Are these Class 7 Mathematics NCERT solutions updated for CBSE 2026-27?

Yes, all solutions are 100% updated and strictly aligned with the latest NCERT rationalized textbooks and CBSE guidelines for academic year 2026-27.

Do these solutions provide step-by-step CBSE marking schemes?

Yes, every question is solved following official CBSE marking rubrics, with explicit marks awarded for given data, formulas, derivations, and final conclusions.

Are competency-based questions (CBQ) and HOTS included?

Yes, every chapter features competency-based case studies, assertion-reason questions, and Higher Order Thinking Skills (HOTS) questions based on NEP 2020.

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NCERT Solutions for Class 7 Maths (CBSE 2026-27) — All Chapters Step-by-Step

1

Chapter 1: Integers

NCERT Exercise 1.1 to 1.3 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Properties Summary

Integers (ℤ (Integers)): The set {..., -3, -2, -1, 0, 1, 2, 3, ...} containing whole numbers and their negatives.
Addition & Subtraction Properties:
- Closure: a + b and a - b are always integers.
- Commutative: a + b = b + a. (Note: a - b ≠ b - a).
- Associative: (a + b) + c = a + (b + c).
- Additive Identity: a + 0 = a = 0 + a. Additive inverse of a is -a.
Multiplication Properties:
- Rules of Signs: (+) × (+) = (+), (-) × (-) = (+), (+) × (-) = (-), (-) × (+) = (-).
- Distributive Property over Addition: a × (b + c) = (a × b) + (a × c).
- Distributive Property over Subtraction: a × (b - c) = (a × b) - (a × c).
- Multiplicative Identity: a × 1 = a. Product with zero: a × 0 = 0.
Division of Integers:
- (-a) ÷ b = a ÷ (-b) = -(a ÷ b) when b ≠ 0.
- (-a) ÷ (-b) = a ÷ b.
- Division by 0 is meaningless (a ÷ 0 is undefined); 0 ÷ a = 0 (a ≠ 0); a ÷ 1 = a.

Exercise 1.1

Q1

Write down a pair of integers whose: (a) sum is −7, (b) difference is −10, (c) sum is 0.

3 Marks

✅ CBSE Standard Answer

(a) Sum is −7:
Let the two integers be −4 and −3.
Verification: (−4) + (−3) = −4 − 3 = −7.
(Another valid pair: −10 and 3, since −10 + 3 = −7).

(b) Difference is −10:
Let the two integers be −12 and −2.
Verification: (−12) − (−2) = −12 + 2 = −10.
(Another valid pair: 2 and 12, since 2 − 12 = −10).

(c) Sum is 0:
Any integer and its additive inverse.
Let the integers be 5 and −5.
Verification: 5 + (−5) = 5 − 5 = 0.

CBSE Marking Scheme 2026-27

Part (a): Stating a valid pair with verification1 Mark

Part (b): Stating a valid pair with subtraction step1 Mark

Part (c): Stating additive inverse pair with sum verification1 Mark

Q2

(a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and a positive integer whose sum is −5.
(c) Write a negative integer and a positive integer whose difference is −3.

3 Marks

✅ CBSE Standard Answer

(a) Pair of negative integers with difference 8:
Let the negative integers be −2 and −10.
Calculation: (−2) − (−10) = −2 + 10 = 8.

(b) Negative integer and positive integer with sum −5:
Let the negative integer be −7 and the positive integer be 2.
Calculation: (−7) + 2 = −5.

(c) Negative integer and positive integer with difference −3:
Let the negative integer be −1 and the positive integer be 2.
Calculation: (−1) − (+2) = −1 − 2 = −3.

CBSE Marking Scheme 2026-27

Part (a): (−2) − (−10) = 8 with signs shown1 Mark

Part (b): (−7) + 2 = −51 Mark

Part (c): (−1) − 2 = −31 Mark

Q3

In a quiz, team A scored −40, 10, 0 and team B scored 10, 0, −40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?

2 Marks

✅ CBSE Standard Answer

Step 1: Total score of Team A:
Total score of Team A = (−40) + 10 + 0 = −30.

Step 2: Total score of Team B:
Total score of Team B = 10 + 0 + (−40) = 10 − 40 = −30.

Step 3: Comparison & Order Property:
Both teams scored −30. Hence, both teams scored equally; neither team scored more.
Yes, we can add integers in any order due to the Commutative and Associative Properties of addition.

CBSE Marking Scheme 2026-27

Calculating both total scores accurately (−30 each)1 Mark

Affirmative answer stating commutative property1 Mark

Q4

Fill in the blanks to make the following statements true:
(i) (−5) + (−8) = (−8) + (.....)
(ii) −53 + ..... = −53
(iii) 17 + ..... = 0
(iv) [13 + (−12)] + (.....) = 13 + [(−12) + (−7)]
(v) (−4) + [15 + (−3)] = [(−4) + 15] + (.....)

3 Marks

✅ CBSE Standard Answer

(i) (−5) + (−8) = (−8) + (−5)   [Commutative property: a + b = b + a]
(ii) −53 + 0 = −53   [Additive identity: a + 0 = a]
(iii) 17 + (−17) = 0   [Additive inverse: a + (-a) = 0]
(iv) [13 + (−12)] + (−7) = 13 + [(−12) + (−7)]   [Associative property]
(v) (−4) + [15 + (−3)] = [(−4) + 15] + (−3)   [Associative property]

CBSE Marking Scheme 2026-27

Each correct fill-in-the-blank with underlying property0.6 Mark each (Total 3 Marks)

Exercise 1.2

Q1

Find each of the following products:
(a) 3 × (−1)
(b) (−1) × 225
(c) (−21) × (−30)
(d) (−316) × (−1)
(e) (−15) × 0 × (−18)
(f) (−12) × (−11) × (10)
(g) 9 × (−3) × (−6)
(h) (−18) × (−5) × (−4)
(i) (−1) × (−2) × (−3) × 4
(j) (−3) × (−6) × (−2) × (−1)

4 Marks

✅ CBSE Standard Answer

(a) 3 × (−1) = −3 [(+) × (−) = (−)]
(b) (−1) × 225 = −225
(c) (−21) × (−30) = +(21 × 30) = 630 [(−) × (−) = (+)]
(d) (−316) × (−1) = 316
(e) (−15) × 0 × (−18) = 0 [Any number multiplied by 0 is 0]
(f) (−12) × (−11) × 10 = 132 × 10 = 1320
(g) 9 × (−3) × (−6) = 9 × 18 = 162
(h) (−18) × (−5) × (−4) = 90 × (−4) = −360 [3 negative signs ⇒ result is negative]
(i) (−1) × (−2) × (−3) × 4 = [2] × [−12] = −24
(j) (−3) × (−6) × (−2) × (−1) = 18 × 2 = 36 [4 negative signs ⇒ result is positive]

CBSE Marking Scheme 2026-27

Sign rule correct application2 Marks

All 10 numerical values accurately computed2 Marks

Q2

Verify the following:
(a) 18 × [7 + (−3)] = [18 × 7] + [18 × (−3)]
(b) (−21) × [(−4) + (−6)] = [(−21) × (−4)] + [(−21) × (−6)]

3 Marks

✅ CBSE Standard Answer

(a) Verification of Distributive Property:
LHS = 18 × [7 + (−3)] = 18 × [7 − 3] = 18 × 4 = 72.
RHS = [18 × 7] + [18 × (−3)] = 126 + (−54) = 126 − 54 = 72.
Since LHS = RHS = 72, the statement is verified.

(b) Verification:
LHS = (−21) × [(−4) + (−6)] = (−21) × [−10] = 210.
RHS = [(−21) × (−4)] + [(−21) × (−6)] = 84 + 126 = 210.
Since LHS = RHS = 210, the statement is verified.

CBSE Marking Scheme 2026-27

Part (a): LHS = 72 and RHS = 72 shown with conclusion1.5 Marks

Part (b): LHS = 210 and RHS = 210 shown with conclusion1.5 Marks

Q3

Find the product using suitable properties:
(a) 26 × (−48) + (−48) × (−36)
(b) 8 × 53 × (−125)
(c) 15 × (−25) × (−4) × (−10)
(d) (−41) × 102
(e) 625 × (−35) + (−625) × 65
(f) 7 × (50 − 2)
(g) (−17) × (−29)
(h) (−57) × (−19) + 57

4 Marks

✅ CBSE Standard Answer

(a) 26 × (−48) + (−48) × (−36):
Take common factor (−48) using Distributive Property:
= (−48) × [26 + (−36)] = (−48) × (−10) = 480.

(b) 8 × 53 × (−125):
Regroup using Associative & Commutative Property:
= [8 × (−125)] × 53 = (−1000) × 53 = −53000.

(c) 15 × (−25) × (−4) × (−10):
Group (−25) × (−4) = 100:
= 15 × 100 × (−10) = 1500 × (−10) = −15000.

(d) (−41) × 102:
Write 102 = 100 + 2:
= (−41) × (100 + 2) = (−41 × 100) + (−41 × 2) = −4100 − 82 = −4182.

(e) 625 × (−35) + (−625) × 65:
Rewrite (−625) × 65 = 625 × (−65):
= 625 × [(−35) + (−65)] = 625 × (−100) = −62500.

(f) 7 × (50 − 2):
= (7 × 50) − (7 × 2) = 350 − 14 = 336.

(g) (−17) × (−29):
Write −29 = −30 + 1:
= (−17) × (−30 + 1) = [(−17) × (−30)] + [(−17) × 1] = 510 − 17 = 493.

(h) (−57) × (−19) + 57:
Note that (−57) × (−19) = 57 × 19 and 57 = 57 × 1:
= 57 × [19 + 1] = 57 × 20 = 1140.

CBSE Marking Scheme 2026-27

Stating property name (Distributive / Regrouping)1.5 Marks

Accurate step-by-step simplification for all parts2.5 Marks

Exercise 1.3

Q1

Evaluate each of the following:
(a) (−30) ÷ 10
(b) 50 ÷ (−5)
(c) (−36) ÷ (−9)
(d) (−49) ÷ (49)
(e) 13 ÷ [(−2) + 1]
(f) 0 ÷ (−12)
(g) (−31) ÷ [(−30) + (−1)]
(h) [(−36) ÷ 12] ÷ 3
(i) [(−6) + 5] ÷ [(−2) + 1]

3 Marks

✅ CBSE Standard Answer

(a) (−30) ÷ 10 = −30/10 = −3
(b) 50 ÷ (−5) = −50/5 = −10
(c) (−36) ÷ (−9) = +36/9 = 4
(d) (−49) ÷ 49 = −1
(e) 13 ÷ [(−2) + 1] = 13 ÷ (−1) = −13
(f) 0 ÷ (−12) = 0 [Zero divided by any non-zero integer is 0]
(g) (−31) ÷ [(−30) + (−1)] = (−31) ÷ (−31) = 1
(h) [(−36) ÷ 12] ÷ 3 = (−3) ÷ 3 = −1
(i) [(−6) + 5] ÷ [(−2) + 1] = (−1) ÷ (−1) = 1

CBSE Marking Scheme 2026-27

Sign rule in division accurately applied1.5 Marks

Correct evaluation of nested brackets1.5 Marks

Q2

Verify that a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) for each of the following values of a, b and c:
(a) a = 12, b = −4, c = 2
(b) a = (−10), b = 1, c = 1

3 Marks

✅ CBSE Standard Answer

(a) When a = 12, b = −4, c = 2:
LHS = a ÷ (b + c) = 12 ÷ [(−4) + 2] = 12 ÷ (−2) = −6.
RHS = (a ÷ b) + (a ÷ c) = [12 ÷ (−4)] + [12 ÷ 2] = (−3) + 6 = 3.
Since −6 ≠ 3, LHS ≠ RHS. Verified that division is not distributive over addition!

(b) When a = −10, b = 1, c = 1:
LHS = (−10) ÷ (1 + 1) = (−10) ÷ 2 = −5.
RHS = [(−10) ÷ 1] + [(−10) ÷ 1] = (−10) + (−10) = −20.
Since −5 ≠ −20, LHS ≠ RHS. Verified!

CBSE Marking Scheme 2026-27

Part (a): LHS = −6, RHS = 3, LHS ≠ RHS1.5 Marks

Part (b): LHS = −5, RHS = −20, LHS ≠ RHS1.5 Marks

Q3

An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach −350 m?

3 Marks

✅ CBSE Standard Answer

Step 1: Determine total distance to descend:
Initial position = +10 m (above ground).
Final position = −350 m (below ground).
Total distance = 10 − (−350) = 10 + 350 = 360 m.

Step 2: Calculate time taken:
Speed of descent = 6 m/min.
Time taken = Distance/Speed = 360/6 = 60 minutes = 1 hour.

Final Answer: The elevator will take 1 hour (60 minutes) to reach −350 m.

CBSE Marking Scheme 2026-27

Distance formulation: 10 − (−350) = 360 m1.5 Marks

Time = 360/6 = 60 min = 1 hour with unit1.5 Marks

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

Case Study: Temperature Variation

Scenario: In a cold storage plant, the temperature at 12:00 noon was 10°C above zero. It decreases at a rate of 2°C per hour until midnight. (i) At what time would the temperature be 8°C below zero? (ii) What would be the temperature at midnight?

(i) Time when temperature is 8°C below zero (−8°C):
Initial temp = +10°C; Target temp = −8°C.
Total drop in temp = 10 − (−8) = 18°C.
Rate of drop = 2°C/hour.
Time taken = 18 ÷ 2 = 9 hours.
Time = 12 noon + 9 hours = 9:00 PM.

(ii) Temperature at midnight (12:00 AM):
Time duration from 12 noon to 12 midnight = 12 hours.
Drop in temperature = 12 × 2°C = 24°C.
Temp at midnight = 10 − 24 = −14°C.

⚡ Assertion & Reason

Assertion (A): The product of an odd number of negative integers is always negative.
Reason (R): Two negative signs pair up to produce a positive sign, leaving one unpaired negative sign when the count is odd.
Choose: (a) Both A and R are true and R is correct explanation of A. (b) Both A and R are true, but R is not correct explanation. (c) A is true, R is false. (d) A is false, R is true.

Answer: (a) Both A and R are true and R is the correct explanation of A.
Explanation: Every pair of negative integers gives a positive product: (−) × (−) = (+). For an odd count 2k + 1, the k pairs give (+), and multiplying with the remaining negative integer yields (+) × (−) = (−).

2

Chapter 2: Fractions and Decimals

NCERT Exercise 2.1 to 2.5 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Formulas Summary

Types of Fractions:
- Proper Fraction: Numerator < Denominator (e.g. 3/5). Value < 1.
- Improper Fraction: Numerator ≥ Denominator (e.g. 7/4). Value ≥ 1. Can be written as Mixed Fraction: 1 3/4.
Multiplication of Fractions: (Numerator₁ × Numerator₂)/(Denominator₁ × Denominator₂) The operator 'of' signifies multiplication: 1/2 of 24 = 1/2 × 24 = 12.
Reciprocal of a Fraction: The non-zero number obtained by interchanging numerator and denominator. Product with original fraction equals 1: a/b × b/a = 1.
Division of Fractions: Multiply by the reciprocal of the divisor: a/b ÷ c/d = a/b × d/c.
Decimal Operations:
- Multiplying by 10, 100, 1000: Shift decimal point to the right by as many places as there are zeros.
- Dividing by 10, 100, 1000: Shift decimal point to the left by as many places as there are zeros.
- Division by decimal: Multiply both dividend and divisor by suitable powers of 10 to make divisor a whole number.

Exercise 2.1

Q1

Solve the following fractions:
(i) 2 − 3/5
(ii) 4 + 7/8
(iii) 3/5 + 2/7
(iv) 9/11 − 4/15
(v) 7/10 + 2/5 + 3/2
(vi) 2⅔ + 3½
(vii) 8½ − 3⅝

4 Marks

✅ CBSE Standard Answer

(i) 2 − 3/5:
= 2/1 − 3/5 = (2 × 5 − 3)/5 = (10 − 3)/5 = 7/5 = 1 2/5.

(ii) 4 + 7/8:
= (4 × 8 + 7)/8 = (32 + 7)/8 = 39/8 = 4 7/8.

(iii) 3/5 + 2/7:
LCM of 5 and 7 is 35.
= (3 × 7 + 2 × 5)/35 = (21 + 10)/35 = 31/35.

(iv) 9/11 − 4/15:
LCM of 11 and 15 is 165.
= (9 × 15 − 4 × 11)/165 = (135 − 44)/165 = 91/165.

(v) 7/10 + 2/5 + 3/2:
LCM of 10, 5, 2 is 10.
= (7 + (2 × 2) + (3 × 5))/10 = (7 + 4 + 15)/10 = 26/10 = 13/5 = 2 3/5.

(vi) 2 2/3 + 3 1/2:
Convert to improper fractions: 8/3 + 7/2. LCM of 3, 2 is 6.
= (8 × 2 + 7 × 3)/6 = (16 + 21)/6 = 37/6 = 6 1/6.

(vii) 8 1/2 − 3 5/8:
Convert to improper fractions: 17/2 − 29/8. LCM of 2, 8 is 8.
= (17 × 4 − 29)/8 = (68 − 29)/8 = 39/8 = 4 7/8.

CBSE Marking Scheme 2026-27

LCM determination and equivalent fractions2 Marks

Correctly evaluated and simplified final fractions2 Marks

Q2

Arrange the following in descending order:
(i) 2/9, 2/3, 8/21
(ii) 1/5, 3/7, 7/10

3 Marks

✅ CBSE Standard Answer

(i) 2/9, 2/3, 8/21:
LCM of denominators 9, 3, 21 is 63.
Convert each to like fractions:
2/9 = (2 × 7)/63 = 14/63
2/3 = (2 × 21)/63 = 42/63
8/21 = (8 × 3)/63 = 24/63
Comparing numerators: 42 > 24 > 14.
Descending order: 2/3 > 8/21 > 2/9.

(ii) 1/5, 3/7, 7/10:
LCM of 5, 7, 10 is 70.
1/5 = (1 × 14)/70 = 14/70
3/7 = (3 × 10)/70 = 30/70
7/10 = (7 × 7)/70 = 49/70
Comparing numerators: 49 > 30 > 14.
Descending order: 7/10 > 3/7 > 1/5.

CBSE Marking Scheme 2026-27

Part (i): LCM = 63 and correct descending order1.5 Marks

Part (ii): LCM = 70 and correct descending order1.5 Marks

Exercise 2.2

Q1

Multiply and reduce to lowest form and convert into a mixed fraction:
(i) 7 × 3/5
(ii) 4 × 1/3
(iii) 2 × 6/7
(iv) 5 × 2/9
(v) 2/3 × 4
(vi) 5/2 × 6
(vii) 11 × 4/7

3 Marks

✅ CBSE Standard Answer

(i) 7 × 3/5 = 21/5 = 4 1/5
(ii) 4 × 1/3 = 4/3 = 1 1/3
(iii) 2 × 6/7 = 12/7 = 1 5/7
(iv) 5 × 2/9 = 10/9 = 1 1/9
(v) 2/3 × 4 = 8/3 = 2 2/3
(vi) 5/2 × 6 = 5 × 3 = 15
(vii) 11 × 4/7 = 44/7 = 6 2/7

CBSE Marking Scheme 2026-27

Numerator multiplication and conversion to mixed fraction3 Marks

Q2

Find: (a) ½ of (i) 24 (ii) 46
(b) ⅔ of (i) 18 (ii) 27
(c) ¾ of (i) 16 (ii) 36

3 Marks

✅ CBSE Standard Answer

(a) ½ of:
(i) 1/2 × 24 = 12
(ii) 1/2 × 46 = 23

(b) ⅔ of:
(i) 2/3 × 18 = 2 × 6 = 12
(ii) 2/3 × 27 = 2 × 9 = 18

(c) ¾ of:
(i) 3/4 × 16 = 3 × 4 = 12
(ii) 3/4 × 36 = 3 × 9 = 27

CBSE Marking Scheme 2026-27

Each correct evaluation with multiplication setup0.5 Mark each (3 Marks total)

Q3

A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2¾ litres of petrol?

2 Marks

✅ CBSE Standard Answer

Step 1: Quantity of petrol given:
2 3/4 litres = (2 × 4 + 3)/4 = 11/4 litres.

Step 2: Calculate total distance:
Distance covered in 1 litre = 16 km.
Total distance = 16 × 11/4 = (16 × 11)/4 = 4 × 11 = 44 km.

Conclusion: The car will cover 44 km using 2¾ litres of petrol.

CBSE Marking Scheme 2026-27

Improper fraction conversion: 11/4 litres0.5 Mark

Multiplication 16 × (11/4) = 44 km with statement1.5 Marks

Exercise 2.3

Q1

Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers:
(i) 3/7 (ii) 5/8 (iii) 9/7 (iv) 6/5 (v) 12/7 (vi) 1/8 (vii) 1/11

3 Marks

✅ CBSE Standard Answer

(i) Reciprocal of 3/7 is 7/3 ⇒ Improper fraction.
(ii) Reciprocal of 5/8 is 8/5 ⇒ Improper fraction.
(iii) Reciprocal of 9/7 is 7/9 ⇒ Proper fraction.
(iv) Reciprocal of 6/5 is 5/6 ⇒ Proper fraction.
(v) Reciprocal of 12/7 is 7/12 ⇒ Proper fraction.
(vi) Reciprocal of 1/8 is 8/1 = 8 ⇒ Whole number.
(vii) Reciprocal of 1/11 is 11/1 = 11 ⇒ Whole number.

CBSE Marking Scheme 2026-27

Reciprocal values correctly found1.5 Marks

Accurate classification (Proper / Improper / Whole)1.5 Marks

Q2

Find: (i) 7/3 ÷ 2, (ii) 4/9 ÷ 5, (iii) 4⅓ ÷ 3, (iv) 2/5 ÷ 1/2, (v) 3½ ÷ 8/7, (vi) 2⅕ ÷ 1⅕.

3 Marks

✅ CBSE Standard Answer

(i) 7/3 ÷ 2 = 7/3 × 1/2 = 7/6 = 1 1/6
(ii) 4/9 ÷ 5 = 4/9 × 1/5 = 4/45
(iii) 4 1/3 ÷ 3 = 13/3 × 1/3 = 13/9 = 1 4/9
(iv) 2/5 ÷ 1/2 = 2/5 × 2/1 = 4/5
(v) 3 1/2 ÷ 8/7 = 7/2 × 7/8 = 49/16 = 3 1/16
(vi) 2 1/5 ÷ 1 1/5 = 11/5 ÷ 6/5 = 11/5 × 5/6 = 11/6 = 1 5/6

CBSE Marking Scheme 2026-27

Reciprocal multiplication rule application1.5 Marks

Accurate simplified fraction values1.5 Marks

Exercise 2.4 & 2.5 (Decimals Operations)

Q1

Find the product:
(i) 0.2 × 6
(ii) 8 × 4.6
(iii) 2.71 × 5
(iv) 20.1 × 4
(v) 0.05 × 7
(vi) 211.02 × 4
(vii) 2 × 0.86

3 Marks

✅ CBSE Standard Answer

(i) 0.2 × 6 = 1.2
(ii) 8 × 4.6 = 36.8
(iii) 2.71 × 5 = 13.55
(iv) 20.1 × 4 = 80.4
(v) 0.05 × 7 = 0.35
(vi) 211.02 × 4 = 844.08
(vii) 2 × 0.86 = 1.72

CBSE Marking Scheme 2026-27

Count of decimal places correctly maintained1.5 Marks

Accurate multiplication arithmetic1.5 Marks

Q2

Find:
(a) 7.75 ÷ 0.25
(b) 76.5 ÷ 0.15
(c) 37.8 ÷ 1.4
(d) 2.7 ÷ 100
(e) 0.5 ÷ 1000
(f) 7 ÷ 3.5

3 Marks

✅ CBSE Standard Answer

(a) 7.75 ÷ 0.25 = 775/25 = 31
(b) 76.5 ÷ 0.15 = 7650/15 = 510
(c) 37.8 ÷ 1.4 = 378/14 = 27
(d) 2.7 ÷ 100 = 0.027 [Shift decimal 2 places left]
(e) 0.5 ÷ 1000 = 0.0005 [Shift decimal 3 places left]
(f) 7 ÷ 3.5 = 70/35 = 2

CBSE Marking Scheme 2026-27

Each correct evaluation with step0.5 Mark each (3 Marks total)

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

Case Study: Uniform Tailoring

Scenario: A school needs 1.8 metres of cloth to stitch one uniform shirt. A roll of fabric contains 45 metres of cloth. (i) How many complete shirts can be tailored from this roll? (ii) What length of fabric will be left? (iii) If each shirt requires 6 buttons, how many buttons will be needed in total?

(i) Number of shirts:
Number of shirts = (Total cloth)/(Cloth per shirt) = 45 ÷ 1.8 = 450/18 = 25 complete shirts.

(ii) Leftover fabric:
Total used = 25 × 1.8 = 45 m. Leftover = 45 − 45 = 0 metres.

(iii) Total buttons needed:
25 × 6 = 150 buttons.

⚡ HOTS Question

Each side of a regular polygon is 2.5 cm in length. The perimeter of the polygon is 12.5 cm. How many sides does the polygon have?

Let number of sides be n.
In a regular polygon, all sides are equal. Perimeter = n × side.
n × 2.5 = 12.5 ⇒ n = 12.5/2.5 = 125/25 = 5.
Therefore, the polygon has 5 sides (it is a regular pentagon).

3

Chapter 3: Data Handling

NCERT Exercise 3.1 to 3.4 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Formulas Summary

Data & Range: Range = Highest Value - Lowest Value
Arithmetic Mean (Average): Mean = (Sum of all observations)/(Number of observations)
Mode: The observation that occurs most frequently in the given data. A data set can have more than one mode (multimodal).
Median: The middle observation when data is arranged in ascending or descending order.
- For odd n: Median = ((n+1)/2)th observation.
Bar Graphs: Representation of numbers using bars of uniform width where lengths/heights are proportional to the frequencies. Double bar graphs compare two sets of data simultaneously.
Probability (P): Measure of chance of an event happening: P(Event) = (Number of favourable outcomes)/(Total number of possible outcomes) 0 ≤ P ≤ 1. Impossible event = 0, Certain event = 1.

Exercise 3.1

Q1

Find the range of heights of any ten students of your class.

2 Marks

✅ CBSE Standard Answer

Sample Data (Heights in cm):
Let the heights of 10 students be: 128, 132, 135, 139, 140, 142, 145, 148, 150, 152 cm.

Calculation of Range:
Highest height = 152 cm
Lowest height = 128 cm
Range = Highest height − Lowest height = 152 − 128 = 24 cm.

CBSE Marking Scheme 2026-27

Listing observations & identifying max/min values1 Mark

Applying Range formula and correct result1 Mark

Q2

Find the mean of the first five whole numbers.

2 Marks

✅ CBSE Standard Answer

Step 1: Identify whole numbers:
The first five whole numbers are: 0, 1, 2, 3, 4.

Step 2: Calculate sum and mean:
Sum = 0 + 1 + 2 + 3 + 4 = 10.
Number of observations = 5.
Mean = (Sum of observations)/(Number of observations) = 10/5 = 2.

Conclusion: The mean of the first five whole numbers is 2.

CBSE Marking Scheme 2026-27

Listing correct numbers starting from 00.5 Mark

Mean formula & final answer 21.5 Marks

Q3

A cricketer scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.

2 Marks

✅ CBSE Standard Answer

Step 1: Sum of runs:
Total runs = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400.

Step 2: Number of innings:
n = 8.

Step 3: Mean score:
Mean = (Total runs)/(Number of innings) = 400/8 = 50.

Conclusion: The mean score of the cricketer is 50 runs.

CBSE Marking Scheme 2026-27

Sum of observations = 4001 Mark

Mean = 400/8 = 50 runs1 Mark

Exercise 3.2

Q1

The scores in mathematics test (out of 25) of 15 students is as follows: 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20. Find the mode and median of this data. Are they same?

3 Marks

✅ CBSE Standard Answer

Step 1: Arrange data in ascending order:
5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25.
Total students, n = 15.

Step 2: Find Mode:
The score 20 occurs 4 times (highest frequency).
Therefore, Mode = 20.

Step 3: Find Median:
Since n = 15 is odd:
Median = ((15+1)/2)th observation = 8th observation = 20.

Conclusion:
Yes, the mode and median are the same (both equal 20).

CBSE Marking Scheme 2026-27

Ascending order + Mode = 201 Mark

Median = 8th value = 201 Mark

Affirmative conclusion stating both are equal1 Mark

Q2

The weights (in kg) of 15 students of a class are: 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47.
(i) Find the mode and median of this data.
(ii) Is there more than one mode?

3 Marks

✅ CBSE Standard Answer

Arranging in ascending order:
32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50.
Total number of observations, n = 15.

(i) Mode and Median:
• Both 38 and 43 occur 3 times each (highest frequency).
Therefore, Modes = 38 kg and 43 kg (bimodal).
• Median = ((15+1)/2)th observation = 8th observation = 40 kg.

(ii) Is there more than one mode?
Yes, there are two modes (38 kg and 43 kg).

CBSE Marking Scheme 2026-27

Modes: 38 and 43 kg1 Mark

Median: 40 kg1 Mark

Yes, with two modes mentioned1 Mark

Exercise 3.3 & 3.4 (Graphs & Probability)

Q1

Consider this data collected from a survey of a colony:
Cricket (Watch: 1240, Participate: 620), Basketball (Watch: 470, Participate: 320), Swimming (Watch: 510, Participate: 320), Hockey (Watch: 430, Participate: 250), Athletics (Watch: 250, Participate: 105).
(i) Which sport is most popular?
(ii) Which is more preferred, watching or participating in sports?

3 Marks

✅ CBSE Standard Answer

(i) Most popular sport:
Total people interested in Cricket = 1240 + 620 = 1860.
Cricket has the highest bar for both watching and participating.
Answer: Cricket is the most popular sport.

(ii) Watching vs. Participating preference:
Total watching = 1240 + 470 + 510 + 430 + 250 = 2900.
Total participating = 620 + 320 + 320 + 250 + 105 = 1615.
Since 2900 > 1615, watching sports is preferred more than participating.

CBSE Marking Scheme 2026-27

Part (i): Identifying Cricket with total count1.5 Marks

Part (ii): Calculating totals (2900 vs 1615) & conclusion1.5 Marks

Q2

There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.
(i) What is the probability of drawing a marble with number 2?
(ii) What is the probability of drawing a marble with number 5?

2 Marks

✅ CBSE Standard Answer

Sample Space:
Total number of marbles = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6.

(i) Probability of drawing marble number 2:
Number 2 appears on only 1 marble ⇒ n(E) = 1.
P(Number 2) = n(E)/n(S) = 1/6.

(ii) Probability of drawing marble number 5:
Number 5 appears on only 1 marble ⇒ n(E) = 1.
P(Number 5) = n(E)/n(S) = 1/6.

CBSE Marking Scheme 2026-27

Formula P = Favourable / Total outcomes1 Mark

Answers 1/6 for both parts1 Mark

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

Case Study: Die Roll Outcomes

Problem: When a standard 6-faced die is thrown once, list the outcomes and find the probability of an event of getting: (i) a prime number, (ii) not a prime number, (iii) a number greater than 5, (iv) a number not greater than 5.

Total possible outcomes = {1, 2, 3, 4, 5, 6}, total = 6.

(i) Prime number: Outcomes are {2, 3, 5} (3 outcomes) ⇒ P = 3/6 = 1/2.

(ii) Not a prime number: Outcomes are {1, 4, 6} (3 outcomes) ⇒ P = 3/6 = 1/2. (Note: 1 is neither prime nor composite).

(iii) Number greater than 5: Only {6} (1 outcome) ⇒ P = 1/6.

(iv) Number not greater than 5: Outcomes {1, 2, 3, 4, 5} (5 outcomes) ⇒ P = 5/6.

4

Chapter 4: Simple Equations

NCERT Exercise 4.1 to 4.4 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Formulas Summary

Equation: A mathematical statement showing that two expressions have equal value (LHS = RHS), involving at least one variable.
Solution of an Equation: The value of the variable for which the equation is satisfied (LHS = RHS).
Rules for Balancing Equations:
- Add or subtract the same number on both sides.
- Multiply or divide both sides by the same non-zero number.
Transposition Method: Moving a term from one side of the equation to the other changes its sign:
- + becomes - | - becomes +
- × becomes ÷ | ÷ becomes ×

Exercise 4.1

Q1

Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
(b) 7n + 5 = 19 (n = −2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p − 3 = 13 (p = 1)
(e) 4p − 3 = 13 (p = −4)
(f) 4p − 3 = 13 (p = 0)

3 Marks

✅ CBSE Standard Answer

(a) n + 5 = 19 for n = 1:
LHS = 1 + 5 = 6 ≠ 19 (RHS) ⇒ No, not a solution.

(b) 7n + 5 = 19 for n = −2:
LHS = 7(−2) + 5 = −14 + 5 = −9 ≠ 19 (RHS) ⇒ No, not a solution.

(c) 7n + 5 = 19 for n = 2:
LHS = 7(2) + 5 = 14 + 5 = 19 = RHS ⇒ Yes, it is a solution.

(d) 4p − 3 = 13 for p = 1:
LHS = 4(1) − 3 = 4 − 3 = 1 ≠ 13 (RHS) ⇒ No, not a solution.

(e) 4p − 3 = 13 for p = −4:
LHS = 4(−4) − 3 = −16 − 3 = −19 ≠ 13 (RHS) ⇒ No, not a solution.

(f) 4p − 3 = 13 for p = 0:
LHS = 4(0) − 3 = −3 ≠ 13 (RHS) ⇒ No, not a solution.

CBSE Marking Scheme 2026-27

LHS vs RHS evaluated for each part0.5 Mark each (3 Marks total)

Q2

Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.

3 Marks

✅ CBSE Standard Answer

(i) x + 4 = 9
(ii) y − 2 = 8
(iii) 10a = 70
(iv) b/5 = 6
(v) 3/4t = 15
(vi) 7m + 7 = 77

CBSE Marking Scheme 2026-27

Each correct algebraic equation translation0.5 Mark each

Exercise 4.2 & 4.3 (Solving Equations)

Q1

Solve the following equations:
(a) 2y + 5/2 = 37/2
(b) 5t + 28 = 10
(c) a/5 + 3 = 2
(d) q/4 + 7 = 5
(e) 5/2 x = −10
(f) 5/2 x = 25/4

3 Marks

✅ CBSE Standard Answer

(a) 2y + 5/2 = 37/2:
Transpose 5/2: 2y = 37/2 − 5/2 = 32/2 = 16 ⇒ y = 16/2 = 8.

(b) 5t + 28 = 10:
5t = 10 − 28 = −18 ⇒ t = -18/5.

(c) a/5 + 3 = 2:
a/5 = 2 − 3 = −1 ⇒ a = −1 × 5 = −5.

(d) q/4 + 7 = 5:
q/4 = 5 − 7 = −2 ⇒ q = −2 × 4 = −8.

(e) 5/2x = −10:
x = −10 × 2/5 = −2 × 2 = −4.

(f) 5/2x = 25/4:
x = 25/4 × 2/5 = 5/2 = 2 1/2.

CBSE Marking Scheme 2026-27

Each correct step-by-step solution0.5 Mark each (3 Marks total)

Q2

Solve the following equations with brackets:
(a) 2(x + 4) = 12
(b) 3(n − 5) = 21
(c) 3(n − 5) = −21
(d) −4(2 + x) = 8
(e) 4(2 − x) = 8

3 Marks

✅ CBSE Standard Answer

(a) 2(x + 4) = 12:
Divide both sides by 2: x + 4 = 6 ⇒ x = 6 − 4 = 2.

(b) 3(n − 5) = 21:
Divide both sides by 3: n − 5 = 7 ⇒ n = 7 + 5 = 12.

(c) 3(n − 5) = −21:
Divide both sides by 3: n − 5 = −7 ⇒ n = −7 + 5 = −2.

(d) −4(2 + x) = 8:
Divide by −4: 2 + x = 8/(−4) = −2 ⇒ x = −2 − 2 = −4.

(e) 4(2 − x) = 8:
Divide by 4: 2 − x = 2 ⇒ −x = 2 − 2 = 0 ⇒ x = 0.

CBSE Marking Scheme 2026-27

Bracket expansion/division step & solution0.6 Mark each

Exercise 4.4 (Applications & Word Problems)

Q1

Set up equations and solve them to find the unknown numbers:
(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.

4 Marks

✅ CBSE Standard Answer

(a) Let the number be x.
Equation: 8x + 4 = 60 ⇒ 8x = 60 − 4 = 56 ⇒ x = 56/8 = 7.

(b) Let the number be y.
Equation: y/5 − 4 = 3 ⇒ y/5 = 7 ⇒ y = 7 × 5 = 35.

(c) Let the number be z.
Equation: 3/4z + 3 = 21 ⇒ 3/4z = 18 ⇒ z = 18 × 4/3 = 6 × 4 = 24.

(d) Let the number be n.
Equation: 2n − 11 = 15 ⇒ 2n = 15 + 11 = 26 ⇒ n = 26/2 = 13.

CBSE Marking Scheme 2026-27

Each part: Equation setup (0.5 Mark) + Solution (0.5 Mark)1 Mark each (4 Marks total)

Q2

Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

3 Marks

✅ CBSE Standard Answer

Step 1: Define variable:
Let Laxmi's present age be y years.

Step 2: Set up equation:
Three times Laxmi's age = 3y.
Father is 4 years older than 3y: 3y + 4.
Given father's age is 49 years:
3y + 4 = 49.

Step 3: Solve for y:
3y = 49 − 4 = 45 ⇒ y = 45/3 = 15.

Conclusion: Laxmi’s age is 15 years.

CBSE Marking Scheme 2026-27

Assuming variable and equation 3y + 4 = 491.5 Marks

Solving y = 15 with final conclusion statement1.5 Marks

Q3

Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?

3 Marks

✅ CBSE Standard Answer

Step 1: Define variable:
Let the number of marbles Parmit has be m.

Step 2: Form equation:
Five times Parmit's marbles = 5m.
7 marbles more than 5m = 5m + 7.
Given Irfan has 37 marbles:
5m + 7 = 37.

Step 3: Solve equation:
5m = 37 − 7 = 30 ⇒ m = 30/5 = 6.

Conclusion: Parmit has 6 marbles.

CBSE Marking Scheme 2026-27

Forming equation 5m + 7 = 371.5 Marks

Solving for m = 6 with statement1.5 Marks

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

Case Study: Cricket Partnership

Scenario: Rahul scored twice as many runs as Sachin. Together, their runs fell two short of a double century (200 runs). (i) Form an equation to represent this situation. (ii) How many runs did each one score?

(i) Equation Formulation:
Let Sachin's score = x runs.
Rahul's score = 2x runs.
Combined runs = x + 2x = 3x.
Two short of a double century = 200 − 2 = 198.
Equation: 3x = 198.

(ii) Solving for individual scores:
x = 198/3 = 66.
Sachin scored = 66 runs.
Rahul scored = 2 × 66 = 132 runs.
Check: 66 + 132 = 198 (verified).

5

Chapter 5: Lines and Angles

NCERT Exercise 5.1 to 5.2 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Angle Properties Summary

Complementary Angles: Two angles whose sum of measures equals 90°. Each angle is called the complement of the other.
Supplementary Angles: Two angles whose sum of measures equals 180°. Each angle is called the supplement of the other.
Adjacent Angles: Angles that have a common vertex, a common arm, and non-common arms on opposite sides of the common arm.
Linear Pair: A pair of adjacent angles whose non-common sides form opposite rays (a straight line). Sum = 180°.
Vertically Opposite Angles: When two lines intersect at a point, vertically opposite angles are formed and are always equal.
Parallel Lines and Transversal: When two parallel lines l ∥ m are cut by a transversal t:
- Corresponding angles are equal (∠ 1 = ∠ 5, etc.).
- Alternate interior angles are equal (∠ 3 = ∠ 5, ∠ 4 = ∠ 6).
- Alternate exterior angles are equal (∠ 1 = ∠ 7, ∠ 2 = ∠ 8).
- Co-interior angles (interior angles on same side of transversal) are supplementary (∠ 4 + ∠ 5 = 180°).

Exercise 5.1

Q1

Find the complement of each of the following angles:
(i) 20°
(ii) 63°
(iii) 57°

3 Marks

✅ CBSE Standard Answer

Two angles are complementary if their sum is 90°. The complement of θ = 90° − θ.

(i) Complement of 20°:
= 90° − 20° = 70°.

(ii) Complement of 63°:
= 90° − 63° = 27°.

(iii) Complement of 57°:
= 90° − 57° = 33°.

CBSE Marking Scheme 2026-27

Each correct complement subtraction1 Mark each (3 Marks total)

Q2

Find the supplement of each of the following angles:
(i) 105°
(ii) 87°
(iii) 154°

3 Marks

✅ CBSE Standard Answer

Two angles are supplementary if their sum is 180°. The supplement of θ = 180° − θ.

(i) Supplement of 105°:
= 180° − 105° = 75°.

(ii) Supplement of 87°:
= 180° − 87° = 93°.

(iii) Supplement of 154°:
= 180° − 154° = 26°.

CBSE Marking Scheme 2026-27

Each correct supplement calculation1 Mark each (3 Marks total)

Q3

Identify which of the following pairs of angles are complementary and which are supplementary:
(i) 65°, 115°
(ii) 63°, 27°
(iii) 112°, 68°
(iv) 130°, 50°
(v) 45°, 45°
(vi) 80°, 10°

3 Marks

✅ CBSE Standard Answer

(i) 65° + 115° = 180°: ⇒ Supplementary.
(ii) 63° + 27° = 90°: ⇒ Complementary.
(iii) 112° + 68° = 180°: ⇒ Supplementary.
(iv) 130° + 50° = 180°: ⇒ Supplementary.
(v) 45° + 45° = 90°: ⇒ Complementary.
(vi) 80° + 10° = 90°: ⇒ Complementary.

CBSE Marking Scheme 2026-27

Each correct sum and classification0.5 Mark each (3 Marks total)

Q4

(a) Find the angle which is equal to its complement.
(b) Find the angle which is equal to its supplement.

2 Marks

✅ CBSE Standard Answer

(a) Angle equal to its complement:
Let the angle be x.
Its complement is 90° − x.
According to question: x = 90° − x ⇒ 2x = 90° ⇒ x = 45°.

(b) Angle equal to its supplement:
Let the angle be y.
Its supplement is 180° − y.
According to question: y = 180° − y ⇒ 2y = 180° ⇒ y = 90°.

CBSE Marking Scheme 2026-27

Part (a): Equation 2x = 90° and x = 45°1 Mark

Part (b): Equation 2y = 180° and y = 90°1 Mark

Q5

Two lines intersect at a point. If one angle is 55°, find the values of all other angles x, y and z.

3 Marks

✅ CBSE Standard Answer

Step 1: Vertically opposite angle:
Angle x is vertically opposite to 55°.
Therefore, x = 55°.

Step 2: Linear pair:
Angle y and 55° form a linear pair on a straight line:
y + 55° = 180° ⇒ y = 180° − 55° = 125°.

Step 3: Vertically opposite to y:
Angle z is vertically opposite to y:
z = y = 125°.

Final Answer: x = 55°, y = 125°, z = 125°.

CBSE Marking Scheme 2026-27

x = 55° with vertically opposite angle reason1 Mark

y = 125° with linear pair equation1 Mark

z = 125° with vertically opposite reason1 Mark

Exercise 5.2 (Parallel Lines & Transversal)

Q1

In a given figure, lines l ∥ m are cut by transversal p. If one exterior angle is 125°, find the measures of remaining angles a, b, c, d, e, f.

3 Marks

✅ CBSE Standard Answer

Step 1: Linear pair with 125°:
e + 125° = 180° ⇒ e = 180° − 125° = 55°.

Step 2: Vertically opposite angles:
f = e = 55°.
Angle opposite to 125° is 125°.

Step 3: Corresponding angles (since l ∥ m):
d = 125° (corresponding to 125°).
a = e = 55° (corresponding to e).
b = d = 125° (vertically opposite to d).
c = a = 55° (vertically opposite to a).

Final Answer: a = 55°, b = 125°, c = 55°, d = 125°, e = 55°, f = 55°.

CBSE Marking Scheme 2026-27

Linear pair to find e = 55°1 Mark

Corresponding angle arguments for d, a1 Mark

Vertically opposite arguments for b, c, f1 Mark

Q2

Find the value of x in each of the following figures if l ∥ m:
(i) One angle is 110° and alternate interior to x with linear pair.
(ii) Co-interior angles are (2x)° and (3x)°.

3 Marks

✅ CBSE Standard Answer

(i) Figure with angle 110°:
The angle adjacent to 110° on the straight line is: 180° − 110° = 70°.
Since l ∥ m, angle x is corresponding (or alternate) to this angle:
x = 180° − 110° = 70°.

(ii) Co-interior angles 2x and 3x:
Interior angles on the same side of a transversal are supplementary:
2x + 3x = 180° ⇒ 5x = 180° ⇒ x = 180°/5 = 36°.
The angles are 2(36°) = 72° and 3(36°) = 108°.

CBSE Marking Scheme 2026-27

Part (i): Linear pair / corresponding reasoning x = 70°1.5 Marks

Part (ii): Co-interior angles sum = 180° and x = 36°1.5 Marks

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

HOTS: Ratio of Supplementary Angles

Problem: Two supplementary angles are in the ratio 4 : 5. Find the measure of both angles.

Let the common ratio multiplier be x.
The two angles are 4x and 5x.
Since they are supplementary:
4x + 5x = 180° ⇒ 9x = 180° ⇒ x = 180°/9 = 20°.
First angle = 4 × 20° = 80°.
Second angle = 5 × 20° = 100°.
Check: 80° + 100° = 180° (verified).

⚡ Assertion & Reason

Assertion (A): If two parallel lines are intersected by a transversal, then each pair of alternate interior angles are equal.
Reason (R): If two parallel lines are intersected by a transversal, the interior angles on the same side of the transversal are supplementary.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, R is false. (d) A is false, R is true.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation: Both statements are fundamental, correct geometric properties of parallel lines and transversals, but Reason (R) states the co-interior angle property, which is not the explanation for why alternate interior angles are equal.

6

Chapter 6: The Triangle and its Properties

NCERT Exercise 6.1 to 6.5 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Theorems Summary

Median of a Triangle: The line segment connecting a vertex to the midpoint of the opposite side. Every triangle has 3 medians that concur at the Centroid.
Altitude of a Triangle: The perpendicular line segment from a vertex to the line containing the opposite side. A triangle has 3 altitudes that concur at the Orthocentre.
Exterior Angle Property: An exterior angle of a triangle is equal to the sum of its two interior opposite angles: ∠ ext = ∠ A + ∠ B
Angle Sum Property: The sum of all three interior angles of a triangle is always 180°: ∠ A + ∠ B + ∠ C = 180°
Triangle Inequality Property: The sum of the lengths of any two sides of a triangle is always greater than the length of the third side: a + b > c, b + c > a, c + a > b
Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: Hypotenuse² = Base² + Perpendicular² ⇒ h² = b² + p²

Exercise 6.1

Q1

In ΔPQR, D is the mid-point of QR. (i) Line segment PM is ..... (ii) Line segment PD is ..... (iii) Is QM = MR?

3 Marks

✅ CBSE Standard Answer

(i) Line segment PM:
Since PM ⊥ QR (makes a right angle 90° with QR), PM is an Altitude of ΔPQR.

(ii) Line segment PD:
Since D is the midpoint of QR, line segment joining vertex P to midpoint D is the Median of ΔPQR.

(iii) Is QM = MR?
No, QM ≠ MR. Because D is the midpoint (so QD = DR), whereas M is the foot of the perpendicular altitude, which is not necessarily the midpoint for a scalene triangle.

CBSE Marking Scheme 2026-27

(i) Altitude1 Mark

(ii) Median1 Mark

(iii) No, with justification1 Mark

Exercise 6.2 (Exterior Angle Property)

Q1

Find the value of the unknown exterior angle x in the following figures:
(i) Interior opposite angles are 50° and 70°
(ii) Interior opposite angles are 65° and 45°
(iii) Interior opposite angles are 30° and 40°

3 Marks

✅ CBSE Standard Answer

By Exterior Angle Property: Exterior angle = Sum of interior opposite angles.

(i) x = 50° + 70° = 120°

(ii) x = 65° + 45° = 110°

(iii) x = 30° + 40° = 70°

CBSE Marking Scheme 2026-27

Exterior angle formula statement1 Mark

Each correct evaluation0.67 Mark each (2 Marks total)

Q2

Find the value of the unknown interior angle x in the following figures:
(i) Exterior angle = 115°, one interior opposite angle = 50°
(ii) Exterior angle = 100°, one interior opposite angle = 70°
(iii) Exterior angle = 125°, right angled at other interior opposite angle (90°)

3 Marks

✅ CBSE Standard Answer

(i) x + 50° = 115° ⇒ x = 115° − 50° = 65°

(ii) x + 70° = 100° ⇒ x = 100° − 70° = 30°

(iii) x + 90° = 125° ⇒ x = 125° − 90° = 35°

CBSE Marking Scheme 2026-27

Each correct subtraction from exterior angle1 Mark each

Exercise 6.3 (Angle Sum Property)

Q1

Find the value of the unknown x in the following triangles:
(i) Two angles are 50° and 60°
(ii) Right-angled triangle with one acute angle 30°
(iii) Two angles are 30° and 110°
(iv) Isosceles triangle with two equal angles x and vertex angle 50°
(v) Equilateral-like triangle with all angles equal to x

3 Marks

✅ CBSE Standard Answer

By Angle Sum Property of a triangle: Sum of all 3 angles = 180°.

(i) x + 50° + 60° = 180° ⇒ x + 110° = 180° ⇒ x = 70°.

(ii) x + 90° + 30° = 180° ⇒ x + 120° = 180° ⇒ x = 60°.

(iii) x + 30° + 110° = 180° ⇒ x + 140° = 180° ⇒ x = 40°.

(iv) x + x + 50° = 180° ⇒ 2x = 180° − 50° = 130° ⇒ x = 130°/2 = 65°.

(v) x + x + x = 180° ⇒ 3x = 180° ⇒ x = 180°/3 = 60°.

CBSE Marking Scheme 2026-27

Stating Angle Sum Property = 180°1 Mark

Accurately computed angle values2 Marks

Exercise 6.4 (Triangle Inequality)

Q1

Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm

3 Marks

✅ CBSE Standard Answer

Property: In any triangle, the sum of lengths of any two sides must be strictly greater than the third side (a + b > c).

(i) 2 cm, 3 cm, 5 cm:
Check sum of smaller sides: 2 + 3 = 5 cm.
Since 5 is not greater than the third side 5 (5 = 5), a triangle is NOT possible.

(ii) 3 cm, 6 cm, 7 cm:
Check all three combinations:
• 3 + 6 = 9 > 7 (True)
• 6 + 7 = 13 > 3 (True)
• 3 + 7 = 10 > 6 (True)
Therefore, a triangle IS possible.

(iii) 6 cm, 3 cm, 2 cm:
Check sum of smaller sides: 3 + 2 = 5 cm.
Since 5 < 6 (less than third side), a triangle is NOT possible.

CBSE Marking Scheme 2026-27

Each part checked against a + b > c with clear decision1 Mark each

Q2

The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

2 Marks

✅ CBSE Standard Answer

Mathematical Principle:
For any triangle with sides a, b, c:
|a − b| < c < (a + b).
That is, the third side must be greater than the difference of the two sides and less than their sum.

Calculation:
Sum of two sides = 15 + 12 = 27 cm.
Difference of two sides = 15 − 12 = 3 cm.

Conclusion: The length of the third side must be greater than 3 cm and less than 27 cm (i.e. between 3 cm and 27 cm).

CBSE Marking Scheme 2026-27

Difference bound: 15 − 12 = 3 cm1 Mark

Sum bound: 15 + 12 = 27 cm with final statement1 Mark

Exercise 6.5 (Pythagoras Theorem)

Q1

PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

3 Marks

✅ CBSE Standard Answer

Step 1: Identify sides of right-angled triangle:
Right angle is at vertex P ⇒ QR is the hypotenuse.
Given legs: PQ = 10 cm, PR = 24 cm.

Step 2: Apply Pythagoras Theorem:
QR² = PQ² + PR²
QR² = 10² + 24² = 100 + 576 = 676
QR = √(676) = 26 cm.

Conclusion: The length of QR is 26 cm.

CBSE Marking Scheme 2026-27

Pythagoras formula: QR² = PQ² + PR²1 Mark

10² + 24² = 6761 Mark

QR = 26 cm with unit1 Mark

Q2

A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

3 Marks

✅ CBSE Standard Answer

Step 1: Geometry setup:
The wall and the horizontal ground meet at 90°.
Ladder acts as hypotenuse: h = 15 m.
Height of window: p = 12 m.
Distance of foot from wall: b = a.

Step 2: Pythagoras Theorem:
a² + 12² = 15²
a² + 144 = 225
a² = 225 − 144 = 81
a = √(81) = 9 m.

Conclusion: The distance of the foot of the ladder from the wall is 9 metres.

CBSE Marking Scheme 2026-27

Pythagoras setup: a² + 12² = 15²1 Mark

a² = 81 evaluation1 Mark

Final answer a = 9 m with unit1 Mark

Q3

A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

3 Marks

✅ CBSE Standard Answer

Step 1: Right-angled triangle model:
Let BC = 5 m be the standing trunk (perpendicular).
Distance from trunk base to top AB = 12 m (base).
The broken fallen trunk AC is the hypotenuse.

Step 2: Calculate hypotenuse AC:
AC² = AB² + BC² = 12² + 5² = 144 + 25 = 169
AC = √(169) = 13 m.

Step 3: Calculate original tree height:
Original height = BC + AC = 5 m + 13 m = 18 m.

Conclusion: The original height of the tree was 18 metres.

CBSE Marking Scheme 2026-27

Pythagoras theorem: AC = 13 m1.5 Marks

Adding 5 m + 13 m = 18 m with conclusion1.5 Marks

Q4

Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

3 Marks

✅ CBSE Standard Answer

Step 1: Find breadth using Pythagoras theorem:
In a rectangle, adjacent sides meet at 90°. Diagonal = 41 cm, Length = 40 cm.
Breadth² + 40² = 41²
b² + 1600 = 1681 ⇒ b² = 1681 − 1600 = 81 ⇒ b = √(81) = 9 cm.

Step 2: Calculate perimeter:
Perimeter = 2(l + b) = 2(40 + 9) = 2 × 49 = 98 cm.

Conclusion: The perimeter of the rectangle is 98 cm.

CBSE Marking Scheme 2026-27

Pythagoras theorem to find breadth b = 9 cm1.5 Marks

Perimeter = 2(40 + 9) = 98 cm1.5 Marks

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

HOTS: Perimeter of Rhombus via Diagonals

Problem: The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Properties of Rhombus:
1. Diagonals bisect each other at right angles (90°).
Half of diagonal 1 = 16/2 = 8 cm.
Half of diagonal 2 = 30/2 = 15 cm.
2. In right-angled triangle formed by half-diagonals and side s:
s² = 8² + 15² = 64 + 225 = 289 ⇒ s = √(289) = 17 cm.
3. Perimeter of rhombus = 4 × s = 4 × 17 = 68 cm.

7

Chapter 7: Comparing Quantities

NCERT Exercise 7.1 to 7.2 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Commercial Mathematics Formulas

Ratio: A comparison of two quantities of the same kind by division (a : b = a/b). Both quantities must have the same unit.
Percentage: Ratios with denominator 100.
- Fraction/Decimal to % ⇒ Multiply by 100%.
- % to Fraction/Decimal ⇒ Divide by 100.
Percentage Change: Percentage Increase / Decrease = (Amount of change)/(Original value) × 100%
Profit and Loss:
- When SP > CP ⇒ Profit = SP - CP; Profit % = Profit/CP × 100%.
- When CP > SP ⇒ Loss = CP - SP; Loss % = Loss/CP × 100%.
Simple Interest: SI = (P × R × T)/100, Total Amount A = P + SI where P = Principal (sum borrowed/invested), R = Rate of interest per annum, T = Time in years.

Exercise 7.1 (Ratios & Percentages)

Q1

Find the ratio of:
(a) ₹5 to 50 paise
(b) 15 kg to 210 g
(c) 9 m to 27 cm
(d) 30 days to 36 hours

3 Marks

✅ CBSE Standard Answer

(a) ₹5 to 50 paise:
Convert to paise: ₹5 = 5 × 100 = 500 paise.
Ratio = 500/50 = 10/1 = 10 : 1.

(b) 15 kg to 210 g:
Convert to grams: 15 kg = 15 × 1000 = 15000 g.
Ratio = 15000/210 = 1500/21 = 500/7 = 500 : 7.

(c) 9 m to 27 cm:
Convert to cm: 9 m = 900 cm.
Ratio = 900/27 = 100/3 = 100 : 3.

(d) 30 days to 36 hours:
Convert to hours: 30 days = 30 × 24 = 720 hours.
Ratio = 720/36 = 20/1 = 20 : 1.

CBSE Marking Scheme 2026-27

Unit conversions done accurately1.5 Marks

All ratios simplified to lowest terms1.5 Marks

Q2

In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?

2 Marks

✅ CBSE Standard Answer

Unitary Method:
Computers required for 6 students = 3.
Computers required for 1 student = 3/6 = 1/2.

Computers required for 24 students = 24 × 1/2 = 12 computers.

Conclusion: Exactly 12 computers will be needed for 24 students.

CBSE Marking Scheme 2026-27

Rate for 1 student = 3/6 = 1/21 Mark

Total computers = 24 × 1/2 = 121 Mark

Q3

Convert the given fractional numbers to per cents:
(a) 1/8
(b) 5/4
(c) 3/40
(d) 2/7

2 Marks

✅ CBSE Standard Answer

(a) 1/8 × 100% = 25/2% = 12.5% = 12 1/2%
(b) 5/4 × 100% = 5 × 25% = 125%
(c) 3/40 × 100% = 15/2% = 7.5%
(d) 2/7 × 100% = 200/7% = 28 4/7%

CBSE Marking Scheme 2026-27

Each fraction converted to %0.5 Mark each

Q4

Find:
(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of ₹2500
(d) 75% of 1 kg

2 Marks

✅ CBSE Standard Answer

(a) 15% of 250 = 15/100 × 250 = (15 × 5)/2 = 75/2 = 37.5
(b) 1% of 1 hour = 1/100 × 60 min = (60 × 60)/100 sec = 36 seconds
(c) 20% of ₹2500 = 20/100 × 2500 = 20 × 25 = ₹500
(d) 75% of 1 kg = 75/100 × 1000 g = 75 × 10 = 750 g = 0.75 kg

CBSE Marking Scheme 2026-27

Each correct percentage evaluation0.5 Mark each

Exercise 7.2 (Profit & Loss, Simple Interest)

Q1

Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case:
(a) Gardening shears bought for ₹250 and sold for ₹325.
(b) A refrigerator bought for ₹12,000 and sold at ₹13,500.
(c) A cupboard bought for ₹2,500 and sold at ₹3,000.
(d) A skirt bought for ₹250 and sold at ₹150.

4 Marks

✅ CBSE Standard Answer

(a) Gardening shears:
CP = ₹250, SP = ₹325. Since SP > CP, there is a profit.
Profit = 325 − 250 = ₹75.
Profit % = 75/250 × 100 = 3/10 × 100 = 30%.

(b) Refrigerator:
CP = ₹12,000, SP = ₹13,500. Profit.
Profit = 13500 − 12000 = ₹1,500.
Profit % = 1500/12000 × 100 = 150/12 = 12.5%.

(c) Cupboard:
CP = ₹2,500, SP = ₹3,000. Profit.
Profit = 3000 − 2500 = ₹500.
Profit % = 500/2500 × 100 = 1/5 × 100 = 20%.

(d) Skirt:
CP = ₹250, SP = ₹150. Since CP > SP, there is a loss.
Loss = 250 − 150 = ₹100.
Loss % = 100/250 × 100 = 2/5 × 100 = 40%.

CBSE Marking Scheme 2026-27

Each part: Profit/Loss calculation (0.5 Mark) + % calculation (0.5 Mark)1 Mark each (4 Marks total)

Q2

Find the amount to be paid at the end of 3 years in each case:
(a) Principal = ₹1,200 at 12% p.a.
(b) Principal = ₹7,500 at 5% p.a.

3 Marks

✅ CBSE Standard Answer

(a) P = ₹1,200, R = 12% p.a., T = 3 years:
SI = (P × R × T)/100 = (1200 × 12 × 3)/100 = 12 × 36 = ₹432.
Amount = P + SI = ₹1200 + ₹432 = ₹1,632.

(b) P = ₹7,500, R = 5% p.a., T = 3 years:
SI = (P × R × T)/100 = (7500 × 5 × 3)/100 = 75 × 15 = ₹1,125.
Amount = P + SI = ₹7500 + ₹1125 = ₹8,625.

CBSE Marking Scheme 2026-27

Part (a): SI = ₹432, Amount = ₹1,6321.5 Marks

Part (b): SI = ₹1,125, Amount = ₹8,6251.5 Marks

Q3

What rate gives ₹280 as interest on a sum of ₹56,000 in 2 years?

2 Marks

✅ CBSE Standard Answer

Given:
SI = ₹280, P = ₹56,000, T = 2 years.

Formula for Rate:
R = (SI × 100)/(P × T) = (280 × 100)/(56000 × 2) = 28000/112000 = 28/112 = 1/4% = 0.25%.

Conclusion: The rate of interest is 0.25% (or ¼%) per annum.

CBSE Marking Scheme 2026-27

Formula R = (SI × 100) / (P × T)1 Mark

R = 0.25% p.a. with statement1 Mark

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

Case Study: Population Growth & Depreciation

Scenario: The population of a city decreased from 25,000 to 24,500. (i) Find the percentage decrease in population. (ii) If the population decreases by the same percentage next year, what will the new population be?

(i) Percentage decrease:
Original population = 25,000.
Decrease = 25,000 − 24,500 = 500.
Percentage Decrease = 500/25000 × 100% = 50/25% = 2%.

(ii) Next year's population:
Decrease next year = 2% of 24,500 = 2/100 × 24500 = 490.
New population = 24,500 − 490 = 24,010.

8

Chapter 8: Rational Numbers

NCERT Exercise 8.1 to 8.2 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Properties Summary

Rational Number: A number that can be written in the form p/q, where p and q are integers and q ≠ 0.
- Positive: Both p and q have the same sign (e.g. 2/3, (-5)/(-7)).
- Negative: p and q have opposite signs (e.g. (-3)/8, 5/(-9)).
- 0 is a rational number since 0 = 0/1.
Standard Form: A rational number p/q is in standard form if q > 0 and p, q have no common factor other than 1.
Equivalent Rational Numbers: Obtained by multiplying or dividing numerator and denominator by the same non-zero integer: p/q = (p × m)/(q × m)
Density Property: Between any two rational numbers, there exist infinitely many rational numbers.
Operations on Rational Numbers:
- Addition/Subtraction: Take LCM of denominators and adjust numerators accordingly.
- Multiplication: a/b × c/d = (a × c)/(b × d).
- Division: a/b ÷ c/d = a/b × d/c (multiplication by reciprocal of divisor).

Exercise 8.1

Q1

List five rational numbers between:
(i) −1 and 0
(ii) −2 and −1
(iii) −4/5 and −2/3
(iv) −1/2 and 2/3

4 Marks

✅ CBSE Standard Answer

(i) Between −1 and 0:
Multiply numerator and denominator by 5 + 1 = 6:
−1 = -6/6, 0 = 0/6.
Five rational numbers: -5/6, -4/6, -3/6, -2/6, -1/6
(Simplified: -5/6, -2/3, -1/2, -1/3, -1/6).

(ii) Between −2 and −1:
Multiply by 6:
−2 = -12/6, −1 = -6/6.
Five rational numbers: -11/6, -10/6, -9/6, -8/6, -7/6.

(iii) Between −4/5 and −2/3:
LCM of 5 and 3 is 15. Make denominators equal to 15 × 3 = 45:
-4/5 = -36/45, -2/3 = -30/45.
Five rational numbers: -35/45, -34/45, -33/45, -32/45, -31/45.

(iv) Between −1/2 and 2/3:
LCM of 2 and 3 is 6. Multiply by 2 ⇒ denominator 12:
-1/2 = -6/12, 2/3 = 8/12.
Five rational numbers: -5/12, -4/12, 0, 1/12, 5/12.

CBSE Marking Scheme 2026-27

Each part: Finding suitable common denominator & listing 5 rationals1 Mark each (4 Marks total)

Q2

Rewrite the following rational numbers in the simplest (standard) form:
(i) −8/6
(ii) 25/45
(iii) −44/72
(iv) −8/10

2 Marks

✅ CBSE Standard Answer

Divide numerator and denominator by their HCF:

(i) -8/6:
HCF of 8 and 6 is 2: -(8 ÷ 2)/(6 ÷ 2) = -4/3.

(ii) 25/45:
HCF of 25 and 45 is 5: (25 ÷ 5)/(45 ÷ 5) = 5/9.

(iii) -44/72:
HCF of 44 and 72 is 4: -(44 ÷ 4)/(72 ÷ 4) = -11/18.

(iv) -8/10:
HCF of 8 and 10 is 2: -(8 ÷ 2)/(10 ÷ 2) = -4/5.

CBSE Marking Scheme 2026-27

Each correct reduction to simplest standard form0.5 Mark each

Q3

Which is greater in each of the following?
(i) 2/3, 5/2
(ii) −5/6, −4/3
(iii) −3/4, 2/−3
(iv) −3 2/7, −3 4/5

3 Marks

✅ CBSE Standard Answer

(i) 2/3 and 5/2:
LCM = 6: 2/3 = 4/6, 5/2 = 15/6.
Since 15 > 4 ⇒ 5/2 > 2/3. (Also 5/2 > 1 while 2/3 < 1).

(ii) −5/6 and −4/3:
LCM = 6: -5/6, -4/3 = -8/6.
On the number line, −5 > −8. Hence, -5/6 > -4/3.

(iii) −3/4 and 2/−3:
In standard form: -3/4 and -2/3.
LCM = 12: -3/4 = -9/12, -2/3 = -8/12.
Since −8 > −9 ⇒ 2/(−3) > -3/4.

(iv) −3 2/7 and −3 4/5:
-3 2/7 = -23/7 = -115/35.
-3 4/5 = -19/5 = -133/35.
Since −115 > −133 ⇒ -3 2/7 > -3 4/5.

CBSE Marking Scheme 2026-27

Converting to common denominator and comparing numerators3 Marks

Exercise 8.2 (Operations on Rational Numbers)

Q1

Find the sum:
(i) 5/4 + (−11/4)
(ii) 5/3 + 3/5
(iii) −9/10 + 22/15
(iv) −3/−11 + 5/9
(v) −8/19 + (−2/57)
(vi) −2⅓ + 4⅗

3 Marks

✅ CBSE Standard Answer

(i) 5/4 + (-11/4) = (5 − 11)/4 = (−6)/4 = -3/2 = -1 1/2

(ii) 5/3 + 3/5: LCM = 15.
= (5 × 5 + 3 × 3)/15 = (25 + 9)/15 = 34/15 = 2 4/15.

(iii) -9/10 + 22/15: LCM = 30.
= (−9 × 3 + 22 × 2)/30 = (−27 + 44)/30 = 17/30.

(iv) (−3)/(−11) + 5/9 = 3/11 + 5/9: LCM = 99.
= (3 × 9 + 5 × 11)/99 = (27 + 55)/99 = 82/99.

(v) -8/19 + (-2/57): LCM = 57.
= (−8 × 3 − 2)/57 = (−24 − 2)/57 = -26/57.

(vi) -2 1/3 + 4 3/5 = -7/3 + 23/5: LCM = 15.
= (−7 × 5 + 23 × 3)/15 = (−35 + 69)/15 = 34/15 = 2 4/15.

CBSE Marking Scheme 2026-27

Each correct fraction addition with LCM steps0.5 Mark each (3 Marks total)

Q2

Find the product:
(i) 9/2 × (−7/4)
(ii) 3/10 × (−9)
(iii) −6/5 × 9/11
(iv) 3/7 × (−2/5)
(v) 3/−5 × (−5/3)

2.5 Marks

✅ CBSE Standard Answer

(i) 9/2 × (-7/4) = (9 × (−7))/(2 × 4) = -63/8 = -7 7/8

(ii) 3/10 × (−9) = (3 × (−9))/10 = -27/10 = -2 7/10

(iii) -6/5 × 9/11 = (−6 × 9)/(5 × 11) = -54/55

(iv) 3/7 × (-2/5) = (3 × (−2))/(7 × 5) = -6/35

(v) 3/(−5) × (-5/3) = (−3)/5 × (−5)/3 = ((−3) × (−5))/(5 × 3) = 15/15 = 1

CBSE Marking Scheme 2026-27

Multiplying numerators and denominators with correct signs0.5 Mark each

Q3

Find the value of:
(i) (−4) ÷ 2/3
(ii) −3/5 ÷ 2
(iii) −4/5 ÷ (−3)
(iv) −1/8 ÷ 3/4
(v) −2/13 ÷ 1/7
(vi) −7/12 ÷ (−2/13)

3 Marks

✅ CBSE Standard Answer

Multiply the first number by the reciprocal of the divisor:

(i) (−4) ÷ 2/3 = (−4) × 3/2 = −2 × 3 = −6

(ii) -3/5 ÷ 2 = -3/5 × 1/2 = -3/10

(iii) -4/5 ÷ (−3) = -4/5 × (-1/3) = 4/15

(iv) -1/8 ÷ 3/4 = -1/8 × 4/3 = -1/6

(v) -2/13 ÷ 1/7 = -2/13 × 7/1 = -14/13 = -1 1/13

(vi) -7/12 ÷ (-2/13) = -7/12 × (-13/2) = 91/24 = 3 19/24

CBSE Marking Scheme 2026-27

Reciprocal inversion & accurate evaluation0.5 Mark each

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

HOTS: Insertion of Rationals

Problem: Is it true that there are only 5 rational numbers between −1 and 0? Explain mathematically how many rational numbers exist between any two distinct rational numbers.

Answer: No, it is not true.

Between any two distinct rational numbers, there are infinitely many rational numbers.

Reasoning: We can multiply the numerator and denominator by any arbitrarily large integer k. For example, choosing k = 100 gives -100/100 and 0/100, yielding 99 rational numbers. Choosing k = 10,000 yields 9,999 rational numbers. By choosing k arbitrarily large, we can insert as many rational numbers as we desire. This fundamental property is called the Density Property of rational numbers.

9

Chapter 9: Perimeter and Area

NCERT Exercise 9.1 to 9.2 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Mensuration Formulas Summary

Square & Rectangle:
- Perimeter of square = 4 × side; Area of square = side².
- Perimeter of rectangle = 2(l + b); Area of rectangle = l × b.
Area of Parallelogram: Area = Base × Height = b × h
Area of Triangle: Area = 1/2 × Base × Height = 1/2 b h (Every triangle can be viewed as half of a parallelogram with the same base and height).
Circle:
- Circumference = 2π r = π d (where π ≈ 22/7 or 3.14).
- Area of circle = π r².
- Area of circular ring / path = π (R² - r²) = π(R + r)(R - r).
Unit Conversions: 1 cm² = 100 mm², 1 m² = 10,000 cm², 1 hectare = 10,000 m².

Exercise 9.1 (Parallelograms & Triangles)

Q1

Find the area of each of the following parallelograms:
(a) base = 7 cm, height = 4 cm
(b) base = 5 cm, height = 3 cm
(c) base = 2.5 cm, height = 3.5 cm
(d) base = 5 cm, height = 4.8 cm
(e) base = 2 cm, height = 4.4 cm

3 Marks

✅ CBSE Standard Answer

Formula: Area of parallelogram = base × height.

(a) Area = 7 cm × 4 cm = 28 cm²

(b) Area = 5 cm × 3 cm = 15 cm²

(c) Area = 2.5 cm × 3.5 cm = 8.75 cm²

(d) Area = 5 cm × 4.8 cm = 24 cm²

(e) Area = 2 cm × 4.4 cm = 8.8 cm²

CBSE Marking Scheme 2026-27

Formula stated: Base × Height0.5 Mark

Each correct evaluation with proper units (cm²)0.5 Mark each (2.5 Marks)

Q2

Find the area of each of the following triangles:
(a) base = 4 cm, height = 3 cm
(b) base = 5 cm, height = 3.2 cm
(c) base = 3 cm, height = 4 cm (right-angled)
(d) base = 3 cm, height = 2 cm (obtuse)

3 Marks

✅ CBSE Standard Answer

Formula: Area of triangle = 1/2 × base × height.

(a) Area = 1/2 × 4 × 3 = 2 × 3 = 6 cm²

(b) Area = 1/2 × 5 × 3.2 = 5 × 1.6 = 8 cm²

(c) Area = 1/2 × 3 × 4 = 3 × 2 = 6 cm²

(d) Area = 1/2 × 3 × 2 = 3 cm²

CBSE Marking Scheme 2026-27

Formula stated: ½ × Base × Height1 Mark

Each correct evaluation with unit (cm²)0.5 Mark each (2 Marks)

Q3

ΔABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of ΔABC. Also find the length of AD.

3 Marks

✅ CBSE Standard Answer

Step 1: Calculate Area of ΔABC using legs AB and AC:
Since right angle is at A, legs AB = 5 cm and AC = 12 cm serve as base and height.
Area = 1/2 × AB × AC = 1/2 × 5 × 12 = 5 × 6 = 30 cm².

Step 2: Find altitude AD using hypotenuse BC as base:
Area = 1/2 × BC × AD
30 = 1/2 × 13 × AD ⇒ AD = (30 × 2)/13 = 60/13 = 4 8/13 cm ≈ 4.62 cm.

Conclusion: Area of ΔABC is 30 cm² and length of AD is 60/13 cm (4.62 cm).

CBSE Marking Scheme 2026-27

Area = ½ × 5 × 12 = 30 cm²1.5 Marks

Equating ½ × 13 × AD = 30 ⇒ AD = 60/13 cm1.5 Marks

Exercise 9.2 (Circles: Circumference & Area)

Q1

Find the circumference of the circles with the following radius (Take π = 22/7):
(a) 14 cm
(b) 28 mm
(c) 21 cm

2 Marks

✅ CBSE Standard Answer

Formula: C = 2π r.

(a) C = 2 × 22/7 × 14 = 2 × 22 × 2 = 88 cm

(b) C = 2 × 22/7 × 28 = 2 × 22 × 4 = 176 mm

(c) C = 2 × 22/7 × 21 = 2 × 22 × 3 = 132 cm

CBSE Marking Scheme 2026-27

Formula 2πr + each evaluation2 Marks

Q2

Find the area of the following circles, given that (Take π = 22/7):
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius = 5 cm

3 Marks

✅ CBSE Standard Answer

Formula: Area = π r².

(a) r = 14 mm:
Area = 22/7 × 14 × 14 = 22 × 2 × 14 = 616 mm².

(b) d = 49 m ⇒ r = 49/2 m:
Area = 22/7 × 49/2 × 49/2 = (11 × 7 × 49)/2 = 3773/2 = 1886.5 m².

(c) r = 5 cm:
Area = 22/7 × 5 × 5 = 550/7 = 78 4/7 cm² ≈ 78.57 cm².

CBSE Marking Scheme 2026-27

Each correct evaluation with formula and units1 Mark each

Q3

A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹4 per metre. (Take π = 22/7)

3 Marks

✅ CBSE Standard Answer

Step 1: Calculate circumference of garden:
Diameter d = 21 m ⇒ r = 21/2 m.
C = 2π r = 2 × 22/7 × 21/2 = 22 × 3 = 66 m.

Step 2: Total length of rope for 2 rounds:
Length of rope = 2 × C = 2 × 66 = 132 m.

Step 3: Calculate total cost:
Cost per metre = ₹4.
Total Cost = 132 × ₹4 = ₹528.

Conclusion: He needs 132 metres of rope and the total cost is ₹528.

CBSE Marking Scheme 2026-27

Circumference C = 66 m1 Mark

Rope length for 2 rounds = 132 m1 Mark

Total cost = 132 × 4 = ₹5281 Mark

Q4

From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

3 Marks

✅ CBSE Standard Answer

Step 1: Area of outer circular sheet (R = 4 cm):
Area₁ = π R² = 3.14 × 4² = 3.14 × 16 = 50.24 cm².

Step 2: Area of removed circle (r = 3 cm):
Area₂ = π r² = 3.14 × 3² = 3.14 × 9 = 28.26 cm².

Step 3: Area of remaining sheet:
Area = Area₁ − Area₂ = 50.24 − 28.26 = 21.98 cm².

Conclusion: The area of the remaining sheet is 21.98 cm².

CBSE Marking Scheme 2026-27

Outer area = 50.24 cm²1 Mark

Inner area = 28.26 cm²1 Mark

Difference = 21.98 cm²1 Mark

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

HOTS: Wire Rebent from Square to Rectangle

Problem: A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 12 cm, find its breadth. Which figure encloses more area and by how much?

1. Length of wire (Perimeter):
Perimeter of square = 4 × side = 4 × 10 = 40 cm.
When rebent into a rectangle, perimeter is still 40 cm:
2(l + b) = 40 ⇒ 2(12 + b) = 40 ⇒ 12 + b = 20 ⇒ b = 20 − 12 = 8 cm.

2. Compare Areas:
Area of square = side² = 10 × 10 = 100 cm².
Area of rectangle = l × b = 12 × 8 = 96 cm².
Difference = 100 − 96 = 4 cm².
Conclusion: The square encloses more area by 4 cm².

10

Chapter 10: Algebraic Expressions

NCERT Exercise 10.1 to 10.2 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Algebraic Rules Summary

Terms and Factors: An algebraic expression is formed by terms. For example, in 4x² - 3xy, the terms are 4x² and -3xy. The factors of 4x² are 4, x, x and of -3xy are -3, x, y.
Coefficients: The numerical factor of a term is called its numerical coefficient (e.g. In -5xy, the coefficient is -5).
Like and Unlike Terms:
- Like terms: Terms having the same algebraic factors with identical exponents (e.g. 7x and -12x, 4x^2y and -9x^2y).
- Unlike terms: Terms having different algebraic factors (e.g. 7x and 7y, 4x² and 4x).
Classification of Expressions:
- Monomial: Exactly one term (e.g. 7xy, -5).
- Binomial: Exactly two terms (e.g. a + b, 4x² - 3).
- Trinomial: Exactly three terms (e.g. a + b + c, x² + 5x + 6).
- Polynomial: An expression with one or more terms having non-negative integral exponents.
Addition and Subtraction: Only like terms can be combined (added or subtracted) by adding their numerical coefficients.
Evaluating Expressions: Substitute the given numerical values of variables into the expression and apply BODMAS.

Exercise 10.1

Q1

Identify the terms and their factors in the following expressions:
(a) x − 3
(b) 1 + x + x²
(c) y − y³
(d) 5xy² + 7x²y
(e) −ab + 2b² − 3a²

3 Marks

✅ CBSE Standard Answer

(a) x − 3:
• Term: x ⇒ Factors: x
• Term: −3 ⇒ Factors: −3 (or −1, 3)

(b) 1 + x + x²:
• Terms: 1, x, x²
• Factors of x²: x, x

(c) y − y³:
• Terms: y and −y³
• Factors of −y³: −1, y, y, y

(d) 5xy² + 7x²y:
• Term: 5xy² ⇒ Factors: 5, x, y, y
• Term: 7x^2y ⇒ Factors: 7, x, x, y

(e) −ab + 2b² − 3a²:
• Term −ab: −1, a, b
• Term 2b²: 2, b, b
• Term −3a²: −3, a, a

CBSE Marking Scheme 2026-27

Identifying terms and breaking into factors3 Marks

Q2

Classify into monomials, binomials and trinomials:
(i) 4y − 7z
(ii) y²
(iii) x + y − xy
(iv) 100
(v) ab − a − b
(vi) 5 − 3t
(vii) 4p²q − 4pq²
(viii) 7mn
(ix) z² − 3z + 8

3 Marks

✅ CBSE Standard Answer

• Monomials (1 term): (ii) y², (iv) 100, (viii) 7mn.
• Binomials (2 terms): (i) 4y − 7z, (vi) 5 − 3t, (vii) 4p^2q − 4pq².
• Trinomials (3 terms): (iii) x + y − xy, (v) ab − a − b, (ix) z² − 3z + 8.

CBSE Marking Scheme 2026-27

Each correct classification0.33 Mark each (3 Marks total)

Q3

Simplify by combining like terms:
(i) 21b − 32 + 7b − 20b
(ii) −z² + 13z² − 5z + 7z³ − 15z
(iii) p − (p − q) − q − (q − p)
(iv) 3a − 2b − ab − (a − b + ab) + 3ab + b − a

4 Marks

✅ CBSE Standard Answer

(i) 21b − 32 + 7b − 20b:
Group terms with variable b:
= (21 + 7 − 20)b − 32 = (28 − 20)b − 32 = 8b − 32.

(ii) −z² + 13z² − 5z + 7z³ − 15z:
Arrange by descending powers of z:
= 7z³ + (−1 + 13)z² + (−5 − 15)z = 7z³ + 12z² − 20z.

(iii) p − (p − q) − q − (q − p):
Open parentheses with sign rules:
= p − p + q − q − q + p = (p − p + p) + (q − q − q) = p − q.

(iv) 3a − 2b − ab − (a − b + ab) + 3ab + b − a:
Open brackets: = 3a − 2b − ab − a + b − ab + 3ab + b − a.
Group like terms:
• Terms in a: (3 − 1 − 1)a = a
• Terms in b: (−2 + 1 + 1)b = 0b = 0
• Terms in ab: (−1 − 1 + 3)ab = ab
= a + ab.

CBSE Marking Scheme 2026-27

Each part: Grouping like terms and simplifying1 Mark each (4 Marks total)

Exercise 10.2 (Evaluation of Expressions)

Q1

If m = 2, find the value of:
(i) m − 2
(ii) 3m − 5
(iii) 9 − 5m
(iv) 3m² − 2m − 7
(v) 5m/2 − 4

3 Marks

✅ CBSE Standard Answer

Substitute m = 2:

(i) m − 2 = 2 − 2 = 0

(ii) 3m − 5 = 3(2) − 5 = 6 − 5 = 1

(iii) 9 − 5m = 9 − 5(2) = 9 − 10 = −1

(iv) 3m² − 2m − 7 = 3(2)² − 2(2) − 7 = 3(4) − 4 − 7 = 12 − 11 = 1

(v) 5m/2 − 4 = 5(2)/2 − 4 = 5 − 4 = 1

CBSE Marking Scheme 2026-27

Each correct numerical evaluation0.6 Mark each (3 Marks total)

Q2

If a = 2, b = −2, find the value of:
(i) a² + b²
(ii) a² + ab + b²
(iii) a² − b²

3 Marks

✅ CBSE Standard Answer

Given: a = 2, b = −2.

(i) a² + b² = (2)² + (−2)² = 4 + 4 = 8

(ii) a² + ab + b² = (2)² + (2)(−2) + (−2)² = 4 − 4 + 4 = 4

(iii) a² − b² = (2)² − (−2)² = 4 − 4 = 0

CBSE Marking Scheme 2026-27

Each part evaluated with square of negative numbers1 Mark each (3 Marks total)

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

HOTS: Constant Determination

Problem: What should be the value of a if the value of 2x² + x − a equals 5, when x = 0?

Given equation condition: 2x² + x − a = 5 when x = 0.

Substitute x = 0 into the expression:
2(0)² + 0 − a = 5
0 + 0 − a = 5 ⇒ −a = 5 ⇒ a = −5.

Conclusion: The value of a is −5.

11

Chapter 11: Exponents and Powers

NCERT Exercise 11.1 to 11.3 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Laws of Exponents Summary

Exponential Form: Repeated multiplication written concisely: a × a × a ... (n times) = aⁿ, where a is the base and n is the exponent.
Laws of Exponents (for non-zero integers a, b and whole numbers m, n):
- Multiplying powers with same base: aᵐ × aⁿ = aᵐ⁺ⁿ.
- Dividing powers with same base: aᵐ ÷ aⁿ = aᵐ⁻ⁿ (m > n).
- Power of a power: (aᵐ)ⁿ = aᵐⁿ.
- Multiplying powers with same exponents: aᵐ × bᵐ = (a × b)ᵐ = (ab)ᵐ.
- Dividing powers with same exponents: aᵐ ÷ bᵐ = (a/b)ᵐ.
- Zero Exponent: a⁰ = 1.
- Powers of -1: (-1)eveⁿ = +1, (-1)odd = -1.
Standard Form (Scientific Notation): Any number can be expressed as k × 10ⁿ, where 1 ≤ k < 10 and n is an integer. Used to write very large numbers compactly.

Exercise 11.1

Q1

Find the value of:
(i) 2⁶
(ii) 9³
(iii) 11²
(iv) 5⁴

2 Marks

✅ CBSE Standard Answer

(i) 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

(ii) 9³ = 9 × 9 × 9 = 81 × 9 = 729

(iii) 11² = 11 × 11 = 121

(iv) 5⁴ = 5 × 5 × 5 × 5 = 25 × 25 = 625

CBSE Marking Scheme 2026-27

Each correct evaluation0.5 Mark each

Q2

Express each of the following as product of powers of their prime factors:
(i) 648
(ii) 405
(iii) 540
(iv) 3,600

3 Marks

✅ CBSE Standard Answer

(i) 648:
648 = 2 × 324 = 2 × 2 × 162 = 2 × 2 × 2 × 81 = 2³ × 3⁴ = 2³ × 3⁴.

(ii) 405:
405 = 5 × 81 = 5 × 3⁴ = 3⁴ × 5.

(iii) 540:
540 = 2 × 270 = 2 × 2 × 135 = 2² × 3 × 45 = 2² × 3³ × 5 = 2² × 3³ × 5.

(iv) 3,600:
3600 = 36 × 100 = (2² × 3²) × (2² × 5²) = 2⁴ × 3² × 5².

CBSE Marking Scheme 2026-27

Prime factorisation steps and final exponential notation3 Marks

Q3

Identify the greater number, wherever possible, in each of the following:
(i) 4³ or 3⁴
(ii) 5³ or 3⁵
(iii) 2⁸ or 8²
(iv) 100² or 2¹⁰⁰

3 Marks

✅ CBSE Standard Answer

(i) 4³ vs 3⁴:
4³ = 4 × 4 × 4 = 64.
3⁴ = 3 × 3 × 3 × 3 = 81.
Since 81 > 64 ⇒ 3⁴ > 4³.

(ii) 5³ vs 3⁵:
5³ = 125.
3⁵ = 243.
Since 243 > 125 ⇒ 3⁵ > 5³.

(iii) 2⁸ vs 8²:
2⁸ = 256.
8² = 64.
Since 256 > 64 ⇒ 2⁸ > 8².

(iv) 100² vs 2¹⁰⁰:
100² = 10,000.
Notice 2¹⁰ = 1024 > 1000 ⇒ 2¹⁰⁰ = (2¹⁰)¹⁰ > (10³)¹⁰ = 10³⁰.
Hence, 2¹⁰⁰ > 100² by an enormous margin.

CBSE Marking Scheme 2026-27

Each correct evaluation & comparison0.75 Mark each

Exercise 11.2 (Laws of Exponents)

Q1

Using laws of exponents, simplify and write the answer in exponential form:
(i) 3² × 3⁴ × 3⁸
(ii) 6¹⁵ ÷ 6¹⁰
(iii) a³ × a²
(iv) 7ˣ × 7²
(v) (5²)³ ÷ 5³
(vi) 2⁵ × 5⁵
(vii) (2²⁰ ÷ 2¹⁵) × 2³
(viii) 8ᵗ ÷ 8²

4 Marks

✅ CBSE Standard Answer

(i) 3² × 3⁴ × 3⁸ = 3² ⁺ ⁴ ⁺ ⁸ = 3¹⁴ [Law: aᵐ × aⁿ = aᵐ⁺ⁿ]
(ii) 6¹⁵ ÷ 6¹⁰ = 6¹⁵ ⁻ ¹⁰ = 6⁵ [Law: aᵐ ÷ aⁿ = aᵐ⁻ⁿ]
(iii) a³ × a² = a³ ⁺ ² = a⁵
(iv) 7ˣ × 7² = 7ˣ⁺²
(v) (5²)³ ÷ 5³ = 5² × ³ ÷ 5³ = 5⁶ ÷ 5³ = 5⁶ ⁻ ³ = 5³
(vi) 2⁵ × 5⁵ = (2 × 5)⁵ = 10⁵ [Law: aᵐ × bᵐ = (ab)ᵐ]
(vii) (2²⁰ ÷ 2¹⁵) × 2³ = 2²⁰ ⁻ ¹⁵ × 2³ = 2⁵ × 2³ = 2⁵ ⁺ ³ = 2⁸
(viii) 8^t ÷ 8² = 8t⁻²

CBSE Marking Scheme 2026-27

Applying exponent law formulas correctly2 Marks

All 8 simplified exponential results2 Marks

Q2

Simplify and express each of the following in exponential form:
(i) (2³ × 3⁴ × 4) / (3 × 32)
(ii) [(5²)³ × 5⁴] ÷ 5⁷
(iii) 25⁴ ÷ 5³
(iv) (3 × 7² × 11⁸) / (21 × 11³)

4 Marks

✅ CBSE Standard Answer

(i) (2³ × 3⁴ × 4)/(3 × 32):
Express in prime factors: 4 = 2², 32 = 2⁵.
= (2³ × 3⁴ × 2²)/(3¹ × 2⁵) = (2³⁺² × 3⁴)/(2⁵ × 3¹) = (2⁵ × 3⁴)/(2⁵ × 3¹) = 2⁵⁻⁵ × 3⁴⁻¹ = 2⁰ × 3³ = 1 × 3³ = 3³.

(ii) [(5²)³ × 5⁴] ÷ 5⁷:
= [5² × ³ × 5⁴] ÷ 5⁷ = [5⁶ × 5⁴] ÷ 5⁷ = 5⁶⁺⁴ ÷ 5⁷ = 5¹⁰ ÷ 5⁷ = 5¹⁰⁻⁷ = 5³.

(iii) 25⁴ ÷ 5³:
Write 25 = 5² ⇒ (5²)⁴ ÷ 5³ = 5⁸ ÷ 5³ = 5⁸⁻³ = 5⁵.

(iv) (3 × 7² × 11⁸)/(21 × 11³):
Write 21 = 3 × 7.
= (3¹ × 7² × 11⁸)/(3¹ × 7¹ × 11³) = 3¹⁻¹ × 7²⁻¹ × 11⁸⁻³ = 3⁰ × 7¹ × 11⁵ = 1 × 7 × 11⁵ = 7 × 11⁵.

CBSE Marking Scheme 2026-27

Each part: Prime base conversion & exponent law steps1 Mark each

Exercise 11.3 (Expanded Form & Scientific Notation)

Q1

Write the following numbers in expanded forms using powers of 10:
(i) 279404
(ii) 3006194
(iii) 2806196

3 Marks

✅ CBSE Standard Answer

(i) 279404:
= 2 × 10⁵ + 7 × 10⁴ + 9 × 10³ + 4 × 10² + 0 × 10¹ + 4 × 10⁰.

(ii) 3006194:
= 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰.

(iii) 2806196:
= 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 6 × 10⁰.

CBSE Marking Scheme 2026-27

Each correct expanded power-of-10 representation1 Mark each

Q2

Express the following in standard scientific form (k × 10ⁿ):
(a) 5,00,00,000
(b) 70,00,000
(c) 3,18,65,00,000
(d) 39087.8
(e) Distance between Sun and Earth = 149,600,000,000 m
(f) Speed of light in vacuum = 300,000,000 m/s

3 Marks

✅ CBSE Standard Answer

Scientific notation format: k × 10ⁿ where 1 ≤ k < 10.

(a) 5,00,00,000 = 5.0 × 10⁷
(b) 70,00,000 = 7.0 × 10⁶
(c) 3,18,65,00,000 = 3.1865 × 10⁹
(d) 39087.8 = 3.90878 × 10⁴
(e) 149,600,000,000 m = 1.496 × 10¹¹ m
(f) 300,000,000 m/s = 3.0 × 10⁸ m/s

CBSE Marking Scheme 2026-27

Each correct standard form representation0.5 Mark each (3 Marks total)

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

HOTS: Finding Unknown Exponent

Problem: Find the value of x in the equation: (2/5)³ × (2/5)⁶ = (2/5)³ˣ.

Using law aᵐ × aⁿ = aᵐ⁺ⁿ on LHS:
(2/5)³ ⁺ ⁶ = (2/5)³ˣ
(2/5)⁹ = (2/5)³ˣ.

Equating exponents since bases are identical:
3x = 9 ⇒ x = 9/3 = 3.

12

Chapter 12: Symmetry

NCERT Exercise 12.1 to 12.3 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & Symmetry Properties Summary

Line Symmetry: A figure has line symmetry if a line can be drawn dividing it into two identical halves that coincide exactly when folded along that line.
Lines of Symmetry for Regular Polygons: Each regular polygon of n sides has exactly n lines of symmetry:
- Equilateral triangle: 3 lines
- Square: 4 lines
- Regular pentagon: 5 lines
- Regular hexagon: 6 lines
- Circle: Infinitely many lines (along any diameter)
Rotational Symmetry: When a figure rotated around a fixed point (centre of rotation) by an angle ≤ 360° looks exactly the same as in its original orientation.
Order of Rotational Symmetry: The number of times a figure fits onto itself in one full 360° rotation: Order = 360°/(Angle of rotation)
Dual Symmetry: Figures possessing both line symmetry and rotational symmetry of order > 1 (e.g. Equilateral triangle, square, rectangle, circle, and English letters H, I, O, X).

Exercise 12.1 (Line Symmetry)

Q1

State the number of lines of symmetry for the following figures:
(a) An equilateral triangle
(b) An isosceles triangle
(c) A scalene triangle
(d) A square
(e) A rectangle
(f) A rhombus
(g) A parallelogram
(h) A quadrilateral
(i) A regular hexagon
(j) A circle

4 Marks

✅ CBSE Standard Answer

(a) Equilateral Triangle: 3 lines of symmetry (each altitude/median).
(b) Isosceles Triangle: 1 line of symmetry (bisector of vertex angle).
(c) Scalene Triangle: 0 lines of symmetry (none of the sides or angles are equal).
(d) Square: 4 lines of symmetry (2 diagonals + 2 perpendicular bisectors of opposite sides).
(e) Rectangle: 2 lines of symmetry (perpendicular bisectors of opposite sides).
(f) Rhombus: 2 lines of symmetry (along the two diagonals).
(g) Parallelogram: 0 lines of symmetry (in general, cannot be folded to coincide).
(h) Quadrilateral: 0 lines of symmetry (in general).
(i) Regular Hexagon: 6 lines of symmetry (3 joining opposite vertices + 3 joining midpoints of opposite sides).
(j) Circle: Infinitely many lines of symmetry (any straight line through the centre / diameter).

CBSE Marking Scheme 2026-27

Each correct geometric figure symmetry count0.4 Mark each (4 Marks total)

Q2

What letters of the English alphabet have reflectional symmetry (i.e. symmetry related to mirror reflection) about:
(a) a vertical mirror
(b) a horizontal mirror
(c) both horizontal and vertical mirrors

3 Marks

✅ CBSE Standard Answer

(a) Vertical Mirror Symmetry (Left-Right reflection):
Letters: A, H, I, M, O, T, U, V, W, X, Y.

(b) Horizontal Mirror Symmetry (Top-Bottom reflection):
Letters: B, C, D, E, H, I, K, O, X.

(c) Both Horizontal and Vertical Mirrors:
Letters in the intersection of both sets: H, I, O, X.

CBSE Marking Scheme 2026-27

Part (a): Vertical symmetry letters1 Mark

Part (b): Horizontal symmetry letters1 Mark

Part (c): Both vertical & horizontal (H, I, O, X)1 Mark

Exercise 12.2 (Rotational Symmetry)

Q1

Give the order of the rotational symmetry and the angle of rotation for each of the following figures:
(a) Square
(b) Rectangle
(c) Rhombus
(d) Equilateral triangle
(e) Regular hexagon
(f) Circle

3 Marks

✅ CBSE Standard Answer

(a) Square:
• Centre of rotation: Intersection of diagonals.
• Angle of rotation: 360°/4 = 90°.
• Order of rotational symmetry = 4.

(b) Rectangle:
• Centre of rotation: Intersection of diagonals.
• Angle of rotation: 360°/2 = 180°.
• Order of rotational symmetry = 2.

(c) Rhombus:
• Centre of rotation: Intersection of diagonals.
• Angle of rotation: 360°/2 = 180°.
• Order of rotational symmetry = 2.

(d) Equilateral Triangle:
• Centre of rotation: Centroid (intersection of medians).
• Angle of rotation: 360°/3 = 120°.
• Order of rotational symmetry = 3.

(e) Regular Hexagon:
• Centre of rotation: Centre of hexagon.
• Angle of rotation: 360°/6 = 60°.
• Order of rotational symmetry = 6.

(f) Circle:
• Centre of rotation: Centre of circle.
• Angle of rotation: Any angle θ.
• Order of rotational symmetry = Infinitely many.

CBSE Marking Scheme 2026-27

Each figure: Order and angle of rotation0.5 Mark each (3 Marks total)

Q2

Can we have a rotational symmetry of order more than 1 whose angle of rotation is:
(i) 45°?
(ii) 17°?

2 Marks

✅ CBSE Standard Answer

Rule: A figure can have rotational symmetry if and only if the angle of rotation divides 360° completely without a remainder.

(i) Angle of rotation = 45°:
360°/45° = 8 (an exact integer).
Yes, we can have a rotational symmetry of order 8 (e.g. a regular octagon).

(ii) Angle of rotation = 17°:
360°/17° = 21.176... (not an integer).
No, since 360° is not divisible by 17°, a rotational symmetry of order more than 1 is not possible.

CBSE Marking Scheme 2026-27

Part (i): Yes with division 360/45 = 81 Mark

Part (ii): No with divisibility justification1 Mark

Exercise 12.3 (Dual Symmetry)

Q1

Name any two figures that have both line symmetry and rotational symmetry.

2 Marks

✅ CBSE Standard Answer

1. A Square:
• Has 4 lines of symmetry.
• Has rotational symmetry of order 4 (angle of rotation = 90°).

2. An Equilateral Triangle:
• Has 3 lines of symmetry.
• Has rotational symmetry of order 3 (angle of rotation = 120°).
(A Circle or Regular Hexagon are also valid examples).

CBSE Marking Scheme 2026-27

Each correct geometric figure with its dual symmetry properties1 Mark each

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

Alphabet Symmetry Analysis

Problem: Name any two letters of the English alphabet that possess: (i) Both line symmetry and rotational symmetry, (ii) Only rotational symmetry of order > 1 but no line symmetry.

(i) Both line and rotational symmetry:
Letters H, I, O, X.
• Letter H has 2 lines of symmetry (1 vertical, 1 horizontal) and rotational symmetry of order 2 (rotates by 180°).
• Letter O has infinitely many lines of symmetry and infinite rotational symmetry.

(ii) Only rotational symmetry (no line symmetry):
Letters S, Z, N.
• Letter S rotates by 180° onto itself (order 2), but cannot be folded along any line to coincide.
• Letter Z rotates by 180° (order 2), but has no line of symmetry.

13

Chapter 13: Visualising Solid Shapes

NCERT Exercise 13.1 to 13.3 — Complete Solutions as per CBSE Marking Scheme 2026-27

📌 Key Concepts & 3-D Solid Properties Summary

3-Dimensional Shapes (Solids): Figures with length, breadth, and height/depth (e.g. Cube, Cuboid, Cylinder, Cone, Sphere, Prism, Pyramid).
Polyhedrons: Solids formed entirely of polygonal planar regions called Faces (F). The line segments where faces meet are Edges (E). The points where edges meet are Vertices (V).
Euler’s Formula: For any convex polyhedron: F + V - E = 2
Net of a Solid: A 2-D outline skeleton on paper that can be folded along its edges to construct the 3-D solid shape.
Visualising Cross-Sections:
- Horizontal cut (parallel to base): Slicing through the height.
- Vertical cut (perpendicular to base): Slicing through the cross-axis.

Exercise 13.1 (Nets of 3-D Solids)

Q1

Identify the nets which can be used to make cubes (cut out copies of the nets and try it). How many faces does a cube have?

2 Marks

✅ CBSE Standard Answer

Properties of a Cube Net:
• A cube has exactly 6 identical square faces, 12 edges, and 8 vertices.
• In a valid cube net, no two folded squares overlap, and opposite faces never share a common edge in the net.
• Standard valid nets include the "T-shape" (1-4-1 pattern), the cross-shape, and the step patterns (2-3-1 pattern).

CBSE Marking Scheme 2026-27

Stating cube has 6 faces1 Mark

Identifying valid non-overlapping foldable nets1 Mark

Q2

Name the solid that each net can make:
(a) A net with 2 identical circular bases and 1 rectangular surface
(b) A net with 1 circular base and 1 sector of a circle
(c) A net with 1 square base and 4 triangular faces
(d) A net with 2 triangular bases and 3 rectangular faces

3 Marks

✅ CBSE Standard Answer

(a) 2 circular bases + 1 rectangular lateral face: Cylinder.

(b) 1 circular base + 1 sector: Cone.

(c) 1 square base + 4 triangular faces meeting at apex: Square Pyramid.

(d) 2 triangular bases + 3 rectangular faces: Triangular Prism.

CBSE Marking Scheme 2026-27

Each correctly identified 3-D solid0.75 Mark each (3 Marks total)

Exercise 13.2 (Isometric Sketches & Dimension Combining)

Q1

Three cubes each with 2 cm edge are placed side by side to form a cuboid. What would be the dimensions (length, breadth, and height) of this cuboid?

2 Marks

✅ CBSE Standard Answer

Geometry Analysis:
Edge of each cube = 2 cm.
When 3 cubes are placed side by side in a row:
• Length (l) = 2 + 2 + 2 = 6 cm.
• Breadth (b) remains unchanged = 2 cm.
• Height (h) remains unchanged = 2 cm.

Conclusion: The dimensions of the resulting cuboid are 6 cm × 2 cm × 2 cm.

CBSE Marking Scheme 2026-27

Length = 2 + 2 + 2 = 6 cm1 Mark

Breadth = 2 cm, Height = 2 cm1 Mark

Exercise 13.3 (Cross-Sections of Solids)

Q1

What cross-section do you get when you give a: (i) vertical cut, (ii) horizontal cut to the following solids?
(a) A brick
(b) A round apple
(c) A die
(d) A circular pipe
(e) An ice cream cone

4 Marks

✅ CBSE Standard Answer

(a) A brick (Cuboid):
(i) Vertical cut: Rectangle
(ii) Horizontal cut: Rectangle

(b) A round apple (Sphere):
(i) Vertical cut: Circle
(ii) Horizontal cut: Circle

(c) A die (Cube):
(i) Vertical cut: Square
(ii) Horizontal cut: Square

(d) A circular pipe (Cylinder):
(i) Vertical cut (along axis): Rectangle
(ii) Horizontal cut (parallel to base): Circular ring (or Circle)

(e) An ice cream cone (Cone):
(i) Vertical cut (through apex): Isosceles Triangle
(ii) Horizontal cut (parallel to base): Circle

CBSE Marking Scheme 2026-27

Each solid: Vertical and horizontal cut cross-section shape0.8 Mark each (4 Marks total)

Q2

Verify Euler’s formula (F + V − E = 2) for:
(a) A Cuboid
(b) A Triangular Prism
(c) A Square Pyramid

3 Marks

✅ CBSE Standard Answer

Euler's formula: F + V − E = 2.

(a) Cuboid:
• Faces (F) = 6
• Vertices (V) = 8
• Edges (E) = 12
Verification: F + V − E = 6 + 8 − 12 = 14 − 12 = 2. Verified!

(b) Triangular Prism:
• Faces (F) = 5 (2 triangular + 3 rectangular)
• Vertices (V) = 6
• Edges (E) = 9
Verification: F + V − E = 5 + 6 − 9 = 11 − 9 = 2. Verified!

(c) Square Pyramid:
• Faces (F) = 5 (1 square + 4 triangular)
• Vertices (V) = 5 (1 apex + 4 base corners)
• Edges (E) = 8 (4 base edges + 4 slant edges)
Verification: F + V − E = 5 + 5 − 8 = 10 − 8 = 2. Verified!

CBSE Marking Scheme 2026-27

Each solid: F, V, E count and verification of F + V − E = 21 Mark each

🎯 Competency-Based Questions (CBQ) & HOTS CBSE Board Exam Pattern

Euler's Formula Polyhedron Test

Problem: Can a polyhedron have 10 faces, 20 edges and 15 vertices? Justify your answer using Euler’s formula.

Given parameters:
Faces, F = 10
Edges, E = 20
Vertices, V = 15.

By Euler's Formula for any polyhedron:
F + V − E = 2.

Substituting given values:
LHS = F + V − E = 10 + 15 − 20 = 25 − 20 = 5.

Since 5 ≠ 2, the given values violate Euler’s formula.

Conclusion: No, a polyhedron cannot have 10 faces, 20 edges, and 15 vertices.