CBSE Class 8 Mathematics Annual Exam Sample Paper 08 with Answers (Old NCERT Syllabus)

Use this Class 8 Maths annual exam sample paper to practise a complete 60-mark exam: 29 questions across sections A to E, all solved step by step. The paper follows the earlier (old) NCERT Class 8 Maths syllabus and suits state-board students as well as anyone revising these chapters. It focuses on direct and inverse proportion in work, food and money problems, negative exponents and standard form, volume and surface area of cylinders and cuboids, adding, subtracting and simplifying algebraic expressions, factorisation and division, and reading a distance–time graph and plotting a cost graph.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 16. What should be subtracted from x⁴ + 2x² − 3x + 7 to get x³ + x² + x − 1?
Answer:
Required expression = (x⁴ + 2x² − 3x + 7) − (x³ + x² + x − 1).
= x⁴ + 2x² − 3x + 7 − x³ − x² − x + 1
= x⁴ − x³ + (2x² − x²) + (−3x − x) + (7 + 1)
= x⁴ − x³ + x² − 4x + 8
Check: (x⁴ + 2x² − 3x + 7) − (x⁴ − x³ + x² − 4x + 8) = x³ + x² + x − 1. ✓
In simple words: The amount to subtract is the starting expression minus the result you want.

Teacher's Note:
1. If A − ? = B, then ? = A − B.
2. Subtracting −1 gives +1, so the constant is 7 + 1 = 8.

Question 17. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Answer:
Total surface area of one suitcase = 2(lb + bh + hl) = 2(80 × 48 + 48 × 24 + 24 × 80).
= 2(3840 + 1152 + 1920) = 2 × 6912 = 13824 cm².
For 100 suitcases: 13824 × 100 = 1382400 cm².
Length of tarpaulin = area ÷ width = 1382400 ÷ 96 = 14400 cm.
= 144 m of tarpaulin.
In simple words: Find the total cloth area needed, then divide by the 96 cm width to get the length.

Teacher's Note:
1. Area of a strip of cloth = length × width, so length = area ÷ width.
2. Change cm to m at the end: 14400 cm = 144 m.

Question 18. Find the value of x for which (−2)³ × (−2)⁻⁶ = −2²ˣ⁻¹.
Answer:
(−2)³ × (−2)⁻⁶ = (−2)³⁺⁽⁻⁶⁾ = (−2)⁻³.
(−2)⁻³ = \( \frac{1}{(-2)^3} \) = \( \frac{-1}{8} \) = −2⁻³.
So −2⁻³ = −2²ˣ⁻¹, which gives 2x − 1 = −3.
2x = −2, so x = −1.
In simple words: Add the powers on the left, write the answer as −2 to a power, and compare the powers.

Teacher's Note:
1. An odd power of a negative number is negative.
2. Check: 2x − 1 = 2(−1) − 1 = −3.

Question 19. Factorise: 2ax² + 4axy + 3bx² + 2ay² + 6bxy + 3by²
Answer:
Group the terms with a and the terms with b:
= (2ax² + 4axy + 2ay²) + (3bx² + 6bxy + 3by²)
= 2a(x² + 2xy + y²) + 3b(x² + 2xy + y²)
= (2a + 3b)(x² + 2xy + y²) = (2a + 3b)(x + y)²
In simple words: Group by a and b, take out the common bracket, and notice it is the perfect square (x + y)².

Teacher's Note:
1. Regrouping is allowed because addition is commutative.
2. Always check whether the remaining bracket is a known identity.

Question 20. In a boys hostel food provision for 50 boys lasts for 40 days. If 30 more boys join the hostel, how long will this provision last?
Answer:
New number of boys = 50 + 30 = 80.
More boys means the food lasts fewer days, so this is inverse variation.
50 × 40 = 80 × x, so x = \( \frac{2000}{80} \) = 25.
The provision will last 25 days.
In simple words: The food is enough for 2000 boy-days; shared by 80 boys it lasts 25 days.

Teacher's Note:
1. Inverse variation: boys × days stays constant.
2. Add the new boys to the original 50 before calculating.

Section C: Short Answer Questions (3 Marks Each)

Question 21. Simplify: \( \frac{49 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}} \) (z ≠ 0)
Answer:
Write 49 = 7².
\( \frac{7^2 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}} = \frac{7^{2-(-3)} \times z^{-3-(-5)}}{10} \).
\( = \frac{7^{5} \times z^{2}}{10} = \frac{16807\,z^2}{10} \).
Answer: \( \frac{7^5 z^2}{10} = \frac{16807 z^2}{10} \).
In simple words: Bring the powers of 7 and of z together and subtract the powers when dividing.

Teacher's Note:
1. Dividing by a negative power is the same as multiplying by a positive power.
2. 7⁵ = 16807.

Question 22. If 52 men can do a piece of work in 35 days:
(a) In how many days 28 men will do it?
(b) How many men will do it in 14 days?
Answer:
The number of men and the number of days vary inversely: men × days = 52 × 35 = 1820.
(a) 28 × x = 1820, so x = \( \frac{1820}{28} \) = 65. 28 men will do it in 65 days.
(b) y × 14 = 1820, so y = \( \frac{1820}{14} \) = 130. 130 men will do it in 14 days.
In simple words: The work needs 1820 man-days; divide by the number of men or by the number of days.

Teacher's Note:
1. Fewer men need more days; fewer days need more men.
2. Check: 28 × 65 = 1820 and 130 × 14 = 1820.

Question 23. The students of Anju's class sold posters to raise money. Anju wanted to create a ratio for finding the amount of money her class would make for different numbers of posters sold. She knew they could raise ₹ 250 for every 60 posters sold.
(a) How much money would Anju's class make for selling 102 posters?
(b) Could Anju's class raise exactly ₹ 2,000? If so, how many posters would they need to sell? If not, why?
Answer:
More posters raise more money, so the money varies directly with the number of posters.
(a) \( \frac{250}{60} = \frac{x}{102} \), so x = \( \frac{250 \times 102}{60} \) = 425.
The class would make ₹ 425.
(b) \( \frac{250}{60} = \frac{2000}{n} \), so n = \( \frac{2000 \times 60}{250} \) = 480.
480 is a whole number of posters, so yes, the class can raise exactly ₹ 2,000 by selling 480 posters.
In simple words: ₹ 250 for 60 posters means ₹ 2000 needs 8 times as many posters: 8 × 60 = 480.

Teacher's Note:
1. In direct variation, the ratio money ÷ posters stays constant.
2. The answer "yes" depends on getting a whole number of posters.

Question 24. Factorise: (i) x² + 19x − 20 (ii) x² − 7x + 10
Answer:
(i) Numbers with product −20 and sum 19: 20 and −1.
x² + 20x − x − 20 = x(x + 20) − 1(x + 20) = (x + 20)(x − 1).
(ii) Numbers with product 10 and sum −7: −5 and −2.
x² − 5x − 2x + 10 = x(x − 5) − 2(x − 5) = (x − 5)(x − 2).
In simple words: Split the middle term into two parts whose product equals the constant term.

Teacher's Note:
1. A negative constant means the two numbers have opposite signs.
2. Check by expanding the brackets.

Section D: Long Answer Questions (5 Marks Each)

Question 25. Simplify: (a) (a + b)(c − d) + (a − b)(c + d) + 2(ac + bd)
(b) (x + y)(2x + y) + (x + 2y)(x − y)
Answer:
(a) (a + b)(c − d) = ac − ad + bc − bd.
(a − b)(c + d) = ac + ad − bc − bd.
Adding: 2ac − 2bd. Then add 2(ac + bd) = 2ac + 2bd.
Total = 2ac − 2bd + 2ac + 2bd = 4ac.
(b) (x + y)(2x + y) = 2x² + xy + 2xy + y² = 2x² + 3xy + y².
(x + 2y)(x − y) = x² − xy + 2xy − 2y² = x² + xy − 2y².
Sum = 3x² + 4xy − y².
In simple words: Expand each product separately, then add and cancel the like terms.

Teacher's Note:
1. In (a), the ad, bc and bd terms all cancel.
2. Expanding each bracket on its own line keeps the work tidy.

Teacher's Note:
1. Cost and quantity are in direct proportion, so the graph is a straight line through (0, 0).
2. Draw the guide lines with a ruler to read the value accurately.

Question 27. Factorise the expressions and divide them as directed.
(a) (5p² − 25p + 20) ÷ (p − 1) (b) 4yz(z² + 6z − 16) ÷ 2y(z + 8)
Answer:
(a) 5p² − 25p + 20 = 5(p² − 5p + 4) = 5(p − 4)(p − 1).
\( \frac{5(p - 4)(p - 1)}{(p - 1)} = 5(p - 4) \). Answer: 5(p − 4).
(b) z² + 6z − 16 = z² + 8z − 2z − 16 = (z + 8)(z − 2).
\( \frac{4yz(z + 8)(z - 2)}{2y(z + 8)} = 2z(z - 2) \). Answer: 2z(z − 2).
In simple words: Factorise the top so that the divisor appears as a factor, then cancel it.

Teacher's Note:
1. Take out the common number 5 first; the quadratic becomes easier.
2. 4yz ÷ 2y = 2z.

Section E: Case Study Based Questions (4 Marks Each)

Question 28. A man started his journey in a car from Location A, travelled to a destination, and then returned to the starting point. The given graph represents his position at different times throughout the journey. The graph provides insights into his travel pattern, including movement, pauses, and speed variations. Understanding such motion graphs is crucial in real-life applications such as traffic analysis, vehicle tracking, and travel time estimation. By analysing the given data, one can determine when the car was moving, when it stopped, and how the speed varied at different intervals.
(a) At what time did he start and end his journey? (1)
(b) What was the total duration of journey? (1)
(c) Which journey, forward or return, was of longer duration? (1)
(d) For how many hours did he not move? (1)
Answer:
(a) The graph leaves 0 km at 5:30 am and comes back to 0 km at 6 pm. He started at 5:30 am and ended at 6 pm.
(b) From 5:30 am to 6 pm is 12 hours 30 minutes.
(c) Forward journey (to the farthest point, 100 km at 2 pm): 5:30 am to 2 pm = 8 h 30 min. Return: 2 pm to 6 pm = 4 h.
So the forward journey took longer.
(d) The graph is flat from 6 am to 9 am (3 h) and from 10 am to 1 pm (3 h). He did not move for 3 + 3 = 6 hours.
In simple words: A rising line means moving away, a falling line means coming back, and a flat line means the car was standing still.

Teacher's Note:
1. On a distance–time graph, a horizontal segment means no movement.
2. Each small square across is half an hour, which helps to read 5:30 am and 1:30 pm.

Question 29. A Vidyalaya organised a competition where students were asked to make and decorate penholders using cardboard. Each penholder was designed in the shape of a cylinder with a base, having a radius of 3 cm and a height of 10.5 cm. The school was responsible for providing the required cardboard to all participants. A total of 35 students participated in the competition. The school administration needs to determine the total amount of cardboard required for the event while ensuring minimum wastage.
(a) How much cardboard was required to be bought for the competition? (2)
(b) Find the total volume of all the penholders. (2)
Answer:
Radius r = 3 cm, height h = 10.5 cm. Each penholder is open at the top, so it has a curved surface and one base.
(a) Cardboard for one penholder = 2πrh + πr² = πr(2h + r) = \( \frac{22}{7} \) × 3 × (21 + 3) = \( \frac{22}{7} \) × 3 × 24 cm².
For 35 penholders: \( \frac{22}{7} \) × 3 × 24 × 35 = 22 × 3 × 24 × 5 = 7920 cm².
Cardboard required = 7920 cm².
(b) Volume of one penholder = πr²h = \( \frac{22}{7} \) × 3 × 3 × 10.5 = 297 cm³.
Total volume = 35 × 297 = 10395 cm³.
In simple words: Each penholder needs its curved side plus one circular base; multiply one penholder's area and volume by 35.

Teacher's Note:
1. There is no top lid, so only one πr² is added.
2. Cancel the 7 in \( \frac{22}{7} \) with 35 early: 35 ÷ 7 = 5.