CLASS 7 MATH Question Paper Set 4 – Higher + Application + Logical Reasoning – TOPICS: A Tale of three intersecting lines, Arithmetic Expressions, Expressions using letter numbers.

🚀 SET 4 (Higher + Application + Logical Reasoning)

TOPICS: A Tale of three intersecting lines, Arithmetic Expressions, Expressions using letter numbers.

SECTION – A

  1. If two sides are 15 cm and 9 cm, which value makes right triangle possible?
    (a) 6 cm
    (b) 12 cm
    (c) 18 cm
    (d) 24 cm
  2. Twin primes differ by 2 and product is 323. Find them.
  3. If (ax² + bx + c) + (2x² – 3x + 4) = (5x² + x + 10), find a, b, c.
  4. Ram’s present age is 4 times his son’s age. After 5 years, Ram will be 3 times his son’s age. Form equation and find ages.
  5. If a and b are consecutive even numbers, prove a + b is divisible by 2.
  6. Assertion–Reason
    Assertion: A triangle with sides 7 cm, 24 cm, 25 cm is right angled.
    Reason: It satisfies Pythagoras theorem.

SECTION – B

  1. Simplify:

(i) 100 – 5(6 – 2) + 3 × 4

(ii) 6x – 4y(3 – 5) – (2x + 3y) + (x – 2y)


  1. Three departments contribute:

A: 5(2a – b + c) – 4(a + b – c)
B: 3(a – 2b + 3c) – 2(2a + b – c)
C: 6(a + b – 2c) – 3(3a – b + c)

Find total and compare with (25a – 10b + 15c).


  1. Perimeter of triangle (4x + 3), (5x – 2), (3x + 7) equals perimeter of rectangle length (6x + 1) width (2x + 4). Find x.

  1. In isosceles triangle, vertex angle is 20° more than twice base angle. Find all angles.

SECTION – C (Advanced Case Study)

  1. A builder wants triangular frame. Two sides are 10 m and 14 m.

(a) Find range of third side.
(b) If third side is 22 m, is triangle possible?
(c) If perimeter must be 40 m, find third side and check validity.

Answer Key

SECTION – A

  1. (b) 12 cm
    (9² + 12² = 15²)
  2. Twin primes: 17 and 19

a = 3
b = 4
c = 6

Son = 5 years
Ram = 20 years

  1. a = 2n
    b = 2n + 2
    a + b = 4n + 2 → divisible by 2 ✔
  2. (a) Both true and correct explanation

SECTION – B

(i) 100 – 5(4) + 12
= 100 – 20 + 12
= 92

(ii) 6x – 4y(-2) – 2x – 3y + x – 2y
= 6x + 8y – 2x – 3y + x – 2y
= 5x + 3y


  1. Total = (4a – 5b + 5c)

Target = (25a – 10b + 15c)

Shortage = (21a – 5b + 10c)


  1. x = 1

  1. Base angle = 40°
    Vertex angle = 100°

SECTION – C

(a) Range:
|14 – 10| < x < 24
4 < x < 24

(b) 22 m ✔ possible

(c) Perimeter = 40
10 + 14 + x = 40
x = 16

Check: 10 + 14 > 16 ✔ valid

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