CLASS 7 – MATH QUESTION PAPER – SET 2

📘 CLASS 7 – MATH QUESTION PAPER – SET 2

ANNUAL EXAMINATION (2025–2026)
SUBJECT: MATHEMATICS

TOPICS:

Ch.2 Arithmetic Expressions
Ch.4 Expressions using Letter-Numbers
Ch.6 Number Play
Ch.7 A Tale of Three Intersecting lines
Part 2:
Ch.1 Geometric Twins
Ch.2 Operations with Integers
Ch.4 Another Peek Beyond the Point


SECTION – A

I. Choose the correct option:

  1. Two sides of a triangle are 12 cm and 8 cm. Which of the following cannot be the third side?
    (a) 7 cm
    (b) 5 cm
    (c) 21 cm
    (d) 10 cm

  1. If p and q are twin primes and p + q = 18, find the value of pq.
    (a) 63
    (b) 80
    (c) 48
    (d) 35

  1. What should be added to (4y² – 3y + 2) to get (7y² + y – 5)?
    (a) 3y² + 4y – 7
    (b) 3y² – 4y – 7
    (c) 3y² + 4y + 7
    (d) 2y² + 4y – 7

  1. Ravi’s age is twice his sister’s age minus 4 years. If sister’s age is x years, express Ravi’s age.
    (a) 2x + 4
    (b) 2x – 4
    (c) x – 4
    (d) 4x – 2

  1. The sum of two odd numbers is always:
    (a) Odd
    (b) Prime
    (c) Even
    (d) Composite

  1. Assertion (A): An exterior angle of a triangle is equal to the sum of its two interior opposite angles.
    Reason (R): The sum of angles of a triangle is 180°.

(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 and R is false
(d) A is false and R is true


SECTION – B

II. Answer the following:

  1. Simplify:

(i) 15 × 4 – 10 + 6

(ii) 5a – (3a – 2b) – b


  1. Three friends contribute money represented by algebraic expressions:

A’s contribution: 3(2p + q – 4r) – (p – 2q + r)
B’s contribution: 2(p – 3q + 2r) + 4q
C’s contribution: 5(p + q – r) – 3(2p – q + r)

a) Find the total contribution.
b) If the target amount is (12p + 5q – 6r), check whether their contribution is sufficient. If not, find how much more is required.


  1. The perimeter of a triangle with sides (3x + 2), (2x – 1), and (x + 5) is equal to the perimeter of a rectangle with length (2x + 4) and width (x + 2).
    Find the value of x.

  1. In an isosceles triangle, the vertex angle is 120°.
    Find the measure of each base angle and all angles of the triangle.

  1. Construct a triangle ABC using ASA criteria with the following measurements. Also write the steps of construction.

AB = 7 cm
∠ ABC = 60°
∠ BAC = 50°


SECTION – C

III. CASE STUDY

  1. Mohan collected three sticks of lengths 4 cm, 5 cm, and 11 cm to form a triangle.

(a) Can he form a triangle? Why?
(b) State the triangle inequality property.
(c) Check whether sticks of lengths 6 cm, 8 cm, and 10 cm can form a triangle. Justify your answer.

📘 ANSWER KEY – SET 2

Subject: Mathematics


SECTION – A

1.

Two sides = 12 cm and 8 cm

For triangle:
|12 − 8| < third side < 12 + 8
4 < third side < 20

21 cm is not possible.

Correct Answer: (c) 21 cm


2.

Twin primes whose sum is 18 → 5 and 7 (sum 12 ❌), 11 and 13 (sum 24 ❌)

No twin primes add up to 18.

No correct option (Question data inconsistent)


3.

Required expression =
(7y² + y – 5) – (4y² – 3y + 2)

= 3y² + 4y – 7

Correct Answer: (a) 3y² + 4y – 7


4.

Ravi’s age = 2x – 4

Correct Answer: (b) 2x – 4


5.

Odd + Odd = Even

Correct Answer: (c) Even


6.

Both statements are true and Reason explains Assertion.

Correct Answer: (a)


SECTION – B

7. Simplify

(i)

15 × 4 – 10 + 6
= 60 – 10 + 6
= 56

(ii)

5a – (3a – 2b) – b
= 5a – 3a + 2b – b
= 2a + b

✅ Answers:
(i) 56
(ii) 2a + b


8. Contributions

A = 3(2p + q – 4r) – (p – 2q + r)
= 6p + 3q – 12r – p + 2q – r
= 5p + 5q – 13r

B = 2(p – 3q + 2r) + 4q
= 2p – 6q + 4r + 4q
= 2p – 2q + 4r

C = 5(p + q – r) – 3(2p – q + r)
= 5p + 5q – 5r – 6p + 3q – 3r
= –p + 8q – 8r

Total Contribution:

(5p + 5q – 13r)

  • (2p – 2q + 4r)
  • (–p + 8q – 8r)

= 6p + 11q – 17r


Target = 12p + 5q – 6r

Difference =
(12p + 5q – 6r) – (6p + 11q – 17r)

= 6p – 6q + 11r

❌ Not sufficient

They need 6p – 6q + 11r more


9. Perimeter Equality

Triangle perimeter:
(3x + 2) + (2x – 1) + (x + 5)
= 6x + 6

Rectangle perimeter:
2(2x + 4 + x + 2)
= 2(3x + 6)
= 6x + 12

Equating:
6x + 6 = 6x + 12

6 = 12 ❌

❗ No solution (Question inconsistent)


10.

Vertex angle = 120°

Remaining angles = 180 – 120 = 60°

Each base angle = 60 ÷ 2 = 30°

✅ Angles are:
30°, 30°, 120°


11. Construction Steps (ASA)

  1. Draw AB = 7 cm
  2. At B construct angle 60°
  3. At A construct angle 50°
  4. Extend rays to meet at C
  5. Join AC and BC

Triangle ABC constructed.


SECTION – C

12. Case Study

(a) 4 + 5 = 9 < 11
Triangle cannot be formed.

(b) Triangle Inequality Property:
Sum of any two sides of a triangle must be greater than the third side.

(c) Check 6, 8, 10

6 + 8 = 14 > 10
8 + 10 = 18 > 6
6 + 10 = 16 > 8

✅ Triangle can be formed.

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