Class VIII Mathematics Annual Examination Solutions 2024-25

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Class VIII Mathematics Annual Examination Solutions 2024-25

DOWNLOAS QUESTION PAPER –CLICK

SECTION – A (Multiple Choice Questions)

1. (i)If 2z – 3 = z + 2, then the value of (z – 4) is:
Solve for z: 2z – z = 2 + 3 ⇒ z = 5.
Find (z – 4): 5 – 4 = 1.

Answer: (a) 1
1. (ii)In the given figure, the measure of angle ‘x’ is:
Sum of angles around a point is 360°.
x + 36° + 64° + 150° = 360°
x + 250° = 360° ⇒ x = 110°.

Answer: (d) 110°
1. (iii) The value of (5° + 6°) × [cite_start]7° is:
Any non-zero number to the power of 0 is 1.
(1 + 1) × 1 = 2 × 1 = 2.

Answer: (c) 2
1. (iv)A square number ends in the digit 4. The unit digit in the square root is:

22 = 4 and 82 = 64.

Answer: (d) 2 or 8
1. (v) Standard exponential form of 384467000 m:

Move decimal 8 places to the left: 3.84467.

Answer: (b) 3.84467 × 108 m
1. (vi) Seema weaves 25 baskets in 35 days. [cite_start]Days for 100 baskets:
Direct proportion: 25/35 = 100/x.
x = (100 × 35) / 25 = 4 × 35 = 140.

Answer: (a) 140
1. (vii) Number of 6 cm cubes in a 216 m3 cuboid:
Volume of cuboid = 216 m3 = 216 × 106 cm3.
Volume of cube = 63 = 216 cm3.
Number = (216 × 106) / 216 = 106.

Answer: (a) 106
1. (ix)The value of ‘x’ in the figure (Polygon Exterior Angles):
Sum of exterior angles = 360°.
Exterior angle at P = 180° – 89° = 91°.
x + 90° + 60° + 91° + 40° = 360°.
x + 281° = 360° ⇒ x = 79°.

Answer: (a) 79°
1. (x) If ∛4096 = 16, then ∛4.096 + ∛4096 is:
∛4.096 = 1.6.
1.6 + 16 = 17.6.

Answer: (c) 17.6
1. (xi) Factors of (49p2 – 36):
Using a2 – b2 = (a-b)(a+b).
(7p)2 – 62 = (7p – 6)(7p + 6).

Answer: (d) (7p – 6)(7p + 6)
1. (xii)If x and y vary inversely (90, 10) and (45, a):
xy = k ⇒ 90 × 10 = 900.
45 × a = 900 ⇒ a = 20.

Answer: (b) 20

SECTION – B (Objective Type Questions)

2. HABD and GCEF are parallelograms. Find ‘x’.
In HABD: ∠A = 130° ⇒ ∠ABD = 50° (Adjacent angles sum to 180°).
In GCEF: ∠F = 30° ⇒ Corresponding angle in the small triangle is 30°.
In the small triangle: x = 180° – (50° + 30°) = 100°.

Answer: 100°
3. If (-2)3 × (-2)-6 = (-2)2m-6, find ‘m’.
Add powers on LHS: 3 + (-6) = -3.
Equate powers: -3 = 2m – 6.
2m = 3 ⇒ m = 1.5.

Answer: m = 1.5
4. Meeta saves Rs 4000 (10% of salary). Find salary.
Let Salary = S. 10% of S = 4000.
S = 4000 × 10 = 40,000.

Answer: Rs 40,000
5. Skates: Discount 20%, Payment Rs 1920. Find Marked Price
Payment is 80% of MP.
0.8 × MP = 1920.
MP = 1920 / 0.8 = 2400.

Answer: Rs 2400
6. What must be added to 2m2 – 3mn + 3n2 to get 5m2 + 2mn + 7n2?
Subtract the first expression from the second.
(5 – 2)m2 + (2 – (-3))mn + (7 – 3)n2.

Answer: 3m2 + 5mn + 4n2
7. Trapezium Area = 480 m2, h = 15m, one side = 20m. Find other side. [cite_start][cite: 2101-2102]
Area = 1/2 × (a + b) × h.
480 = 1/2 × (20 + x) × 15.
64 = 20 + x ⇒ x = 44.

Answer: 44 m

SECTION – C (Short Answer Questions)

8. Find the least number to subtract from 4000 to get a perfect square. [cite_start][cite: 2106-2107]
632 = 3969 and 642 = 4096.
Nearest smaller square is 3969.
Subtract: 4000 – 3969 = 31.

Answer: 31 (Square root: 63)
9. Simplify: (a+b)(2a-3b+c) – (2a-3b)c
Expand: 2a2 – 3ab + ac + 2ab – 3b2 + bc – (2ac – 3bc).
Combine like terms: 2a2 – ab – 3b2 + ac + bc – 2ac + 3bc.

Answer: 2a2 – 3b2 – ab – ac + 4bc
10. 2 persons take 3 days to fit windows.
(i) 1 person falls ill. Work = 2 × 3 = 6 man-days. Remaining person takes 6 days.
(ii) To complete in 1 day: 6 persons needed.

Answer: (i) 6 days, (ii) 6 persons

SECTION – D (Long Answer Questions)

11. Divide 44(y4 – 5y3 – 24y2) by 11y(y – 8).
Factorize numerator: 44y2(y2 – 5y – 24).
Split middle term: y2 – 8y + 3y – 24 = (y-8)(y+3).
Expression: [44y2(y-8)(y+3)] / [11y(y-8)].

Answer: 4y(y + 3)
12. Solve: 5x – 2(2x – 7) = 2(3x – 1) + 7/2.
5x – 4x + 14 = 6x – 2 + 3.5.
x + 14 = 6x + 1.5.
12.5 = 5x ⇒ x = 2.5.

Answer: x = 2.5
13. Line Graph for Car Distance.
(a) Suitable scale: 1 unit = 40 km.
(b) Distance in last 2 hours (7:30 to 9:30): 160 – 80 = 80 km.

Answer: (a) 1 unit = 40km, (b) 80 km

SECTION – E (Case Study Questions)

14. Ravi’s Loan of Rs 7000 for 2 years.
(i) Simple Interest: 7000 × 0.05 × 2 = Rs 700.
(ii) Total Amount (Scheme I): 7000 + 700 = Rs 7700.
(iii) Compound Amount (Scheme II): 7000(1.05)2 = 7000(1.1025) = Rs 7717.5.

Conclusion: Scheme I is better.
15. Straws of length 10cm, 8cm, 6cm, 3cm.
(i) Scalene triangles.
(ii) 62 + 82 = 36 + 64 = 100 = 102. They form a Right-Angled Triangle.
(iii) Akash’s straws: 10, 6, 3 (Because 6 + 3 < 10, no triangle possible).

OR: Sonia’s straws: 6, 8, 10.
16. Community Place Pillars (r=0.7m, h=5m).
(i) Area of 1 pillar (CSA): 2πrh = 2 × (22/7) × 0.7 × 5 = 22 m2.
(ii) Cans required: 22 / 10 = 2.2 ⇒ 3 cans.
(iii) Total Cost: 4 pillars × 22 m2 × Rs 20 = Rs 1760.
OR (Roof): Volume = 10 × 10 × 0.25 = 25 m3. Cost = 25 × 60 = Rs 1500.

 

Class 8 Math – Optional Question Solutions

 

SECTION-B

Q4 (OR): Cost Price of Toy Car

[cite_start]

Question: Selling price of a toy car is ₹ 540. If the profit made by the shopkeeper is 20%, what is the cost price of this toy? [cite: 434-435]

Solution:
Given: Selling Price (SP) = ₹ 540, Profit = 20%
Formula: CP = (SP × 100) / (100 + Profit%)

Substituting values:
CP = (540 × 100) / (100 + 20)
CP = (540 × 100) / 120
CP = 4.5 × 100 = 450

Answer: The Cost Price is ₹ 450.

SECTION-C

Q9 (OR): Subtraction of Algebraic Expressions

[cite_start]

Question: Subtract 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c). [cite: 516-517]

Solution:

  1. Simplify the term to subtract from (First Term):
    4c(-a + b + c) = -4ac + 4bc + 4c2
  2. Simplify the term to be subtracted (Second Term):
    3a(a + b + c) – 2b(a – b + c)
    = (3a2 + 3ab + 3ac) – (2ab – 2b2 + 2bc)
    = 3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc
    Combine like terms (3ab – 2ab):
    = 3a2 + 2b2 + ab + 3ac – 2bc
  3. Perform Subtraction (First Term – Second Term):
    (-4ac + 4bc + 4c2) – (3a2 + 2b2 + ab + 3ac – 2bc)
    Change signs of the second polynomial:
    -4ac + 4bc + 4c2 – 3a2 – 2b2 – ab – 3ac + 2bc
  4. Arranging like terms:
    -3a2 – 2b2 + 4c2 – ab + (4bc + 2bc) + (-4ac – 3ac)

Answer: -3a2 – 2b2 + 4c2 – ab + 6bc – 7ac

SECTION-D

Q11 (OR): Factorisation

[cite_start]

(i) p4 – 81 [cite: 531]

Solution:
Write as difference of squares:
= (p2)2 – (9)2
= (p2 – 9)(p2 + 9)
Further factorise (p2 – 9):
= (p2 – 32)(p2 + 9)
Answer: (p – 3)(p + 3)(p2 + 9)

[cite_start]

(ii) 25a2 – 4b2 + 28bc – 49c2 [cite: 532]

Solution:
Group the last three terms and factor out the negative sign:
= 25a2 – (4b2 – 28bc + 49c2)
The bracket is a perfect square (2b – 7c)2:
= (5a)2 – (2b – 7c)2
Apply identity a2 – b2 = (a-b)(a+b):
= [5a – (2b – 7c)] [5a + (2b – 7c)]
Answer: (5a – 2b + 7c)(5a + 2b – 7c)

Q13 (OR): Data Handling (Line Graph)

[cite_start]

Context: Graph showing marks obtained by a student in two tests. [cite: 576-586]

  • [cite_start](i) In which subject did the student score full marks? [cite: 588]
    Looking at the graph, the solid line (Test I) reaches 10 in Maths.
    Answer: Maths
  • [cite_start](ii) In which subject(s) did the student score the least marks in Test II? [cite: 590]
    Looking at the dotted line (Test II), the lowest point is at 6 marks.
    Answer: English and Hindi
  • (iii) In which test was the student’s performance better? [cite_start]Explain. [cite: 592]
    Total Marks Test I: 7 (Eng) + 8 (Hin) + 10 (Mat) + 6 (Sci) + 5 (S.St) = 36
    Total Marks Test II: 6 (Eng) + 6 (Hin) + 8 (Mat) + 9 (Sci) + 8 (S.St) = 37
    Answer: Test II was better because the total score (37) is higher than Test I (36).

SECTION-E

Q14 (OR): Comparing Interest Schemes

Context: Ravi borrows ₹7000 for 2 years. Scheme I is 5% Simple Interest. [cite_start]Scheme II is 5% Compound Interest. [cite: 633-636]

Scheme I (Simple Interest):
Interest = (P × R × T) / 100 = (7000 × 5 × 2) / 100 = ₹ 700

Scheme II (Compound Interest):
Amount = P(1 + R/100)T = 7000(1 + 5/100)2
= 7000(1.05)2 = 7000 × 1.1025 = ₹ 7717.5
Interest = Amount – Principal = 7717.5 – 7000 = ₹ 717.5

Answer: Scheme I is better for Ravi because he has to pay less interest (₹700 vs ₹717.5).

Q15 (OR): Triangles Property

Question: Sonia chose three straws and formed a right-angled triangle. [cite_start]Identify the straws and give a reason. [cite: 658-660]

Solution:
Available straws: A=10cm, B=8cm, C=6cm, D=3cm.
For a right-angled triangle, Pythagoras theorem must hold (Hypotenuse2 = Base2 + Height2).
Checking straws A, B, and C:
62 + 82 = 36 + 64 = 100
102 = 100
Since LHS = RHS, these form a right-angled triangle.
Answer: Straws A (10cm), B (8cm), and C (6cm).

Q16 (OR): Construction Cost

[cite_start]

Question: Calculate the amount to be paid to the contractor for the construction of the roof. [cite: 699]

Solution:
[cite_start]Dimensions of roof = 10m × 10m × 25cm [cite: 687]
Convert height to meters: 25 cm = 0.25 m
Volume of roof = Length × Breadth × Height
Volume = 10 × 10 × 0.25 = 25 m3
[cite_start]Rate of construction = ₹ 60 per cubic meter [cite: 688]
Total Cost = Volume × Rate = 25 × 60
Answer: ₹ 1,500

 

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