Official CBSE Practice Papers for Class 8 Mathematics
Review targeted exam resources with the CBSE Class 8 Mathematics Annual Exam Sample Paper 07 with Answers (Old NCERT Syllabus). Built according to official CBSE standards for the 2026-27 academic year, these downloadable Class 8 Mathematics sample papers support effective revision and performance tracking.
Solved Model Papers for Mathematics
Access the complete sample paper PDF for Class 8 Mathematics below. Regular practice with these targeted mock exams builds familiarity with expected question patterns and chapter weightage to help secure higher marks.
This Class 8 Maths annual exam sample paper is a full 60-mark paper with 29 questions in sections A to E, and every question has a worked, step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, which makes it useful for state-board students and for annual exam revision of these topics. It checks areas of parallelograms, trapeziums and rhombuses, surface area of cuboids and cylinders, direct and inverse proportion word problems, laws of exponents and standard form, multiplying and factorising polynomials, and reading a line graph and plotting a simple interest graph.
Section A: Multiple Choice Questions (1 Mark Each)
(a) 12 cm
(b) 6 cm
(c) 4 cm
(d) 2 cm
Show Answer & Explanation
Answer: (a) 12 cm
Explanation:
1. Area of a parallelogram = base × corresponding altitude.
2. 60 = base × 5, so base = 60 ÷ 5 = 12 cm.
3. So the corresponding side is 12 cm.
Teacher's Note:
1. The altitude must be the one drawn to that particular side.
2. Check: 12 × 5 = 60 cm².
(a) 1 : 4
(b) 1 : 8
(c) 1 : 16
(d) 1 : 32
Show Answer & Explanation
Answer: (c) 1 : 16
Explanation:
1. Let the edges be a and b. Then a³ : b³ = 1 : 64.
2. Taking cube roots, a : b = 1 : 4.
3. Area of a face is a² and b², so the ratio is 1² : 4² = 1 : 16.
Teacher's Note:
1. Volume ratio → cube root → edge ratio → square → area ratio.
2. 1 : 4 is the ratio of the edges, not of the areas.
(a) 14 cm
(b) 18 cm
(c) 22 cm
(d) 12 cm
Show Answer & Explanation
Answer: (c) 22 cm
Explanation:
1. Area = \( \frac{1}{2} \) × (sum of parallel sides) × height. Let the other side be x cm.
2. 180 = \( \frac{1}{2} \) × (8 + x) × 12 = 6(8 + x).
3. 8 + x = 180 ÷ 6 = 30, so x = 30 − 8 = 22.
4. So the other parallel side is 22 cm. Check: \( \frac{1}{2} \) × (8 + 22) × 12 = 180 cm².
Teacher's Note:
1. Half of the height (12 ÷ 2 = 6) can be multiplied first to make the equation simple.
2. Always substitute your answer back into the formula to check it.
(a) 10
(b) 12
(c) 16
(d) 6
Show Answer & Explanation
Answer: (b) 12
Explanation:
1. To finish in less time we need more pumps, so this is inverse variation: pumps × hours is constant.
2. 8 × 24 = x × 16.
3. x = \( \frac{8 \times 24}{16} \) = 12.
4. So 12 pumps are needed.
Teacher's Note:
1. Inverse variation: the product of the two quantities stays the same.
2. The answer must be more than 8 because the time is shorter.
(a) 250
(b) 300
(c) 350
(d) 400
Show Answer & Explanation
Answer: (b) 300
Explanation:
1. In the same time, more machines produce more units, so this is direct variation.
2. \( \frac{200}{10} \) = \( \frac{x}{15} \), so x = \( \frac{200 \times 15}{10} \) = 300.
3. So 15 machines produce 300 units.
Teacher's Note:
1. The time (5 hours) is the same for both, so it does not affect the answer.
2. One machine makes 20 units in 5 hours; 15 machines make 15 × 20 = 300.
(a) \( \frac{4}{3} \)
(b) \( \frac{3}{4} \)
(c) \( \frac{16}{9} \)
(d) \( \frac{9}{16} \)
Show Answer & Explanation
Answer: (a) \( \frac{4}{3} \)
Explanation:
1. A negative power flips the fraction: \( \left(\frac{9}{16}\right)^{-\frac{1}{2}} = \left(\frac{16}{9}\right)^{\frac{1}{2}} \).
2. The power \( \frac{1}{2} \) means the square root: \( \sqrt{\frac{16}{9}} = \frac{4}{3} \).
3. So the value is \( \frac{4}{3} \).
Teacher's Note:
1. Deal with the minus sign (reciprocal) and the \( \frac{1}{2} \) (square root) one at a time.
2. \( \frac{3}{4} \) would be the value of \( \left(\frac{9}{16}\right)^{\frac{1}{2}} \), without the minus sign.
(a) 7.2 × 10⁵
(b) 7.2 × 10⁻⁶
(c) 7.2 × 10⁻⁷
(d) 72 × 10⁻⁶
Show Answer & Explanation
Answer: (b) 7.2 × 10⁻⁶
Explanation:
1. Move the decimal point to the right until it is just after the 7: 0.0000072 → 7.2.
2. The point moved 6 places to the right, so the power of 10 is −6.
3. So 0.0000072 = 7.2 × 10⁻⁶. (Check: 7.2 ÷ 1000000 = 0.0000072.)
Teacher's Note:
1. Count the places moved, not the number of zeros: here there are 5 zeros but 6 moves.
2. 72 × 10⁻⁶ is not in standard form because 72 is greater than 10.
(a) 0.00091
(b) 0.0091
(c) 0.091
(d) 910
Show Answer & Explanation
Answer: (b) 0.0091
Explanation:
1. 10⁻³ means divide by 1000, so move the decimal point 3 places to the left.
2. 9.1 → 0.91 → 0.091 → 0.0091.
3. So 9.1 × 10⁻³ = 0.0091.
Teacher's Note:
1. A negative power of 10 makes the number smaller than 1.
2. Check: 9.1 ÷ 1000 = 0.0091.
(a) 16
(b) 6
(c) 4
(d) 8
Show Answer & Explanation
Answer: (b) 6
Explanation:
1. For the same base, aᵐ ÷ aⁿ = aᵐ⁻ⁿ.
2. y⁸ ÷ y² = y⁸⁻² = y⁶.
3. So n = 6.
Teacher's Note:
1. Subtract the powers when dividing; do not divide them (8 ÷ 2 = 4 is a common mistake).
2. 16 would come from multiplying the powers, which is used for a power of a power.
(a) 2a² + 4a + 1
(b) −2a² + 10a + 1
(c) 2a² + 4a − 1
(d) 2a² − 4a + 1
Show Answer & Explanation
Answer: (a) 2a² + 4a + 1
Explanation:
1. a² terms: 4a² − 2a² = 2a².
2. a terms: −3a + 7a = 4a; constants: 5 − 4 = 1.
3. So the sum is 2a² + 4a + 1.
Teacher's Note:
1. Group like terms before adding.
2. Keep the sign attached to each term.
(a) 2x² − 5x + 11
(b) 2x² − 5x + 3
(c) 2x² − x + 3
(d) 8x² − x + 11
Show Answer & Explanation
Answer: (a) 2x² − 5x + 11
Explanation:
1. (5x² − 3x + 7) − (3x² + 2x − 4) = 5x² − 3x + 7 − 3x² − 2x + 4.
2. x² terms: 2x²; x terms: −3x − 2x = −5x; constants: 7 + 4 = 11.
3. So the difference is 2x² − 5x + 11.
Teacher's Note:
1. Subtracting −4 is the same as adding 4, so the constant is 7 + 4 = 11.
2. Option (b) comes from forgetting to change the sign of −4.
(a) x² + x − 6
(b) x² − x − 6
(c) x² + 5x − 6
(d) x² − 5x − 6
Show Answer & Explanation
Answer: (a) x² + x − 6
Explanation:
1. (x − 2)(x + 3) = x² + 3x − 2x − 6.
2. = x² + x − 6.
3. So the product is x² + x − 6.
Teacher's Note:
1. Use (x + a)(x + b) = x² + (a + b)x + ab with a = −2, b = 3.
2. a + b = 1 and ab = −6.
(a) (a − b − c)(a − b + c)
(b) (a + b − c)(a + b + c)
(c) (a − b + c)(a − b + c)
(d) none of these
Show Answer & Explanation
Answer: (a) (a − b − c)(a − b + c)
Explanation:
1. a² − 2ab + b² = (a − b)², so the expression is (a − b)² − c².
2. Using x² − y² = (x − y)(x + y) with x = a − b and y = c:
3. = (a − b − c)(a − b + c).
Teacher's Note:
1. Look for a perfect square among the first three terms.
2. Then the whole expression becomes a difference of two squares.
Assertion (A): A cylinder with a radius of 7 cm and a height of 10 cm has a lateral surface area of 440 cm².
Reason (R): The formula for the lateral surface area of a cylinder is 2πrh.
(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, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Lateral (curved) surface area = 2πrh = 2 × \( \frac{22}{7} \) × 7 × 10 = 440 cm², so A is true.
2. R gives the correct formula, so R is true.
3. A is obtained by using the formula in R, so R explains A. The answer is (a).
Teacher's Note:
1. Lateral surface area does not include the two circular ends.
2. Total surface area would be 2πr(h + r).
Assertion (A): If 8 men take 15 days to complete a task, then 10 men will take fewer days to complete the same task.
Reason (R): The number of men and the time taken to complete a task are inversely proportional.
(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, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Men and days are inversely proportional, so R is true.
2. 10 men take \( \frac{8 \times 15}{10} \) = 12 days, which is fewer than 15, so A is true.
3. A follows directly from R, so the answer is (a).
Teacher's Note:
1. More workers finish the same job in less time.
2. Use men × days = constant to find the actual number of days.
Section B: Short Answer Questions (2 Marks Each)
Question 16. Subtract: 3t⁴ − 4t³ + 2t² − 6t + 6 from −4t⁴ + 8t³ − 4t² − 2t + 11
Answer:
(−4t⁴ + 8t³ − 4t² − 2t + 11) − (3t⁴ − 4t³ + 2t² − 6t + 6)
= −4t⁴ + 8t³ − 4t² − 2t + 11 − 3t⁴ + 4t³ − 2t² + 6t − 6
= (−4t⁴ − 3t⁴) + (8t³ + 4t³) + (−4t² − 2t²) + (−2t + 6t) + (11 − 6)
= −7t⁴ + 12t³ − 6t² + 4t + 5
In simple words: Write the second expression first, change every sign of the one being subtracted, then add like terms.
Teacher's Note:
1. "Subtract A from B" means B − A.
2. Arrange the answer in descending powers of t.
Question 17. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is ₹ 4.
Answer:
Area of one rhombus tile = \( \frac{1}{2} \) × d₁ × d₂ = \( \frac{1}{2} \) × 45 × 30 = 675 cm².
Area of 3000 tiles = 675 × 3000 = 2025000 cm².
1 m² = 10000 cm², so area = 2025000 ÷ 10000 = 202.5 m².
Cost = 202.5 × ₹ 4 = ₹ 810.
In simple words: Find the area of one tile, multiply by 3000, change cm² to m², then multiply by ₹ 4.
Teacher's Note:
1. Area of a rhombus is half the product of its diagonals.
2. Divide by 10000 (not 100) to change cm² into m².
Question 18. Express the following numbers in the usual form:
(a) 3.02 × 10⁻⁶ (b) 1.0001 × 10⁹
Answer:
(a) 10⁻⁶ means move the decimal point 6 places to the left: 3.02 × 10⁻⁶ = 0.00000302.
(b) 10⁹ means move the decimal point 9 places to the right: 1.0001 × 10⁹ = 1000100000.
In simple words: Negative power: move the point left; positive power: move it right, filling gaps with zeros.
Teacher's Note:
1. In (b), the four digits after the point use up 4 of the 9 moves; add 5 more zeros.
2. Check the size: 10⁹ is a billion, so (b) should be just over 1000000000.
Question 19. Factorise: (i) 15pq + 15 + 9q + 25p (ii) z − 7 + 7xy − xyz
Answer:
(i) Rearrange: 15pq + 25p + 9q + 15.
= 5p(3q + 5) + 3(3q + 5) = (3q + 5)(5p + 3).
(ii) z − 7 + 7xy − xyz = (z − 7) − xy(z − 7).
= (z − 7)(1 − xy).
In simple words: Group the terms in pairs that share a factor, so the same bracket appears twice.
Teacher's Note:
1. In (ii), taking out −xy from 7xy − xyz gives −xy(z − 7).
2. Check by multiplying the factors back.
Question 20. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of children is reduced by 4?
Answer:
Total sweets = 24 × 5 = 120 (this does not change).
New number of children = 24 − 4 = 20.
Fewer children get more sweets each (inverse variation): each gets 120 ÷ 20 = 6.
Each child would get 6 sweets.
In simple words: There are 120 sweets in all; shared among 20 children, each gets 6.
Teacher's Note:
1. In inverse variation, children × sweets each stays constant.
2. The answer should be more than 5 because there are fewer children.
Section C: Short Answer Questions (3 Marks Each)
Question 21. Simplify: (a) \( \left(\frac{1}{4}\right)^{-2} + \left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2} \) (b) \( \left[\left\{\left(\frac{-3}{5}\right)^{-2}\right\}^{2}\right]^{-1} \)
Answer:
(a) \( \left(\frac{1}{a}\right)^{-2} = a^2 \), so the expression = 4² + 2² + 3².
= 16 + 4 + 9 = 29.
(b) Using (aᵐ)ⁿ = aᵐⁿ, the powers multiply: (−2) × 2 × (−1) = 4.
\( \left(\frac{-3}{5}\right)^{4} = \frac{(-3)^4}{5^4} = \frac{81}{625} \).
Answer: \( \frac{81}{625} \).
In simple words: A negative power flips a fraction; a power of a power multiplies the indices.
Teacher's Note:
1. An even power of a negative number is positive: (−3)⁴ = 81.
2. Multiply all three indices in (b) at once to save steps.
Answer:
5 hours = 5 × 60 = 300 minutes. Let the distance be x km.
| Distance (in km) | 14 | x |
| Time (in minutes) | 25 | 300 |
At the same speed, distance and time vary directly: \( \frac{14}{25} = \frac{x}{300} \).
x = \( \frac{14 \times 300}{25} \) = 168.
The truck can travel 168 km in 5 hours.
In simple words: 300 minutes is 12 times 25 minutes, so the truck goes 12 × 14 = 168 km.
Teacher's Note:
1. Change both times to the same unit (minutes) first.
2. Direct variation: the ratio distance ÷ time stays the same.
Question 23. Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?
Answer:
The number of persons and the number of days are in inverse proportion: persons × days = 2 × 3 = 6.
(i) Now only 1 person works: 1 × x = 6, so x = 6. The job would take 6 days.
(ii) To finish in 1 day: y × 1 = 6, so y = 6. 6 persons would be needed.
In simple words: The job needs 6 person-days of work, so 1 person takes 6 days and 6 persons take 1 day.
Teacher's Note:
1. Fewer workers take more days; more workers take fewer days.
2. Find the fixed product first; then each part is one division.
Question 24. Factorise: (a) p² + 6p + 8 (b) q² − 10q + 21
Answer:
(a) Numbers with product 8 and sum 6: 2 and 4.
p² + 6p + 8 = p² + 2p + 4p + 8 = p(p + 2) + 4(p + 2) = (p + 2)(p + 4).
(b) Numbers with product 21 and sum −10: −7 and −3.
q² − 10q + 21 = q² − 7q − 3q + 21 = q(q − 7) − 3(q − 7) = (q − 7)(q − 3).
In simple words: Split the middle term using two numbers whose product is the constant and whose sum is the middle coefficient.
Teacher's Note:
1. A positive constant with a negative middle term means both numbers are negative.
2. Multiply the factors back to check.
Section D: Long Answer Questions (5 Marks Each)
Question 25. Multiply (i) (3x² + 4x − 8) and (2x² − 4x + 3) (ii) (2x + 2y − 3) and (x + y + 5)
Answer:
(i) (3x² + 4x − 8)(2x² − 4x + 3)
= 3x²(2x² − 4x + 3) + 4x(2x² − 4x + 3) − 8(2x² − 4x + 3)
= 6x⁴ − 12x³ + 9x² + 8x³ − 16x² + 12x − 16x² + 32x − 24
= 6x⁴ − 4x³ − 23x² + 44x − 24
(ii) (2x + 2y − 3)(x + y + 5)
= 2x(x + y + 5) + 2y(x + y + 5) − 3(x + y + 5)
= 2x² + 2xy + 10x + 2xy + 2y² + 10y − 3x − 3y − 15
= 2x² + 4xy + 2y² + 7x + 7y − 15
In simple words: Multiply every term of the first bracket by the whole second bracket, then collect like terms.
Teacher's Note:
1. Line up like powers in columns to avoid missing a term.
2. Check (ii) with x = y = 1: (1)(7) = 7 and 2 + 4 + 2 + 7 + 7 − 15 = 7.
| Deposit (in ₹) | 1000 | 2000 | 3000 | 4000 | 5000 |
|---|---|---|---|---|---|
| Simple Interest (in ₹) | 80 | 160 | 240 | 320 | 400 |
(ii) Use the graph to find the interest on ₹ 2500 for a year.
(iii) To get an interest of ₹ 280 per year, how much money should be deposited?
Answer:
Take deposit on the x-axis (1 big square = ₹ 1000) and simple interest on the y-axis (1 big square = ₹ 80).
Plot (1000, 80), (2000, 160), (3000, 240), (4000, 320), (5000, 400) and join them. They lie on a straight line.
(i) Extending the line backwards, it passes through (0, 0). Yes, the graph passes through the origin (no deposit, no interest).
(ii) From 2500 on the x-axis go up to the line and across to the y-axis: the interest is ₹ 200.
(iii) From 280 on the y-axis go across to the line and down to the x-axis: the deposit is ₹ 3500.
In simple words: The interest is always 8% of the deposit, so the graph is a straight line through the origin.
Teacher's Note:
1. A straight line through the origin shows direct proportion.
2. Check (ii) and (iii): 8% of 2500 = 200 and 280 ÷ 0.08 = 3500.
Question 27. Factorise (i) x⁴ − (x − z)⁴ (ii) a⁴ − 2a²b² + b⁴
Answer:
(i) x⁴ − (x − z)⁴ = (x²)² − [(x − z)²]² = [x² − (x − z)²][x² + (x − z)²].
x² − (x − z)² = [x − (x − z)][x + (x − z)] = z(2x − z).
x² + (x − z)² = x² + x² − 2xz + z² = 2x² − 2xz + z².
So x⁴ − (x − z)⁴ = z(2x − z)(2x² − 2xz + z²).
(ii) a⁴ − 2a²b² + b⁴ = (a²)² − 2a²b² + (b²)² = (a² − b²)².
= [(a + b)(a − b)]² = (a + b)²(a − b)².
In simple words: Use the difference of squares twice in (i), and the identity (x − y)² in (ii), then factorise a² − b².
Teacher's Note:
1. Treat (x − z)² as a single quantity when using a² − b².
2. Check (i) with x = 2, z = 1: 16 − 1 = 15 and 1 × 3 × 5 = 15.
Section E: Case Study Based Questions (4 Marks Each)
Question 28. A retail shop owner wants to analyse customer footfall patterns to optimise store operations and staffing. The following graph shows the number of people present at the shop at different times of the day. By studying these trends, the owner aims to determine peak hours, plan staff shifts efficiently, and enhance customer service.
(a) What type of a graph is this? (1)
(b) What is the busiest time of day at the shop? (1)
(c) How many people are there in the shop at 2:30 pm? (1)
(d) About how many people are there in the shop at 1:30 pm? (1)
Answer:
(a) The points are joined by line segments, so it is a line graph.
(b) The highest point of the graph is 25 people at 1 pm, so 1 pm is the busiest time.
(c) At 2:30 pm (halfway between 2 pm and 3 pm) the graph is at 5, so there are 5 people.
(d) At 1:30 pm the graph is halfway between 25 (1 pm) and 15 (2 pm), so there are about 20 people.
In simple words: Find the time on the horizontal axis, go up to the graph, and read the number on the vertical axis.
Teacher's Note:
1. Each small square across is half an hour and each small square up is 2.5 people.
2. The highest point of a line graph shows the maximum value.
Question 29. A small indoor greenhouse (herbarium) is constructed entirely from glass panes, including the base, and held together with tape. It has dimensions of 30 cm in length, 25 cm in width, and 25 cm in height.
The greenhouse is used to grow small herbs and plants in a controlled environment. To ensure proper insulation and durability, the total surface area of the glass panes must be calculated to determine the amount of material required. Additionally, knowing the volume of the greenhouse helps in estimating the available space for plant growth.
(i) What is the area of the glass? (2)
(ii) How much of tape is needed for all the 12 edges? (2)
Answer:
Length l = 30 cm, breadth b = 25 cm, height h = 25 cm.
(i) All six faces are glass, so area = total surface area = 2(lb + bh + hl).
= 2(30 × 25 + 25 × 25 + 25 × 30) = 2(750 + 625 + 750) = 2 × 2125 = 4250 cm².
Area of the glass = 4250 cm².
(ii) A cuboid has 12 edges: 4 lengths, 4 breadths and 4 heights.
Tape = 4(l + b + h) = 4(30 + 25 + 25) = 4 × 80 = 320 cm.
In simple words: Add the areas of all six glass faces for (i); add all twelve edges for (ii).
Teacher's Note:
1. Total surface area is used because the base is glass too.
2. Edges are lengths (cm), while faces are areas (cm²).
Free study material for Mathematics
Download CBSE Sample Papers: Class 8 Mathematics
Get Started with CBSE Class 8 Mathematics Annual Exam Sample Paper 07 with Answers (Old NCERT Syllabus) (CBSE)
Explore downloadable sample sets for Class 8 Mathematics. Utilizing the CBSE Class 8 Mathematics Annual Exam Sample Paper 07 with Answers (Old NCERT Syllabus) allows learners to gauge exam readiness and master official CBSE assessment structures.
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- Concept Reinforcement: Consult the official NCERT book for Class 8 Mathematics when stuck before re-attempting problems.
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