Answer key- class 7 annual practice paper

 Answer key

SECTION A

1. 3x − 5 = 2x + 7 ⇒ 3x − 2x = 7 + 5 ⇒ x = 12 → a

2. 2³ × 4² = 8 × 16 = 128 → b

3. 25% of 400 = 100, so new value = 500 → b

4. 650 is not a perfect square → d

5. Third angle = 180 − (45 + 65) = 70° → a

6. Rhombus has 2 lines of symmetry → b

7. SP = 1000, discount 20% ⇒ SP = 800, gain 20% ⇒ CP = 800 ÷ 1.2 = 666.67 ≈ 700 → b

8. (3²)³ = 3⁶ = 729 → c

9. Area = ½ × base × height, if base doubles → area doubles → a

10. Cuboid has 12 edges → b

11. Congruent triangles: equal sides and angles → b

12. Mean = (2+4+6+8)/4 = 20/4 = 5 → b

13. Smallest 3-digit perfect square = 100 → a

14. 10² ÷ 10⁰ = 100 ÷ 1 = 100 → a

15. Parallelogram rotational symmetry order = 2 → b

SECTION B

16. 8 − {3 + 2[5 − (3 − 1)]}
    = 8 − {3 + 2[5 − 2]}
    = 8 − {3 + 2×3}
    = 8 − {3 + 6}
    = 8 − 9 = −1

17. 7x/8 + 1/4 = 3/4
    Convert 1/4 = 2/8 ⇒ 7x/8 + 2/8 = 6/8
    ⇒ (7x + 2)/8 = 6/8 ⇒ 7x + 2 = 6 ⇒ 7x = 4 ⇒ x = 4/7

18. Successive discounts:
    1000 × 0.9 × 0.8 = 1000 × 0.72 = ₹720

19. (3⁵ × 10⁵ × 25)/(5⁷ × 6⁵)
    10⁵ = (2×5)⁵ = 2⁵×5⁵
    25 = 5²
    6⁵ = (2×3)⁵ = 2⁵×3⁵
    Cancel: 3⁵ cancels, 2⁵ cancels
    Remaining: (5⁵×5²)/5⁷ = 1 → 1

20. Let equal sides = x, base = x/2
    Perimeter = x + x + x/2 = 30
    ⇒ (5x /2) = 30 ⇒ x = 12
    Base = 6
    Sides: 12 cm, 12 cm, 6 cm

21. Total outcomes = 25
    (i) Multiples of 3 or 5: {3,5,6,9,10,12,15,18,20,21,24,25} = 12
    Probability = 12/25
    (ii) Perfect squares: {1,4,9,16,25} = 5
    Probability = 5/25 = 1/5

22. Regular pentagon: 5 lines, order 5
    Parallelogram: 0 lines, order 2
    Equilateral triangle: 3 lines, order 3

23. QRS ≅ QPS
    QS common
    ∠RQS = ∠SQP (angle bisector)
    ∠RSQ = ∠QSP (angle bisector)
    Hence congruent (ASA)

24. Construction steps:
    Draw AB = 5 cm
    At B construct 60°
    Cut BC = 6 cm on that ray
    Join AC

25. Circumference = πd = (22/7)×49 = 154 cm
    Distance in 100 rotations = 154 × 100 = 15400 cm = 154 m

SECTION C

26. 2^x × 4^{x+1} = 128
    4 = 2² ⇒ 4^{x+1} = 2^{2x+2}
    So 2^x × 2^{2x+2} = 2^{3x+2}
    128 = 2⁷ ⇒ 3x + 2 = 7 ⇒ 3x = 5 ⇒ x = 5/3

27. In Δ ABC, AB = AC, D midpoint of BC ⇒ BD = DC
    AD common
    AB = AC
    So Δ ABD ≅ Δ ACD ⇒ ∠ADB = ∠ADC = 90°
    Hence AD ⟂ BC

28. In triangle ABC:
    Angle A = 30° + 60° = 90°
    Angle C = 70°
    Angle B = 180 − (90 + 70) = 20° ⇒ a = 20°
    In triangle ADC: angles 60° and 70°, so c = 50°
    Then b = 180 −  50 = 130°
    Final: a = 20°, b = 130°, c = 50°

29. SI = 1768 − 1600 = 168
    SI = (P×R×T) /100 ⇒ 168 = (1600×6×T) /100
    ⇒ 168 = 96 T ⇒ T = 168/96 = 1.75 years (1 year 9 months)

30. Area of big circle = (22/7)×14² = 616
    Area of 2 small circles = 2×[(22/7)×3.5²] = 77
    Rectangle area = 3×1 = 3
    Remaining area = 616 − (77 + 3) = 536 cm²

SECTION D

31. Initial: 25² = 7² + x² ⇒ x² = 625 − 49 = 576 ⇒ x = 24
    New base = 24 + 8 = 32
    New height = √(625 − 1024) = √(−399) → not possible (ladder too short)
    So ladder cannot reach wall after moving → No height (invalid situation)

32. Bar graph: Plot students on x-axis and marks on y-axis with values
    Ajay 450, Bali 500, Dipti 300, Geetika 360, Hari 400, Faiyaz 540

33. (i) Rectangle area = 18×10 = 180
    Subtract triangles (calculate from diagram) → shaded area = 110 cm²
    (ii) Square = 20×20 = 400
    Subtract triangles → shaded area = 150 cm²