CBSE Class 8 Mathematics Annual Exam Sample Paper 04 with Answers (Old NCERT Syllabus)

This Class 8 Maths annual exam sample paper carries 60 marks and has 29 questions in five sections (A to E), with a step-by-step answer for every question. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it is useful for state-board students and for revising these chapters before the final exam. The paper tests like terms and products of algebraic expressions, area of a trapezium and rhombus, surface area of cylinders, direct and inverse proportion, negative exponents and standard form, factorisation and division of polynomials, drawing the graph of y = 6x, and two case studies on a cyclist's distance–time graph and the cardboard needed for cylindrical penholders.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 16. Subtract: 3l(l − 4m + 5n) from 4l(10n − 3m + 2l)
Answer:
Required expression = 4l(10n − 3m + 2l) − 3l(l − 4m + 5n)
First expand: 4l(10n − 3m + 2l) = 40ln − 12lm + 8l² and 3l(l − 4m + 5n) = 3l² − 12lm + 15ln
= (40ln − 12lm + 8l²) − (3l² − 12lm + 15ln)
= 40ln − 12lm + 8l² − 3l² + 12lm − 15ln
= 5l² + 25ln
In simple words: Open both brackets, change the signs of the expression being subtracted, and combine like terms.

Teacher's Note:
1. 'Subtract A from B' means B − A, so the second expression is written first.
2. The −12lm and +12lm cancel out completely.

Question 17. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.
Answer:
A rhombus is a parallelogram, so Area = base × altitude = 5 × 4.8 = 24 cm².
Also, Area of rhombus = \( \frac{1}{2} \) × d₁ × d₂.
24 = \( \frac{1}{2} \) × 8 × d₂ ⇒ 24 = 4d₂ ⇒ d₂ = 6
The other diagonal is 6 cm.
In simple words: Find the area with base × height, then use half the product of diagonals to get the missing diagonal.

Teacher's Note:
1. A rhombus has two area formulas; this question links them.
2. Check: half-diagonals 4 cm and 3 cm give side √(16 + 9) = 5 cm, as stated.

Question 18. Find the value of x for which 2^x ÷ 2⁻⁴ = 4⁵
Answer:
Write 4 as 2²: 4⁵ = (2²)⁵ = 2¹⁰
Left side: 2^x ÷ 2⁻⁴ = 2^{x − (−4)} = 2^{x + 4}
So 2^{x + 4} = 2¹⁰ ⇒ x + 4 = 10
x = 6
In simple words: Make both sides powers of 2, then the exponents must be equal.

Teacher's Note:
1. Subtracting a negative exponent means adding it: x − (−4) = x + 4.
2. Equal bases allow you to equate the exponents.

Question 19. Factorise: ab(x² + y²) − xy(a² + b²)
Answer:
Expand: abx² + aby² − a²xy − b²xy
Regroup: (abx² − a²xy) + (aby² − b²xy)
= ax(bx − ay) + by(ay − bx)
= ax(bx − ay) − by(bx − ay)
= (bx − ay)(ax − by)
In simple words: Open the brackets, pair the terms so each pair has a common factor, then take out the common bracket.

Teacher's Note:
1. (ay − bx) = −(bx − ay); changing the sign this way gives a common bracket.
2. Multiply the factors back to check your answer.

Question 20. A car can finish a certain journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance?
Answer:
For a fixed distance, speed and time are in inverse proportion, so speed × time is constant.
Let the new speed be (48 + x) km/h. Then 48 × 10 = (48 + x) × 8
480 = 384 + 8x ⇒ 8x = 96 ⇒ x = 12
The speed should be increased by 12 km/h (new speed 60 km/h).
In simple words: The distance is 480 km; covering it in 8 hours needs 60 km/h, which is 12 km/h more.

Teacher's Note:
1. The question asks for the increase, not the new speed.
2. Less time needs more speed, so the relation is inverse.

Section C: Short Answer Questions (3 Marks Each)

Question 21. Simplify: (a) \( \left[\left(\frac{1}{2}\right)^2 - \left(\frac{1}{4}\right)^3\right] \times 2^3 \)    (b) \( (3^2 - 2^2) \div \left(\frac{1}{5}\right)^2 \)
Answer:
(a) \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \) and \( \left(\frac{1}{4}\right)^3 = \frac{1}{64} \)
\( \left[\frac{1}{4} - \frac{1}{64}\right] \times 8 = \frac{16 - 1}{64} \times 8 = \frac{15}{64} \times 8 = \frac{15}{8} \)
Answer: \( \frac{15}{8} \)
(b) 3² − 2² = 9 − 4 = 5 and \( \left(\frac{1}{5}\right)^2 = \frac{1}{25} \)
5 ÷ \( \frac{1}{25} \) = 5 × 25 = 125
In simple words: Work out each power first, then do the subtraction inside the bracket, and finally multiply or divide.

Teacher's Note:
1. (1/4)³ means 1/(4 × 4 × 4) = 1/64, not 3/4.
2. Dividing by a fraction is the same as multiplying by its reciprocal.

Question 22. Which of the following quantities vary inversely as each other?
(i) The number of x men hired to construct a wall and the time y taken to finish the job.
(ii) The length x of a journey by bus and price y of the ticket.
(iii) Journey (x km) undertaken by a car and the petrol (y litres) consumed by it.
Answer:
(i) More men finish the wall in less time; as x increases, y decreases. This is inverse variation.
(ii) A longer journey costs more; as x increases, y also increases. Not inverse.
(iii) A longer journey uses more petrol; x and y increase together. Not inverse.
Hence only (i) shows inverse variation.
In simple words: Ask: if one quantity goes up, does the other go down? Only men and time behave that way.

Teacher's Note:
1. Inverse variation: one increases while the other decreases, keeping their product constant.
2. (ii) and (iii) are examples of direct variation.

Question 23. If x and y vary inversely as each other and
(i) x = 3 when y = 8, find y when x = 4
(ii) x = 5 when y = 15, find x when y = 12
(iii) x = 30, find y when constant of variation = 900.
Answer:
For inverse variation, xy = k (constant).
(i) k = 3 × 8 = 24. When x = 4: 4y = 24 ⇒ y = 6
(ii) k = 5 × 15 = 75. When y = 12: 12x = 75 ⇒ x = \( \frac{75}{12} \) = \( \frac{25}{4} \) (6.25)
(iii) k = 900. 30y = 900 ⇒ y = 30
In simple words: Find the constant product xy first, then divide it by the known value to get the unknown.

Teacher's Note:
1. The constant of variation in inverse proportion is the product xy.
2. Leave answers like 75/12 in lowest terms: 25/4.

Question 24. Factorise: (i) y² + 10y + 24   (ii) x² + 5x + 6
Answer:
(i) Find two numbers with sum 10 and product 24: 6 and 4.
y² + 6y + 4y + 24 = y(y + 6) + 4(y + 6) = (y + 6)(y + 4)
(ii) Two numbers with sum 5 and product 6: 3 and 2.
x² + 3x + 2x + 6 = x(x + 3) + 2(x + 3) = (x + 3)(x + 2)
In simple words: Split the middle term using two numbers that add to the middle coefficient and multiply to the constant.

Teacher's Note:
1. Check by expanding: (y + 6)(y + 4) = y² + 10y + 24.
2. This splitting method works whenever the coefficient of the square term is 1.

Section D: Long Answer Questions (5 Marks Each)

Question 25. Multiply (i) (x − 4) and (2x + 3)   (ii) (x − y) and (3x + 5y)
Answer:
(i) (x − 4)(2x + 3) = x(2x + 3) − 4(2x + 3)
= 2x² + 3x − 8x − 12
= 2x² − 5x − 12
(ii) (x − y)(3x + 5y) = x(3x + 5y) − y(3x + 5y)
= 3x² + 5xy − 3xy − 5y²
= 3x² + 2xy − 5y²
In simple words: Multiply every term of the first bracket by every term of the second, then add the like terms.

Teacher's Note:
1. xy and yx are like terms, so 5xy − 3xy = 2xy.
2. Watch the sign when the second term of the first bracket is negative.

Teacher's Note:
1. A relation of the form y = kx always gives a straight line through the origin.
2. Choose a scale on the y-axis so that the largest value fits on the sheet.

Question 27. Factorise the expressions and divide them as directed.
(i) 4yz(z² + 6z − 16) ÷ 2y(z + 8)   (ii) 12xy(9x² − 16y²) ÷ 4xy(3x + 4y)
Answer:
(i) z² + 6z − 16 = z² + 8z − 2z − 16 = z(z + 8) − 2(z + 8) = (z + 8)(z − 2)
So 4yz(z² + 6z − 16) = 4yz(z + 8)(z − 2)
4yz(z + 8)(z − 2) ÷ 2y(z + 8) = \( \frac{4yz(z + 8)(z - 2)}{2y(z + 8)} \) = 2z(z − 2)
(ii) 9x² − 16y² = (3x)² − (4y)² = (3x + 4y)(3x − 4y)
So 12xy(9x² − 16y²) = 12xy(3x + 4y)(3x − 4y)
12xy(3x + 4y)(3x − 4y) ÷ 4xy(3x + 4y) = 3(3x − 4y)
In simple words: Factorise the dividend fully, then cancel the factors it shares with the divisor.

Teacher's Note:
1. Look for a common factor or the identity a² − b² = (a + b)(a − b) before dividing.
2. The answer 3(3x − 4y) can also be written as 9x − 12y.

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

Question 28. The following graph represents the journey of a cyclist over a period of 6 hours, starting from an initial point. The x-axis represents time in hours, while the y-axis represents the distance of the cyclist from the starting point in kilometres. The journey commenced at 9:00 AM, and the graph illustrates variations in speed, possible breaks, and overall progress. Analyze the graph carefully to understand the cyclist's speed patterns, rest periods, and total distance covered during the journey.
Observe the graph carefully and answer the following questions:
(a) What was happening between B and C? (1)
(b) During which of the periods AB, BC, CD and DE did the cyclist maintain his greatest speed? (2)
(c) What was average speed from A to E? (1)
Answer:
From the graph: A (9:00, 0 km), B (10:00, 20 km), C (10:30, 20 km), D (12:00, 40 km), E (15:00, 0 km).
(a) From B to C the distance stays at 20 km for half an hour, so the cyclist was at rest (taking a break).
(b) AB: 20 km in 1 h = 20 km/h. BC: 0 km in \( \frac{1}{2} \) h = 0 km/h.
CD: (40 − 20) km in \( \frac{3}{2} \) h = \( \frac{40}{3} \) ≈ 13.33 km/h. DE: 40 km (coming back) in 3 h = \( \frac{40}{3} \) ≈ 13.33 km/h.
The greatest speed was during AB (20 km/h).
(c) Total distance travelled = 20 + 0 + 20 + 40 = 80 km; total time = 6 h.
Average speed = \( \frac{80}{6} \) = \( \frac{40}{3} \) ≈ 13.33 km/h.
In simple words: A flat part of a distance–time graph means no movement, and the steepest part shows the fastest speed.

Teacher's Note:
1. The steeper the line, the greater the speed.
2. For average speed, use the total distance actually travelled (80 km), including the return, and the total time including rest.

Question 29. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. There were 35 competitors to participate in this competition. The cost of cardboard is Rs. 20 per 10 square cm.
Based on the above situation, answer the following questions:
(a) Find the lateral surface area of a pen holder. (1)
(b) Find the total surface area of a penholder. (1)
(c) Find the area of cardboard sheet used by 35 competitors. (1)
(d) What is the cost of cardboard? (1)
Answer:
(a) Lateral (curved) surface area = 2πrh = 2 × \( \frac{22}{7} \) × 3 × 10.5 = 198 cm²
(b) A penholder is open at the top, so it has the curved surface and one base.
Area of base = πr² = \( \frac{22}{7} \) × 9 = \( \frac{198}{7} \) cm²
Total surface area = 198 + \( \frac{198}{7} \) = \( \frac{1584}{7} \) cm² ≈ 226.29 cm²
(c) Cardboard for 35 penholders = 35 × \( \frac{1584}{7} \) = 7920 cm²
(d) Cost = \( \frac{7920}{10} \) × Rs. 20 = 792 × 20 = Rs. 15840
In simple words: Find the cardboard for one open cylinder (side plus bottom), multiply by 35, then charge Rs. 20 for every 10 cm².

Teacher's Note:
1. A penholder has no top lid, so only one circular base is added.
2. The original answer key gave Rs. 1584 for (d); the correct cost is 792 × Rs. 20 = Rs. 15840.