 1 square, 2 squares, 3 squares, and 4 squares.
(ii) Write the general rule for the number of matchsticks required to make n squares.
(iii) How many matchsticks are needed to make a row of 25 connected squares?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Count of Matchsticks:
- For 1 square: 4 matchsticks.
- For 2 connected squares: 4 + 3 = 7 matchsticks (1 edge shared).
- For 3 connected squares: 7 + 3 = 10 matchsticks.
- For 4 connected squares: 10 + 3 = 13 matchsticks.
- For 1 square: 4 matchsticks.
- For 2 connected squares: 4 + 3 = 7 matchsticks (1 edge shared).
- For 3 connected squares: 7 + 3 = 10 matchsticks.
- For 4 connected squares: 10 + 3 = 13 matchsticks.
(ii) Formulation of General Rule:
Notice: 4 = 3(1) + 1, 7 = 3(2) + 1, 10 = 3(3) + 1, 13 = 3(4) + 1.
Therefore, for $n$ connected squares, the general rule is:
Number of matchsticks = 3n + 1.
Notice: 4 = 3(1) + 1, 7 = 3(2) + 1, 10 = 3(3) + 1, 13 = 3(4) + 1.
Therefore, for $n$ connected squares, the general rule is:
Number of matchsticks = 3n + 1.
(iii) Calculation for n = 25:
Matchsticks = 3(25) + 1 = 75 + 1 = 76 matchsticks.
Matchsticks = 3(25) + 1 = 75 + 1 = 76 matchsticks.
CBSE Board Marking Scheme 2026-27
Part (i): Tabulating stick counts (4, 7, 10, 13)1 Mark
Part (ii): Deriving formula 3n + 11 Mark
Part (iii): Correct calculation yielding 761 Mark
Q4
A series of connected equilateral triangles is formed using matchsticks.
(i) How many matchsticks are required for 1, 2, 3, and 4 triangles?
(ii) Find the rule for the number of matchsticks needed for n triangles.
(iii) If 41 matchsticks are used, how many triangles are formed in the row?
(i) How many matchsticks are required for 1, 2, 3, and 4 triangles?
(ii) Find the rule for the number of matchsticks needed for n triangles.
(iii) If 41 matchsticks are used, how many triangles are formed in the row?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Count of Matchsticks for Triangles:
- 1 triangle requires 3 matchsticks.
- 2 connected triangles share 1 common side: 3 + 2 = 5 matchsticks.
- 3 connected triangles: 5 + 2 = 7 matchsticks.
- 4 connected triangles: 7 + 2 = 9 matchsticks.
- 1 triangle requires 3 matchsticks.
- 2 connected triangles share 1 common side: 3 + 2 = 5 matchsticks.
- 3 connected triangles: 5 + 2 = 7 matchsticks.
- 4 connected triangles: 7 + 2 = 9 matchsticks.
(ii) General Rule:
Observe: 3 = 2(1) + 1, 5 = 2(2) + 1, 7 = 2(3) + 1, 9 = 2(4) + 1.
Therefore, for $n$ triangles: Matchsticks = 2n + 1.
Observe: 3 = 2(1) + 1, 5 = 2(2) + 1, 7 = 2(3) + 1, 9 = 2(4) + 1.
Therefore, for $n$ triangles: Matchsticks = 2n + 1.
(iii) Finding n when 41 matchsticks are used:
$2n + 1 = 41 \implies 2n = 40 \implies n = 20$.
Therefore, 20 triangles are formed.
$2n + 1 = 41 \implies 2n = 40 \implies n = 20$.
Therefore, 20 triangles are formed.
CBSE Board Marking Scheme 2026-27
Part (i): Correct counts (3, 5, 7, 9)1 Mark
Part (ii): Writing general rule 2n + 11 Mark
Part (iii): Solving 2n + 1 = 41 to get n = 201 Mark
Q5
Observe the pattern in the following number pyramid:
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
Write down the next two lines of this pattern without actual multiplication, and describe the rule.
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
Write down the next two lines of this pattern without actual multiplication, and describe the rule.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Pattern Rule Analysis:
- In the first factor, digits 1, 2, 3, ... are appended in ascending sequence.
- The number added is the next natural number (1, 2, 3, 4, 5, 6).
- The product digits decrease consecutively starting from 9.
Line 5: 12345 × 8 + 5 = 98765
Line 6: 123456 × 8 + 6 = 987654
- In the first factor, digits 1, 2, 3, ... are appended in ascending sequence.
- The number added is the next natural number (1, 2, 3, 4, 5, 6).
- The product digits decrease consecutively starting from 9.
Line 5: 12345 × 8 + 5 = 98765
Line 6: 123456 × 8 + 6 = 987654
CBSE Board Marking Scheme 2026-27
Line 5 correct: 12345 × 8 + 5 = 987651 Mark
Line 6 correct: 123456 × 8 + 6 = 9876541 Mark
Q6
A sequence is defined by the rule: "Multiply the term number by 4 and then subtract 3".
(i) Write the first 5 terms of this sequence.
(ii) Find the 20ᵗʰ term.
(iii) Is the number 85 a term in this sequence? Justify your answer.
(i) Write the first 5 terms of this sequence.
(ii) Find the 20ᵗʰ term.
(iii) Is the number 85 a term in this sequence? Justify your answer.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) First 5 Terms:
Formula: $a_n = 4n - 3$.
- For $n = 1$: $4(1) - 3 = 1$
- For $n = 2$: $4(2) - 3 = 5$
- For $n = 3$: $4(3) - 3 = 9$
- For $n = 4$: $4(4) - 3 = 13$
- For $n = 5$: $4(5) - 3 = 17$
The first 5 terms are 1, 5, 9, 13, 17.
Formula: $a_n = 4n - 3$.
- For $n = 1$: $4(1) - 3 = 1$
- For $n = 2$: $4(2) - 3 = 5$
- For $n = 3$: $4(3) - 3 = 9$
- For $n = 4$: $4(4) - 3 = 13$
- For $n = 5$: $4(5) - 3 = 17$
The first 5 terms are 1, 5, 9, 13, 17.
(ii) 20ᵗʰ Term:
$a_{20} = 4(20) - 3 = 80 - 3 = \mathbf{77}$.
$a_{20} = 4(20) - 3 = 80 - 3 = \mathbf{77}$.
(iii) Checking if 85 is in the sequence:
Set $4n - 3 = 85 \implies 4n = 88 \implies n = 22$.
Since $n = 22$ is a whole number, Yes, 85 is the 22ⁿᵈ term.
Set $4n - 3 = 85 \implies 4n = 88 \implies n = 22$.
Since $n = 22$ is a whole number, Yes, 85 is the 22ⁿᵈ term.
CBSE Board Marking Scheme 2026-27
Part (i): Computing first 5 terms (1, 5, 9, 13, 17)1 Mark
Part (ii): Finding a₂₀ = 771 Mark
Part (iii): Solving 4n - 3 = 85 and concluding n = 221 Mark
Q7
Observe the dot pattern for square numbers: 1, 4, 9, 16, 25, 36, ...
(i) Express 25 as the sum of consecutive odd natural numbers.
(ii) Express 36 as the sum of two consecutive triangular numbers.
(iii) Express 49 as the sum of consecutive odd natural numbers.
(i) Express 25 as the sum of consecutive odd natural numbers.
(ii) Express 36 as the sum of two consecutive triangular numbers.
(iii) Express 49 as the sum of consecutive odd natural numbers.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Expressing 25 as sum of odd numbers:
$25 = 5^2$. It is the sum of the first 5 odd natural numbers:
$25 = \mathbf{1 + 3 + 5 + 7 + 9}$.
$25 = 5^2$. It is the sum of the first 5 odd natural numbers:
$25 = \mathbf{1 + 3 + 5 + 7 + 9}$.
(ii) Expressing 36 as sum of two consecutive triangular numbers:
$36 = 6^2$. A square number $n^2$ is the sum of the $(n-1)$-th and $n$-th triangular numbers.
$T_5 = \frac{5 \times 6}{2} = 15$; $T_6 = \frac{6 \times 7}{2} = 21$.
$36 = \mathbf{15 + 21}$.
$36 = 6^2$. A square number $n^2$ is the sum of the $(n-1)$-th and $n$-th triangular numbers.
$T_5 = \frac{5 \times 6}{2} = 15$; $T_6 = \frac{6 \times 7}{2} = 21$.
$36 = \mathbf{15 + 21}$.
(iii) Expressing 49 as sum of odd numbers:
$49 = 7^2$. It is the sum of the first 7 odd natural numbers:
$49 = \mathbf{1 + 3 + 5 + 7 + 9 + 11 + 13}$.
$49 = 7^2$. It is the sum of the first 7 odd natural numbers:
$49 = \mathbf{1 + 3 + 5 + 7 + 9 + 11 + 13}$.
CBSE Board Marking Scheme 2026-27
Part (i): Writing 1 + 3 + 5 + 7 + 9 = 251 Mark
Part (ii): Writing 15 + 21 = 361 Mark
Part (iii): Writing 1 + 3 + 5 + 7 + 9 + 11 + 13 = 491 Mark
Q8
Observe the palindromic multiplication pattern:
11 × 11 = 121
111 × 111 = 12321
1111 × 1111 = 1234321
(i) Without actual calculation, predict the value of 11111 × 11111.
(ii) What is the square root of 12345654321?
11 × 11 = 121
111 × 111 = 12321
1111 × 1111 = 1234321
(i) Without actual calculation, predict the value of 11111 × 11111.
(ii) What is the square root of 12345654321?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Value of 11111 × 11111:
Since there are 5 ones, digits increase from 1 up to 5 and then decrease back to 1:
$11111 \times 11111 = \mathbf{123454321}$.
Since there are 5 ones, digits increase from 1 up to 5 and then decrease back to 1:
$11111 \times 11111 = \mathbf{123454321}$.
(ii) Square root of 12345654321:
The peak middle digit is 6, which indicates the number is the square of a 6-digit number consisting of all 1s:
$\sqrt{12345654321} = \mathbf{111111}$.
The peak middle digit is 6, which indicates the number is the square of a 6-digit number consisting of all 1s:
$\sqrt{12345654321} = \mathbf{111111}$.
CBSE Board Marking Scheme 2026-27
Part (i): Stating 123454321 correctly1 Mark
Part (ii): Identifying 1111111 Mark
Q9
A calendar page displays a 3 × 3 square of dates as follows:
4 5 6
11 12 13
18 19 20
(i) Find the sum of all 9 numbers in this square.
(ii) Relate this total sum to the central number (12).
(iii) Does this rule hold true for any 3 × 3 date block on a calendar? Give a mathematical reason.
4 5 6
11 12 13
18 19 20
(i) Find the sum of all 9 numbers in this square.
(ii) Relate this total sum to the central number (12).
(iii) Does this rule hold true for any 3 × 3 date block on a calendar? Give a mathematical reason.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Sum of all 9 numbers:
$4 + 5 + 6 + 11 + 12 + 13 + 18 + 19 + 20 = 108$.
$4 + 5 + 6 + 11 + 12 + 13 + 18 + 19 + 20 = 108$.
(ii) Relationship to central number:
Notice that $108 = 9 \times 12$.
The total sum of all 9 numbers equals 9 times the central number.
Notice that $108 = 9 \times 12$.
The total sum of all 9 numbers equals 9 times the central number.
(iii) Mathematical Justification:
Let the central number be $x$. The 8 surrounding numbers are symmetric around $x$:
$(x-8) + (x-7) + (x-6) + (x-1) + x + (x+1) + (x+6) + (x+7) + (x+8) = 9x$.
The positive and negative offsets cancel out completely. Hence, the rule holds true for any 3 × 3 square on any calendar month.
Let the central number be $x$. The 8 surrounding numbers are symmetric around $x$:
$(x-8) + (x-7) + (x-6) + (x-1) + x + (x+1) + (x+6) + (x+7) + (x+8) = 9x$.
The positive and negative offsets cancel out completely. Hence, the rule holds true for any 3 × 3 square on any calendar month.
CBSE Board Marking Scheme 2026-27
Part (i): Calculating total sum 1081 Mark
Part (ii): Stating Sum = 9 × Central Number (9 × 12 = 108)1 Mark
Part (iii): Algebraic proof with cancellation of symmetric offsets1 Mark
Q10
A pattern of "matchstick houses" is built in a row. The first house requires 6 matchsticks. Each subsequent connected house requires 5 additional matchsticks.
(i) Write the rule for the number of matchsticks required for h houses.
(ii) Find the number of matchsticks needed to build 12 houses.
(iii) If a student has 106 matchsticks, how many complete houses can be built?
(i) Write the rule for the number of matchsticks required for h houses.
(ii) Find the number of matchsticks needed to build 12 houses.
(iii) If a student has 106 matchsticks, how many complete houses can be built?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Rule Formulation:
For 1 house: 6 matchsticks.
For 2 houses: 6 + 5 = 11 matchsticks.
For 3 houses: 11 + 5 = 16 matchsticks.
Rule for $h$ houses: $6 + 5(h - 1) = \mathbf{5h + 1}$.
For 1 house: 6 matchsticks.
For 2 houses: 6 + 5 = 11 matchsticks.
For 3 houses: 11 + 5 = 16 matchsticks.
Rule for $h$ houses: $6 + 5(h - 1) = \mathbf{5h + 1}$.
(ii) Matchsticks for 12 houses:
$5(12) + 1 = 60 + 1 = \mathbf{61\text{ matchsticks}}$.
$5(12) + 1 = 60 + 1 = \mathbf{61\text{ matchsticks}}$.
(iii) Houses from 106 matchsticks:
$5h + 1 = 106 \implies 5h = 105 \implies h = 21$.
Thus, 21 complete houses can be built.
$5h + 1 = 106 \implies 5h = 105 \implies h = 21$.
Thus, 21 complete houses can be built.
CBSE Board Marking Scheme 2026-27
Part (i): Deriving rule 5h + 11 Mark
Part (ii): Finding 61 matchsticks1 Mark
Part (iii): Solving for h = 211 Mark
Q11
Consider the Fibonacci-like pattern: 1, 1, 2, 3, 5, 8, 13, 21, ...
(i) Explain the rule used to generate each subsequent term.
(ii) Write the next two terms of this sequence.
(iii) What is the sum of the first 6 terms? Verify that it equals the 8ᵗʰ term minus 1.
(i) Explain the rule used to generate each subsequent term.
(ii) Write the next two terms of this sequence.
(iii) What is the sum of the first 6 terms? Verify that it equals the 8ᵗʰ term minus 1.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Generation Rule:
Each term after the first two is obtained by adding the two immediately preceding terms:
$1 + 1 = 2$, $1 + 2 = 3$, $2 + 3 = 5$, $3 + 5 = 8$, $5 + 8 = 13$, $8 + 13 = 21$.
Each term after the first two is obtained by adding the two immediately preceding terms:
$1 + 1 = 2$, $1 + 2 = 3$, $2 + 3 = 5$, $3 + 5 = 8$, $5 + 8 = 13$, $8 + 13 = 21$.
(ii) Next Two Terms:
- 9ᵗʰ term: $13 + 21 = \mathbf{34}$
- 10ᵗʰ term: $21 + 34 = \mathbf{55}$
- 9ᵗʰ term: $13 + 21 = \mathbf{34}$
- 10ᵗʰ term: $21 + 34 = \mathbf{55}$
(iii) Sum and Verification:
Sum of first 6 terms = $1 + 1 + 2 + 3 + 5 + 8 = 20$.
8ᵗʰ term = 21.
Verification: $21 - 1 = 20$. Hence verified.
Sum of first 6 terms = $1 + 1 + 2 + 3 + 5 + 8 = 20$.
8ᵗʰ term = 21.
Verification: $21 - 1 = 20$. Hence verified.
CBSE Board Marking Scheme 2026-27
Part (i): Stating addition of previous two terms1 Mark
Part (ii): Calculating 34 and 551 Mark
Part (iii): Verifying Sum = 20 = 21 - 11 Mark
CBSE Competency-Based Question (Real-Life Case Study)
Case Study: School Auditorium Seating Arrangement:
In a newly constructed school auditorium, the chairs in the rows are arranged in a triangular stage pattern. Row 1 has 3 chairs, Row 2 has 5 chairs, Row 3 has 7 chairs, and so on.
(a) Formulate the algebraic rule for the number of chairs in Row n.
(b) How many chairs are in the 15ᵗʰ row?
(c) If the last row contains 41 chairs, find the total number of rows in the auditorium.
(d) Calculate the total seating capacity of the first 10 rows.
In a newly constructed school auditorium, the chairs in the rows are arranged in a triangular stage pattern. Row 1 has 3 chairs, Row 2 has 5 chairs, Row 3 has 7 chairs, and so on.
(a) Formulate the algebraic rule for the number of chairs in Row n.
(b) How many chairs are in the 15ᵗʰ row?
(c) If the last row contains 41 chairs, find the total number of rows in the auditorium.
(d) Calculate the total seating capacity of the first 10 rows.
Detailed Analytical Solution & Marking Scheme
(a) Algebraic Rule:
The number of chairs in consecutive rows is 3, 5, 7, 9, ...
Common difference $d = 2$. For Row 1: $2(1) + 1 = 3$.
Therefore, for Row $n$: Chairs = 2n + 1.
(b) Chairs in 15ᵗʰ Row:
Substitute $n = 15$: Chairs = $2(15) + 1 = 30 + 1 = \mathbf{31\text{ chairs}}$.
(c) Total Number of Rows when Last Row has 41 Chairs:
$2n + 1 = 41 \implies 2n = 40 \implies n = \mathbf{20\text{ rows}}$.
(d) Total Seating Capacity of First 10 Rows:
Sum = $3 + 5 + 7 + \dots + 21$.
Notice this is $(1 + 3 + 5 + \dots + 21) - 1 = 11^2 - 1 = 121 - 1 = \mathbf{120\text{ seats}}$.
2
Chapter 2: Lines and Angles
NCERT Ganita Prakash (Class 6) — Points, Lines, Rays, Angle Classifications, Clock Revolutions, Protractor Measurement, Linear Pairs & Intersecting Lines | CBSE 2026-27
📌 Key Geometric Concepts & Angle Classifications
- Point, Line, Ray & Line Segment: A point marks a fixed position with zero dimensions. A line extends infinitely in both directions ($overleftrightarrow{AB}$). A line segment has two fixed endpoints ($overline{AB}$). A ray has one initial point and extends indefinitely in one direction ($ ec{AB}$).
- Angle Classifications:
- Acute Angle: Greater than 0° and less than 90°.
- Right Angle: Exactly equal to 90° ($\frac{1}{4}$ of a full revolution).
- Obtuse Angle: Greater than 90° and less than 180°.
- Straight Angle: Exactly equal to 180° ($\frac{1}{2}$ of a full revolution).
- Reflex Angle: Greater than 180° and less than 360°.
- Complete Angle: Exactly equal to 360° (1 full revolution).
- Complementary & Supplementary Angles: Two angles are complementary if their sum is 90°. Two angles are supplementary if their sum is 180°.
- Linear Pair & Vertically Opposite Angles: When two lines intersect, adjacent angles forming a straight line add to 180° (linear pair), and vertically opposite angles are equal.
Figure 2.1: Geometric Representation of Acute, Right, Obtuse, and Straight Angles
NCERT Exercise 2.1: Angle Types & Identification
Q1
Classify each of the following angles into acute, obtuse, right, straight, or reflex:
(i) 35° (ii) 90° (iii) 142° (iv) 180° (v) 245° (vi) 89°
(i) 35° (ii) 90° (iii) 142° (iv) 180° (v) 245° (vi) 89°
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) 35°: Since 35° < 90°, it is an Acute Angle.
(ii) 90°: Exactly 90°, it is a Right Angle.
(iii) 142°: Since 90° < 142° < 180°, it is an Obtuse Angle.
(iv) 180°: Exactly 180°, it is a Straight Angle.
(v) 245°: Since 180° < 245° < 360°, it is a Reflex Angle.
(vi) 89°: Since 89° < 90°, it is an Acute Angle.
CBSE Board Marking Scheme 2026-27
Each correct classification with reason0.5 Mark each (Total 3 Marks)
Figure 2.2: Clock Face Revolutions: Quarter and Half Turns
NCERT Exercise 2.2: Clock Angles & Direction Revolutions
Q2
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from:
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Formula: A clock has 12 equal hours. 1 full revolution = 12 hours = 360°.
Fraction of revolution = $\frac{\text{Number of hours passed}}{12}$.
(a) From 3 to 9: Hours passed = $9 - 3 = 6$. Fraction = $\frac{6}{12} = \mathbf{\frac{1}{2}}$ revolution (180°).
(b) From 4 to 7: Hours passed = $7 - 4 = 3$. Fraction = $\frac{3}{12} = \mathbf{\frac{1}{4}}$ revolution (90°).
(c) From 7 to 10: Hours passed = $10 - 7 = 3$. Fraction = $\frac{3}{12} = \mathbf{\frac{1}{4}}$ revolution (90°).
(d) From 12 to 9 (clockwise): Hours passed = 9. Fraction = $\frac{9}{12} = \mathbf{\frac{3}{4}}$ revolution (270°).
Fraction of revolution = $\frac{\text{Number of hours passed}}{12}$.
(a) From 3 to 9: Hours passed = $9 - 3 = 6$. Fraction = $\frac{6}{12} = \mathbf{\frac{1}{2}}$ revolution (180°).
(b) From 4 to 7: Hours passed = $7 - 4 = 3$. Fraction = $\frac{3}{12} = \mathbf{\frac{1}{4}}$ revolution (90°).
(c) From 7 to 10: Hours passed = $10 - 7 = 3$. Fraction = $\frac{3}{12} = \mathbf{\frac{1}{4}}$ revolution (90°).
(d) From 12 to 9 (clockwise): Hours passed = 9. Fraction = $\frac{9}{12} = \mathbf{\frac{3}{4}}$ revolution (270°).
CBSE Board Marking Scheme 2026-27
Formula statement: Hours / 120.5 Mark
Parts (a), (b), (c), (d) correct simplified fractions2.5 Marks
Q3
Where will the hand of a clock stop if it:
(a) Starts at 12 and makes 1/2 of a revolution, clockwise?
(b) Starts at 2 and makes 1/2 of a revolution, clockwise?
(c) Starts at 5 and makes 1/4 of a revolution, clockwise?
(d) Starts at 8 and makes 3/4 of a revolution, clockwise?
(a) Starts at 12 and makes 1/2 of a revolution, clockwise?
(b) Starts at 2 and makes 1/2 of a revolution, clockwise?
(c) Starts at 5 and makes 1/4 of a revolution, clockwise?
(d) Starts at 8 and makes 3/4 of a revolution, clockwise?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Starts at 12, 1/2 revolution:
Hours moved = $\frac{1}{2} \times 12 = 6$ hours. $12 + 6 = 18 \equiv 6$. Stops at 6.
Hours moved = $\frac{1}{2} \times 12 = 6$ hours. $12 + 6 = 18 \equiv 6$. Stops at 6.
(b) Starts at 2, 1/2 revolution:
Hours moved = $\frac{1}{2} \times 12 = 6$ hours. $2 + 6 = 8$. Stops at 8.
Hours moved = $\frac{1}{2} \times 12 = 6$ hours. $2 + 6 = 8$. Stops at 8.
(c) Starts at 5, 1/4 revolution:
Hours moved = $\frac{1}{4} \times 12 = 3$ hours. $5 + 3 = 8$. Stops at 8.
Hours moved = $\frac{1}{4} \times 12 = 3$ hours. $5 + 3 = 8$. Stops at 8.
(d) Starts at 8, 3/4 revolution:
Hours moved = $\frac{3}{4} \times 12 = 9$ hours. $8 + 9 = 17 \equiv 5$. Stops at 5.
Hours moved = $\frac{3}{4} \times 12 = 9$ hours. $8 + 9 = 17 \equiv 5$. Stops at 5.
CBSE Board Marking Scheme 2026-27
Each correct position with clear calculation0.75 Mark each (Total 3 Marks)
Q4
In Figure 2.3, two lines AB and CD intersect at point O. If ∠AOC = 50°, find the measures of ∠BOD, ∠AOD, and ∠BOC. Give reasons for each step.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Step 1: Vertically Opposite Angle:
$\angle BOD$ and $\angle AOC$ are vertically opposite angles.
Therefore, $\angle BOD = \angle AOC = \mathbf{50^\circ}$.
$\angle BOD$ and $\angle AOC$ are vertically opposite angles.
Therefore, $\angle BOD = \angle AOC = \mathbf{50^\circ}$.
Step 2: Linear Pair with ∠AOC:
$AB$ is a straight line, so ray $OC$ stands on it.
$\angle AOC + \angle BOC = 180^\circ$ (Linear pair property).
$50^\circ + \angle BOC = 180^\circ \implies \angle BOC = 180^\circ - 50^\circ = \mathbf{130^\circ}$.
$AB$ is a straight line, so ray $OC$ stands on it.
$\angle AOC + \angle BOC = 180^\circ$ (Linear pair property).
$50^\circ + \angle BOC = 180^\circ \implies \angle BOC = 180^\circ - 50^\circ = \mathbf{130^\circ}$.
Step 3: Vertically Opposite Angle for ∠AOD:
$\angle AOD = \angle BOC = \mathbf{130^\circ}$ (Vertically opposite angles).
Summary: $\angle BOD = 50^\circ$, $\angle AOD = 130^\circ$, $\angle BOC = 130^\circ$.
$\angle AOD = \angle BOC = \mathbf{130^\circ}$ (Vertically opposite angles).
Summary: $\angle BOD = 50^\circ$, $\angle AOD = 130^\circ$, $\angle BOC = 130^\circ$.
CBSE Board Marking Scheme 2026-27
∠BOD = 50° with vertically opposite angles reason1 Mark
∠BOC = 130° using linear pair property1 Mark
∠AOD = 130° with proper geometrical reason1 Mark
Q5
Explain the step-by-step procedure to measure an angle ∠PQR using a protractor. Also describe how to avoid parallax error during reading.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Procedure for Protractor Measurement:
1. Place the centre point of the protractor directly on the vertex $Q$ of the angle.
2. Adjust the protractor so that the base line (zero-edge) lies precisely along one arm, $QR$.
3. Look at the scale where the zero mark coincides with arm $QR$.
4. Read the mark on that scale where the second arm, $QP$, crosses the curved edge.
Avoiding Parallax Error: Position your eye directly perpendicular above the scale reading mark rather than viewing at an angle.
1. Place the centre point of the protractor directly on the vertex $Q$ of the angle.
2. Adjust the protractor so that the base line (zero-edge) lies precisely along one arm, $QR$.
3. Look at the scale where the zero mark coincides with arm $QR$.
4. Read the mark on that scale where the second arm, $QP$, crosses the curved edge.
Avoiding Parallax Error: Position your eye directly perpendicular above the scale reading mark rather than viewing at an angle.
CBSE Board Marking Scheme 2026-27
Accurate 4-step protractor alignment procedure1.5 Marks
Explanation of perpendicular eye position for zero parallax0.5 Mark
Q6
(a) Find the complement of: (i) 28° (ii) 63°
(b) Find the supplement of: (i) 105° (ii) 82°
(c) An angle is equal to its supplement. Find the measure of this angle.
(b) Find the supplement of: (i) 105° (ii) 82°
(c) An angle is equal to its supplement. Find the measure of this angle.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Complementary Angles (Sum = 90°):
(i) Complement of 28° = $90^\circ - 28^\circ = \mathbf{62^\circ}$.
(ii) Complement of 63° = $90^\circ - 63^\circ = \mathbf{27^\circ}$.
(i) Complement of 28° = $90^\circ - 28^\circ = \mathbf{62^\circ}$.
(ii) Complement of 63° = $90^\circ - 63^\circ = \mathbf{27^\circ}$.
(b) Supplementary Angles (Sum = 180°):
(i) Supplement of 105° = $180^\circ - 105^\circ = \mathbf{75^\circ}$.
(ii) Supplement of 82° = $180^\circ - 82^\circ = \mathbf{98^\circ}$.
(i) Supplement of 105° = $180^\circ - 105^\circ = \mathbf{75^\circ}$.
(ii) Supplement of 82° = $180^\circ - 82^\circ = \mathbf{98^\circ}$.
(c) Angle equal to its supplement:
Let the angle be $x$. $x + x = 180^\circ \implies 2x = 180^\circ \implies x = \mathbf{90^\circ}$.
Let the angle be $x$. $x + x = 180^\circ \implies 2x = 180^\circ \implies x = \mathbf{90^\circ}$.
CBSE Board Marking Scheme 2026-27
Part (a): Both complementary calculations correct1 Mark
Part (b): Both supplementary calculations correct1 Mark
Part (c): Equation 2x = 180° and x = 90°1 Mark
Q7
You are standing facing North.
(a) If you turn clockwise to face East, what fraction of a revolution have you turned? What angle is this?
(b) If you turn anti-clockwise from North to South, what angle have you turned?
(c) If you face West and turn 270° clockwise, which direction do you face?
(a) If you turn clockwise to face East, what fraction of a revolution have you turned? What angle is this?
(b) If you turn anti-clockwise from North to South, what angle have you turned?
(c) If you face West and turn 270° clockwise, which direction do you face?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) North to East (Clockwise):
Turn = 1 right turn = $90^\circ$. Fraction = $\frac{90}{360} = \mathbf{\frac{1}{4}}$ revolution.
Turn = 1 right turn = $90^\circ$. Fraction = $\frac{90}{360} = \mathbf{\frac{1}{4}}$ revolution.
(b) North to South (Anti-clockwise):
North to South is a straight half turn = $\mathbf{180^\circ}$ (Straight angle = $\frac{1}{2}$ revolution).
North to South is a straight half turn = $\mathbf{180^\circ}$ (Straight angle = $\frac{1}{2}$ revolution).
(c) West turning 270° clockwise:
- West to North = 90°
- North to East = 90° (cumulative 180°)
- East to South = 90° (cumulative 270°)
Therefore, you will be facing South.
- West to North = 90°
- North to East = 90° (cumulative 180°)
- East to South = 90° (cumulative 270°)
Therefore, you will be facing South.
CBSE Board Marking Scheme 2026-27
Part (a): 1/4 revolution and 90°1 Mark
Part (b): 180° straight angle1 Mark
Part (c): Step-by-step turn leading to South1 Mark
Q8
Two angles form a linear pair. If one angle is 36° more than the other angle, find the measure of both angles.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Let the smaller angle be $x$.
Then the larger angle is $(x + 36^\circ)$.
Since they form a linear pair, their sum is $180^\circ$:
$x + (x + 36^\circ) = 180^\circ$
$2x + 36^\circ = 180^\circ \implies 2x = 180^\circ - 36^\circ = 144^\circ$
$x = \frac{144^\circ}{2} = 72^\circ$.
Larger angle = $72^\circ + 36^\circ = 108^\circ$.
Answer: The two angles are 72° and 108°.
Then the larger angle is $(x + 36^\circ)$.
Since they form a linear pair, their sum is $180^\circ$:
$x + (x + 36^\circ) = 180^\circ$
$2x + 36^\circ = 180^\circ \implies 2x = 180^\circ - 36^\circ = 144^\circ$
$x = \frac{144^\circ}{2} = 72^\circ$.
Larger angle = $72^\circ + 36^\circ = 108^\circ$.
Answer: The two angles are 72° and 108°.
CBSE Board Marking Scheme 2026-27
Formulating equation: x + (x + 36°) = 180°1 Mark
Solving to get 72° and 108°1 Mark
Q9
Four rays meet at a common point O, forming four angles: ∠AOB = 110°, ∠BOC = 80°, ∠COD = 95°, and ∠DOA = x. Find the value of x. Give the geometrical property used.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Geometrical Property: The sum of all angles formed around a single point equals a complete angle ($360^\circ$).
$\angle AOB + \angle BOC + \angle COD + \angle DOA = 360^\circ$
$110^\circ + 80^\circ + 95^\circ + x = 360^\circ$
$285^\circ + x = 360^\circ \implies x = 360^\circ - 285^\circ = \mathbf{75^\circ}$.
$\angle AOB + \angle BOC + \angle COD + \angle DOA = 360^\circ$
$110^\circ + 80^\circ + 95^\circ + x = 360^\circ$
$285^\circ + x = 360^\circ \implies x = 360^\circ - 285^\circ = \mathbf{75^\circ}$.
CBSE Board Marking Scheme 2026-27
Stating property: Sum of angles around a point is 360°1 Mark
Computing x = 75° accurately1 Mark
Q10
Differentiate clearly between a line, a ray, and a line segment in terms of: (i) Number of endpoints, (ii) Definite length, (iii) Standard symbolic notation.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
| Property | Line Segment | Ray | Line |
|---|---|---|---|
| Endpoints | 2 fixed endpoints | 1 initial endpoint | 0 endpoints (extends infinitely) |
| Definite Length | Yes, measurable | No, infinite length | No, infinite length |
| Symbolic Notation | $\overline{AB}$ | $\vec{AB}$ | $\overleftrightarrow{AB}$ |
CBSE Board Marking Scheme 2026-27
1 Mark per correct comparison row3 Marks
Q11
Identify the types of angles formed by the hour hand and minute hand of a clock at the following exact times:
(a) 3:00 (b) 6:00 (c) 1:00 (d) 5:00
(a) 3:00 (b) 6:00 (c) 1:00 (d) 5:00
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Each 1-hour interval on a clock represents $\frac{360^\circ}{12} = 30^\circ$.
(a) At 3:00: Hands are 3 hours apart: $3 \times 30^\circ = 90^\circ$ $\implies$ Right Angle.
(b) At 6:00: Hands are 6 hours apart: $6 \times 30^\circ = 180^\circ$ $\implies$ Straight Angle.
(c) At 1:00: Hands are 1 hour apart: $1 \times 30^\circ = 30^\circ$ $\implies$ Acute Angle.
(d) At 5:00: Hands are 5 hours apart: $5 \times 30^\circ = 150^\circ$ $\implies$ Obtuse Angle.
(a) At 3:00: Hands are 3 hours apart: $3 \times 30^\circ = 90^\circ$ $\implies$ Right Angle.
(b) At 6:00: Hands are 6 hours apart: $6 \times 30^\circ = 180^\circ$ $\implies$ Straight Angle.
(c) At 1:00: Hands are 1 hour apart: $1 \times 30^\circ = 30^\circ$ $\implies$ Acute Angle.
(d) At 5:00: Hands are 5 hours apart: $5 \times 30^\circ = 150^\circ$ $\implies$ Obtuse Angle.
CBSE Board Marking Scheme 2026-27
Each correct time classification with angle calculation0.5 Mark each (Total 2 Marks)
Q12
Define the following geometric terms with an illustrative sketch:
(i) Collinear Points
(ii) Concurrent Lines
(i) Collinear Points
(ii) Concurrent Lines
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Collinear Points: Three or more points that lie on the same straight line are called collinear points. If no single straight line can pass through all of them, they are non-collinear.
(ii) Concurrent Lines: Three or more lines that intersect at a single common point are called concurrent lines. The common point of intersection is called the point of concurrence.
(ii) Concurrent Lines: Three or more lines that intersect at a single common point are called concurrent lines. The common point of intersection is called the point of concurrence.
CBSE Board Marking Scheme 2026-27
Definition of Collinear Points1 Mark
Definition of Concurrent Lines with point of concurrence1 Mark
CBSE Competency-Based Question (Real-World Application)
Case Study: Ship Navigation Compass & Lighthouse Bearings:
A mariner's magnetic compass has four cardinal directions (North, South, East, West) and four intercardinal directions (NE, SE, SW, NW).
(a) A ship sails facing North. What angle does the ship turn through if it turns clockwise to face South-East (SE)? What type of angle is this?
(b) If the ship is initially facing West and turns anti-clockwise through 270°, in which direction is it now facing?
(c) Two ships leave harbour along bearings of North-East and South-East. What is the angle between their routes?
A mariner's magnetic compass has four cardinal directions (North, South, East, West) and four intercardinal directions (NE, SE, SW, NW).
(a) A ship sails facing North. What angle does the ship turn through if it turns clockwise to face South-East (SE)? What type of angle is this?
(b) If the ship is initially facing West and turns anti-clockwise through 270°, in which direction is it now facing?
(c) Two ships leave harbour along bearings of North-East and South-East. What is the angle between their routes?
Detailed Analytical Solution & Marking Scheme
(a) North to South-East (Clockwise):
- North to East = 90°
- East to South-East = 45°
- Total angle = $90^\circ + 45^\circ = \mathbf{135^\circ}$.
Since $90^\circ < 135^\circ < 180^\circ$, it is an Obtuse Angle.
(b) West turning anti-clockwise through 270°:
- West to South = 90°
- South to East = 90° (cumulative 180°)
- East to North = 90° (cumulative 270°)
The ship is now facing North.
(c) Angle between NE and SE routes:
NE is 45° north of East. SE is 45° south of East.
Angle = $45^\circ + 45^\circ = \mathbf{90^\circ}$ (A Right Angle).
3
Chapter 3: Number Play
NCERT Ganita Prakash (Class 6) — Place Value Systems, Large Numbers, Estimation, Brackets, Roman Numerals & Magic Squares | CBSE 2026-27
📌 Key Number Concepts & Place Value Systems
- Indian System of Numeration: Commas after Hundreds (3 digits), then after every 2 digits: Crores, Ten Lakhs, Lakhs, Ten Thousands, Thousands, Hundreds, Tens, Ones. (e.g., $7,34,56,819$).
- International System of Numeration: Commas after every 3 digits from right: Billions, Millions (Hundred, Ten, One), Thousands (Hundred, Ten, One), Hundreds, Tens, Ones. (e.g., $73,456,819$).
- Estimation (Rounding Off): To round off to nearest 10, look at ones digit (< 5 round down, $\ge$ 5 round up). To nearest 100, look at tens digit; to nearest 1000, look at hundreds digit.
- Use of Brackets & Order of Operations: Brackets indicate precedence. Rule: $\text{Bar} \to \text{Round}() \to \text{Curly}\{\} \to \text{Square}[]$. Within brackets, follow DMAS (Division, Multiplication, Addition, Subtraction).
- Roman Numerals: $I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000$. Symbol cannot be repeated more than 3 times; $V, L, D$ are never repeated or subtracted.
- Magic Squares: A square grid where numbers in each row, column, and both diagonals sum to the exact same constant (Magic Constant). For a $3 \times 3$ grid using numbers $1$ to $9$, Magic Sum $= 15$.
NCERT Exercise 3.1: Large Numbers & Numeration Systems
Q1
Write the number 8,75,95,762 according to:
(i) Indian System of Numeration with commas and in words.
(ii) International System of Numeration with commas and in words.
(i) Indian System of Numeration with commas and in words.
(ii) International System of Numeration with commas and in words.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Indian System of Numeration:
Placement of commas: 8,75,95,762
In words: Eight crore, seventy-five lakh, ninety-five thousand, seven hundred sixty-two.
Placement of commas: 8,75,95,762
In words: Eight crore, seventy-five lakh, ninety-five thousand, seven hundred sixty-two.
(ii) International System of Numeration:
Re-grouping in sets of 3 digits from right: 87,595,762
In words: Eighty-seven million, five hundred ninety-five thousand, seven hundred sixty-two.
Re-grouping in sets of 3 digits from right: 87,595,762
In words: Eighty-seven million, five hundred ninety-five thousand, seven hundred sixty-two.
CBSE Board Marking Scheme 2026-27
Part (i): Correct commas and words in Indian system1.5 Marks
Part (ii): Correct commas and words in International system1.5 Marks
Q2
Find the difference between the place value and the face value of the digit 7 in the number 5,78,49,236.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Given number: 5,78,49,236.
The digit 7 is at the Ten Lakhs position.
- Place Value of 7 = $7 \times 10,00,000 = \mathbf{70,00,000}$.
- Face Value of 7 = $\mathbf{7}$.
Difference = $70,00,000 - 7 = \mathbf{69,99,993}$.
The digit 7 is at the Ten Lakhs position.
- Place Value of 7 = $7 \times 10,00,000 = \mathbf{70,00,000}$.
- Face Value of 7 = $\mathbf{7}$.
Difference = $70,00,000 - 7 = \mathbf{69,99,993}$.
CBSE Board Marking Scheme 2026-27
Identifying place value (70,00,000) and face value (7)1 Mark
Subtracting to get 69,99,9931 Mark
Q3
Use the digits 4, 0, 7, 1, 9, 5 to form:
(i) The greatest 6-digit number using each digit only once.
(ii) The smallest 6-digit number using each digit only once.
(iii) Find the difference between the greatest and smallest numbers formed.
(i) The greatest 6-digit number using each digit only once.
(ii) The smallest 6-digit number using each digit only once.
(iii) Find the difference between the greatest and smallest numbers formed.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Greatest 6-digit number:
Arrange digits in descending order: 9, 7, 5, 4, 1, 0 $\implies \mathbf{9,75,410}$.
Arrange digits in descending order: 9, 7, 5, 4, 1, 0 $\implies \mathbf{9,75,410}$.
(ii) Smallest 6-digit number:
Since 0 cannot be the leading digit, start with the smallest non-zero digit 1, followed by 0, then remaining digits in ascending order: 1, 0, 4, 5, 7, 9 $\implies \mathbf{1,04,579}$.
Since 0 cannot be the leading digit, start with the smallest non-zero digit 1, followed by 0, then remaining digits in ascending order: 1, 0, 4, 5, 7, 9 $\implies \mathbf{1,04,579}$.
(iii) Difference:
$9,75,410 - 1,04,579 = \mathbf{8,70,831}$.
$9,75,410 - 1,04,579 = \mathbf{8,70,831}$.
CBSE Board Marking Scheme 2026-27
Part (i): Forming 9,75,4101 Mark
Part (ii): Forming 1,04,579 (noting 0 cannot lead)1 Mark
Part (iii): Accurate difference 8,70,8311 Mark
NCERT Exercise 3.2: Estimation & Rounding Off
Q4
Round off the number 73,485 to the nearest:
(a) Tens
(b) Hundreds
(c) Thousands
(d) Ten thousands
(a) Tens
(b) Hundreds
(c) Thousands
(d) Ten thousands
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Nearest 10: Ones digit is 5 $\implies$ round up $\implies \mathbf{73,490}$.
(b) Nearest 100: Tens digit is 8 (> 5) $\implies$ round up $\implies \mathbf{73,500}$.
(c) Nearest 1000: Hundreds digit is 4 (< 5) $\implies$ round down $\implies \mathbf{73,000}$.
(d) Nearest 10,000: Thousands digit is 3 (< 5) $\implies$ round down $\implies \mathbf{70,000}$.
(b) Nearest 100: Tens digit is 8 (> 5) $\implies$ round up $\implies \mathbf{73,500}$.
(c) Nearest 1000: Hundreds digit is 4 (< 5) $\implies$ round down $\implies \mathbf{73,000}$.
(d) Nearest 10,000: Thousands digit is 3 (< 5) $\implies$ round down $\implies \mathbf{70,000}$.
CBSE Board Marking Scheme 2026-27
Each correct rounding with stated reason0.5 Mark each (Total 2 Marks)
Q5
Estimate the following using the general rule:
(a) 730 + 998
(b) 7,961 - 428
(c) 578 × 161
(a) 730 + 998
(b) 7,961 - 428
(c) 578 × 161
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
General Rule: Round off each number to its greatest place value.
(a) 730 + 998:
- 730 rounds off to nearest 100 $\implies$ 700.
- 998 rounds off to nearest 100 $\implies$ 1,000.
Estimated Sum = $700 + 1,000 = \mathbf{1,700}$.
(a) 730 + 998:
- 730 rounds off to nearest 100 $\implies$ 700.
- 998 rounds off to nearest 100 $\implies$ 1,000.
Estimated Sum = $700 + 1,000 = \mathbf{1,700}$.
(b) 7,961 - 428:
- 7,961 rounds to nearest 1,000 $\implies$ 8,000.
- 428 rounds to nearest 100 $\implies$ 400.
Estimated Difference = $8,000 - 400 = \mathbf{7,600}$.
- 7,961 rounds to nearest 1,000 $\implies$ 8,000.
- 428 rounds to nearest 100 $\implies$ 400.
Estimated Difference = $8,000 - 400 = \mathbf{7,600}$.
(c) 578 × 161:
- 578 rounds to nearest 100 $\implies$ 600.
- 161 rounds to nearest 100 $\implies$ 200.
Estimated Product = $600 \times 200 = \mathbf{1,20,000}$.
- 578 rounds to nearest 100 $\implies$ 600.
- 161 rounds to nearest 100 $\implies$ 200.
Estimated Product = $600 \times 200 = \mathbf{1,20,000}$.
CBSE Board Marking Scheme 2026-27
Part (a): 1,7001 Mark
Part (b): 7,6001 Mark
Part (c): 1,20,0001 Mark
NCERT Exercise 3.3: Roman Numerals & Brackets
Q6
Write the following in Roman numerals:
(a) 69 (b) 98 (c) 144 (d) 490
(a) 69 (b) 98 (c) 144 (d) 490
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) 69: $60 + 9 = (50 + 10) + (10 - 1) = \text{LX} + \text{IX} = \mathbf{LXIX}$.
(b) 98: $90 + 8 = (100 - 10) + 8 = \text{XC} + \text{VIII} = \mathbf{XCVIII}$.
(c) 144: $100 + 40 + 4 = \text{C} + (50 - 10) + (5 - 1) = \text{C} + \text{XL} + \text{IV} = \mathbf{CXLIV}$.
(d) 490: $400 + 90 = (500 - 100) + (100 - 10) = \text{CD} + \text{XC} = \mathbf{CDXC}$.
(b) 98: $90 + 8 = (100 - 10) + 8 = \text{XC} + \text{VIII} = \mathbf{XCVIII}$.
(c) 144: $100 + 40 + 4 = \text{C} + (50 - 10) + (5 - 1) = \text{C} + \text{XL} + \text{IV} = \mathbf{CXLIV}$.
(d) 490: $400 + 90 = (500 - 100) + (100 - 10) = \text{CD} + \text{XC} = \mathbf{CDXC}$.
CBSE Board Marking Scheme 2026-27
Each correct Roman numeral with expansion0.5 Mark each (Total 2 Marks)
Q7
Simplify the following expressions using brackets and order of operations:
(i) 15 + (12 - 3) × 4
(ii) 45 - [38 - {60 ÷ 3 - (6 - 9 ÷ 3)}]
(i) 15 + (12 - 3) × 4
(ii) 45 - [38 - {60 ÷ 3 - (6 - 9 ÷ 3)}]
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) 15 + (12 - 3) × 4:
Step 1 (Inside bracket): $12 - 3 = 9$.
Step 2 (Multiplication): $9 \times 4 = 36$.
Step 3 (Addition): $15 + 36 = \mathbf{51}$.
Step 1 (Inside bracket): $12 - 3 = 9$.
Step 2 (Multiplication): $9 \times 4 = 36$.
Step 3 (Addition): $15 + 36 = \mathbf{51}$.
(ii) 45 - [38 - {60 ÷ 3 - (6 - 9 ÷ 3)}]:
Innermost round bracket $(6 - 9 \div 3)$: $9 \div 3 = 3$, so $6 - 3 = 3$.
Next curly bracket: $\{60 \div 3 - 3\} = \{20 - 3\} = 17$.
Next square bracket: $[38 - 17] = 21$.
Final subtraction: $45 - 21 = \mathbf{24}$.
Innermost round bracket $(6 - 9 \div 3)$: $9 \div 3 = 3$, so $6 - 3 = 3$.
Next curly bracket: $\{60 \div 3 - 3\} = \{20 - 3\} = 17$.
Next square bracket: $[38 - 17] = 21$.
Final subtraction: $45 - 21 = \mathbf{24}$.
CBSE Board Marking Scheme 2026-27
Part (i): Step-by-step simplification to 511 Mark
Part (ii): Nested bracket hierarchy correctly executed to 242 Marks
NCERT Exercise 3.4: Magic Squares & Word Problems
Q8
A 3 × 3 magic square contains numbers 1 to 9 without repetition. The sum of numbers in every row, column, and diagonal is 15.
(i) Explain why the central number must be 5.
(ii) Construct the complete 3 × 3 magic square.
(i) Explain why the central number must be 5.
(ii) Construct the complete 3 × 3 magic square.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Proof that Central Number is 5:
The sum of numbers 1 to 9 is: $\frac{9 \times 10}{2} = 45$.
Since there are 3 rows, Magic Sum = $\frac{45}{3} = 15$.
The center cell belongs to 4 lines (row 2, col 2, and both diagonals). Adding these 4 lines:
$4 \times 15 = 60$.
In this sum, all 9 cells are covered, plus the center cell is counted 3 extra times:
$45 + 3 \times (\text{Center}) = 60 \implies 3 \times (\text{Center}) = 15 \implies \mathbf{\text{Center} = 5}$.
The sum of numbers 1 to 9 is: $\frac{9 \times 10}{2} = 45$.
Since there are 3 rows, Magic Sum = $\frac{45}{3} = 15$.
The center cell belongs to 4 lines (row 2, col 2, and both diagonals). Adding these 4 lines:
$4 \times 15 = 60$.
In this sum, all 9 cells are covered, plus the center cell is counted 3 extra times:
$45 + 3 \times (\text{Center}) = 60 \implies 3 \times (\text{Center}) = 15 \implies \mathbf{\text{Center} = 5}$.
(ii) Completed 3 × 3 Magic Square:
Row sums: $8+1+6=15$; $3+5+7=15$; $4+9+2=15$. Diagonals: $8+5+2=15$; $6+5+4=15$.
8 | 1 | 6
---+---+---
3 | 5 | 7
---+---+---
4 | 9 | 2
---+---+---
3 | 5 | 7
---+---+---
4 | 9 | 2
Row sums: $8+1+6=15$; $3+5+7=15$; $4+9+2=15$. Diagonals: $8+5+2=15$; $6+5+4=15$.
CBSE Board Marking Scheme 2026-27
Part (i): Mathematical proof that Center = 51.5 Marks
Part (ii): Correct 3 × 3 grid verification1.5 Marks
Q9
A medicine manufacturing company packs tablets in boxes. Each box contains 2,00,000 tablets, and each tablet weighs 20 mg. Find the total weight of all tablets in the box in:
(i) Milligrams (ii) Grams (iii) Kilograms
(i) Milligrams (ii) Grams (iii) Kilograms
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Total weight in milligrams:
Weight = $2,00,000 \times 20\text{ mg} = \mathbf{40,00,000\text{ mg}}$.
Weight = $2,00,000 \times 20\text{ mg} = \mathbf{40,00,000\text{ mg}}$.
(ii) Total weight in grams:
Since $1\text{ g} = 1,000\text{ mg}$:
Weight = $\frac{40,00,000}{1,000} = \mathbf{4,000\text{ grams}}$.
Since $1\text{ g} = 1,000\text{ mg}$:
Weight = $\frac{40,00,000}{1,000} = \mathbf{4,000\text{ grams}}$.
(iii) Total weight in kilograms:
Since $1\text{ kg} = 1,000\text{ g}$:
Weight = $\frac{4,000}{1,000} = \mathbf{4\text{ kilograms}}$.
Since $1\text{ kg} = 1,000\text{ g}$:
Weight = $\frac{4,000}{1,000} = \mathbf{4\text{ kilograms}}$.
CBSE Board Marking Scheme 2026-27
Part (i): 40,00,000 mg1 Mark
Part (ii): 4,000 g1 Mark
Part (iii): 4 kg1 Mark
Q10
A merchant had ₹ 78,592 with her. She placed an order for purchasing 40 radio sets at ₹ 1,200 each. How much money will remain with her after the purchase?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Total money with merchant = ₹ 78,592.
Cost of 1 radio set = ₹ 1,200.
Cost of 40 radio sets = $40 \times 1,200 = \text{₹ } 48,000$.
Money remaining = ₹ 78,592 - ₹ 48,000 = ₹ 30,592.
Cost of 1 radio set = ₹ 1,200.
Cost of 40 radio sets = $40 \times 1,200 = \text{₹ } 48,000$.
Money remaining = ₹ 78,592 - ₹ 48,000 = ₹ 30,592.
CBSE Board Marking Scheme 2026-27
Calculating total expenditure ₹ 48,0001 Mark
Subtracting to get ₹ 30,5921 Mark
Q11
A vessel has 4 litres and 500 mL of curd. In how many glasses, each of 25 mL capacity, can it be filled?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Total capacity of vessel = 4 L 500 mL.
Convert to millilitres: $4 \times 1,000\text{ mL} + 500\text{ mL} = 4,500\text{ mL}$.
Capacity of each glass = 25 mL.
Number of glasses = $\frac{4,500}{25} = \mathbf{180\text{ glasses}}$.
Convert to millilitres: $4 \times 1,000\text{ mL} + 500\text{ mL} = 4,500\text{ mL}$.
Capacity of each glass = 25 mL.
Number of glasses = $\frac{4,500}{25} = \mathbf{180\text{ glasses}}$.
CBSE Board Marking Scheme 2026-27
Conversion: 4 L 500 mL = 4,500 mL1 Mark
Division yielding 180 glasses1 Mark
Q12
To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Convert measurements into centimetres:
- Cloth for 1 shirt = $2\text{ m } 15\text{ cm} = 200 + 15 = 215\text{ cm}$.
- Total available cloth = $40\text{ m} = 40 \times 100 = 4,000\text{ cm}$.
Dividing 4,000 by 215:
$4,000 = 215 \times 18 + 130$.
Quotient = 18; Remainder = 130 cm = 1 m 30 cm.
Answer: 18 shirts can be stitched and 1 m 30 cm cloth will remain.
- Cloth for 1 shirt = $2\text{ m } 15\text{ cm} = 200 + 15 = 215\text{ cm}$.
- Total available cloth = $40\text{ m} = 40 \times 100 = 4,000\text{ cm}$.
Dividing 4,000 by 215:
$4,000 = 215 \times 18 + 130$.
Quotient = 18; Remainder = 130 cm = 1 m 30 cm.
Answer: 18 shirts can be stitched and 1 m 30 cm cloth will remain.
CBSE Board Marking Scheme 2026-27
Conversion to centimetres (215 cm and 4,000 cm)1 Mark
Concluding 18 shirts and 1 m 30 cm cloth left1 Mark
CBSE Competency-Based Question (Electoral Data Case Study)
Case Study: Assembly Election Voter Turnout Analysis:
In an assembly election, the winning candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. A third independent candidate secured 42,350 votes, while 7,450 votes were declared invalid.
(a) By what margin of votes did the successful candidate win the election?
(b) Find the total number of votes polled in the constituency.
(c) Express the total votes polled in the International System of Numeration.
In an assembly election, the winning candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. A third independent candidate secured 42,350 votes, while 7,450 votes were declared invalid.
(a) By what margin of votes did the successful candidate win the election?
(b) Find the total number of votes polled in the constituency.
(c) Express the total votes polled in the International System of Numeration.
Detailed Analytical Solution & Marking Scheme
(a) Victory Margin:
Margin = Winning votes - Nearest rival votes
Margin = $5,77,500 - 3,48,700 = \mathbf{2,28,800\text{ votes}}$.
(b) Total Votes Polled:
Total votes = $5,77,500 + 3,48,700 + 42,350 + 7,450 = \mathbf{9,76,000\text{ votes}}$.
(c) International System Notation:
Grouping by sets of three digits: 976,000.
In words: Nine hundred seventy-six thousand.
4
Chapter 4: Data Handling and Presentation
NCERT Ganita Prakash (Class 6) — Tally Marks, Frequency Tables, Pictographs with Scales, Bar Graphs & Statistical Interpretation | CBSE 2026-27
📌 Key Statistical Concepts & Representations
- Data: A collection of numbers gathered to give information. Raw data is unorganized data.
- Tally Marks: Marks used for counting in groups of five: $\text{I, II, III, IIII, } \cancel{\text{IIII}}$ (fifth mark crosses the four vertical strokes).
- Pictograph: Represents data through pictures or symbols of objects. A scale or key must indicate what each symbol represents (e.g., $\bigstar = 10\text{ books}$).
- Bar Graph: A visual representation of numbers using bars of uniform width drawn horizontally or vertically with equal spacing between them. The length of each bar represents the given quantity according to a chosen scale on the axis.
NCERT Exercise 4.1: Tally Marks & Frequency Distribution
Q1
In a class of 30 students, the shoe sizes recorded are:
5, 4, 6, 5, 7, 6, 5, 4, 5, 6, 7, 5, 4, 6, 5, 5, 6, 7, 4, 5, 6, 5, 4, 6, 7, 5, 6, 5, 4, 5.
(i) Organize the data in a frequency table using tally marks.
(ii) Which shoe size is worn by the maximum number of students?
(iii) Which shoe size is worn by the minimum number of students?
5, 4, 6, 5, 7, 6, 5, 4, 5, 6, 7, 5, 4, 6, 5, 5, 6, 7, 4, 5, 6, 5, 4, 6, 7, 5, 6, 5, 4, 5.
(i) Organize the data in a frequency table using tally marks.
(ii) Which shoe size is worn by the maximum number of students?
(iii) Which shoe size is worn by the minimum number of students?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Frequency Distribution Table:
| Shoe Size | Tally Marks | Number of Students (Frequency) |
|---|---|---|
| 4 | 卌 | 6 |
| 5 | 卌 卌 II | 12 |
| 6 | 卌 III | 8 |
| 7 | IIII | 4 |
| Total | 30 |
(ii) Maximum shoe size: Size 5 (worn by 12 students).
(iii) Minimum shoe size: Size 7 (worn by 4 students).
(iii) Minimum shoe size: Size 7 (worn by 4 students).
CBSE Board Marking Scheme 2026-27
Part (i): Accurate tally marks table summing to 301.5 Marks
Part (ii): Size 5 (Maximum)0.75 Mark
Part (iii): Size 7 (Minimum)0.75 Mark
NCERT Exercise 4.2: Pictograph Interpretation & Construction
Q2
The following pictograph shows the number of tractors in 5 villages:
Village A: 🚜 🚜 🚜 🚜 🚜 🚜
Village B: 🚜 🚜 🚜 🚜 🚜
Village C: 🚜 🚜 🚜 🚜 🚜 🚜 🚜 🚜
Village D: 🚜 🚜 🚜
Village E: 🚜 🚜 🚜 🚜 🚜 🚜
(Scale: 1 🚜 symbol = 1 tractor)
(i) Which village has the minimum number of tractors?
(ii) Which village has the maximum number of tractors?
(iii) How many more tractors does Village C have than Village B?
(iv) What is the total number of tractors in all 5 villages?
Village A: 🚜 🚜 🚜 🚜 🚜 🚜
Village B: 🚜 🚜 🚜 🚜 🚜
Village C: 🚜 🚜 🚜 🚜 🚜 🚜 🚜 🚜
Village D: 🚜 🚜 🚜
Village E: 🚜 🚜 🚜 🚜 🚜 🚜
(Scale: 1 🚜 symbol = 1 tractor)
(i) Which village has the minimum number of tractors?
(ii) Which village has the maximum number of tractors?
(iii) How many more tractors does Village C have than Village B?
(iv) What is the total number of tractors in all 5 villages?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Data from Pictograph:
- Village A = 6 tractors
- Village B = 5 tractors
- Village C = 8 tractors
- Village D = 3 tractors
- Village E = 6 tractors
(i) Minimum tractors: Village D (3 tractors).
(ii) Maximum tractors: Village C (8 tractors).
(iii) Difference (C vs B): $8 - 5 = \mathbf{3\text{ more tractors}}$.
(iv) Total tractors: $6 + 5 + 8 + 3 + 6 = \mathbf{28\text{ tractors}}$.
- Village A = 6 tractors
- Village B = 5 tractors
- Village C = 8 tractors
- Village D = 3 tractors
- Village E = 6 tractors
(i) Minimum tractors: Village D (3 tractors).
(ii) Maximum tractors: Village C (8 tractors).
(iii) Difference (C vs B): $8 - 5 = \mathbf{3\text{ more tractors}}$.
(iv) Total tractors: $6 + 5 + 8 + 3 + 6 = \mathbf{28\text{ tractors}}$.
CBSE Board Marking Scheme 2026-27
Each sub-part correct0.75 Mark each (Total 3 Marks)
Q3
The total number of animals in five villages are as follows:
Village A: 80, Village B: 120, Village C: 90, Village D: 40, Village E: 60.
(i) Prepare a pictograph of these animals using one symbol ⨂ to represent 10 animals.
(ii) Which village has the maximum number of animals?
(iii) Which village has more animals: Village A or Village C?
Village A: 80, Village B: 120, Village C: 90, Village D: 40, Village E: 60.
(i) Prepare a pictograph of these animals using one symbol ⨂ to represent 10 animals.
(ii) Which village has the maximum number of animals?
(iii) Which village has more animals: Village A or Village C?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Scale: 1 symbol ⨂ = 10 animals.
- Village A (80 animals): $\frac{80}{10} = 8$ symbols: ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂
- Village B (120 animals): $\frac{120}{10} = 12$ symbols: ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂
- Village C (90 animals): $\frac{90}{10} = 9$ symbols: ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂
- Village D (40 animals): $\frac{40}{10} = 4$ symbols: ⨂ ⨂ ⨂ ⨂
- Village E (60 animals): $\frac{60}{10} = 6$ symbols: ⨂ ⨂ ⨂ ⨂ ⨂ ⨂
- Village A (80 animals): $\frac{80}{10} = 8$ symbols: ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂
- Village B (120 animals): $\frac{120}{10} = 12$ symbols: ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂
- Village C (90 animals): $\frac{90}{10} = 9$ symbols: ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂ ⨂
- Village D (40 animals): $\frac{40}{10} = 4$ symbols: ⨂ ⨂ ⨂ ⨂
- Village E (60 animals): $\frac{60}{10} = 6$ symbols: ⨂ ⨂ ⨂ ⨂ ⨂ ⨂
(ii) Maximum animals: Village B (120 animals).
(iii) Comparison: Village C has 90 animals, Village A has 80 animals. Therefore, Village C has more animals.
(iii) Comparison: Village C has 90 animals, Village A has 80 animals. Therefore, Village C has more animals.
CBSE Board Marking Scheme 2026-27
Part (i): Pictograph with chosen scale symbol1.5 Marks
Part (ii): Village B0.75 Mark
Part (iii): Village C0.75 Mark
NCERT Exercise 4.3: Bar Graph Interpretation & Drawing
Figure 4.1: Bar Graph Representation of Student Examination Marks
Q4
Observe the bar graph given in Figure 4.1 and answer the following questions:
(a) What information does the bar graph give?
(b) In which subject has the student scored maximum marks?
(c) In which subject has the student scored minimum marks?
(d) State the marks obtained in Science and Social Studies.
(a) What information does the bar graph give?
(b) In which subject has the student scored maximum marks?
(c) In which subject has the student scored minimum marks?
(d) State the marks obtained in Science and Social Studies.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Information given: The bar graph represents the marks obtained by a student in 5 different school subjects in an examination.
(b) Maximum marks: Mathematics (75 marks).
(c) Minimum marks: Hindi (45 marks).
(d) Marks: In Science = 65 marks; in Social Studies = 50 marks.
(b) Maximum marks: Mathematics (75 marks).
(c) Minimum marks: Hindi (45 marks).
(d) Marks: In Science = 65 marks; in Social Studies = 50 marks.
CBSE Board Marking Scheme 2026-27
Each sub-part answered accurately0.75 Mark each (Total 3 Marks)
Q5
The number of mathematics books sold by a shopkeeper on six consecutive days is shown below:
Sunday: 65, Monday: 40, Tuesday: 30, Wednesday: 50, Thursday: 20, Friday: 70.
(i) Draw a vertical bar graph to represent this data with a scale of 1 unit length = 10 books.
(ii) On which day were the maximum number of books sold?
(iii) What was the total number of books sold during these 6 days?
Sunday: 65, Monday: 40, Tuesday: 30, Wednesday: 50, Thursday: 20, Friday: 70.
(i) Draw a vertical bar graph to represent this data with a scale of 1 unit length = 10 books.
(ii) On which day were the maximum number of books sold?
(iii) What was the total number of books sold during these 6 days?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) Bar Graph Specifications:
- Horizontal Axis (X-axis): Days of the week.
- Vertical Axis (Y-axis): Number of books sold.
- Scale: 1 unit length = 10 books.
- Bar heights: Sunday = 6.5 units; Monday = 4 units; Tuesday = 3 units; Wednesday = 5 units; Thursday = 2 units; Friday = 7 units.
- Horizontal Axis (X-axis): Days of the week.
- Vertical Axis (Y-axis): Number of books sold.
- Scale: 1 unit length = 10 books.
- Bar heights: Sunday = 6.5 units; Monday = 4 units; Tuesday = 3 units; Wednesday = 5 units; Thursday = 2 units; Friday = 7 units.
(ii) Day of maximum sales: Friday (70 books).
(iii) Total books sold: $65 + 40 + 30 + 50 + 20 + 70 = \mathbf{275\text{ books}}$.
(iii) Total books sold: $65 + 40 + 30 + 50 + 20 + 70 = \mathbf{275\text{ books}}$.
CBSE Board Marking Scheme 2026-27
Part (i): Scale selection and accurate bar heights1.5 Marks
Part (ii): Friday0.75 Mark
Part (iii): Total 275 books0.75 Mark
Q6
Explain the essential differences between a pictograph and a bar graph. What are the advantages of using a bar graph over a pictograph for large quantities?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Differences & Advantages:
1. Representation: A pictograph uses visual symbols or drawings, whereas a bar graph uses rectangular bars of uniform width and varying heights.
2. Precision for Large Quantities: In pictographs, fractional or intermediate values (e.g. 73 when 1 symbol = 10) require difficult partial drawings. In bar graphs, any precise value can be accurately marked on a calibrated continuous axis.
3. Speed of Construction: Drawing numerous detailed icons is time-consuming; bars can be drawn swiftly using a ruler.
1. Representation: A pictograph uses visual symbols or drawings, whereas a bar graph uses rectangular bars of uniform width and varying heights.
2. Precision for Large Quantities: In pictographs, fractional or intermediate values (e.g. 73 when 1 symbol = 10) require difficult partial drawings. In bar graphs, any precise value can be accurately marked on a calibrated continuous axis.
3. Speed of Construction: Drawing numerous detailed icons is time-consuming; bars can be drawn swiftly using a ruler.
CBSE Board Marking Scheme 2026-27
Precise structural distinction1 Mark
Stating advantages of scale precision and speed1 Mark
Q7
The scores of 7 students in a science quiz out of 25 are: 18, 22, 14, 25, 19, 22, 20.
(a) Find the highest and lowest scores.
(b) Calculate the range of the scores.
(c) Find the average (mean) score of the group.
(a) Find the highest and lowest scores.
(b) Calculate the range of the scores.
(c) Find the average (mean) score of the group.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Highest and Lowest:
- Highest score = 25
- Lowest score = 14
- Highest score = 25
- Lowest score = 14
(b) Range:
$\text{Range} = \text{Highest Score} - \text{Lowest Score} = 25 - 14 = \mathbf{11}$.
$\text{Range} = \text{Highest Score} - \text{Lowest Score} = 25 - 14 = \mathbf{11}$.
(c) Mean Score:
$\text{Mean} = \frac{\text{Sum of scores}}{\text{Number of students}} = \frac{18 + 22 + 14 + 25 + 19 + 22 + 20}{7} = \frac{140}{7} = \mathbf{20}$.
$\text{Mean} = \frac{\text{Sum of scores}}{\text{Number of students}} = \frac{18 + 22 + 14 + 25 + 19 + 22 + 20}{7} = \frac{140}{7} = \mathbf{20}$.
CBSE Board Marking Scheme 2026-27
Part (a): Highest (25) and Lowest (14)1 Mark
Part (b): Range = 111 Mark
Part (c): Mean = 20 with formula1 Mark
Q8
What is meant by "scale" on a bar graph? Why is choosing an appropriate scale critical when representing large statistical datasets?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Definition of Scale:
The scale of a bar graph is the ratio that compares the unit length on the axis to the actual quantity represented (e.g., 1 cm = 100 people).
Importance:
If the scale chosen is too small (e.g. 1 unit = 1 person for 50,000 people), the graph will not fit on paper. If the scale is too large (e.g. 1 unit = 20,000 for values differing by 50), the bars will look identical and differences cannot be distinguished.
The scale of a bar graph is the ratio that compares the unit length on the axis to the actual quantity represented (e.g., 1 cm = 100 people).
Importance:
If the scale chosen is too small (e.g. 1 unit = 1 person for 50,000 people), the graph will not fit on paper. If the scale is too large (e.g. 1 unit = 20,000 for values differing by 50), the bars will look identical and differences cannot be distinguished.
CBSE Board Marking Scheme 2026-27
Definition of scale1 Mark
Importance for legibility and graphical proportion1 Mark
Q9
A researcher records modes of transport used by 250 school students:
School Bus: 100, Walking: 60, Bicycle: 50, Car: 25, Public Bus: 15.
(i) What fraction of students travel by School Bus?
(ii) What percentage of students walk to school?
(iii) Which mode of transport is used least?
School Bus: 100, Walking: 60, Bicycle: 50, Car: 25, Public Bus: 15.
(i) What fraction of students travel by School Bus?
(ii) What percentage of students walk to school?
(iii) Which mode of transport is used least?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Total students = 250.
(i) School Bus fraction: $\frac{100}{250} = \mathbf{\frac{2}{5}}$ (or 40%).
(ii) Walking percentage: $\frac{60}{250} \times 100 = \frac{600}{25} = \mathbf{24\%}$.
(iii) Least used mode: Public Bus (used by only 15 students).
(i) School Bus fraction: $\frac{100}{250} = \mathbf{\frac{2}{5}}$ (or 40%).
(ii) Walking percentage: $\frac{60}{250} \times 100 = \frac{600}{25} = \mathbf{24\%}$.
(iii) Least used mode: Public Bus (used by only 15 students).
CBSE Board Marking Scheme 2026-27
Part (i): Fraction 2/50.75 Mark
Part (ii): 24%0.75 Mark
Part (iii): Public Bus0.5 Mark
Q10
State whether each of the following statements is True or False. Justify your answer:
(a) In a bar graph, the widths of the bars can vary according to preference.
(b) The spaces between adjacent bars must always be uniform.
(c) A pictograph can only be drawn horizontally.
(a) In a bar graph, the widths of the bars can vary according to preference.
(b) The spaces between adjacent bars must always be uniform.
(c) A pictograph can only be drawn horizontally.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) False: The widths of all bars must be strictly uniform. Width carries no numerical meaning, only the bar length or height represents data.
(b) True: Equal spacing between bars is mandatory to maintain visual clarity and fair comparison.
(c) False: Pictographs can be drawn either horizontally (rows) or vertically (columns).
(b) True: Equal spacing between bars is mandatory to maintain visual clarity and fair comparison.
(c) False: Pictographs can be drawn either horizontally (rows) or vertically (columns).
CBSE Board Marking Scheme 2026-27
Each statement correctly evaluated with justification0.7 Mark each (Total 2 Marks)
Q11
Distinguish between qualitative data and quantitative data with two practical examples of each from school records.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Qualitative Data: Descriptive data that characterizes attributes or qualities without numerical measurement.
Examples: Blood group (A, B, AB, O), Grade awarded in art (A, B, C).
Quantitative Data: Numerical data that can be counted or measured directly on a scale.
Examples: Height of students in cm (142 cm, 150 cm), Marks obtained in mathematics (85/100).
Examples: Blood group (A, B, AB, O), Grade awarded in art (A, B, C).
Quantitative Data: Numerical data that can be counted or measured directly on a scale.
Examples: Height of students in cm (142 cm, 150 cm), Marks obtained in mathematics (85/100).
CBSE Board Marking Scheme 2026-27
Definition and examples of qualitative data1 Mark
Definition and examples of quantitative data1 Mark
CBSE Competency-Based Question (School Survey Case Study)
Case Study: School Canteen Healthy Fruit Preference Survey:
The school wellness committee surveyed 400 middle-school students regarding their preferred fruit choice for the subsidized canteen menu. The recorded preferences were:
- Apple: 120 students
- Banana: 100 students
- Orange: 80 students
- Guava: 60 students
- Papaya: 40 students
(a) If you choose a scale of 1 unit length = 20 students, what will be the heights of the bars for Apple and Orange?
(b) What percentage of the surveyed students selected either Apple or Banana?
(c) Formulate a recommendation for the canteen manager based on this data.
The school wellness committee surveyed 400 middle-school students regarding their preferred fruit choice for the subsidized canteen menu. The recorded preferences were:
- Apple: 120 students
- Banana: 100 students
- Orange: 80 students
- Guava: 60 students
- Papaya: 40 students
(a) If you choose a scale of 1 unit length = 20 students, what will be the heights of the bars for Apple and Orange?
(b) What percentage of the surveyed students selected either Apple or Banana?
(c) Formulate a recommendation for the canteen manager based on this data.
Detailed Analytical Solution & Marking Scheme
(a) Bar Heights (Scale: 1 unit = 20 students):
- Height for Apple = $\frac{120}{20} = \mathbf{6\text{ units}}$.
- Height for Orange = $\frac{80}{20} = \mathbf{4\text{ units}}$.
(b) Percentage for Apple or Banana:
Total choosing Apple or Banana = $120 + 100 = 220$ students.
Percentage = $\frac{220}{400} \times 100 = \mathbf{55\%}$.
More than half of the student population prefers Apple or Banana.
(c) Canteen Recommendation:
The canteen manager should allocate at least 55% of the fruit procurement budget to Apples and Bananas to avoid food wastage, while stocking moderate quantities of Guava and Papaya.
5
Chapter 5: Prime Time
NCERT Ganita Prakash (Class 6) — Factors, Multiples, Prime & Composite Numbers, Divisibility Rules, Prime Factor Tree, HCF & LCM | CBSE 2026-27
📌 Key Number Theory & Divisibility Rules
- Factors & Multiples: A factor of a number divides it exactly without a remainder. A multiple is obtained by multiplying the number by integers.
- Prime & Composite: A prime number has exactly two distinct factors: 1 and itself (e.g., $2, 3, 5, 7, 11$). 2 is the only even prime. 1 is neither prime nor composite.
- Co-prime Numbers: Two numbers having only 1 as a common factor are called co-primes (e.g., 4 and 9).
- Divisibility Tests:
- By 2, 5, 10: Last digit is even (for 2); 0 or 5 (for 5); 0 (for 10).
- By 3 & 9: Sum of digits is divisible by 3 (or 9).
- By 4 & 8: Last 2 digits divisible by 4; last 3 digits divisible by 8.
- By 6: Divisible by both 2 and 3.
- By 11: Difference between the sum of digits at odd places and sum of digits at even places is 0 or divisible by 11.
- HCF & LCM: Highest Common Factor (HCF) is the greatest common divisor. Lowest Common Multiple (LCM) is the smallest common positive multiple. For two numbers $a$ and $b$: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$.
Figure 5.1: Prime Factor Tree Decomposition for 72
NCERT Exercise 5.1: Factors, Multiples & Divisibility Tests
Q1
Write all the factors of the following numbers:
(a) 24 (b) 36 (c) 48
(a) 24 (b) 36 (c) 48
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Factors of 24:
$1 \times 24 = 24$, $2 \times 12 = 24$, $3 \times 8 = 24$, $4 \times 6 = 24$.
Factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
$1 \times 24 = 24$, $2 \times 12 = 24$, $3 \times 8 = 24$, $4 \times 6 = 24$.
Factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
(b) Factors of 36:
$1 \times 36 = 36$, $2 \times 18 = 36$, $3 \times 12 = 36$, $4 \times 9 = 36$, $6 \times 6 = 36$.
Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
$1 \times 36 = 36$, $2 \times 18 = 36$, $3 \times 12 = 36$, $4 \times 9 = 36$, $6 \times 6 = 36$.
Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
(c) Factors of 48:
$1 \times 48 = 48$, $2 \times 24 = 48$, $3 \times 16 = 48$, $4 \times 12 = 48$, $6 \times 8 = 48$.
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.
$1 \times 48 = 48$, $2 \times 24 = 48$, $3 \times 16 = 48$, $4 \times 12 = 48$, $6 \times 8 = 48$.
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.
CBSE Board Marking Scheme 2026-27
1 Mark per part listing all factors completely3 Marks
Q2
Using divisibility tests, determine which of the following numbers are divisible by 6:
(a) 2,97,144
(b) 1,258
(c) 4,335
(a) 2,97,144
(b) 1,258
(c) 4,335
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Divisibility Rule for 6: A number is divisible by 6 if it is divisible by both 2 and 3.
(a) 2,97,144:
- Last digit is 4 (even) $\implies$ Divisible by 2.
- Sum of digits = $2 + 9 + 7 + 1 + 4 + 4 = 27$. Since 27 is divisible by 3 $\implies$ Divisible by 3.
Since it is divisible by both 2 and 3, 2,97,144 is divisible by 6.
(a) 2,97,144:
- Last digit is 4 (even) $\implies$ Divisible by 2.
- Sum of digits = $2 + 9 + 7 + 1 + 4 + 4 = 27$. Since 27 is divisible by 3 $\implies$ Divisible by 3.
Since it is divisible by both 2 and 3, 2,97,144 is divisible by 6.
(b) 1,258:
- Last digit is 8 (even) $\implies$ Divisible by 2.
- Sum of digits = $1 + 2 + 5 + 8 = 16$. 16 is NOT divisible by 3.
Therefore, 1,258 is NOT divisible by 6.
- Last digit is 8 (even) $\implies$ Divisible by 2.
- Sum of digits = $1 + 2 + 5 + 8 = 16$. 16 is NOT divisible by 3.
Therefore, 1,258 is NOT divisible by 6.
(c) 4,335:
- Last digit is 5 (odd) $\implies$ NOT divisible by 2.
Therefore, 4,335 is NOT divisible by 6.
- Last digit is 5 (odd) $\implies$ NOT divisible by 2.
Therefore, 4,335 is NOT divisible by 6.
CBSE Board Marking Scheme 2026-27
Part (a): Verified for 2 and 3 -> Divisible by 61 Mark
Part (b): Fails test for 3 -> Not divisible1 Mark
Part (c): Fails test for 2 -> Not divisible1 Mark
Q3
Using divisibility tests, determine whether the number 71,38,965 is divisible by 11.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Divisibility Test for 11:
Digits at odd places (from right): 5, 9, 3, 7.
Sum of odd-placed digits = $5 + 9 + 3 + 7 = 24$.
Digits at even places (from right): 6, 8, 1.
Sum of even-placed digits = $6 + 8 + 1 = 15$.
Difference = $24 - 15 = 9$.
Since 9 is neither 0 nor divisible by 11, the number 71,38,965 is NOT divisible by 11.
Digits at odd places (from right): 5, 9, 3, 7.
Sum of odd-placed digits = $5 + 9 + 3 + 7 = 24$.
Digits at even places (from right): 6, 8, 1.
Sum of even-placed digits = $6 + 8 + 1 = 15$.
Difference = $24 - 15 = 9$.
Since 9 is neither 0 nor divisible by 11, the number 71,38,965 is NOT divisible by 11.
CBSE Board Marking Scheme 2026-27
Finding sum of odd places (24) and even places (15)1 Mark
Computing difference 9 and concluding not divisible1 Mark
NCERT Exercise 5.2: Prime Factorization & HCF
Q4
Find the prime factorization of 980 by drawing a factor tree or using the repeated division method.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Let us divide 980 by consecutive prime numbers:
$980 \div 2 = 490$
$490 \div 2 = 245$
$245 \div 5 = 49$
$49 \div 7 = 7$
$7 \div 7 = 1$
Therefore, the prime factorization of 980 is:
980 = 2 × 2 × 5 × 7 × 7 = 2² × 5 × 7².
$980 \div 2 = 490$
$490 \div 2 = 245$
$245 \div 5 = 49$
$49 \div 7 = 7$
$7 \div 7 = 1$
Therefore, the prime factorization of 980 is:
980 = 2 × 2 × 5 × 7 × 7 = 2² × 5 × 7².
CBSE Board Marking Scheme 2026-27
Step-by-step prime factor division1 Mark
Final product expression 2² × 5 × 7²1 Mark
Q5
Find the Highest Common Factor (HCF) of 144, 180, and 192 using the prime factorization method.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Prime Factorization of each number:
- $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$
- $180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5$
- $192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1$
- $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$
- $180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5$
- $192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1$
Common Prime Factors with Lowest Powers:
- Common factor 2: lowest power is $2^2 = 4$
- Common factor 3: lowest power is $3^1 = 3$
- Factor 5 is not common to all three.
$\text{HCF} = 2^2 \times 3^1 = 4 \times 3 = \mathbf{12}$.
- Common factor 2: lowest power is $2^2 = 4$
- Common factor 3: lowest power is $3^1 = 3$
- Factor 5 is not common to all three.
$\text{HCF} = 2^2 \times 3^1 = 4 \times 3 = \mathbf{12}$.
CBSE Board Marking Scheme 2026-27
Prime factorization of all three numbers1.5 Marks
Extracting common factors to get HCF = 121.5 Marks
NCERT Exercise 5.3: LCM & Practical Applications
Q6
Find the Lowest Common Multiple (LCM) of 20, 25, and 30 using the common division method.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Common Division Method:
Divide by primes (2, 2, 3, 5, 5):
- Divide by 2: (10, 25, 15)
- Divide by 2: (5, 25, 15)
- Divide by 3: (5, 25, 5)
- Divide by 5: (1, 5, 1)
- Divide by 5: (1, 1, 1)
$\text{LCM} = 2 \times 2 \times 3 \times 5 \times 5 = 4 \times 3 \times 25 = 12 \times 25 = \mathbf{300}$.
Divide by primes (2, 2, 3, 5, 5):
- Divide by 2: (10, 25, 15)
- Divide by 2: (5, 25, 15)
- Divide by 3: (5, 25, 5)
- Divide by 5: (1, 5, 1)
- Divide by 5: (1, 1, 1)
$\text{LCM} = 2 \times 2 \times 3 \times 5 \times 5 = 4 \times 3 \times 25 = 12 \times 25 = \mathbf{300}$.
CBSE Board Marking Scheme 2026-27
Division grid showing prime factors1.5 Marks
Multiplying factors yielding LCM = 3001.5 Marks
Q7
The HCF and LCM of two numbers are 16 and 240 respectively. If one of the numbers is 48, find the other number.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Mathematical Property:
$\text{Product of two numbers} = \text{HCF} \times \text{LCM}$
$48 \times \text{Other Number} = 16 \times 240$
$\text{Other Number} = \frac{16 \times 240}{48} = \frac{1 \times 240}{3} = \mathbf{80}$.
The other number is 80.
$\text{Product of two numbers} = \text{HCF} \times \text{LCM}$
$48 \times \text{Other Number} = 16 \times 240$
$\text{Other Number} = \frac{16 \times 240}{48} = \frac{1 \times 240}{3} = \mathbf{80}$.
The other number is 80.
CBSE Board Marking Scheme 2026-27
Stating formula HCF × LCM = Product of numbers1 Mark
Evaluation yielding 801 Mark
Q8
Two tankers contain 850 litres and 680 litres of kerosene oil respectively. Find the maximum capacity of a container which can measure the kerosene of both tankers when used an exact number of times.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
To find the maximum capacity of the measuring container, we must calculate the HCF of 850 and 680.
- Prime factorization of $850 = 2 \times 5 \times 5 \times 17 = 2 \times 5^2 \times 17$
- Prime factorization of $680 = 2 \times 2 \times 2 \times 5 \times 17 = 2^3 \times 5 \times 17$
Common prime factors: $2^1 \times 5^1 \times 17^1 = 10 \times 17 = \mathbf{170}$.
Therefore, the maximum capacity of the container is 170 litres.
- Prime factorization of $850 = 2 \times 5 \times 5 \times 17 = 2 \times 5^2 \times 17$
- Prime factorization of $680 = 2 \times 2 \times 2 \times 5 \times 17 = 2^3 \times 5 \times 17$
Common prime factors: $2^1 \times 5^1 \times 17^1 = 10 \times 17 = \mathbf{170}$.
Therefore, the maximum capacity of the container is 170 litres.
CBSE Board Marking Scheme 2026-27
Identifying problem requires HCF calculation1 Mark
Prime factorization of 850 and 6801 Mark
Answer: 170 litres with units1 Mark
Q9
Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm, and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
To find the minimum common distance, we must compute the LCM of 63, 70, and 77.
- Prime factors of $63 = 3^2 \times 7$
- Prime factors of $70 = 2 \times 5 \times 7$
- Prime factors of $77 = 7 \times 11$
$\text{LCM} = 2 \times 3^2 \times 5 \times 7 \times 11 = 2 \times 9 \times 5 \times 7 \times 11 = 90 \times 77 = \mathbf{6,930\text{ cm}}$.
In metres: $6,930\text{ cm} = \mathbf{69\text{ m } 30\text{ cm}}$.
- Prime factors of $63 = 3^2 \times 7$
- Prime factors of $70 = 2 \times 5 \times 7$
- Prime factors of $77 = 7 \times 11$
$\text{LCM} = 2 \times 3^2 \times 5 \times 7 \times 11 = 2 \times 9 \times 5 \times 7 \times 11 = 90 \times 77 = \mathbf{6,930\text{ cm}}$.
In metres: $6,930\text{ cm} = \mathbf{69\text{ m } 30\text{ cm}}$.
CBSE Board Marking Scheme 2026-27
Recognizing need for LCM1 Mark
Calculating LCM = 6,930 cm1.5 Marks
Final answer in cm / m with units0.5 Mark
Q10
Define twin prime numbers and co-prime numbers. Give two examples of each.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Twin Primes: Two prime numbers whose difference is 2 are called twin primes.
Examples: (3, 5), (5, 7), (11, 13), (17, 19).
Co-prime Numbers: Two numbers that have only 1 as their common factor are called co-prime numbers. They do not need to be prime individually.
Examples: (4, 9), (8, 15), (7, 10).
Examples: (3, 5), (5, 7), (11, 13), (17, 19).
Co-prime Numbers: Two numbers that have only 1 as their common factor are called co-prime numbers. They do not need to be prime individually.
Examples: (4, 9), (8, 15), (7, 10).
CBSE Board Marking Scheme 2026-27
Definition and examples of Twin Primes1 Mark
Definition and examples of Co-primes1 Mark
Q11
What is a "perfect number"? Show that 6 and 28 are perfect numbers by finding the sum of all their factors.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Definition: A number for which the sum of all its factors is equal to twice the number is called a Perfect Number.
- For 6: Factors are 1, 2, 3, 6.
Sum = $1 + 2 + 3 + 6 = 12 = 2 \times 6$. Hence, 6 is a perfect number.
- For 28: Factors are 1, 2, 4, 7, 14, 28.
Sum = $1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 \times 28$. Hence, 28 is a perfect number.
- For 6: Factors are 1, 2, 3, 6.
Sum = $1 + 2 + 3 + 6 = 12 = 2 \times 6$. Hence, 6 is a perfect number.
- For 28: Factors are 1, 2, 4, 7, 14, 28.
Sum = $1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 \times 28$. Hence, 28 is a perfect number.
CBSE Board Marking Scheme 2026-27
Definition of perfect number1 Mark
Verifying sums for 6 (12) and 28 (56)1 Mark
Q12
Find the smallest 4-digit number which is divisible by 18, 24, and 32.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Step 1: Find the LCM of 18, 24, and 32.
- $18 = 2 \times 3^2$
- $24 = 2^3 \times 3$
- $32 = 2^5$
$\text{LCM} = 2^5 \times 3^2 = 32 \times 9 = 288$.
Step 2: Smallest 4-digit number is 1,000.
Divide 1,000 by 288: $1,000 = 288 \times 3 + 136$.
Required number = $1,000 + (288 - 136) = 1,000 + 152 = \mathbf{1,152}$.
Verification: $1,152 = 288 \times 4$.
- $18 = 2 \times 3^2$
- $24 = 2^3 \times 3$
- $32 = 2^5$
$\text{LCM} = 2^5 \times 3^2 = 32 \times 9 = 288$.
Step 2: Smallest 4-digit number is 1,000.
Divide 1,000 by 288: $1,000 = 288 \times 3 + 136$.
Required number = $1,000 + (288 - 136) = 1,000 + 152 = \mathbf{1,152}$.
Verification: $1,152 = 288 \times 4$.
CBSE Board Marking Scheme 2026-27
Calculating LCM(18, 24, 32) = 2881.5 Marks
Finding required smallest 4-digit multiple 1,1521.5 Marks
CBSE Competency-Based Question (Traffic Signal Synchronization)
Case Study: Smart City Traffic Signal Synchronization:
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds, and 108 seconds respectively.
(a) If they change simultaneously at 7:00:00 a.m., after how much time will they change simultaneously again?
(b) At what exact clock time will they change simultaneously next?
(c) How many times will they change simultaneously in 1 hour?
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds, and 108 seconds respectively.
(a) If they change simultaneously at 7:00:00 a.m., after how much time will they change simultaneously again?
(b) At what exact clock time will they change simultaneously next?
(c) How many times will they change simultaneously in 1 hour?
Detailed Analytical Solution & Marking Scheme
(a) Time Interval (LCM of 48, 72, 108):
- $48 = 2^4 \times 3$
- $72 = 2^3 \times 3^2$
- $108 = 2^2 \times 3^3$
$\text{LCM} = 2^4 \times 3^3 = 16 \times 27 = \mathbf{432\text{ seconds}}$.
Convert to minutes and seconds: $432 = 7 \times 60 + 12 = \mathbf{7\text{ minutes } 12\text{ seconds}}$.
(b) Exact Next Clock Time:
7:00:00 a.m. + 7 min 12 sec = 7:07:12 a.m.
(c) Occurrences in 1 hour (3600 seconds):
$\frac{3600}{432} \approx 8.33$.
They will change simultaneously 8 times in the next hour.
6
Chapter 6: Perimeter and Area
NCERT Ganita Prakash (Class 6) — Perimeter of Polygons, Area of Rectangles & Squares, Grid Estimation & Mensuration Applications | CBSE 2026-27
📌 Key Mensuration Formulas & Units
- Perimeter: The distance covered along the boundary forming a closed figure when you go around the figure once. Units: cm, m, km.
- Perimeter Formulas:
- Rectangle: $P = 2 \times (\text{Length} + \text{Breadth}) = 2(l + b)$.
- Square: $P = 4 \times \text{Side} = 4s$.
- Equilateral Triangle: $P = 3 \times \text{Side} = 3s$.
- Regular Polygon of $n$ sides: $P = n \times \text{Side}$.
- Area: The amount of surface enclosed by a closed figure. Units: sq cm ($ ext{cm}^2$), sq m ($ ext{m}^2$).
- Area Formulas:
- Rectangle: $\text{Area} = \text{Length} \times \text{Breadth} = l \times b$.
- Square: $\text{Area} = \text{Side} \times \text{Side} = s^2$.
- Estimating Area on Squared Paper: Count fully covered squares as 1, more than half as 1, exactly half as $\frac{1}{2}$, and ignore less than half.
NCERT Exercise 6.1: Perimeter of Polygons
Q1
Find the perimeter of each of the following figures:
(a) A rectangle of sides 15 cm and 9 cm.
(b) A square of side 8.5 cm.
(c) An equilateral triangle of side 12 cm.
(d) A regular hexagon of side 6 cm.
(a) A rectangle of sides 15 cm and 9 cm.
(b) A square of side 8.5 cm.
(c) An equilateral triangle of side 12 cm.
(d) A regular hexagon of side 6 cm.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Rectangle: $P = 2(l + b) = 2(15 + 9) = 2(24) = \mathbf{48\text{ cm}}$.
(b) Square: $P = 4s = 4 \times 8.5 = \mathbf{34\text{ cm}}$.
(c) Equilateral Triangle: $P = 3s = 3 \times 12 = \mathbf{36\text{ cm}}$.
(d) Regular Hexagon: $P = 6s = 6 \times 6 = \mathbf{36\text{ cm}}$.
(b) Square: $P = 4s = 4 \times 8.5 = \mathbf{34\text{ cm}}$.
(c) Equilateral Triangle: $P = 3s = 3 \times 12 = \mathbf{36\text{ cm}}$.
(d) Regular Hexagon: $P = 6s = 6 \times 6 = \mathbf{36\text{ cm}}$.
CBSE Board Marking Scheme 2026-27
Each correct perimeter calculation with units0.75 Mark each (Total 3 Marks)
Q2
A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) A square?
(b) An equilateral triangle?
(c) A regular hexagon?
(a) A square?
(b) An equilateral triangle?
(c) A regular hexagon?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
The perimeter of each figure equals the total string length = 30 cm.
(a) Square (4 equal sides): Side = $\frac{30}{4} = \mathbf{7.5\text{ cm}}$.
(b) Equilateral Triangle (3 equal sides): Side = $\frac{30}{3} = \mathbf{10\text{ cm}}$.
(c) Regular Hexagon (6 equal sides): Side = $\frac{30}{6} = \mathbf{5\text{ cm}}$.
(a) Square (4 equal sides): Side = $\frac{30}{4} = \mathbf{7.5\text{ cm}}$.
(b) Equilateral Triangle (3 equal sides): Side = $\frac{30}{3} = \mathbf{10\text{ cm}}$.
(c) Regular Hexagon (6 equal sides): Side = $\frac{30}{6} = \mathbf{5\text{ cm}}$.
CBSE Board Marking Scheme 2026-27
1 Mark per part with division and units3 Marks
Q3
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the length of its third side?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Perimeter of triangle = Sum of all three sides.
$12 + 14 + \text{Third Side} = 36$
$26 + \text{Third Side} = 36$
$\text{Third Side} = 36 - 26 = \mathbf{10\text{ cm}}$.
$12 + 14 + \text{Third Side} = 36$
$26 + \text{Third Side} = 36$
$\text{Third Side} = 36 - 26 = \mathbf{10\text{ cm}}$.
CBSE Board Marking Scheme 2026-27
Setting up equation 26 + Third Side = 361 Mark
Answer 10 cm1 Mark
Q4
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹ 12 per metre.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Step 1: Find perimeter of rectangular park:
$P = 2(l + b) = 2(175 + 125) = 2(300) = 600\text{ m}$.
Step 2: Cost of fencing:
$\text{Cost} = 600 \times \text{₹ } 12 = \mathbf{\text{₹ } 7,200}$.
$P = 2(l + b) = 2(175 + 125) = 2(300) = 600\text{ m}$.
Step 2: Cost of fencing:
$\text{Cost} = 600 \times \text{₹ } 12 = \mathbf{\text{₹ } 7,200}$.
CBSE Board Marking Scheme 2026-27
Perimeter calculation 600 m1 Mark
Cost calculation ₹ 7,2001 Mark
NCERT Exercise 6.2: Area of Rectangles & Squares
Q5
Find the area of a rectangle whose length and breadth are 12 cm and 4 cm respectively. Also find the area of a square of side 7 cm.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Rectangle: $\text{Area} = l \times b = 12\text{ cm} \times 4\text{ cm} = \mathbf{48\text{ cm}^2}$.
Square: $\text{Area} = s^2 = 7\text{ cm} \times 7\text{ cm} = \mathbf{49\text{ cm}^2}$.
Square: $\text{Area} = s^2 = 7\text{ cm} \times 7\text{ cm} = \mathbf{49\text{ cm}^2}$.
CBSE Board Marking Scheme 2026-27
Rectangle area: 48 cm²1 Mark
Square area: 49 cm²1 Mark
Q6
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
$\text{Area} = \text{Length} \times \text{Width}$
$300 = 50 \times \text{Width}$
$\text{Width} = \frac{300}{50} = \mathbf{6\text{ metres}}$.
$300 = 50 \times \text{Width}$
$\text{Width} = \frac{300}{50} = \mathbf{6\text{ metres}}$.
CBSE Board Marking Scheme 2026-27
Formula substitution: 300 = 50 × Width1 Mark
Answer: 6 metres1 Mark
Q7
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Step 1: Calculate area of plot:
$\text{Area} = 500\text{ m} \times 200\text{ m} = 1,00,000\text{ m}^2$.
Step 2: Cost rate is ₹ 8 per 100 sq m:
Number of hundreds = $\frac{1,00,000}{100} = 1,000$.
Total Cost = $1,000 \times \text{₹ } 8 = \mathbf{\text{₹ } 8,000}$.
$\text{Area} = 500\text{ m} \times 200\text{ m} = 1,00,000\text{ m}^2$.
Step 2: Cost rate is ₹ 8 per 100 sq m:
Number of hundreds = $\frac{1,00,000}{100} = 1,000$.
Total Cost = $1,000 \times \text{₹ } 8 = \mathbf{\text{₹ } 8,000}$.
CBSE Board Marking Scheme 2026-27
Area calculation: 1,00,000 m²1.5 Marks
Cost computation: ₹ 8,0001.5 Marks
Q8
A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Length $l = 4\text{ m}$.
Breadth $b = 3\text{ m } 50\text{ cm} = 3.5\text{ m}$.
$\text{Area of floor} = l \times b = 4 \times 3.5 = \mathbf{14\text{ sq metres}}$.
Therefore, 14 sq m of carpet is needed.
Breadth $b = 3\text{ m } 50\text{ cm} = 3.5\text{ m}$.
$\text{Area of floor} = l \times b = 4 \times 3.5 = \mathbf{14\text{ sq metres}}$.
Therefore, 14 sq m of carpet is needed.
CBSE Board Marking Scheme 2026-27
Converting breadth to 3.5 m1 Mark
Calculating area 14 m²1 Mark
Q9
How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are 100 cm and 144 cm?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Step 1: Area of the rectangular region = $100\text{ cm} \times 144\text{ cm} = 14,400\text{ cm}^2$.
Step 2: Area of one tile = $12\text{ cm} \times 5\text{ cm} = 60\text{ cm}^2$.
Step 3: Number of tiles needed = $\frac{\text{Area of region}}{\text{Area of one tile}} = \frac{14,400}{60} = \mathbf{240\text{ tiles}}$.
Step 2: Area of one tile = $12\text{ cm} \times 5\text{ cm} = 60\text{ cm}^2$.
Step 3: Number of tiles needed = $\frac{\text{Area of region}}{\text{Area of one tile}} = \frac{14,400}{60} = \mathbf{240\text{ tiles}}$.
CBSE Board Marking Scheme 2026-27
Area of region (14,400) and tile (60)1.5 Marks
Division yielding 240 tiles1.5 Marks
Q10
Can two different rectangles have the same perimeter but different areas? Justify your answer with two concrete numerical examples.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Yes.
Consider two rectangles with perimeter = 20 cm ($l + b = 10$):
- Rectangle 1: $l = 6\text{ cm}, b = 4\text{ cm}$.
Perimeter = $2(6+4) = 20\text{ cm}$. Area = $6 \times 4 = \mathbf{24\text{ cm}^2}$.
- Rectangle 2: $l = 8\text{ cm}, b = 2\text{ cm}$.
Perimeter = $2(8+2) = 20\text{ cm}$. Area = $8 \times 2 = \mathbf{16\text{ cm}^2}$.
Both have perimeter 20 cm, but their areas are 24 cm² and 16 cm².
Consider two rectangles with perimeter = 20 cm ($l + b = 10$):
- Rectangle 1: $l = 6\text{ cm}, b = 4\text{ cm}$.
Perimeter = $2(6+4) = 20\text{ cm}$. Area = $6 \times 4 = \mathbf{24\text{ cm}^2}$.
- Rectangle 2: $l = 8\text{ cm}, b = 2\text{ cm}$.
Perimeter = $2(8+2) = 20\text{ cm}$. Area = $8 \times 2 = \mathbf{16\text{ cm}^2}$.
Both have perimeter 20 cm, but their areas are 24 cm² and 16 cm².
CBSE Board Marking Scheme 2026-27
Affirmative statement "Yes"0.5 Mark
Two numerical examples with equal perimeter and distinct areas1.5 Marks
Q11
A wire of length 36 cm is bent in the form of a square. If the same wire is straightened and rebent into a rectangle of breadth 6 cm, find the length of the rectangle and compare the areas of both shapes.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Square: $4s = 36 \implies s = 9\text{ cm}$.
Area of square = $9 \times 9 = \mathbf{81\text{ cm}^2}$.
Rectangle: $2(l + 6) = 36 \implies l + 6 = 18 \implies l = \mathbf{12\text{ cm}}$.
Area of rectangle = $12 \times 6 = \mathbf{72\text{ cm}^2}$.
Comparison: The square encloses more area than the rectangle ($81\text{ cm}^2 > 72\text{ cm}^2$).
Area of square = $9 \times 9 = \mathbf{81\text{ cm}^2}$.
Rectangle: $2(l + 6) = 36 \implies l + 6 = 18 \implies l = \mathbf{12\text{ cm}}$.
Area of rectangle = $12 \times 6 = \mathbf{72\text{ cm}^2}$.
Comparison: The square encloses more area than the rectangle ($81\text{ cm}^2 > 72\text{ cm}^2$).
CBSE Board Marking Scheme 2026-27
Square side 9 cm and area 81 cm²1 Mark
Rectangle length 12 cm and area 72 cm²1 Mark
Comparison conclusion1 Mark
CBSE Competency-Based Question (Sports Field Budgeting)
Case Study: School Football Ground & Running Track:
A standard school football pitch is 100 m long and 64 m wide. Surrounding the pitch is an athletics running track path of uniform width 3 m.
(a) Find the playing area of the football pitch.
(b) Find the outer perimeter of the running track.
(c) Calculate the total area of the running track path.
A standard school football pitch is 100 m long and 64 m wide. Surrounding the pitch is an athletics running track path of uniform width 3 m.
(a) Find the playing area of the football pitch.
(b) Find the outer perimeter of the running track.
(c) Calculate the total area of the running track path.
Detailed Analytical Solution & Marking Scheme
(a) Playing Area of Football Pitch:
$\text{Area} = 100\text{ m} \times 64\text{ m} = \mathbf{6,400\text{ m}^2}$.
(b) Outer Dimensions & Perimeter:
- Outer length = $100 + 2(3) = 106\text{ m}$.
- Outer breadth = $64 + 2(3) = 70\text{ m}$.
Outer Perimeter = $2(106 + 70) = 2(176) = \mathbf{352\text{ metres}}$.
(c) Area of the Running Track Path:
Outer Area = $106 \times 70 = 7,420\text{ m}^2$.
Inner Area = $6,400\text{ m}^2$.
$\text{Path Area} = 7,420 - 6,400 = \mathbf{1,020\text{ m}^2}$.
7
Chapter 7: Fractions
NCERT Ganita Prakash (Class 6) — Proper, Improper & Mixed Fractions, Number Line Models, Equivalent Fractions, Simplest Form & Arithmetic | CBSE 2026-27
📌 Key Fraction Concepts & Operation Rules
- Fraction: A number representing a part of a whole. In $\frac{a}{b}$, $a$ is numerator and $b$ is denominator ($b \neq 0$).
- Types of Fractions:
- Proper Fraction: Numerator < Denominator (value < 1), e.g., $\frac{3}{7}$.
- Improper Fraction: Numerator $\ge$ Denominator (value $\ge$ 1), e.g., $\frac{11}{4}$.
- Mixed Fraction: Combination of a whole number and a proper fraction, e.g., $2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{11}{4}$.
- Equivalent Fractions: Fractions having identical values, obtained by multiplying or dividing numerator and denominator by the same non-zero integer.
- Simplest Form: A fraction $\frac{a}{b}$ is in simplest form if $\text{HCF}(a, b) = 1$.
- Addition & Subtraction: Convert unlike fractions to like fractions using the LCM of their denominators before adding or subtracting.
NCERT Exercise 7.1: Visual Fractions & Number Line
Q1
Write the fraction representing the shaded portion in the following figures:
(a) A circle divided into 8 equal parts with 3 parts shaded.
(b) A rectangle divided into 6 equal parts with 5 parts shaded.
(c) What fraction of a day is 8 hours?
(d) What fraction of an hour is 40 minutes?
(a) A circle divided into 8 equal parts with 3 parts shaded.
(b) A rectangle divided into 6 equal parts with 5 parts shaded.
(c) What fraction of a day is 8 hours?
(d) What fraction of an hour is 40 minutes?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Circle: Shaded parts = 3, Total parts = 8 $\implies \mathbf{\frac{3}{8}}$.
(b) Rectangle: Shaded parts = 5, Total parts = 6 $\implies \mathbf{\frac{5}{6}}$.
(c) 8 hours of a day (24 hours): $\frac{8}{24} = \mathbf{\frac{1}{3}}$.
(d) 40 minutes of an hour (60 minutes): $\frac{40}{60} = \mathbf{\frac{2}{3}}$.
(b) Rectangle: Shaded parts = 5, Total parts = 6 $\implies \mathbf{\frac{5}{6}}$.
(c) 8 hours of a day (24 hours): $\frac{8}{24} = \mathbf{\frac{1}{3}}$.
(d) 40 minutes of an hour (60 minutes): $\frac{40}{60} = \mathbf{\frac{2}{3}}$.
CBSE Board Marking Scheme 2026-27
Each correct fraction in lowest form0.5 Mark each (Total 2 Marks)
Q2
Represent the following fractions on a number line:
(a) 2/5 (b) 3/5 (c) 7/5
(a) 2/5 (b) 3/5 (c) 7/5
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Construction on Number Line:
1. Draw a straight line and mark 0, 1, and 2.
2. Divide the unit interval between 0 and 1 into 5 equal parts (each step is $\frac{1}{5}$).
3. Point at 2 steps from 0 marks 2/5.
4. Point at 3 steps from 0 marks 3/5.
5. Divide the interval between 1 and 2 into 5 equal parts. Point at 2 steps beyond 1 ($1 + \frac{2}{5} = \frac{7}{5}$) marks 7/5.
1. Draw a straight line and mark 0, 1, and 2.
2. Divide the unit interval between 0 and 1 into 5 equal parts (each step is $\frac{1}{5}$).
3. Point at 2 steps from 0 marks 2/5.
4. Point at 3 steps from 0 marks 3/5.
5. Divide the interval between 1 and 2 into 5 equal parts. Point at 2 steps beyond 1 ($1 + \frac{2}{5} = \frac{7}{5}$) marks 7/5.
CBSE Board Marking Scheme 2026-27
Dividing unit segment into 5 equal intervals1 Mark
Correctly locating 2/5, 3/5, and 7/51 Mark
NCERT Exercise 7.2: Types of Fractions & Conversions
Q3
(a) Express the following as mixed fractions: (i) 20/3 (ii) 27/5
(b) Express the following as improper fractions: (i) 7 3/4 (ii) 5 6/7
(b) Express the following as improper fractions: (i) 7 3/4 (ii) 5 6/7
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Mixed Fractions:
(i) $20 \div 3 = 6$ (quotient) with remainder 2 $\implies \mathbf{6\frac{2}{3}}$.
(ii) $27 \div 5 = 5$ (quotient) with remainder 2 $\implies \mathbf{5\frac{2}{5}}$.
(i) $20 \div 3 = 6$ (quotient) with remainder 2 $\implies \mathbf{6\frac{2}{3}}$.
(ii) $27 \div 5 = 5$ (quotient) with remainder 2 $\implies \mathbf{5\frac{2}{5}}$.
(b) Improper Fractions:
(i) $7\frac{3}{4} = \frac{7 \times 4 + 3}{4} = \frac{28 + 3}{4} = \mathbf{\frac{31}{4}}$.
(ii) $5\frac{6}{7} = \frac{5 \times 7 + 6}{7} = \frac{35 + 6}{7} = \mathbf{\frac{41}{7}}$.
(i) $7\frac{3}{4} = \frac{7 \times 4 + 3}{4} = \frac{28 + 3}{4} = \mathbf{\frac{31}{4}}$.
(ii) $5\frac{6}{7} = \frac{5 \times 7 + 6}{7} = \frac{35 + 6}{7} = \mathbf{\frac{41}{7}}$.
CBSE Board Marking Scheme 2026-27
Each conversion correct0.5 Mark each (Total 2 Marks)
NCERT Exercise 7.3: Equivalent Fractions & Simplest Form
Q4
Find the equivalent fraction of 3/5 having:
(a) Denominator 20
(b) Numerator 9
(c) Denominator 30
(a) Denominator 20
(b) Numerator 9
(c) Denominator 30
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Denominator 20: Multiply numerator and denominator by 4:
$\frac{3 \times 4}{5 \times 4} = \mathbf{\frac{12}{20}}$.
$\frac{3 \times 4}{5 \times 4} = \mathbf{\frac{12}{20}}$.
(b) Numerator 9: Multiply numerator and denominator by 3:
$\frac{3 \times 3}{5 \times 3} = \mathbf{\frac{9}{15}}$.
$\frac{3 \times 3}{5 \times 3} = \mathbf{\frac{9}{15}}$.
(c) Denominator 30: Multiply numerator and denominator by 6:
$\frac{3 \times 6}{5 \times 6} = \mathbf{\frac{18}{30}}$.
$\frac{3 \times 6}{5 \times 6} = \mathbf{\frac{18}{30}}$.
CBSE Board Marking Scheme 2026-27
1 Mark per correct equivalent fraction3 Marks
Q5
Reduce the following fractions to their simplest (lowest) form by dividing by HCF:
(a) 48/60 (b) 84/98 (c) 12/52
(a) 48/60 (b) 84/98 (c) 12/52
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) 48/60: $\text{HCF}(48, 60) = 12$.
$\frac{48 \div 12}{60 \div 12} = \mathbf{\frac{4}{5}}$.
$\frac{48 \div 12}{60 \div 12} = \mathbf{\frac{4}{5}}$.
(b) 84/98: $\text{HCF}(84, 98) = 14$.
$\frac{84 \div 14}{98 \div 14} = \mathbf{\frac{6}{7}}$.
$\frac{84 \div 14}{98 \div 14} = \mathbf{\frac{6}{7}}$.
(c) 12/52: $\text{HCF}(12, 52) = 4$.
$\frac{12 \div 4}{52 \div 4} = \mathbf{\frac{3}{13}}$.
$\frac{12 \div 4}{52 \div 4} = \mathbf{\frac{3}{13}}$.
CBSE Board Marking Scheme 2026-27
1 Mark per reduction with HCF step3 Marks
NCERT Exercise 7.4: Fraction Comparison & Ordering
Q6
Compare the following pairs of fractions and place appropriate sign (<, =, >):
(a) 3/6 and 5/6
(b) 1/4 and 1/7
(c) 4/5 and 5/6
(a) 3/6 and 5/6
(b) 1/4 and 1/7
(c) 4/5 and 5/6
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) 3/6 and 5/6: Like fractions; since $3 < 5$, $\mathbf{\frac{3}{6} < \frac{5}{6}}$.
(b) 1/4 and 1/7: Same numerator; the fraction with the smaller denominator is larger: $\mathbf{\frac{1}{4} > \frac{1}{7}}$.
(c) 4/5 and 5/6: Unlike fractions. $\text{LCM}(5, 6) = 30$.
$\frac{4}{5} = \frac{24}{30}$ and $\frac{5}{6} = \frac{25}{30}$.
Since $24 < 25$, $\mathbf{\frac{4}{5} < \frac{5}{6}}$.
$\frac{4}{5} = \frac{24}{30}$ and $\frac{5}{6} = \frac{25}{30}$.
Since $24 < 25$, $\mathbf{\frac{4}{5} < \frac{5}{6}}$.
CBSE Board Marking Scheme 2026-27
1 Mark per comparison with clear reasoning3 Marks
NCERT Exercise 7.5: Addition & Subtraction of Fractions
Q7
Solve the following:
(a) 1/18 + 1/18
(b) 8/15 + 3/15
(c) 7/7 - 5/7
(d) 1 - 2/3
(a) 1/18 + 1/18
(b) 8/15 + 3/15
(c) 7/7 - 5/7
(d) 1 - 2/3
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) $\frac{1}{18} + \frac{1}{18} = \frac{2}{18} = \mathbf{\frac{1}{9}}$.
(b) $\frac{8}{15} + \frac{3}{15} = \mathbf{\frac{11}{15}}$.
(c) $\frac{7}{7} - \frac{5}{7} = \mathbf{\frac{2}{7}}$.
(d) $1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \mathbf{\frac{1}{3}}$.
(b) $\frac{8}{15} + \frac{3}{15} = \mathbf{\frac{11}{15}}$.
(c) $\frac{7}{7} - \frac{5}{7} = \mathbf{\frac{2}{7}}$.
(d) $1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \mathbf{\frac{1}{3}}$.
CBSE Board Marking Scheme 2026-27
Each correct evaluation in simplest form0.5 Mark each (Total 2 Marks)
Q8
Find the sum: 2/3 + 3/4 + 1/2.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Step 1: Find LCM of denominators 3, 4, and 2:
$\text{LCM}(3, 4, 2) = 12$.
Step 2: Convert to like fractions:
$\frac{2}{3} = \frac{8}{12}$; $\frac{3}{4} = \frac{9}{12}$; $\frac{1}{2} = \frac{6}{12}$.
Step 3: Add numerators:
$\text{Sum} = \frac{8 + 9 + 6}{12} = \frac{23}{12} = \mathbf{1\frac{11}{12}}$.
$\text{LCM}(3, 4, 2) = 12$.
Step 2: Convert to like fractions:
$\frac{2}{3} = \frac{8}{12}$; $\frac{3}{4} = \frac{9}{12}$; $\frac{1}{2} = \frac{6}{12}$.
Step 3: Add numerators:
$\text{Sum} = \frac{8 + 9 + 6}{12} = \frac{23}{12} = \mathbf{1\frac{11}{12}}$.
CBSE Board Marking Scheme 2026-27
LCM = 12 and conversion to like fractions1 Mark
Sum = 23/12 or 1 11/121 Mark
Q9
Subtract 2 1/5 from 4 2/3.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Convert mixed numbers to improper fractions:
$4\frac{2}{3} = \frac{14}{3}$; $2\frac{1}{5} = \frac{11}{5}$.
$\text{LCM}(3, 5) = 15$.
$\frac{14}{3} = \frac{70}{15}$; $\frac{11}{5} = \frac{33}{15}$.
Difference = $\frac{70 - 33}{15} = \frac{37}{15} = \mathbf{2\frac{7}{15}}$.
$4\frac{2}{3} = \frac{14}{3}$; $2\frac{1}{5} = \frac{11}{5}$.
$\text{LCM}(3, 5) = 15$.
$\frac{14}{3} = \frac{70}{15}$; $\frac{11}{5} = \frac{33}{15}$.
Difference = $\frac{70 - 33}{15} = \frac{37}{15} = \mathbf{2\frac{7}{15}}$.
CBSE Board Marking Scheme 2026-27
Improper conversion and LCM = 151 Mark
Answer 37/15 or 2 7/151 Mark
Q10
Sarita bought 2/5 metre of ribbon and Lalita 3/4 metre of ribbon. What is the total length of the ribbon they bought?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Total length = $\frac{2}{5} + \frac{3}{4}\text{ metres}$.
$\text{LCM}(5, 4) = 20$.
$\frac{2}{5} = \frac{8}{20}$; $\frac{3}{4} = \frac{15}{20}$.
Total = $\frac{8 + 15}{20} = \frac{23}{20} = \mathbf{1\frac{3}{20}\text{ metres}}$ (or 1.15 m).
$\text{LCM}(5, 4) = 20$.
$\frac{2}{5} = \frac{8}{20}$; $\frac{3}{4} = \frac{15}{20}$.
Total = $\frac{8 + 15}{20} = \frac{23}{20} = \mathbf{1\frac{3}{20}\text{ metres}}$ (or 1.15 m).
CBSE Board Marking Scheme 2026-27
Formulating sum with LCM = 201 Mark
Answer 23/20 m or 1 3/20 m1 Mark
Q11
A piece of wire 7/8 metre long broke into two pieces. One piece was 1/4 metre long. How long is the other piece?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Length of other piece = Total length - First piece
$= \frac{7}{8} - \frac{1}{4}$.
$\text{LCM}(8, 4) = 8$.
$\frac{1}{4} = \frac{2}{8}$.
Length = $\frac{7 - 2}{8} = \mathbf{\frac{5}{8}\text{ metre}}$.
$= \frac{7}{8} - \frac{1}{4}$.
$\text{LCM}(8, 4) = 8$.
$\frac{1}{4} = \frac{2}{8}$.
Length = $\frac{7 - 2}{8} = \mathbf{\frac{5}{8}\text{ metre}}$.
CBSE Board Marking Scheme 2026-27
Subtraction setup with LCM1 Mark
Answer: 5/8 metre1 Mark
CBSE Competency-Based Question (Culinary Recipe Fractions)
Case Study: Culinary Recipe Scaling & Nutritional Fractions:
A chef is baking a traditional carrot cake. The original recipe requires $2\frac{1}{2}$ cups of flour, $\frac{3}{4}$ cup of brown sugar, and $1\frac{1}{4}$ cups of grated carrots.
(a) If the chef wants to bake double the batch (2 times), how much flour and brown sugar will be required?
(b) If the chef has only 3 cups of flour available, how much more flour is needed for the double batch?
(c) Express the total liquid sweetener and vegetable ratio as a combined fraction.
A chef is baking a traditional carrot cake. The original recipe requires $2\frac{1}{2}$ cups of flour, $\frac{3}{4}$ cup of brown sugar, and $1\frac{1}{4}$ cups of grated carrots.
(a) If the chef wants to bake double the batch (2 times), how much flour and brown sugar will be required?
(b) If the chef has only 3 cups of flour available, how much more flour is needed for the double batch?
(c) Express the total liquid sweetener and vegetable ratio as a combined fraction.
Detailed Analytical Solution & Marking Scheme
(a) Doubling the Ingredients:
- Flour = $2 \times 2\frac{1}{2} = 2 \times \frac{5}{2} = \mathbf{5\text{ cups}}$.
- Sugar = $2 \times \frac{3}{4} = \frac{6}{4} = \frac{3}{2} = \mathbf{1\frac{1}{2}\text{ cups}}$.
(b) Extra Flour Needed:
$5\text{ cups} - 3\text{ cups} = \mathbf{2\text{ cups of flour}}$.
(c) Combined Sugar and Carrots:
$\frac{3}{4} + 1\frac{1}{4} = \frac{3}{4} + \frac{5}{4} = \frac{8}{4} = \mathbf{2\text{ cups}}$.
8
Chapter 8: Playing with Constructions
NCERT Ganita Prakash (Class 6) — Compass & Ruler Constructions, Perpendicular Bisectors, Angle Bisectors & Standard Angle Geometry | CBSE 2026-27
📌 Key Geometrical Construction Instruments & Rules
- Instruments: Ruler (unmarked straightedge / calibrated scale in cm/mm), Compass (drawing arcs and circles of precise radii), Divider (comparing lengths), Set Squares (30°-60°-90° and 45°-45°-90°), Protractor (measuring/drawing angles).
- Perpendicular Bisector: A line that divides a given line segment into two equal halves at right angles (90°). Every point on the perpendicular bisector is equidistant from the two endpoints.
- Angle Bisector: A ray that divides an angle into two equal angles.
- Standard Angles with Compass & Ruler:
- 60°: First arc cut from initial point on circumference of same radius.
- 120°: Second consecutive arc cut along circumference.
- 90°: Bisecting the interval between 60° and 120°.
- 30°: Bisecting 60°.
- 45°: Bisecting 90°.
NCERT Exercise 8.1: Circle & Line Segment Constructions
Q1
Draw a circle of radius 3.5 cm using a ruler and compass. Mark its centre O, a radius OA, a diameter AB, and a chord CD.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Steps of Construction:
1. Open the compass to measure exactly 3.5 cm on the ruler scale.
2. Mark a point $O$ on paper with a sharp pencil as the centre.
3. Place the compass needle at $O$ and rotate smoothly through 360° to draw the circle.
4. Draw segment from $O$ to any point on circle $A$ $\implies$ Radius $OA = 3.5\text{ cm}$.
5. Extend $AO$ in a straight line through $O$ to point $B$ on the circle $\implies$ Diameter $AB = 7\text{ cm}$.
6. Join any two points $C$ and $D$ on the circumference not passing through $O$ $\implies$ Chord $CD$.
1. Open the compass to measure exactly 3.5 cm on the ruler scale.
2. Mark a point $O$ on paper with a sharp pencil as the centre.
3. Place the compass needle at $O$ and rotate smoothly through 360° to draw the circle.
4. Draw segment from $O$ to any point on circle $A$ $\implies$ Radius $OA = 3.5\text{ cm}$.
5. Extend $AO$ in a straight line through $O$ to point $B$ on the circle $\implies$ Diameter $AB = 7\text{ cm}$.
6. Join any two points $C$ and $D$ on the circumference not passing through $O$ $\implies$ Chord $CD$.
CBSE Board Marking Scheme 2026-27
Drawing circle of radius 3.5 cm1 Mark
Accurately labeling centre O, radius OA, diameter AB, and chord CD1 Mark
Q2
Draw a line segment AB of length 7.3 cm. Using ruler and compass, find its axis of symmetry (perpendicular bisector). Measure each half.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Steps of Construction:
1. Draw a line segment $AB = 7.3\text{ cm}$ using a calibrated ruler.
2. With $A$ as centre and radius more than half of $AB$ (approx. 4.5 cm), draw two arcs, one above and one below $AB$.
3. With $B$ as centre and the identical radius, draw two arcs intersecting the previous arcs at points $P$ and $Q$.
4. Join $P$ and $Q$ with a straight line. Let $PQ$ intersect $AB$ at point $M$.
5. Line $PQ$ is the required perpendicular bisector (axis of symmetry).
Measurement: $AM = MB = \frac{7.3}{2} = \mathbf{3.65\text{ cm}}$.
1. Draw a line segment $AB = 7.3\text{ cm}$ using a calibrated ruler.
2. With $A$ as centre and radius more than half of $AB$ (approx. 4.5 cm), draw two arcs, one above and one below $AB$.
3. With $B$ as centre and the identical radius, draw two arcs intersecting the previous arcs at points $P$ and $Q$.
4. Join $P$ and $Q$ with a straight line. Let $PQ$ intersect $AB$ at point $M$.
5. Line $PQ$ is the required perpendicular bisector (axis of symmetry).
Measurement: $AM = MB = \frac{7.3}{2} = \mathbf{3.65\text{ cm}}$.
CBSE Board Marking Scheme 2026-27
Drawing segment AB = 7.3 cm1 Mark
Constructing intersecting arcs above and below1 Mark
Drawing line PQ and measuring AM = MB = 3.65 cm1 Mark
NCERT Exercise 8.2: Perpendiculars & Angle Constructions
Q3
Draw an angle of measure 80° using a protractor and construct its angle bisector using a compass.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Steps of Construction:
1. Draw a ray $OA$. Using a protractor, draw ray $OB$ such that $\angle AOB = 80^\circ$.
2. With $O$ as centre and any convenient radius, draw an arc cutting ray $OA$ at $P$ and ray $OB$ at $Q$.
3. With $P$ as centre and radius more than $\frac{1}{2}PQ$, draw an arc inside the angle.
4. With $Q$ as centre and the same radius, draw another arc intersecting the previous arc at point $R$.
5. Join $O$ and $R$ and produce it to form ray $OR$.
Ray $OR$ is the bisector of $\angle AOB$.
Verification: $\angle AOR = \angle ROB = \frac{80^\circ}{2} = \mathbf{40^\circ}$.
1. Draw a ray $OA$. Using a protractor, draw ray $OB$ such that $\angle AOB = 80^\circ$.
2. With $O$ as centre and any convenient radius, draw an arc cutting ray $OA$ at $P$ and ray $OB$ at $Q$.
3. With $P$ as centre and radius more than $\frac{1}{2}PQ$, draw an arc inside the angle.
4. With $Q$ as centre and the same radius, draw another arc intersecting the previous arc at point $R$.
5. Join $O$ and $R$ and produce it to form ray $OR$.
Ray $OR$ is the bisector of $\angle AOB$.
Verification: $\angle AOR = \angle ROB = \frac{80^\circ}{2} = \mathbf{40^\circ}$.
CBSE Board Marking Scheme 2026-27
Drawing ∠AOB = 80°1 Mark
Arc intersection steps for bisector1 Mark
Drawing ray OR and verifying 40°1 Mark
Q4
Construct an angle of 60° using only a ruler and a compass. Explain the geometric rationale for why this angle is exactly 60°.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Steps of Construction:
1. Draw a straight line and mark a point $O$ on it, with ray $OA$.
2. With $O$ as centre and any convenient radius $r$, draw an arc cutting ray $OA$ at point $P$.
3. With $P$ as centre and the same radius $r$, draw an arc cutting the first arc at point $Q$.
4. Join $O$ and $Q$ and extend it to form ray $OB$.
Angle $\angle AOB$ is exactly 60°.
Geometric Rationale: Points $O, P, Q$ form an equilateral triangle because $OP = OQ = PQ = r$. In an equilateral triangle, each interior angle measures $180^\circ / 3 = 60^\circ$.
1. Draw a straight line and mark a point $O$ on it, with ray $OA$.
2. With $O$ as centre and any convenient radius $r$, draw an arc cutting ray $OA$ at point $P$.
3. With $P$ as centre and the same radius $r$, draw an arc cutting the first arc at point $Q$.
4. Join $O$ and $Q$ and extend it to form ray $OB$.
Angle $\angle AOB$ is exactly 60°.
Geometric Rationale: Points $O, P, Q$ form an equilateral triangle because $OP = OQ = PQ = r$. In an equilateral triangle, each interior angle measures $180^\circ / 3 = 60^\circ$.
CBSE Board Marking Scheme 2026-27
Accurate construction steps with compass1.5 Marks
Geometric explanation using equilateral triangle property1.5 Marks
Q5
Construct an angle of 90° at the initial point of a given ray using only a ruler and compass. Then construct its bisector to obtain a 45° angle.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Steps of Construction:
1. Draw ray $OA$. With $O$ as centre, draw a semicircle cutting $OA$ at $X$.
2. With $X$ as centre and same radius, cut the arc at $P$ (60°).
3. With $P$ as centre and same radius, cut the arc at $Q$ (120°).
4. From $P$ and $Q$, draw arcs of equal radius intersecting each other at $R$.
5. Join $OR$. Line $OR$ is perpendicular to $OA$ ($\angle AOR = \mathbf{90^\circ}$).
6. To obtain 45°, bisect $\angle AOR$: with $X$ and the intersection point on $OR$ as centres, draw intersecting arcs and join to $O$. Resulting angle = $\frac{90^\circ}{2} = \mathbf{45^\circ}$.
1. Draw ray $OA$. With $O$ as centre, draw a semicircle cutting $OA$ at $X$.
2. With $X$ as centre and same radius, cut the arc at $P$ (60°).
3. With $P$ as centre and same radius, cut the arc at $Q$ (120°).
4. From $P$ and $Q$, draw arcs of equal radius intersecting each other at $R$.
5. Join $OR$. Line $OR$ is perpendicular to $OA$ ($\angle AOR = \mathbf{90^\circ}$).
6. To obtain 45°, bisect $\angle AOR$: with $X$ and the intersection point on $OR$ as centres, draw intersecting arcs and join to $O$. Resulting angle = $\frac{90^\circ}{2} = \mathbf{45^\circ}$.
CBSE Board Marking Scheme 2026-27
Constructing 90° using 60° and 120° arcs1.5 Marks
Bisecting 90° to construct 45°1.5 Marks
Q6
Explain how you would construct an angle of 30° using only a ruler and compass.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
1. Draw ray $OA$.
2. Construct an angle of 60° ($\angle AOB$) using ruler and compass as described in Q4.
3. Draw the angle bisector of $\angle AOB$ by taking arcs from points on arms $OA$ and $OB$.
4. The bisector ray divides 60° into two equal angles of $\frac{60^\circ}{2} = \mathbf{30^\circ}$.
2. Construct an angle of 60° ($\angle AOB$) using ruler and compass as described in Q4.
3. Draw the angle bisector of $\angle AOB$ by taking arcs from points on arms $OA$ and $OB$.
4. The bisector ray divides 60° into two equal angles of $\frac{60^\circ}{2} = \mathbf{30^\circ}$.
CBSE Board Marking Scheme 2026-27
Stating 60° construction step1 Mark
Bisection step yielding 30°1 Mark
Q7
Draw a circle of radius 4.5 cm with centre O. Mark points P, Q, and R such that:
(a) P lies inside the circle.
(b) Q lies on the circumference.
(c) R lies outside the circle.
Compare the lengths OP, OQ, and OR with the radius 4.5 cm.
(a) P lies inside the circle.
(b) Q lies on the circumference.
(c) R lies outside the circle.
Compare the lengths OP, OQ, and OR with the radius 4.5 cm.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Steps of Construction:
1. Open the compass to measure exactly 4.5 cm on a ruler.
2. Mark a point O on paper as the centre. Keeping compass needle at O, draw the circle.
3. Mark point P inside, point Q on the boundary, and point R in the exterior region.
1. Open the compass to measure exactly 4.5 cm on a ruler.
2. Mark a point O on paper as the centre. Keeping compass needle at O, draw the circle.
3. Mark point P inside, point Q on the boundary, and point R in the exterior region.
Geometric Relationships with Radius (r = 4.5 cm):
• (a) Inside point P: Distance $OP < 4.5\text{ cm}$ (strictly less than radius).
• (b) Boundary point Q: Distance $OQ = 4.5\text{ cm}$ (exactly equal to radius).
• (c) Exterior point R: Distance $OR > 4.5\text{ cm}$ (strictly greater than radius).
• (a) Inside point P: Distance $OP < 4.5\text{ cm}$ (strictly less than radius).
• (b) Boundary point Q: Distance $OQ = 4.5\text{ cm}$ (exactly equal to radius).
• (c) Exterior point R: Distance $OR > 4.5\text{ cm}$ (strictly greater than radius).
CBSE Board Marking Scheme 2026-27
Accurate circle drawing of radius 4.5 cm with marked points1.5 Marks
Correct mathematical distance comparisons (OP < r, OQ = r, OR > r)1.5 Marks
Q8
Construct an angle of 45° using only a ruler and a pair of compasses. Write the step-by-step procedure.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Geometric Principle: A 45° angle is the angle bisector of a 90° right angle ($45° = \frac{90°}{2}$).
Steps of Construction:
1. Draw a base ray OA with initial point O.
2. With O as centre and any convenient radius, draw a semicircle intersecting OA at point X.
3. From X, mark an arc of 60° (point P) and from P mark an arc of 120° (point Q).
4. Bisect the arc between P and Q to construct ray OB perpendicular to OA ($angle AOB = 90°$).
5. Ray OB intersects the initial arc at point Y. With centres X and Y, draw intersecting arcs to locate ray OC.
6. Ray OC bisects $angle AOB$. Therefore, $mathbf{angle AOC = 45°}$.
1. Draw a base ray OA with initial point O.
2. With O as centre and any convenient radius, draw a semicircle intersecting OA at point X.
3. From X, mark an arc of 60° (point P) and from P mark an arc of 120° (point Q).
4. Bisect the arc between P and Q to construct ray OB perpendicular to OA ($angle AOB = 90°$).
5. Ray OB intersects the initial arc at point Y. With centres X and Y, draw intersecting arcs to locate ray OC.
6. Ray OC bisects $angle AOB$. Therefore, $mathbf{angle AOC = 45°}$.
CBSE Board Marking Scheme 2026-27
Construction of 90° angle using 60° and 120° arcs1.5 Marks
Accurate bisection of 90° yielding 45° ray1.5 Marks
Q9
Construct an angle of 30° using ruler and compass only. Show all construction arcs clearly.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Geometric Principle: An angle of 30° is obtained by bisecting a standard 60° angle ($30° = \frac{60°}{2}$).
Steps of Construction:
1. Draw a ray OA.
2. With O as centre and any radius, draw an arc cutting OA at point P.
3. With P as centre and the same radius, draw an arc intersecting the first arc at point Q ($angle QOA = 60°$).
4. With P and Q as centres and radius greater than half of PQ, draw two arcs intersecting at point R.
5. Join OR. Ray OR bisects $angle QOA$.
6. Thus, $mathbf{angle ROA = 30°}$.
1. Draw a ray OA.
2. With O as centre and any radius, draw an arc cutting OA at point P.
3. With P as centre and the same radius, draw an arc intersecting the first arc at point Q ($angle QOA = 60°$).
4. With P and Q as centres and radius greater than half of PQ, draw two arcs intersecting at point R.
5. Join OR. Ray OR bisects $angle QOA$.
6. Thus, $mathbf{angle ROA = 30°}$.
CBSE Board Marking Scheme 2026-27
Constructing standard 60° arc with unchanged radius1.5 Marks
Bisection of 60° to obtain 30° ray with steps1.5 Marks
Q10
Draw a line segment AB of length 7.2 cm. Construct its perpendicular bisector using ruler and compass. Verify by measurement that the bisector divides AB into two equal segments of 3.6 cm each.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Steps of Construction:
1. Draw line segment $AB = 7.2\text{ cm}$ using a millimeter ruler.
2. With centre A and radius greater than half of AB ($r > 3.6\text{ cm}$), draw arcs above and below AB.
3. With centre B and the exact same radius, draw arcs cutting the previous arcs at points X and Y.
4. Draw line XY intersecting segment AB at point M.
5. Line XY is the perpendicular bisector ($XY \perp AB$).
1. Draw line segment $AB = 7.2\text{ cm}$ using a millimeter ruler.
2. With centre A and radius greater than half of AB ($r > 3.6\text{ cm}$), draw arcs above and below AB.
3. With centre B and the exact same radius, draw arcs cutting the previous arcs at points X and Y.
4. Draw line XY intersecting segment AB at point M.
5. Line XY is the perpendicular bisector ($XY \perp AB$).
Verification:
Measuring AM and MB with a ruler gives $AM = 3.6\text{ cm}$ and $MB = 3.6\text{ cm}$.
Also $angle XMA = 90°$. Thus, verified.
Measuring AM and MB with a ruler gives $AM = 3.6\text{ cm}$ and $MB = 3.6\text{ cm}$.
Also $angle XMA = 90°$. Thus, verified.
CBSE Board Marking Scheme 2026-27
Correct segment AB = 7.2 cm and intersecting compass arcs1.5 Marks
Drawing bisector XY and confirming AM = MB = 3.6 cm1.5 Marks
Q11
Draw an angle of 120° using compass and ruler. Divide it into four equal parts using repeated angle bisection. What is the measure of each resulting part?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Mathematical Breakdown:
When an angle of 120° is divided into 4 equal parts, each part has measure:
$\text{Each angle} = \frac{120°}{4} = \mathbf{30°}$.
When an angle of 120° is divided into 4 equal parts, each part has measure:
$\text{Each angle} = \frac{120°}{4} = \mathbf{30°}$.
Construction Method:
1. Construct ray OA and an angle of 120° ($angle AOB = 120°$) using two consecutive 60° arcs on the base semicircle.
2. First Bisection: Bisect $angle AOB$ with ray OC, creating two equal angles of $60°$ ($angle AOC = angle COB = 60°$).
3. Second Bisection: Bisect $angle AOC$ with ray OD to get $angle AOD = 30°$.
4. Third Bisection: Bisect $angle COB$ with ray OE to get $angle COE = 30°$.
5. The four equal angles are each $mathbf{30°}$.
1. Construct ray OA and an angle of 120° ($angle AOB = 120°$) using two consecutive 60° arcs on the base semicircle.
2. First Bisection: Bisect $angle AOB$ with ray OC, creating two equal angles of $60°$ ($angle AOC = angle COB = 60°$).
3. Second Bisection: Bisect $angle AOC$ with ray OD to get $angle AOD = 30°$.
4. Third Bisection: Bisect $angle COB$ with ray OE to get $angle COE = 30°$.
5. The four equal angles are each $mathbf{30°}$.
CBSE Board Marking Scheme 2026-27
Accurate 120° angle construction1 Mark
Repeated bisections producing 4 parts of 30°2 Marks
Q12
Two concentric circles are drawn with common centre O. The radius of the inner circle is 3 cm and the radius of the outer circle is 5 cm. A chord AB of the outer circle touches the inner circle at point P.
(a) What is the angle $angle OPB$?
(b) State the geometric property demonstrated.
(a) What is the angle $angle OPB$?
(b) State the geometric property demonstrated.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Measure of $angle OPB$:
Since OP is the radius of the inner circle and line AB touches the inner circle at P, OP is perpendicular to AB.
Therefore, $mathbf{angle OPB = 90°}$.
Since OP is the radius of the inner circle and line AB touches the inner circle at P, OP is perpendicular to AB.
Therefore, $mathbf{angle OPB = 90°}$.
(b) Geometric Property:
• The radius drawn to the point of contact of a tangent to a circle is perpendicular to the tangent line.
• Furthermore, the perpendicular from the centre to a chord bisects the chord, meaning $AP = PB = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$. Total chord $AB = 8\text{ cm}$.
• The radius drawn to the point of contact of a tangent to a circle is perpendicular to the tangent line.
• Furthermore, the perpendicular from the centre to a chord bisects the chord, meaning $AP = PB = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$. Total chord $AB = 8\text{ cm}$.
CBSE Board Marking Scheme 2026-27
Identifying right angle $angle OPB = 90°$1 Mark
Stating radius-tangent perpendicularity property1 Mark
CBSE Competency-Based Question (Pergola Pergola Blueprint)
Case Study: Botanical Garden Pergola Blueprint:
A landscape architect is designing a semi-circular wooden pergola for a school botanical garden.
(a) The foundation requires two support pillars placed at endpoints A and B of a diameter of length 8 m. What radius must be used for the semi-circular arc?
(b) A central decorative arch must stand along the perpendicular bisector of AB. Describe the geometric steps to establish this axis on the ground using two equal measuring ropes.
(c) What angle does the central pillar make with ground line AB?
A landscape architect is designing a semi-circular wooden pergola for a school botanical garden.
(a) The foundation requires two support pillars placed at endpoints A and B of a diameter of length 8 m. What radius must be used for the semi-circular arc?
(b) A central decorative arch must stand along the perpendicular bisector of AB. Describe the geometric steps to establish this axis on the ground using two equal measuring ropes.
(c) What angle does the central pillar make with ground line AB?
Detailed Analytical Solution & Marking Scheme
(a) Radius Required:
$\text{Radius} = \frac{\text{Diameter}}{2} = \frac{8\text{ m}}{2} = \mathbf{4\text{ metres}}$.
(b) Ground Perpendicular Bisector Construction:
1. Fix pegs at endpoints $A$ and $B$.
2. Take two equal ropes of length greater than 4 m (e.g., 6 m). Anchor one at $A$ and the other at $B$.
3. Pull the ropes taut so their ends meet at a point $P$ on one side of $AB$, and peg point $P$.
4. Repeat on the other side to establish point $Q$.
5. The straight line string between $P$ and $Q$ defines the exact perpendicular bisector axis.
(c) Angle with Ground Line:
The central axis makes an angle of exactly 90° (Right Angle) with ground line $AB$.
9
Chapter 9: Symmetry
NCERT Ganita Prakash (Class 6) — Lines of Symmetry, Reflectional Symmetry, Mirror Inversion & Completing Symmetric Figures | CBSE 2026-27
📌 Key Symmetry Concepts & Properties
- Line of Symmetry (Axis of Symmetry): An imaginary line that divides a figure into two identical halves such that folding along the line makes both halves coincide exactly.
- Lines of Symmetry in Regular Polygons: A regular polygon with $n$ sides has exactly $n$ lines of symmetry:
- Equilateral Triangle: 3 lines of symmetry.
- Square: 4 lines of symmetry (2 along medians, 2 along diagonals).
- Regular Pentagon: 5 lines of symmetry.
- Regular Hexagon: 6 lines of symmetry.
- Circle: Infinitely many lines of symmetry (every diameter is an axis of symmetry).
- Reflection and Symmetry: The object and its mirror reflection are symmetrical about the mirror line. The distance of an object point from the mirror line equals the distance of its image point.
Figure 9.1: Lines of Symmetry in Standard Geometric Polygons and Circle
NCERT Exercise 9.1: Identifying Lines of Symmetry
Q1
Find the number of lines of symmetry for each of the following shapes:
(a) An equilateral triangle
(b) An isosceles triangle
(c) A scalene triangle
(d) A square
(e) A rectangle
(f) A rhombus
(a) An equilateral triangle
(b) An isosceles triangle
(c) A scalene triangle
(d) A square
(e) A rectangle
(f) A rhombus
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Equilateral Triangle: 3 lines of symmetry (along the altitudes / medians from each vertex).
(b) Isosceles Triangle: 1 line of symmetry (from the vertex angle bisecting the non-equal base).
(c) Scalene Triangle: 0 lines of symmetry (no two sides or angles are equal).
(d) Square: 4 lines of symmetry (2 along the midpoints of opposite sides, 2 along diagonals).
(e) Rectangle: 2 lines of symmetry (along the lines joining midpoints of opposite sides). Diagonals are NOT lines of symmetry.
(f) Rhombus: 2 lines of symmetry (along its two diagonals).
(b) Isosceles Triangle: 1 line of symmetry (from the vertex angle bisecting the non-equal base).
(c) Scalene Triangle: 0 lines of symmetry (no two sides or angles are equal).
(d) Square: 4 lines of symmetry (2 along the midpoints of opposite sides, 2 along diagonals).
(e) Rectangle: 2 lines of symmetry (along the lines joining midpoints of opposite sides). Diagonals are NOT lines of symmetry.
(f) Rhombus: 2 lines of symmetry (along its two diagonals).
CBSE Board Marking Scheme 2026-27
Each correct shape symmetry count0.5 Mark each (Total 3 Marks)
Q2
Consider the letters of the English alphabet from A to Z.
(a) List five letters having vertical lines of symmetry.
(b) List five letters having horizontal lines of symmetry.
(c) List three letters having both vertical and horizontal lines of symmetry.
(d) List three letters having no lines of symmetry.
(a) List five letters having vertical lines of symmetry.
(b) List five letters having horizontal lines of symmetry.
(c) List three letters having both vertical and horizontal lines of symmetry.
(d) List three letters having no lines of symmetry.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Vertical line of symmetry (reflection across vertical line):
A, M, T, U, V, W, Y.
A, M, T, U, V, W, Y.
(b) Horizontal line of symmetry (reflection across horizontal line):
B, C, D, E, K.
B, C, D, E, K.
(c) Both vertical and horizontal lines of symmetry:
H, I, O, X.
H, I, O, X.
(d) No line of symmetry:
F, G, J, L, N, P, Q, R, S, Z.
F, G, J, L, N, P, Q, R, S, Z.
CBSE Board Marking Scheme 2026-27
Each sub-part accurately categorized0.75 Mark each (Total 3 Marks)
Q3
Why is a diagonal of a rectangle not a line of symmetry? Explain using a paper-folding argument.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Explanation:
When a rectangular sheet of paper is folded along one of its diagonals, the two opposite halves do not coincide exactly.
The corner of one half overlaps outside the boundary of the other half. For a line to be an axis of symmetry, the two folded parts must superimpose exactly and cover each other completely without any overhang. Hence, a diagonal of a non-square rectangle is not a line of symmetry.
When a rectangular sheet of paper is folded along one of its diagonals, the two opposite halves do not coincide exactly.
The corner of one half overlaps outside the boundary of the other half. For a line to be an axis of symmetry, the two folded parts must superimpose exactly and cover each other completely without any overhang. Hence, a diagonal of a non-square rectangle is not a line of symmetry.
CBSE Board Marking Scheme 2026-27
Clear explanation of non-coincidence on folding1 Mark
Correct geometric conclusion1 Mark
Q4
State the number of lines of symmetry for:
(a) A regular pentagon (b) A regular hexagon (c) A regular octagon (d) A circle
(a) A regular pentagon (b) A regular hexagon (c) A regular octagon (d) A circle
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
General Rule: A regular polygon of $n$ sides has exactly $n$ lines of symmetry.
(a) Regular Pentagon (5 equal sides): 5 lines of symmetry.
(b) Regular Hexagon (6 equal sides): 6 lines of symmetry.
(c) Regular Octagon (8 equal sides): 8 lines of symmetry.
(d) Circle: Infinitely many lines of symmetry (any straight line passing through its centre is an axis of symmetry).
(a) Regular Pentagon (5 equal sides): 5 lines of symmetry.
(b) Regular Hexagon (6 equal sides): 6 lines of symmetry.
(c) Regular Octagon (8 equal sides): 8 lines of symmetry.
(d) Circle: Infinitely many lines of symmetry (any straight line passing through its centre is an axis of symmetry).
CBSE Board Marking Scheme 2026-27
Each correct symmetry count0.5 Mark each (Total 2 Marks)
Q5
What is reflection symmetry? How is the distance of an object point from the mirror line related to the distance of its image point?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Reflection Symmetry:
Reflection symmetry occurs when an object and its reflected image are symmetrical with respect to a mirror line (the axis of reflection).
Distance Property:
The distance of any object point from the mirror line is strictly equal to the distance of its corresponding image point from the mirror line ($d_1 = d_2$). Furthermore, the line segment joining the object and its image is perpendicular to the mirror line.
Reflection symmetry occurs when an object and its reflected image are symmetrical with respect to a mirror line (the axis of reflection).
Distance Property:
The distance of any object point from the mirror line is strictly equal to the distance of its corresponding image point from the mirror line ($d_1 = d_2$). Furthermore, the line segment joining the object and its image is perpendicular to the mirror line.
CBSE Board Marking Scheme 2026-27
Definition of reflection symmetry1 Mark
Equal distance and perpendicularity property1 Mark
Q6
Determine the number of lines of symmetry for each of the following regular polygons:
(a) Equilateral triangle
(b) Square
(c) Regular pentagon
(d) Regular hexagon
State the general rule relating the number of sides of a regular polygon to its lines of symmetry.
(a) Equilateral triangle
(b) Square
(c) Regular pentagon
(d) Regular hexagon
State the general rule relating the number of sides of a regular polygon to its lines of symmetry.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Polygon Lines of Symmetry:
• (a) Equilateral Triangle (3 sides): 3 lines of symmetry (connecting each vertex to midpoint of opposite side).
• (b) Square (4 sides): 4 lines of symmetry (2 along diagonals + 2 perpendicular bisectors of opposite sides).
• (c) Regular Pentagon (5 sides): 5 lines of symmetry (connecting each vertex to midpoint of opposite side).
• (d) Regular Hexagon (6 sides): 6 lines of symmetry (3 along main diagonals + 3 bisecting opposite pairs of sides).
• (a) Equilateral Triangle (3 sides): 3 lines of symmetry (connecting each vertex to midpoint of opposite side).
• (b) Square (4 sides): 4 lines of symmetry (2 along diagonals + 2 perpendicular bisectors of opposite sides).
• (c) Regular Pentagon (5 sides): 5 lines of symmetry (connecting each vertex to midpoint of opposite side).
• (d) Regular Hexagon (6 sides): 6 lines of symmetry (3 along main diagonals + 3 bisecting opposite pairs of sides).
General Rule: A regular polygon with $n$ sides possesses exactly n lines of symmetry.
CBSE Board Marking Scheme 2026-27
Correct line counts for all 4 polygons (3, 4, 5, 6)2 Marks
Stating general rule (n sides = n lines of symmetry)1 Mark
Q7
Which of the English capital letters have:
(a) Vertical line of symmetry only?
(b) Horizontal line of symmetry only?
(c) Both vertical and horizontal lines of symmetry?
(d) No line of symmetry?
(a) Vertical line of symmetry only?
(b) Horizontal line of symmetry only?
(c) Both vertical and horizontal lines of symmetry?
(d) No line of symmetry?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Alphabet Symmetry Classification:
• (a) Vertical line of symmetry only: A, M, T, U, V, W, Y.
• (b) Horizontal line of symmetry only: B, C, D, E, K.
• (c) Both vertical and horizontal lines of symmetry: H, I, O, X.
• (d) No line of symmetry (Asymmetric): F, G, J, L, N, P, Q, R, S, Z.
• (a) Vertical line of symmetry only: A, M, T, U, V, W, Y.
• (b) Horizontal line of symmetry only: B, C, D, E, K.
• (c) Both vertical and horizontal lines of symmetry: H, I, O, X.
• (d) No line of symmetry (Asymmetric): F, G, J, L, N, P, Q, R, S, Z.
CBSE Board Marking Scheme 2026-27
Correct letters for vertical, horizontal, both, and none categories3 Marks (0.75 M each)
Q8
A circle has an infinite number of lines of symmetry. Explain why this statement is mathematically true using the concept of diameters.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Mathematical Explanation:
1. Every line passing through the centre of a circle is a line of symmetry because it divides the circle into two identical halves called semicircles.
2. When a circle is folded along any diameter, the two semicircles coincide point-for-point.
3. Since infinitely many diameters can be drawn through the centre of a circle, a circle has infinitely many lines of symmetry.
1. Every line passing through the centre of a circle is a line of symmetry because it divides the circle into two identical halves called semicircles.
2. When a circle is folded along any diameter, the two semicircles coincide point-for-point.
3. Since infinitely many diameters can be drawn through the centre of a circle, a circle has infinitely many lines of symmetry.
CBSE Board Marking Scheme 2026-27
Explaining that every diameter is an axis of symmetry1 Mark
Deducing infinite lines of symmetry from infinite diameters1 Mark
Q9
Can you draw a triangle which has:
(a) Exactly one line of symmetry?
(b) Exactly two lines of symmetry?
(c) Exactly three lines of symmetry?
(d) No lines of symmetry?
Give the specific geometric name for each case.
(a) Exactly one line of symmetry?
(b) Exactly two lines of symmetry?
(c) Exactly three lines of symmetry?
(d) No lines of symmetry?
Give the specific geometric name for each case.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Geometric Triangle Analysis:
• (a) Exactly 1 line of symmetry: Yes, an Isosceles Triangle (line joins apex to midpoint of unequal base).
• (b) Exactly 2 lines of symmetry: No, impossible. If two sides are equal, the third side being equal makes it 3 lines, or if unequal it remains 1 line. No triangle can have exactly 2 lines of symmetry.
• (c) Exactly 3 lines of symmetry: Yes, an Equilateral Triangle (all 3 sides and angles are equal).
• (d) No line of symmetry: Yes, a Scalene Triangle (all three sides have different lengths).
• (a) Exactly 1 line of symmetry: Yes, an Isosceles Triangle (line joins apex to midpoint of unequal base).
• (b) Exactly 2 lines of symmetry: No, impossible. If two sides are equal, the third side being equal makes it 3 lines, or if unequal it remains 1 line. No triangle can have exactly 2 lines of symmetry.
• (c) Exactly 3 lines of symmetry: Yes, an Equilateral Triangle (all 3 sides and angles are equal).
• (d) No line of symmetry: Yes, a Scalene Triangle (all three sides have different lengths).
CBSE Board Marking Scheme 2026-27
Isosceles (1 line), Equilateral (3 lines), Scalene (0 lines)2 Marks
Correctly proving impossibility of 2 lines in a triangle1 Mark
Q10
Complete the mirror reflection for the given coordinates across the y-axis: A(3, 4), B(5, 1), C(2, -3). What is the rule for reflecting a point (x, y) across:
(i) The y-axis?
(ii) The x-axis?
(i) The y-axis?
(ii) The x-axis?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Coordinate Reflection Rules:
• Reflection across the y-axis changes $(x, y) o (-x, y)$ (x-coordinate negates, y remains same).
- $A(3, 4) o A'(-3, 4)$
- $B(5, 1) o B'(-5, 1)$
- $C(2, -3) o C'(-2, -3)$
• Reflection across the x-axis changes $(x, y) o (x, -y)$ (y-coordinate negates, x remains same).
• Reflection across the y-axis changes $(x, y) o (-x, y)$ (x-coordinate negates, y remains same).
- $A(3, 4) o A'(-3, 4)$
- $B(5, 1) o B'(-5, 1)$
- $C(2, -3) o C'(-2, -3)$
• Reflection across the x-axis changes $(x, y) o (x, -y)$ (y-coordinate negates, x remains same).
CBSE Board Marking Scheme 2026-27
Reflected coordinates A', B', C'1 Mark
General reflection rules for x-axis and y-axis1 Mark
Q11
A rhombus has diagonals of lengths 12 cm and 16 cm.
(a) How many lines of symmetry does a rhombus have?
(b) Are the diagonals the lines of symmetry? Explain.
(a) How many lines of symmetry does a rhombus have?
(b) Are the diagonals the lines of symmetry? Explain.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Lines of Symmetry: A rhombus has exactly 2 lines of symmetry.
(b) Explanation:
Yes, the two diagonals of a rhombus are its exact lines of symmetry.
Because all four sides of a rhombus are equal in length and diagonals bisect each other at right angles ($90°$), folding the rhombus along either diagonal causes the opposite halves to coincide perfectly.
Yes, the two diagonals of a rhombus are its exact lines of symmetry.
Because all four sides of a rhombus are equal in length and diagonals bisect each other at right angles ($90°$), folding the rhombus along either diagonal causes the opposite halves to coincide perfectly.
CBSE Board Marking Scheme 2026-27
Stating 2 lines of symmetry1 Mark
Explaining diagonal reflection symmetry1 Mark
Q12
In a traditional Indian Rangoli pattern, an artist creates a 4-fold symmetrical design using rotational and reflectional symmetry. If one quadrant contains 7 circular motifs and 3 petal curves, calculate the total number of motifs and curves in the complete design.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Symmetry Multiplier Calculation:
A 4-fold symmetrical design replicates the master quadrant across all 4 quadrants.
• Total circular motifs = $4 imes 7 = \mathbf{28\text{ motifs}}$.
• Total petal curves = $4 imes 3 = \mathbf{12\text{ curves}}$.
• Total design elements = $28 + 12 = \mathbf{40\text{ elements}}$.
A 4-fold symmetrical design replicates the master quadrant across all 4 quadrants.
• Total circular motifs = $4 imes 7 = \mathbf{28\text{ motifs}}$.
• Total petal curves = $4 imes 3 = \mathbf{12\text{ curves}}$.
• Total design elements = $28 + 12 = \mathbf{40\text{ elements}}$.
CBSE Board Marking Scheme 2026-27
Calculation of motifs (28) and curves (12)1 Mark
Total elements count = 401 Mark
CBSE Competency-Based Question (Ceramic Tile Mosaic Design)
Case Study: Architectural Ceramic Tile Mosaic:
An interior designer creates a square floor tile of side 40 cm featuring a decorative star pattern.
(a) If the star pattern inside the square has 8-fold symmetry (4 lines of symmetry), what are the orientations of these 4 axes?
(b) If a single motif is painted at coordinates (10 cm, 5 cm) from the center, at what other three coordinates must identical motifs be placed to maintain reflection symmetry across both horizontal and vertical axes?
(c) Explain why symmetrical tiles simplify tiling patterns in buildings.
An interior designer creates a square floor tile of side 40 cm featuring a decorative star pattern.
(a) If the star pattern inside the square has 8-fold symmetry (4 lines of symmetry), what are the orientations of these 4 axes?
(b) If a single motif is painted at coordinates (10 cm, 5 cm) from the center, at what other three coordinates must identical motifs be placed to maintain reflection symmetry across both horizontal and vertical axes?
(c) Explain why symmetrical tiles simplify tiling patterns in buildings.
Detailed Analytical Solution & Marking Scheme
(a) Orientations of 4 Axes:
The 4 axes of symmetry of a square are: the vertical axis (y-axis), the horizontal axis (x-axis), and the two diagonal axes (sloping at 45° and 135°).
(b) Coordinates of Symmetrical Motifs:
Taking the center as origin $(0, 0)$:
- Original motif: $(10, 5)$
- Horizontal reflection (across y-axis): $(-10, 5)$
- Vertical reflection (across x-axis): $(10, -5)$
- Double reflection (across origin / both axes): $(-10, -5)$.
(c) Advantage in Architecture:
Symmetrical tiles can be rotated in any 90° orientation during laying without breaking the pattern continuity, significantly reducing installation errors and labor costs.
10
Chapter 10: The Other Side of Zero
NCERT Ganita Prakash (Class 6) — Negative Numbers, Integers, Number Line Jumps, Additive Inverses & Integer Arithmetic | CBSE 2026-27
📌 Key Integer Concepts & Sign Rules
- Integers ($mathbb{Z}$): The collection of positive natural numbers, zero, and negative natural numbers: $dots, -4, -3, -2, -1, 0, 1, 2, 3, 4, dots$
- Ordering on the Number Line: Numbers to the right are always greater than numbers to the left. For any positive integer $a$, $-a < 0 < a$. As negative numbers move further left, their value decreases (e.g., $-10 < -2$).
- Absolute Value: The numerical value of an integer regardless of its sign, denoted by $|x|$. $|-5| = 5$, $|5| = 5$.
- Rules for Addition & Subtraction:
- Same sign: Add absolute values and assign the common sign: $(+5) + (+3) = +8$; $(-5) + (-3) = -8$.
- Different signs: Subtract smaller absolute value from larger and assign the sign of the larger absolute value: $(+7) + (-4) = +3$; $(-9) + (+2) = -7$.
- Subtraction: To subtract an integer, add its additive inverse: $a - b = a + (-b)$.
NCERT Exercise 10.1: Integer Concepts & Number Line
Q1
Write integers with appropriate signs (+ or -) for the following real-life situations:
(a) An increase of 8 marks
(b) A loss of ₹ 450
(c) A submarine cruising 800 m below sea level
(d) Temperature of 15°C below 0°C
(e) Deposit of ₹ 2,000 in a bank account
(a) An increase of 8 marks
(b) A loss of ₹ 450
(c) A submarine cruising 800 m below sea level
(d) Temperature of 15°C below 0°C
(e) Deposit of ₹ 2,000 in a bank account
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) Increase of 8 marks: +8
(b) Loss of ₹ 450: -₹ 450 (or -450)
(c) 800 m below sea level: -800 m (or -800)
(d) 15°C below 0°C: -15°C (or -15)
(e) Deposit of ₹ 2,000: +₹ 2,000 (or +2000)
(b) Loss of ₹ 450: -₹ 450 (or -450)
(c) 800 m below sea level: -800 m (or -800)
(d) 15°C below 0°C: -15°C (or -15)
(e) Deposit of ₹ 2,000: +₹ 2,000 (or +2000)
CBSE Board Marking Scheme 2026-27
Each integer representation correct with sign0.6 Mark each (Total 3 Marks)
Q2
Represent the following integers on a number line:
(a) +5 (b) -8 (c) -1 (d) +4
(a) +5 (b) -8 (c) -1 (d) +4
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Number Line Representation:
1. Draw a horizontal straight line with arrows at both ends.
2. Mark a point in the center as 0 (reference origin).
3. Mark equal unit intervals to the right as positive integers (+1, +2, +3, +4, +5).
4. Mark equal unit intervals to the left as negative integers (-1, -2, -3, ..., -8).
5. Circle and label points: +5 (5 units right), -8 (8 units left), -1 (1 unit left), +4 (4 units right).
1. Draw a horizontal straight line with arrows at both ends.
2. Mark a point in the center as 0 (reference origin).
3. Mark equal unit intervals to the right as positive integers (+1, +2, +3, +4, +5).
4. Mark equal unit intervals to the left as negative integers (-1, -2, -3, ..., -8).
5. Circle and label points: +5 (5 units right), -8 (8 units left), -1 (1 unit left), +4 (4 units right).
CBSE Board Marking Scheme 2026-27
Accurately calibrated number line with origin 01 Mark
Correctly marking all 4 points1 Mark
Q3
Compare the following pairs of integers using > or <:
(a) 0 and -8
(b) -5 and -11
(c) -100 and -101
(d) 7 and -7
(a) 0 and -8
(b) -5 and -11
(c) -100 and -101
(d) 7 and -7
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) 0 and -8: Zero is to the right of every negative integer $\implies \mathbf{0 > -8}$.
(b) -5 and -11: -5 is to the right of -11 on the number line $\implies \mathbf{-5 > -11}$.
(c) -100 and -101: -100 is closer to 0 than -101 $\implies \mathbf{-100 > -101}$.
(d) 7 and -7: Any positive integer is greater than any negative integer $\implies \mathbf{7 > -7}$.
(b) -5 and -11: -5 is to the right of -11 on the number line $\implies \mathbf{-5 > -11}$.
(c) -100 and -101: -100 is closer to 0 than -101 $\implies \mathbf{-100 > -101}$.
(d) 7 and -7: Any positive integer is greater than any negative integer $\implies \mathbf{7 > -7}$.
CBSE Board Marking Scheme 2026-27
Each correct inequality comparison0.5 Mark each (Total 2 Marks)
NCERT Exercise 10.2: Integer Addition & Subtraction
Q4
Using a number line, find the integer which is:
(a) 4 more than -1
(b) 5 less than 3
(c) 3 less than -2
(a) 4 more than -1
(b) 5 less than 3
(c) 3 less than -2
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) 4 more than -1:
Start at -1 on the number line. Move 4 steps to the right (addition):
$-1 + 4 = \mathbf{+3}$.
Start at -1 on the number line. Move 4 steps to the right (addition):
$-1 + 4 = \mathbf{+3}$.
(b) 5 less than 3:
Start at +3 on the number line. Move 5 steps to the left (subtraction):
$3 - 5 = \mathbf{-2}$.
Start at +3 on the number line. Move 5 steps to the left (subtraction):
$3 - 5 = \mathbf{-2}$.
(c) 3 less than -2:
Start at -2 on the number line. Move 3 steps to the left:
$-2 - 3 = \mathbf{-5}$.
Start at -2 on the number line. Move 3 steps to the left:
$-2 - 3 = \mathbf{-5}$.
CBSE Board Marking Scheme 2026-27
1 Mark per correct number line step with answer3 Marks
Q5
Find the sum of:
(a) (-11) + (-12)
(b) (+37) + (-20)
(c) (-312) + 39 + 192
(a) (-11) + (-12)
(b) (+37) + (-20)
(c) (-312) + 39 + 192
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) (-11) + (-12):
Both have same negative signs: $-(11 + 12) = \mathbf{-23}$.
Both have same negative signs: $-(11 + 12) = \mathbf{-23}$.
(b) (+37) + (-20):
Different signs: Subtract absolute values: $37 - 20 = 17$.
Assign sign of larger absolute value: $\mathbf{+17}$.
Different signs: Subtract absolute values: $37 - 20 = 17$.
Assign sign of larger absolute value: $\mathbf{+17}$.
(c) (-312) + 39 + 192:
Add positive integers first: $39 + 192 = 231$.
Now calculate $(-312) + 231 = -(312 - 231) = \mathbf{-81}$.
Add positive integers first: $39 + 192 = 231$.
Now calculate $(-312) + 231 = -(312 - 231) = \mathbf{-81}$.
CBSE Board Marking Scheme 2026-27
1 Mark per sub-part with clear sign steps3 Marks
Q6
Find the value of:
(a) 35 - (20)
(b) (-15) - (-18)
(c) (-20) - (13)
(d) 23 - (-12)
(a) 35 - (20)
(b) (-15) - (-18)
(c) (-20) - (13)
(d) 23 - (-12)
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Rule: $a - b = a + (\text{additive inverse of } b)$.
(a) $35 - 20 = \mathbf{15}$.
(b) $(-15) - (-18) = -15 + 18 = \mathbf{+3}$.
(c) $(-20) - 13 = -20 + (-13) = \mathbf{-33}$.
(d) $23 - (-12) = 23 + 12 = \mathbf{35}$.
(a) $35 - 20 = \mathbf{15}$.
(b) $(-15) - (-18) = -15 + 18 = \mathbf{+3}$.
(c) $(-20) - 13 = -20 + (-13) = \mathbf{-33}$.
(d) $23 - (-12) = 23 + 12 = \mathbf{35}$.
CBSE Board Marking Scheme 2026-27
Each correct evaluation0.5 Mark each (Total 2 Marks)
Q7
Fill in the blanks with >, <, or = sign:
(a) (-3) + (-6) ______ (-3) - (-6)
(b) (-21) - (-10) ______ (-31) + (-11)
(a) (-3) + (-6) ______ (-3) - (-6)
(b) (-21) - (-10) ______ (-31) + (-11)
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a):
- LHS: $(-3) + (-6) = -9$
- RHS: $(-3) - (-6) = -3 + 6 = +3$
Since $-9 < +3$, the answer is <.
- LHS: $(-3) + (-6) = -9$
- RHS: $(-3) - (-6) = -3 + 6 = +3$
Since $-9 < +3$, the answer is <.
(b):
- LHS: $(-21) - (-10) = -21 + 10 = -11$
- RHS: $(-31) + (-11) = -42$
Since $-11 > -42$, the answer is >.
- LHS: $(-21) - (-10) = -21 + 10 = -11$
- RHS: $(-31) + (-11) = -42$
Since $-11 > -42$, the answer is >.
CBSE Board Marking Scheme 2026-27
Part (a): LHS = -9, RHS = +3, sign <1 Mark
Part (b): LHS = -11, RHS = -42, sign >1 Mark
Q8
Write the opposite of each of the following statements using positive or negative integers:
(a) Increase in weight by 5 kg.
(b) 30 km north.
(c) A deposit of ₹2,000.
(d) 100 m below sea level.
(e) Temperature of 8°C below freezing point (0°C).
(a) Increase in weight by 5 kg.
(b) 30 km north.
(c) A deposit of ₹2,000.
(d) 100 m below sea level.
(e) Temperature of 8°C below freezing point (0°C).
[2.5 Marks]
✅ CBSE Standard Step-by-Step Solution
• (a) Increase of 5 kg: Opposite is Decrease of 5 kg (represented by $-5\text{ kg}$).
• (b) 30 km north: Opposite is 30 km south ($-30\text{ km}$).
• (c) A deposit of ₹2,000: Opposite is A withdrawal of ₹2,000 ($-₹2000$).
• (d) 100 m below sea level: Opposite is 100 m above sea level ($+100\text{ m}$).
• (e) 8°C below 0°C: Opposite is 8°C above 0°C ($+8°\text{C}$).
• (b) 30 km north: Opposite is 30 km south ($-30\text{ km}$).
• (c) A deposit of ₹2,000: Opposite is A withdrawal of ₹2,000 ($-₹2000$).
• (d) 100 m below sea level: Opposite is 100 m above sea level ($+100\text{ m}$).
• (e) 8°C below 0°C: Opposite is 8°C above 0°C ($+8°\text{C}$).
CBSE Board Marking Scheme 2026-27
Each correct opposite statement with appropriate sign0.5 Mark each (Total 2.5 M)
Q9
Find the values of the following using the rules of integers:
(a) $(-13) + (+18)
(b) (-10) + (-19)
(c) (+25) - (-15)
(d) (-32) - (-40)
(a) $(-13) + (+18)
(b) (-10) + (-19)
(c) (+25) - (-15)
(d) (-32) - (-40)
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (a) $(-13) + 18 = 18 - 13 = \mathbf{+5}$.
• (b) $(-10) + (-19) = -(10 + 19) = \mathbf{-29}$.
• (c) $(+25) - (-15) = 25 + 15 = \mathbf{+40}$ (Subtracting a negative is adding a positive).
• (d) $(-32) - (-40) = -32 + 40 = 40 - 32 = \mathbf{+8}$.
• (b) $(-10) + (-19) = -(10 + 19) = \mathbf{-29}$.
• (c) $(+25) - (-15) = 25 + 15 = \mathbf{+40}$ (Subtracting a negative is adding a positive).
• (d) $(-32) - (-40) = -32 + 40 = 40 - 32 = \mathbf{+8}$.
CBSE Board Marking Scheme 2026-27
Correct integer arithmetic with signs0.5 Mark each (Total 2 Marks)
Q10
Write all the integers between -5 and +3 in ascending order. Find their sum.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Integers between -5 and +3:
In ascending order: -4, -3, -2, -1, 0, 1, 2.
In ascending order: -4, -3, -2, -1, 0, 1, 2.
Sum Calculation:
$(-4) + (-3) + (-2) + (-1) + 0 + 1 + 2$
Notice: $(-2 + 2) = 0$ and $(-1 + 1) = 0$.
Sum = $(-4) + (-3) + 0 = \mathbf{-7}$.
$(-4) + (-3) + (-2) + (-1) + 0 + 1 + 2$
Notice: $(-2 + 2) = 0$ and $(-1 + 1) = 0$.
Sum = $(-4) + (-3) + 0 = \mathbf{-7}$.
CBSE Board Marking Scheme 2026-27
Listing integers in ascending order1 Mark
Calculating correct sum = -71 Mark
Q11
At Srinagar, the temperature was -2°C on Monday and then it dropped by 3°C on Tuesday. On Wednesday, it rose by 4°C. What was the temperature of Srinagar on (a) Tuesday and (b) Wednesday?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• Monday temperature = $-2°\text{C}$.
• (a) Tuesday Temperature: Temperature drops by 3°C:
$\text{Temp} = -2°\text{C} - 3°\text{C} = \mathbf{-5°\text{C}}$.
• (b) Wednesday Temperature: Temperature rises by 4°C from Tuesday:
$\text{Temp} = -5°\text{C} + 4°\text{C} = \mathbf{-1°\text{C}}$.
• (a) Tuesday Temperature: Temperature drops by 3°C:
$\text{Temp} = -2°\text{C} - 3°\text{C} = \mathbf{-5°\text{C}}$.
• (b) Wednesday Temperature: Temperature rises by 4°C from Tuesday:
$\text{Temp} = -5°\text{C} + 4°\text{C} = \mathbf{-1°\text{C}}$.
CBSE Board Marking Scheme 2026-27
Tuesday temperature calculation (-5°C)1 Mark
Wednesday temperature calculation (-1°C)1 Mark
Q12
A submarine is situated at 800 m below the sea level. If it ascends 250 m, what is its new position relative to the sea level? Write in integer notation.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• Initial position = $-800\text{ m}$ (below sea level).
• Ascending distance = $+250\text{ m}$ (moving upwards).
• New position = $-800 + 250 = \mathbf{-550\text{ m}}$.
In words: The submarine is now 550 m below sea level.
• Ascending distance = $+250\text{ m}$ (moving upwards).
• New position = $-800 + 250 = \mathbf{-550\text{ m}}$.
In words: The submarine is now 550 m below sea level.
CBSE Board Marking Scheme 2026-27
Formulating integer addition -800 + 2501 Mark
Result -550 m with physical interpretation1 Mark
CBSE Competency-Based Question (Siachen Glacier Climate Data)
Case Study: Siachen Glacier Temperature & Altitude Profile:
At a high-altitude base camp on the Siachen glacier, the midnight temperature was recorded at $-18^\circ\text{C}$. By 2:00 p.m. the following afternoon, due to clear sunshine, the temperature rose by $11^\circ\text{C}$. By midnight, a cold blizzard caused the temperature to drop by $9^\circ\text{C}$.
(a) What was the temperature recorded at 2:00 p.m.?
(b) What was the final temperature recorded at midnight?
(c) An army helicopter ascends from an altitude of 3,200 m above sea level to a ridge 5,400 m above sea level. Express the vertical altitude gain as an integer.
At a high-altitude base camp on the Siachen glacier, the midnight temperature was recorded at $-18^\circ\text{C}$. By 2:00 p.m. the following afternoon, due to clear sunshine, the temperature rose by $11^\circ\text{C}$. By midnight, a cold blizzard caused the temperature to drop by $9^\circ\text{C}$.
(a) What was the temperature recorded at 2:00 p.m.?
(b) What was the final temperature recorded at midnight?
(c) An army helicopter ascends from an altitude of 3,200 m above sea level to a ridge 5,400 m above sea level. Express the vertical altitude gain as an integer.
Detailed Analytical Solution & Marking Scheme
(a) Temperature at 2:00 p.m.:
$\text{Temp} = -18^\circ\text{C} + 11^\circ\text{C} = \mathbf{-7^\circ\text{C}}$.
(b) Temperature at Midnight:
$\text{Temp} = -7^\circ\text{C} - 9^\circ\text{C} = \mathbf{-16^\circ\text{C}}$.
(c) Altitude Gain:
$\text{Gain} = +5,400 - (+3,200) = \mathbf{+2,200\text{ metres}}$.
11
Chapter 11: Introduction to Algebra
NCERT Ganita Prakash (Class 6) — Variables, Forming Algebraic Expressions, Evaluating Expressions & Solving Simple Equations | CBSE 2026-27
📌 Key Algebraic Concepts
- Variable: An entity that can take different numerical values, denoted by letters like $x, y, z, m, n, p$.
- Constant: A quantity having a fixed numerical value (e.g., $5, -8, \frac{1}{2}$).
- Algebraic Expression: A combination of variables and constants connected by arithmetic operations ($+, -, \times, \div$), such as $2x + 7$ or $3y - 5$.
- Equation: A mathematical condition on a variable stating that two expressions are equal, containing an equality sign ($=$).
- Transposition: Changing the side of a term across the equality sign with inverse operation ($+ \leftrightarrow -, \times \leftrightarrow \div$).
NCERT Exercise 11.1: Matchstick Rules & Expressions
Q1
Find the rule which gives the number of matchsticks required to make the following letter patterns:
(a) A pattern of letter T as ⊤
(b) A pattern of letter Z as Z
(c) A pattern of letter U as ⨆
(d) A pattern of letter V as V
(a) A pattern of letter T as ⊤
(b) A pattern of letter Z as Z
(c) A pattern of letter U as ⨆
(d) A pattern of letter V as V
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Let $n$ be the number of letters formed.
(a) Letter T: 2 matchsticks per letter $\implies$ Rule = 2n.
(b) Letter Z: 3 matchsticks per letter $\implies$ Rule = 3n.
(c) Letter U: 3 matchsticks per letter $\implies$ Rule = 3n.
(d) Letter V: 2 matchsticks per letter $\implies$ Rule = 2n.
(a) Letter T: 2 matchsticks per letter $\implies$ Rule = 2n.
(b) Letter Z: 3 matchsticks per letter $\implies$ Rule = 3n.
(c) Letter U: 3 matchsticks per letter $\implies$ Rule = 3n.
(d) Letter V: 2 matchsticks per letter $\implies$ Rule = 2n.
CBSE Board Marking Scheme 2026-27
Each correct rule with variable n0.5 Mark each (Total 2 Marks)
Q2
Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows).
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Number of cadets in 1 row = 5.
Let the number of rows be $n$.
Total number of cadets = Number of cadets in a row $\times$ Number of rows
$= 5 \times n = \mathbf{5n}$.
Let the number of rows be $n$.
Total number of cadets = Number of cadets in a row $\times$ Number of rows
$= 5 \times n = \mathbf{5n}$.
CBSE Board Marking Scheme 2026-27
Identifying multiplication by 51 Mark
Writing correct expression 5n1 Mark
Q3
Write algebraic expressions for each of the following statements:
(a) 7 added to p
(b) 7 subtracted from -m
(c) y is multiplied by -5 and the result is added to 16
(d) One-fifth of x added to 3 times y
(a) 7 added to p
(b) 7 subtracted from -m
(c) y is multiplied by -5 and the result is added to 16
(d) One-fifth of x added to 3 times y
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) 7 added to p: $\mathbf{p + 7}$
(b) 7 subtracted from -m: $\mathbf{-m - 7}$
(c) y multiplied by -5 and added to 16: $(-5 \times y) + 16 = \mathbf{16 - 5y}$
(d) One-fifth of x added to 3 times y: $\mathbf{\frac{x}{5} + 3y}$
(b) 7 subtracted from -m: $\mathbf{-m - 7}$
(c) y multiplied by -5 and added to 16: $(-5 \times y) + 16 = \mathbf{16 - 5y}$
(d) One-fifth of x added to 3 times y: $\mathbf{\frac{x}{5} + 3y}$
CBSE Board Marking Scheme 2026-27
Each correct algebraic expression0.5 Mark each (Total 2 Marks)
NCERT Exercise 11.2: Evaluating Expressions & Formulas
Q4
The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l. Also express the perimeter of a regular hexagon of side s using s.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Equilateral Triangle: Has 3 equal sides of length $l$.
Perimeter = $l + l + l = \mathbf{3l}$.
Regular Hexagon: Has 6 equal sides of length $s$.
Perimeter = $6 \times s = \mathbf{6s}$.
Perimeter = $l + l + l = \mathbf{3l}$.
Regular Hexagon: Has 6 equal sides of length $s$.
Perimeter = $6 \times s = \mathbf{6s}$.
CBSE Board Marking Scheme 2026-27
Triangle perimeter: 3l1 Mark
Hexagon perimeter: 6s1 Mark
Q5
Evaluate the expression 3x² - 5x + 8 when:
(a) x = 2
(b) x = -1
(a) x = 2
(b) x = -1
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) When x = 2:
$3(2)^2 - 5(2) + 8 = 3(4) - 10 + 8 = 12 - 10 + 8 = \mathbf{10}$.
$3(2)^2 - 5(2) + 8 = 3(4) - 10 + 8 = 12 - 10 + 8 = \mathbf{10}$.
(b) When x = -1:
$3(-1)^2 - 5(-1) + 8 = 3(1) - (-5) + 8 = 3 + 5 + 8 = \mathbf{16}$.
$3(-1)^2 - 5(-1) + 8 = 3(1) - (-5) + 8 = 3 + 5 + 8 = \mathbf{16}$.
CBSE Board Marking Scheme 2026-27
Part (a): Substitution and evaluation to 101.5 Marks
Part (b): Handling negative signs to get 161.5 Marks
NCERT Exercise 11.3: Solving Simple Equations
Q6
Solve the following simple linear equations using inverse operations:
(i) x + 10 = 25
(ii) p - 8 = 17
(iii) 5m = 45
(iv) q/4 = 7
(i) x + 10 = 25
(ii) p - 8 = 17
(iii) 5m = 45
(iv) q/4 = 7
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(i) x + 10 = 25: Transpose +10 to RHS: $x = 25 - 10 \implies \mathbf{x = 15}$.
(ii) p - 8 = 17: Transpose -8 to RHS: $p = 17 + 8 \implies \mathbf{p = 25}$.
(iii) 5m = 45: Divide both sides by 5: $m = \frac{45}{5} \implies \mathbf{m = 9}$.
(iv) q/4 = 7: Multiply both sides by 4: $q = 7 \times 4 \implies \mathbf{q = 28}$.
(ii) p - 8 = 17: Transpose -8 to RHS: $p = 17 + 8 \implies \mathbf{p = 25}$.
(iii) 5m = 45: Divide both sides by 5: $m = \frac{45}{5} \implies \mathbf{m = 9}$.
(iv) q/4 = 7: Multiply both sides by 4: $q = 7 \times 4 \implies \mathbf{q = 28}$.
CBSE Board Marking Scheme 2026-27
Each correct equation solution with transposition step0.75 Mark each (Total 3 Marks)
Q7
Solve the equation 2x + 7 = 19 and verify your solution.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Solving:
$2x + 7 = 19$
$2x = 19 - 7 = 12$
$x = \frac{12}{2} = \mathbf{6}$.
Verification:
$\text{LHS} = 2(6) + 7 = 12 + 7 = 19 = \text{RHS}$. Hence verified.
$2x + 7 = 19$
$2x = 19 - 7 = 12$
$x = \frac{12}{2} = \mathbf{6}$.
Verification:
$\text{LHS} = 2(6) + 7 = 12 + 7 = 19 = \text{RHS}$. Hence verified.
CBSE Board Marking Scheme 2026-27
Solution x = 61 Mark
LHS = RHS verification1 Mark
Q8
Change the following algebraic expressions into statements in ordinary English language:
(a) $n + 7$
(b) $3p - 5$
(c) $\frac{y}{4}$
(d) $2x + 11$
(a) $n + 7$
(b) $3p - 5$
(c) $\frac{y}{4}$
(d) $2x + 11$
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (a) $n + 7$: 7 more than a number $n$ (or $n$ increased by 7).
• (b) $3p - 5$: 5 subtracted from thrice of a number $p$.
• (c) $\frac{y}{4}$: One-fourth of a number $y$ (or $y$ divided by 4).
• (d) $2x + 11$: 11 added to twice of a number $x$.
• (b) $3p - 5$: 5 subtracted from thrice of a number $p$.
• (c) $\frac{y}{4}$: One-fourth of a number $y$ (or $y$ divided by 4).
• (d) $2x + 11$: 11 added to twice of a number $x$.
CBSE Board Marking Scheme 2026-27
Correct verbal statements for all 4 expressions0.5 Mark each (Total 2 Marks)
Q9
Sarita's present age is x years.
(i) What will be her age 5 years from now?
(ii) What was her age 3 years ago?
(iii) Sarita's grandfather is 6 times her age. What is her grandfather's age?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother's age?
(i) What will be her age 5 years from now?
(ii) What was her age 3 years ago?
(iii) Sarita's grandfather is 6 times her age. What is her grandfather's age?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother's age?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Given present age = $x$ years.
• (i) Age 5 years from now: $(x + 5)$ years.
• (ii) Age 3 years ago: $(x - 3)$ years.
• (iii) Grandfather's age: $6 \times x = \mathbf{6x\text{ years}}$.
• (iv) Grandmother's age: $(6x - 2)$ years.
• (i) Age 5 years from now: $(x + 5)$ years.
• (ii) Age 3 years ago: $(x - 3)$ years.
• (iii) Grandfather's age: $6 \times x = \mathbf{6x\text{ years}}$.
• (iv) Grandmother's age: $(6x - 2)$ years.
CBSE Board Marking Scheme 2026-27
Parts (i), (ii), (iii), (iv) correct algebraic formulations3 Marks (0.75 M each)
Q10
Solve the following simple linear equations by trial and error method:
(a) $x + 6 = 15$
(b) $4m = 28$
(c) $\frac{y}{3} = 5$
(a) $x + 6 = 15$
(b) $4m = 28$
(c) $\frac{y}{3} = 5$
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
• (a) $x + 6 = 15$:
Try $x = 8$: $8 + 6 = 14 eq 15$.
Try $x = 9$: $9 + 6 = 15$ (LHS = RHS).
Therefore, x = 9 is the solution.
• (b) $4m = 28$:
Try $m = 6$: $4 \times 6 = 24 eq 28$.
Try $m = 7$: $4 \times 7 = 28$ (LHS = RHS).
Therefore, m = 7 is the solution.
• (c) $\frac{y}{3} = 5$:
$y = 5 \times 3 = \mathbf{15}$. (Verification: $15 / 3 = 5$).
Try $x = 8$: $8 + 6 = 14 eq 15$.
Try $x = 9$: $9 + 6 = 15$ (LHS = RHS).
Therefore, x = 9 is the solution.
• (b) $4m = 28$:
Try $m = 6$: $4 \times 6 = 24 eq 28$.
Try $m = 7$: $4 \times 7 = 28$ (LHS = RHS).
Therefore, m = 7 is the solution.
• (c) $\frac{y}{3} = 5$:
$y = 5 \times 3 = \mathbf{15}$. (Verification: $15 / 3 = 5$).
CBSE Board Marking Scheme 2026-27
Trial and error working and correct solution for each equation1 Mark each (Total 3 M)
Q11
The perimeter of a regular hexagon is p. Express the side length of the hexagon in terms of p. If p = 54 cm, what is the length of each side?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• A regular hexagon has 6 equal sides ($s$).
• Perimeter $p = 6 \times s \implies s = \mathbf{\frac{p}{6}}$.
• Given $p = 54\text{ cm}$:
Side $s = \frac{54}{6} = \mathbf{9\text{ cm}}$.
• Perimeter $p = 6 \times s \implies s = \mathbf{\frac{p}{6}}$.
• Given $p = 54\text{ cm}$:
Side $s = \frac{54}{6} = \mathbf{9\text{ cm}}$.
CBSE Board Marking Scheme 2026-27
Algebraic formula s = p / 61 Mark
Side calculation: 9 cm1 Mark
Q12
A bus travels at v km/h. It has travelled for 5 hours and still has 20 km left to reach its destination. What is the total distance between the starting point and the destination?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• Speed = $v\text{ km/h}$, Time travelled = $5\text{ hours}$.
• Distance covered = $\text{Speed} \times \text{Time} = 5v\text{ km}$.
• Distance remaining = $20\text{ km}$.
• Total Distance = Distance covered + Distance left = $\mathbf{(5v + 20)\text{ km}}$.
• Distance covered = $\text{Speed} \times \text{Time} = 5v\text{ km}$.
• Distance remaining = $20\text{ km}$.
• Total Distance = Distance covered + Distance left = $\mathbf{(5v + 20)\text{ km}}$.
CBSE Board Marking Scheme 2026-27
Distance covered = 5v km1 Mark
Total distance expression = (5v + 20) km1 Mark
CBSE Competency-Based Question (Algebraic Age Problems)
Case Study: Family Age Relationships:
Sarita's present age is y years.
(a) What will be her age 5 years from now?
(b) What was her age 3 years ago?
(c) Sarita's grandfather is 6 times her age. What is her grandfather's age?
(d) Grandmother is 2 years younger than grandfather. What is grandmother's age?
(e) Sarita's father's age is 5 years more than 3 times Sarita's age. Express father's age.
Sarita's present age is y years.
(a) What will be her age 5 years from now?
(b) What was her age 3 years ago?
(c) Sarita's grandfather is 6 times her age. What is her grandfather's age?
(d) Grandmother is 2 years younger than grandfather. What is grandmother's age?
(e) Sarita's father's age is 5 years more than 3 times Sarita's age. Express father's age.
Detailed Analytical Solution & Marking Scheme
(a) Age in 5 years: $(y + 5)$ years.
(b) Age 3 years ago: $(y - 3)$ years.
(c) Grandfather's age: $6 \times y = \mathbf{6y\text{ years}}$.
(d) Grandmother's age: $(6y - 2)$ years.
(e) Father's age: $(3y + 5)$ years.
12
Chapter 12: Ratio and Proportion
NCERT Ganita Prakash (Class 6) — Simplest Form of Ratios, Proportions, Extreme & Mean Terms & The Unitary Method | CBSE 2026-27
📌 Key Ratio, Proportion & Unitary Rules
- Ratio: Comparing two quantities of the same kind and with the same unit by division. Denoted by $a : b$ (read as "$a$ is to $b$"). Ratio has no units.
- Simplest Form of Ratio: Divide both terms by their HCF: $\frac{a}{b} = \frac{a \div \text{HCF}}{b \div \text{HCF}}$.
- Proportion: If two ratios are equal, they are in proportion: $a : b :: c : d$ (or $\frac{a}{b} = \frac{c}{d}$).
- $a$ and $d$ are called extreme terms (extremes).
- $b$ and $c$ are called middle terms (means).
- Rule: $\text{Product of Extremes} = \text{Product of Means} \implies a \times d = b \times c$.
- Unitary Method: The method in which first we find the value of one unit by division, and then the value of the required number of units by multiplication.
NCERT Exercise 12.1: Ratios in Simplest Form
Q1
There are 20 girls and 15 boys in a class.
(a) What is the ratio of number of girls to the number of boys?
(b) What is the ratio of number of girls to the total number of students in the class?
(a) What is the ratio of number of girls to the number of boys?
(b) What is the ratio of number of girls to the total number of students in the class?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Number of girls = 20; Number of boys = 15.
Total number of students = $20 + 15 = 35$.
(a) Girls to Boys:
$\text{Ratio} = \frac{20}{15} = \frac{20 \div 5}{15 \div 5} = \frac{4}{3} = \mathbf{4 : 3}$.
Total number of students = $20 + 15 = 35$.
(a) Girls to Boys:
$\text{Ratio} = \frac{20}{15} = \frac{20 \div 5}{15 \div 5} = \frac{4}{3} = \mathbf{4 : 3}$.
(b) Girls to Total Students:
$\text{Ratio} = \frac{20}{35} = \frac{20 \div 5}{35 \div 5} = \frac{4}{7} = \mathbf{4 : 7}$.
$\text{Ratio} = \frac{20}{35} = \frac{20 \div 5}{35 \div 5} = \frac{4}{7} = \mathbf{4 : 7}$.
CBSE Board Marking Scheme 2026-27
Part (a): 4 : 3 in simplest form1.5 Marks
Part (b): 4 : 7 in simplest form1.5 Marks
Q2
Find the ratio of the following in simplest form:
(a) 30 minutes to 1.5 hours
(b) 40 cm to 1.5 m
(c) 55 paise to ₹ 1
(a) 30 minutes to 1.5 hours
(b) 40 cm to 1.5 m
(c) 55 paise to ₹ 1
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
(a) 30 minutes to 1.5 hours:
Convert hours to minutes: $1.5 \times 60 = 90\text{ minutes}$.
$\text{Ratio} = \frac{30}{90} = \frac{1}{3} = \mathbf{1 : 3}$.
Convert hours to minutes: $1.5 \times 60 = 90\text{ minutes}$.
$\text{Ratio} = \frac{30}{90} = \frac{1}{3} = \mathbf{1 : 3}$.
(b) 40 cm to 1.5 m:
Convert metres to cm: $1.5 \times 100 = 150\text{ cm}$.
$\text{Ratio} = \frac{40}{150} = \frac{4}{15} = \mathbf{4 : 15}$.
Convert metres to cm: $1.5 \times 100 = 150\text{ cm}$.
$\text{Ratio} = \frac{40}{150} = \frac{4}{15} = \mathbf{4 : 15}$.
(c) 55 paise to ₹ 1:
Convert rupees to paise: ₹ 1 = 100 paise.
$\text{Ratio} = \frac{55}{100} = \frac{11}{20} = \mathbf{11 : 20}$.
Convert rupees to paise: ₹ 1 = 100 paise.
$\text{Ratio} = \frac{55}{100} = \frac{11}{20} = \mathbf{11 : 20}$.
CBSE Board Marking Scheme 2026-27
1 Mark per part showing unit conversion and simplification3 Marks
Q3
Divide ₹ 60 in the ratio 1 : 2 between Kriti and Kiran.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Total parts = $1 + 2 = 3$ parts.
- Kriti's share = $\frac{1}{3} \times 60 = \mathbf{\text{₹ } 20}$.
- Kiran's share = $\frac{2}{3} \times 60 = \mathbf{\text{₹ } 40}$.
Verification: ₹ 20 + ₹ 40 = ₹ 60.
- Kriti's share = $\frac{1}{3} \times 60 = \mathbf{\text{₹ } 20}$.
- Kiran's share = $\frac{2}{3} \times 60 = \mathbf{\text{₹ } 40}$.
Verification: ₹ 20 + ₹ 40 = ₹ 60.
CBSE Board Marking Scheme 2026-27
Finding total ratio parts (1 + 2 = 3)1 Mark
Kriti = ₹ 20, Kiran = ₹ 401 Mark
NCERT Exercise 12.2: Proportions & Cross-Products
Q4
Determine if the following numbers are in proportion:
(a) 15, 45, 40, 120
(b) 33, 121, 9, 96
(a) 15, 45, 40, 120
(b) 33, 121, 9, 96
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Rule: Numbers $a, b, c, d$ are in proportion if $\frac{a}{b} = \frac{c}{d}$ (or $a \times d = b \times c$).
(a) 15, 45, 40, 120:
- Ratio 1: $\frac{15}{45} = \frac{1}{3}$
- Ratio 2: $\frac{40}{120} = \frac{1}{3}$
Since $\frac{1}{3} = \frac{1}{3}$, Yes, they are in proportion ($15 : 45 :: 40 : 120$).
(a) 15, 45, 40, 120:
- Ratio 1: $\frac{15}{45} = \frac{1}{3}$
- Ratio 2: $\frac{40}{120} = \frac{1}{3}$
Since $\frac{1}{3} = \frac{1}{3}$, Yes, they are in proportion ($15 : 45 :: 40 : 120$).
(b) 33, 121, 9, 96:
- Ratio 1: $\frac{33}{121} = \frac{3}{11}$
- Ratio 2: $\frac{9}{96} = \frac{3}{32}$
Since $\frac{3}{11} \neq \frac{3}{32}$, No, they are NOT in proportion.
- Ratio 1: $\frac{33}{121} = \frac{3}{11}$
- Ratio 2: $\frac{9}{96} = \frac{3}{32}$
Since $\frac{3}{11} \neq \frac{3}{32}$, No, they are NOT in proportion.
CBSE Board Marking Scheme 2026-27
Part (a): Yes with ratios 1/3 = 1/31.5 Marks
Part (b): No with ratios 3/11 != 3/321.5 Marks
Q5
Find the missing value x in the proportion: 25 : 30 :: 40 : x.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
In a proportion, $\text{Product of Extremes} = \text{Product of Means}$.
$25 \times x = 30 \times 40$
$25x = 1,200$
$x = \frac{1,200}{25} = \mathbf{48}$.
$25 \times x = 30 \times 40$
$25x = 1,200$
$x = \frac{1,200}{25} = \mathbf{48}$.
CBSE Board Marking Scheme 2026-27
Equation: 25 × x = 30 × 401 Mark
Answer: x = 481 Mark
NCERT Exercise 12.3: The Unitary Method
Q6
If the cost of 7 m of cloth is ₹ 1,470, find the cost of 5 m of cloth.
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Step 1: Cost of 1 metre of cloth:
$\text{Cost of 1 m} = \frac{1,470}{7} = \text{₹ } 210$.
Step 2: Cost of 5 metres of cloth:
$\text{Cost of 5 m} = 5 \times \text{₹ } 210 = \mathbf{\text{₹ } 1,050}$.
$\text{Cost of 1 m} = \frac{1,470}{7} = \text{₹ } 210$.
Step 2: Cost of 5 metres of cloth:
$\text{Cost of 5 m} = 5 \times \text{₹ } 210 = \mathbf{\text{₹ } 1,050}$.
CBSE Board Marking Scheme 2026-27
Cost of 1 m = ₹ 2101 Mark
Cost of 5 m = ₹ 1,0501 Mark
Q7
Ekta earns ₹ 3,000 in 10 days. How much will she earn in 30 days?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
Earnings in 10 days = ₹ 3,000.
Earnings in 1 day = $\frac{3,000}{10} = \text{₹ } 300$.
Earnings in 30 days = $30 \times \text{₹ } 300 = \mathbf{\text{₹ } 9,000}$.
Earnings in 1 day = $\frac{3,000}{10} = \text{₹ } 300$.
Earnings in 30 days = $30 \times \text{₹ } 300 = \mathbf{\text{₹ } 9,000}$.
CBSE Board Marking Scheme 2026-27
Daily earnings = ₹ 3001 Mark
Monthly earnings = ₹ 9,0001 Mark
Q8
A car travels 90 km in 2 1/2 hours.
(a) How much time is required to cover 30 km with the same speed?
(b) Find the distance covered in 2 hours with the same speed.
(a) How much time is required to cover 30 km with the same speed?
(b) Find the distance covered in 2 hours with the same speed.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Speed of car = $\frac{\text{Distance}}{\text{Time}} = \frac{90}{2.5} = 36\text{ km/h}$.
(a) Time to cover 30 km:
$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{30}{36} = \frac{5}{6}\text{ hours} = \frac{5}{6} \times 60 = \mathbf{50\text{ minutes}}$.
(a) Time to cover 30 km:
$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{30}{36} = \frac{5}{6}\text{ hours} = \frac{5}{6} \times 60 = \mathbf{50\text{ minutes}}$.
(b) Distance in 2 hours:
$\text{Distance} = \text{Speed} \times \text{Time} = 36 \times 2 = \mathbf{72\text{ km}}$.
$\text{Distance} = \text{Speed} \times \text{Time} = 36 \times 2 = \mathbf{72\text{ km}}$.
CBSE Board Marking Scheme 2026-27
Calculating uniform speed 36 km/h1 Mark
Part (a): 50 minutes (or 5/6 h)1 Mark
Part (b): 72 km1 Mark
Q9
Find the ratio of the following in simplest form:
(a) 30 minutes to 1.5 hours.
(b) 40 cm to 1.5 m.
(c) 55 paise to ₹1.
(d) 500 mL to 2 litres.
(a) 30 minutes to 1.5 hours.
(b) 40 cm to 1.5 m.
(c) 55 paise to ₹1.
(d) 500 mL to 2 litres.
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Rule: Both quantities must be converted to the same unit before forming a ratio.
• (a) 30 min to 1.5 hours: $1.5\text{ h} = 1.5 \times 60 = 90\text{ min}$. Ratio = $\frac{30}{90} = \mathbf{1 : 3}$.
• (b) 40 cm to 1.5 m: $1.5\text{ m} = 150\text{ cm}$. Ratio = $\frac{40}{150} = \mathbf{4 : 15}$.
• (c) 55 paise to ₹1: $₹1 = 100\text{ paise}$. Ratio = $\frac{55}{100} = \frac{11}{20} = \mathbf{11 : 20}$.
• (d) 500 mL to 2 litres: $2\text{ L} = 2000\text{ mL}$. Ratio = $\frac{500}{2000} = \mathbf{1 : 4}$.
• (a) 30 min to 1.5 hours: $1.5\text{ h} = 1.5 \times 60 = 90\text{ min}$. Ratio = $\frac{30}{90} = \mathbf{1 : 3}$.
• (b) 40 cm to 1.5 m: $1.5\text{ m} = 150\text{ cm}$. Ratio = $\frac{40}{150} = \mathbf{4 : 15}$.
• (c) 55 paise to ₹1: $₹1 = 100\text{ paise}$. Ratio = $\frac{55}{100} = \frac{11}{20} = \mathbf{11 : 20}$.
• (d) 500 mL to 2 litres: $2\text{ L} = 2000\text{ mL}$. Ratio = $\frac{500}{2000} = \mathbf{1 : 4}$.
CBSE Board Marking Scheme 2026-27
Unit conversions and simplified ratios for all 4 parts3 Marks (0.75 M each)
Q10
Determine if the following ratios form a proportion:
(a) 25 cm : 1 m and ₹40 : ₹160
(b) 39 litres : 65 litres and 6 bottles : 10 bottles
(a) 25 cm : 1 m and ₹40 : ₹160
(b) 39 litres : 65 litres and 6 bottles : 10 bottles
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• (a) 25 cm : 1 m and ₹40 : ₹160:
$1\text{ m} = 100\text{ cm} \implies \frac{25}{100} = \frac{1}{4}$.
$\frac{40}{160} = \frac{1}{4}$.
Since $\frac{1}{4} = \frac{1}{4}$, Yes, they form a proportion ($25\text{ cm} : 1\text{ m} :: ₹40 : ₹160$).
• (b) 39 L : 65 L and 6 bottles : 10 bottles:
$\frac{39}{65} = \frac{3 \times 13}{5 \times 13} = \frac{3}{5}$.
$\frac{6}{10} = \frac{3}{5}$.
Since $\frac{3}{5} = \frac{3}{5}$, Yes, they form a proportion.
$1\text{ m} = 100\text{ cm} \implies \frac{25}{100} = \frac{1}{4}$.
$\frac{40}{160} = \frac{1}{4}$.
Since $\frac{1}{4} = \frac{1}{4}$, Yes, they form a proportion ($25\text{ cm} : 1\text{ m} :: ₹40 : ₹160$).
• (b) 39 L : 65 L and 6 bottles : 10 bottles:
$\frac{39}{65} = \frac{3 \times 13}{5 \times 13} = \frac{3}{5}$.
$\frac{6}{10} = \frac{3}{5}$.
Since $\frac{3}{5} = \frac{3}{5}$, Yes, they form a proportion.
CBSE Board Marking Scheme 2026-27
Simplifying both ratios and verifying equivalence1 Mark each (Total 2 Marks)
Q11
Cost of 5 kg of wheat is ₹91.50.
(a) What will be the cost of 8 kg of wheat?
(b) What quantity of wheat can be purchased in ₹183?
(a) What will be the cost of 8 kg of wheat?
(b) What quantity of wheat can be purchased in ₹183?
[3 Marks]
✅ CBSE Standard Step-by-Step Solution
Unitary Method Step 1: Cost of 1 kg of wheat:
$\text{Cost of 1 kg} = \frac{₹91.50}{5} = \mathbf{₹18.30}$.
$\text{Cost of 1 kg} = \frac{₹91.50}{5} = \mathbf{₹18.30}$.
• (a) Cost of 8 kg:
$\text{Cost} = 8 \times ₹18.30 = \mathbf{₹146.40}$.
• (b) Quantity in ₹183:
$\text{Quantity} = \frac{₹183}{₹18.30} = \mathbf{10\text{ kg}}$.
$\text{Cost} = 8 \times ₹18.30 = \mathbf{₹146.40}$.
• (b) Quantity in ₹183:
$\text{Quantity} = \frac{₹183}{₹18.30} = \mathbf{10\text{ kg}}$.
CBSE Board Marking Scheme 2026-27
Unit cost calculation ₹18.30 / kg1 Mark
Part (a): ₹146.40 and Part (b): 10 kg2 Marks
Q12
Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?
[2 Marks]
✅ CBSE Standard Step-by-Step Solution
• Anish's run rate = $\frac{42\text{ runs}}{6\text{ overs}} = \mathbf{7\text{ runs per over}}$.
• Anup's run rate = $\frac{63\text{ runs}}{7\text{ overs}} = \mathbf{9\text{ runs per over}}$.
• Comparing: $9 > 7$.
Therefore, Anup made more runs per over.
• Anup's run rate = $\frac{63\text{ runs}}{7\text{ overs}} = \mathbf{9\text{ runs per over}}$.
• Comparing: $9 > 7$.
Therefore, Anup made more runs per over.
CBSE Board Marking Scheme 2026-27
Calculating both run rates (7 and 9 runs/over)1 Mark
Comparison and stating Anup1 Mark
CBSE Competency-Based Question (Industrial Manufacturing Scale)
Case Study: Solar Power Station Generation Ratio:
A solar farm generates 180 kilowatt-hours (kWh) of clean electricity every 4 hours during peak sunlight.
(a) Formulate the unit rate of energy generation per hour.
(b) How many kilowatt-hours will the solar farm generate in 7.5 hours of continuous sunshine?
(c) If a nearby village requires 720 kWh daily, how many hours of peak sunlight generation are needed to meet the full demand?
A solar farm generates 180 kilowatt-hours (kWh) of clean electricity every 4 hours during peak sunlight.
(a) Formulate the unit rate of energy generation per hour.
(b) How many kilowatt-hours will the solar farm generate in 7.5 hours of continuous sunshine?
(c) If a nearby village requires 720 kWh daily, how many hours of peak sunlight generation are needed to meet the full demand?
Detailed Analytical Solution & Marking Scheme
(a) Unit Rate:
Rate = $\frac{180\text{ kWh}}{4\text{ h}} = \mathbf{45\text{ kWh per hour}}$.
(b) Generation in 7.5 hours:
Energy = $45 \times 7.5 = \mathbf{337.5\text{ kWh}}$.
(c) Hours needed for 720 kWh:
Hours = $\frac{720}{45} = \mathbf{16\text{ hours of peak sunlight}}$.