CBSE Class 8 End Term Maths Sample Paper with Solutions (2024-25)

If you are a Class 8 CBSE student preparing for your end-term maths exam, solving a full-length sample paper is one of the most effective things you can do. It helps you understand the question pattern, manage your time across sections, and identify topics where you need extra revision.

This sample paper follows the CBSE 2024-25 pattern for Class 8 Mathematics. It covers the complete syllabus — from Rational Numbers and Linear Equations to Mensuration and Factorisation. Every question is accompanied by a clear, step-by-step solution so you can learn the method, not just the answer.

Whether you are aiming for full marks or just want to pass comfortably, working through this paper will sharpen your problem-solving speed and boost your confidence.

The CBSE Class 8 End Term Maths paper is structured into 5 sections with a total of 80 marks and a time limit of 3 hours.

Section A (20 marks): 20 questions — MCQs (Q1-14, 1 mark each), Fill-in-the-blanks (Q15-16, 1 mark each), True/False (Q17-20, 1 mark each). No internal choice.

Section B (14 marks): 7 questions of 2 marks each (Q21-27). Short-answer type.

Section C (18 marks): 6 questions of 3 marks each (Q28-33). Application-based problems.

Section D (20 marks): 4 questions of 5 marks each (Q34-37). Long-answer type with detailed working.

Section E (8 marks): 3 case-study questions (Q38-40). Each case study has sub-parts.

All questions are compulsory. Use of calculators is not permitted.

Q1. A train 150 m long is running at 108 km/h. It crosses a bridge 250 m long. How long does it take?

Solution: Convert speed: $108 \times \frac{5}{18} = 30$ m/s. Total distance = train + bridge $= 150 + 250 = 400$ m. Time $= \frac{400}{30} = 13.\overline{3}$ s.

Answer: (b) 13.3 s

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Q2. Rs 2,00,000 is invested at 10% p.a. compounded semi-annually for 2 years. Find the amount.

Solution: Rate per half-year $= \frac{10}{2} = 5\%$, number of periods $= 2 \times 2 = 4$.
$A = 2{,}00{,}000 \times (1.05)^4 = 2{,}00{,}000 \times 1.21550625 = \text{Rs } 2{,}43{,}101$

Answer: (b) Rs 2,43,101

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Q3. The diagonals of a rhombus are in the ratio 3:4. If the longer diagonal is 24 cm, find the side.

Solution: Shorter diagonal $= \frac{3}{4} \times 24 = 18$ cm. Half-diagonals are 12 cm and 9 cm. Each side $= \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15$ cm.

Answer: (c) 15 cm

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Q4. A wire 88 cm long is bent into a circle. Find the area.

Solution: Circumference $= 2\pi r = 88$, so $r = \frac{88}{2\pi} = \frac{88 \times 7}{2 \times 22} = 14$ cm. Area $= \pi r^2 = \frac{22}{7} \times 196 = 616$ cm$^2$.

**Answer: (a) 616 cm$^2$**

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Q5. A cuboid has dimensions 8 cm $\times$ 6 cm $\times$ 5 cm. Find the ratio of TSA to LSA.

Solution: TSA $= 2(8 \times 6 + 6 \times 5 + 8 \times 5) = 2(48 + 30 + 40) = 236$ cm$^2$. LSA $= 2(8 + 6) \times 5 = 140$ cm$^2$. Ratio $= 236 : 140 = 59 : 35$.

Answer: (a) 236 : 140

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Q6. A and B together finish work in 12 days. B alone takes 20 days. B works 5 days alone, then A and B finish the rest together. How many more days?

Solution: B's 5-day work $= \frac{5}{20} = \frac{1}{4}$. Remaining $= \frac{3}{4}$. A+B rate $= \frac{1}{12}$/day. Time for remaining $= \frac{3/4}{1/12} = 9$ days.

Answer: (b) 9 days

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Q7. CP = Rs 100, marked price = Rs 140, discount = 15%. Find profit %.

Solution: SP $= 140 \times 0.85 = \text{Rs } 119$. Profit $= 119 - 100 = 19$. Profit % $= 19\%$.

Answer: (a) 19%

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Q8. A tank is 3 m $\times$ 2 m $\times$ 1.5 m. Water flows in at 0.5 m$^3$/min. How long to fill it?

Solution: Volume $= 3 \times 2 \times 1.5 = 9$ m$^3$. Time $= \frac{9}{0.5} = 18$ min.

Answer: (c) 18 min

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Q9. A rhombus has side 10 cm and one diagonal is 3 times the other. Find the shorter diagonal.

Solution: Let shorter diagonal $= d$, longer $= 3d$. Half-diagonals are $\frac{d}{2}$ and $\frac{3d}{2}$. Using Pythagoras:
$\left(\frac{d}{2}\right)^2 + \left(\frac{3d}{2}\right)^2 = 10^2$
$\frac{d^2 + 9d^2}{4} = 100 \implies 10d^2 = 400 \implies d = 2\sqrt{10}$ cm.

**Answer: (c) $2\sqrt{10}$ cm**

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Q10. A sum of money triples itself in 4 years at CI. In how many years will it become 27 times?

Solution: $(1+r)^4 = 3$. For 27 times: $(1+r)^n = 27 = 3^3 = ((1+r)^4)^3$, so $n = 12$ years.

Answer: (b) 12 years

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Q11. A cuboid 12 cm $\times$ 8 cm $\times$ 6 cm is cut into 3 equal pieces along the longest edge. Find the TSA of each piece.

Solution: Each piece is $4 \times 8 \times 6$ cm. TSA $= 2(4 \times 8 + 8 \times 6 + 4 \times 6) = 2(32 + 48 + 24) = 208$ cm$^2$.

**Answer: (b) 208 cm$^2$**

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Q12. Under simple interest, in what condition does SI equal the principal?

Solution: SI $= P$ means $\frac{P \times R \times T}{100} = P$, giving $R \times T = 100$.

**Answer: (a) $RT = 100$**

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Q13. Which of the following is NOT a linear equation in one variable? (a) $2x + 3 = 7$ (b) $3y - 5 = 10$ (c) $x^2 + y = 5$

Solution: $x^2 + y = 5$ contains $x^2$ (degree 2), so it is not linear.

**Answer: (c) $x^2 + y = 5$**

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Q14. In rhombus PQRS, $\angle QPS = 70°$. Find $\angle PQR$.

Solution: In a rhombus, adjacent angles are supplementary. $\angle PQR = 180° - 70° = 110°$.

Answer: (c) 110°

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Q15. Solve: $\frac{4x}{3} = 7$.

Solution: $4x = 21$, so $x = \frac{21}{4}$.

**Answer: $\frac{21}{4}$**

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Q16. A parallelogram has area 96 cm$^2$ and base 12 cm. Find the altitude.

Solution: Area $=$ base $\times$ height. $96 = 12 \times h$, so $h = 8$ cm.

Answer: 8 cm

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Q17. In a parallelogram, one angle is 70°. The adjacent angle is 110°.

Solution: Adjacent angles in a parallelogram are supplementary: $180° - 70° = 110°$.

Answer: 110°

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Q18. "A linear equation in one variable always has exactly one solution." True or False?

Solution: False. A linear equation can have 0 solutions (inconsistent), exactly 1 solution, or infinitely many solutions (identity like $2x = 2x$).

Answer: False

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Q19. "In a rectangle PQRS, PO = OQ where O is the intersection of diagonals." True or False?

Solution: True. Diagonals of a rectangle are equal and bisect each other, so $PO = OQ$.

Answer: True

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Q20. "The equation $3x + 2y^2 = 10$ is a linear equation." True or False?

Solution: False. The term $y^2$ makes the equation non-linear (degree 2).

Answer: False

Q21. Find the simple interest on Rs 4,500 for 7 months at 8.5% p.a. Also find the amount.

Solution:
$\text{SI} = \frac{P \times R \times T}{100} = \frac{4500 \times 8.5 \times 7}{100 \times 12}$
$= \frac{4500 \times 8.5 \times 7}{1200} = \frac{2{,}67{,}750}{1200} \approx \text{Rs } 223.13$

Amount $= 4500 + 223.13 = \text{Rs } 4{,}723.13$

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Q22. Solve: $\frac{2x+1}{3} - \frac{3x-2}{4} = 1$.

Solution: Multiply both sides by LCM of 3 and 4, which is 12:
$4(2x+1) - 3(3x-2) = 12$
$8x + 4 - 9x + 6 = 12$
$-x + 10 = 12$
$-x = 2 \implies x = -2$

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Q23. The area of a rhombus is 120 cm$^2$ and one diagonal is 15 cm. Find the other diagonal.

Solution: Area of rhombus $= \frac{1}{2} \times d_1 \times d_2$.
$120 = \frac{1}{2} \times 15 \times d_2$
$d_2 = \frac{120 \times 2}{15} = 16$ cm.

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Q24. In the figure, two lines intersect at O. $\angle AOC = 115°$ and $\angle DAO = 45°$. Find $\angle ADO$.

Solution: $\angle AOD = 180° - 115° = 65°$ (linear pair with $\angle AOC$). In triangle AOD:
$\angle ADO = 180° - \angle DAO - \angle AOD = 180° - 45° - 65° = 70°$.

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Q25. $y$ varies inversely as $(2x + 1)$. If $x = 2$ when $y = 6$, find $y$ when $x = 5$.

Solution: $y = \frac{k}{2x+1}$. When $x=2, y=6$: $6 = \frac{k}{5}$, so $k = 30$.
When $x = 5$: $y = \frac{30}{2(5)+1} = \frac{30}{11}$.

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Q26. Factorise: $3x^4 - 48x^2$.

Solution: Take out the common factor:
$3x^4 - 48x^2 = 3x^2(x^2 - 16) = 3x^2(x+4)(x-4)$

(Using the identity $a^2 - b^2 = (a+b)(a-b)$.)

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Q27. A room is 12 m long, 8 m wide, and 4 m high. It has 2 doors (each 3 m$^2$) and 3 windows (each 1.5 m$^2$). Find the cost of painting the walls and ceiling at Rs 12/m$^2$.

Solution: Walls $= 2(12 + 8) \times 4 = 160$ m$^2$. Ceiling $= 12 \times 8 = 96$ m$^2$. Total $= 256$ m$^2$.
Subtract openings: $2(3) + 3(1.5) = 6 + 4.5 = 10.5$ m$^2$.
Paintable area $= 256 - 10.5 = 245.5$ m$^2$. Cost $= 245.5 \times 12 = \text{Rs } 2{,}946$.

Q28. P, Q, and R can finish a job together in 10 days. Q and R together take 15 days. P and R together take 12 days. Find how long each takes alone. Who is most efficient?

Solution: Let rates be $\frac{1}{P}, \frac{1}{Q}, \frac{1}{R}$.
$\frac{1}{P} + \frac{1}{Q} + \frac{1}{R} = \frac{1}{10}$ ... (i)
$\frac{1}{Q} + \frac{1}{R} = \frac{1}{15}$ ... (ii)
$\frac{1}{P} + \frac{1}{R} = \frac{1}{12}$ ... (iii)

From (i) - (ii): $\frac{1}{P} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30}$ $\Rightarrow$ P takes 30 days.
From (iii): $\frac{1}{R} = \frac{1}{12} - \frac{1}{30} = \frac{5-2}{60} = \frac{3}{60} = \frac{1}{20}$ $\Rightarrow$ R takes 20 days.
From (ii): $\frac{1}{Q} = \frac{1}{15} - \frac{1}{20} = \frac{4-3}{60} = \frac{1}{60}$ $\Rightarrow$ Q takes 60 days.

R is most efficient (finishes alone in the fewest days).

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Q29. Find the compound interest on Rs 8,000 for 3 years if the rates are 5%, 6%, and 10% for successive years.

Solution:
After Year 1: $8000 \times 1.05 = 8{,}400$
After Year 2: $8400 \times 1.06 = 8{,}904$
After Year 3: $8904 \times 1.10 = 9{,}794.40$

CI $= 9{,}794.40 - 8{,}000 = \text{Rs } 1{,}794.40$

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Q30. A floor is paved with 1,800 triangular tiles, each with base 45 cm and height 30 cm. Find the cost of polishing the floor at Rs 6/m$^2$.

Solution: Area of one tile $= \frac{1}{2} \times 45 \times 30 = 675$ cm$^2$.
Total area $= 1800 \times 675 = 12{,}15{,}000$ cm$^2 = 121.5$ m$^2$.
Cost $= 121.5 \times 6 = \text{Rs } 729$.

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Q31. (a) Factorise $2x^4 - 32$. (b) Evaluate $998^2 - 4 \times 998 + 4$ using identities.

Solution:
(a) $2x^4 - 32 = 2(x^4 - 16) = 2(x^2 + 4)(x^2 - 4) = 2(x^2 + 4)(x + 2)(x - 2)$

(b) $998^2 - 4 \times 998 + 4 = 998^2 - 2(2)(998) + 2^2 = (998 - 2)^2 = 996^2 = 9{,}92{,}016$

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Q32. Simplify: $\sqrt{\frac{507}{3}}$.

Solution: $\frac{507}{3} = 169$. So $\sqrt{169} = 13$.

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Q33. A warehouse is 15 m $\times$ 12 m $\times$ 5 m. Only 80% of the space can be used. How many boxes of size 50 cm $\times$ 40 cm $\times$ 80 cm can be stored?

Solution: Warehouse volume $= 15 \times 12 \times 5 = 900$ m$^3$. Usable $= 0.8 \times 900 = 720$ m$^3$.
Box volume $= 0.5 \times 0.4 \times 0.8 = 0.16$ m$^3$. Number of boxes $= \frac{720}{0.16} = 4{,}500$.

Q34. A man borrows a sum of money and agrees to repay in 4 annual instalments forming a geometric series with common ratio $\frac{2}{3}$. If the total repayment is Rs 4,422 and the sum of ratios across 4 years works out to $\frac{211}{81}$, find the first instalment.

Solution: Let the first instalment be $P_1$. The 4 instalments form a GP with $r = \frac{2}{3}$.
Sum $= P_1\left(1 + \frac{2}{3} + \frac{4}{9} + \frac{8}{27}\right) = P_1 \times \frac{27 + 18 + 12 + 8}{27} = P_1 \times \frac{65}{27}$

So $P_1 \times \frac{65}{27} = 4422$, giving $P_1 = \frac{4422 \times 27}{65} \approx \text{Rs } 1{,}836.34$.

(If the problem states the factor is $\frac{211}{81}$: $P_1 = \frac{4422 \times 81}{211} \approx \text{Rs } 1{,}698$.)

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Q35. Two trains start at the same time from two stations 480 km apart and travel towards each other at 80 km/h and 60 km/h. When do they meet?

Solution: Relative speed (moving towards each other) $= 80 + 60 = 140$ km/h.
Time to meet $= \frac{480}{140} = \frac{24}{7} \approx 3.43$ hours $= 3$ hours $25$ minutes $43$ seconds.

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Q36. The length of a rectangular field is $(2x + 5)$ m and the breadth is $(x + 2)$ m. Its area is 96 m$^2$. Find $x$.

Solution:
$(2x + 5)(x + 2) = 96$
$2x^2 + 4x + 5x + 10 = 96$
$2x^2 + 9x - 86 = 0$

Using the quadratic formula:
$x = \frac{-9 \pm \sqrt{81 + 688}}{4} = \frac{-9 \pm \sqrt{769}}{4}$
$\sqrt{769} \approx 27.73$, so $x = \frac{-9 + 27.73}{4} \approx 4.68$ m (taking positive value).

Length $\approx 14.36$ m, Breadth $\approx 6.68$ m.

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Q37. Prove that the diagonals of a rhombus are perpendicular to each other.

Solution: Let ABCD be a rhombus with diagonals AC and BD intersecting at O. In a rhombus, all sides are equal: $AB = BC = CD = DA$.

Consider triangles AOB and COB:
- $AB = CB$ (sides of rhombus)
- $OB = OB$ (common)
- $AO = OC$ (diagonals of a parallelogram bisect each other)

By SSS congruence, $\triangle AOB \cong \triangle COB$.
Therefore $\angle AOB = \angle COB$.
Since $\angle AOB + \angle COB = 180°$ (linear pair), we get $\angle AOB = \angle COB = 90°$.

Hence the diagonals are perpendicular.

Q38. Case Study — Aquarium (Open Top)

An aquarium is 60 cm long, 35 cm wide, and 40 cm high (open top).

(a) Find the total area of glass needed.

Solution:
Base $= 60 \times 35 = 2{,}100$ cm$^2$
2 long walls $= 2 \times 60 \times 40 = 4{,}800$ cm$^2$
2 short walls $= 2 \times 35 \times 40 = 2{,}800$ cm$^2$
**Total glass $= 2{,}100 + 4{,}800 + 2{,}800 = 9{,}700$ cm$^2$**

(b) Find the total length of all edges.

Solution: A cuboid has 12 edges: 4 lengths + 4 widths + 4 heights.
Total $= 4(60) + 4(35) + 4(40) = 240 + 140 + 160 = 540$ cm.

(c) If edge tape costs Rs 5/cm, find the total cost.

Solution: Cost $= 540 \times 5 = \text{Rs } 2{,}700$.

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Q39. Case Study — Investment Comparison

Rs 25,000 is to be invested for 3 years. Option A: Simple Interest at 9% p.a. Option B: Compound Interest at 8% p.a.

(a) Find the amount under Option A.

Solution: SI $= \frac{25000 \times 9 \times 3}{100} = \text{Rs } 6{,}750$. Amount $= 25{,}000 + 6{,}750 = \text{Rs } 31{,}750$.

(b) Find the amount under Option B.

Solution: $A = 25000 \times (1.08)^3 = 25000 \times 1.259712 = \text{Rs } 31{,}492.80$.
CI $= 31{,}492.80 - 25{,}000 = \text{Rs } 6{,}492.80$.

(c) Which option is better and by how much?

Solution: Option A gives Rs 31,750 while Option B gives Rs 31,492.80. Option A is better by Rs 257.20.

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Q40. Case Study — Land Comparison

Two plots: one is a trapezium with parallel sides 80 m and 50 m and height 45 m. The other is a circle with diameter 80 m.

(a) Find the area of the trapezium.

Solution: Area $= \frac{1}{2}(80 + 50) \times 45 = \frac{1}{2} \times 130 \times 45 = 2{,}925$ m$^2$.

(b) Find the area of the circular plot.

Solution: Radius $= 40$ m. Area $= \pi r^2 = 3.14 \times 1600 = 5{,}024$ m$^2$.

(c) Which plot is larger and by how much?

Solution: Circular plot is larger by $5{,}024 - 2{,}925 = 2{,}099$ m$^2$.

Here are some tips to get the most out of this paper:

1. Simulate real exam conditions. Set a timer for 3 hours, sit at a desk, and attempt the full paper without looking at the solutions.

2. Write out every step. Do not skip working — CBSE awards step marks, so showing your method matters even if the final answer has an error.

3. Mark and review. After completing the paper, check your answers against the solutions above. Highlight the topics where you made mistakes and revise those chapters.

4. Focus on weak areas. If you struggled with mensuration, go back to the formulas for surface area and volume. If algebra tripped you up, practise more factorisation and linear equation problems.

5. Try the online version. Taking this same paper as a timed online test gives you a different kind of practice — you get instant feedback and an email report.

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