CBSE Class 8 Mathematics Annual Exam Sample Paper 06 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), each with a clear step-by-step answer. 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 adding, subtracting and multiplying algebraic expressions, factorisation and division of polynomials, negative exponents and standard form, direct and inverse proportion, area, volume and surface area in mensuration, and drawing and reading line graphs.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 16. Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz.
Answer:
(5xy − 2yz − 2zx + 10xyz) − (3xy + 5yz − 7zx)
= 5xy − 2yz − 2zx + 10xyz − 3xy − 5yz + 7zx
= (5xy − 3xy) + (−2yz − 5yz) + (−2zx + 7zx) + 10xyz
= 2xy − 7yz + 5zx + 10xyz
In simple words: Change the sign of every term being subtracted, then add the like terms.

Teacher's Note:
1. "Subtract A from B" means B − A, so the second expression is written first.
2. 10xyz has no like term, so it is carried down unchanged.

Question 17. Find the side of a cube whose surface area is 600 cm².
Answer:
Surface area of a cube = 6 × (side)².
6 × (side)² = 600, so (side)² = 100.
side = √100 = 10 cm.
So the side of the cube is 10 cm.
In simple words: Divide 600 by the 6 faces to get 100 cm² per face; the side is √100 = 10 cm.

Teacher's Note:
1. A cube has 6 equal square faces.
2. Check: 6 × 10 × 10 = 600 cm².

Question 18. Express the number appearing in the following statements in standard form.
(i) Charge of an electron is 0.000 000 000 000 000 000 16 coulomb.
(ii) Size of bacteria is 0.0000005 m.
Answer:
(i) 0.000 000 000 000 000 000 16 has 18 zeros after the decimal point before 16.
Moving the point 19 places to the right gives 1.6, so the charge = 1.6 × 10⁻¹⁹ coulomb.
(ii) 0.0000005: moving the point 7 places to the right gives 5.
So the size of bacteria = 5 × 10⁻⁷ m.
In simple words: Move the decimal point to just after the first non-zero digit and count the moves; a move to the right gives a negative power.

Teacher's Note:
1. Count carefully: write the digits in groups of three to avoid errors.
2. In standard form the number before × 10 is at least 1 and less than 10.

Question 19. Factorise: 16x⁴ − 625y⁴
Answer:
16x⁴ − 625y⁴ = (4x²)² − (25y²)².
Using a² − b² = (a + b)(a − b): = (4x² + 25y²)(4x² − 25y²).
4x² − 25y² = (2x)² − (5y)² = (2x + 5y)(2x − 5y).
So 16x⁴ − 625y⁴ = (4x² + 25y²)(2x + 5y)(2x − 5y).
In simple words: Use the difference of squares twice; the sum of squares 4x² + 25y² cannot be split further.

Teacher's Note:
1. Always check whether a factor can be factorised again.
2. A sum of two squares like 4x² + 25y² has no real factors.

Question 20. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?
Answer:
The model is made to scale, so lengths in the model vary directly with lengths of the ship.
Let the length of the model ship be x cm.
\( \frac{\text{model length}}{\text{ship length}} \) = \( \frac{\text{model mast}}{\text{ship mast}} \): \( \frac{x}{28} = \frac{9}{12} \).
x = \( \frac{9 \times 28}{12} \) = 21.
The model mast is in cm, so the model ship is 21 cm long.
In simple words: The model is 9 cm for every 12 m of the real ship, so 28 m of ship becomes 21 cm in the model.

Teacher's Note:
1. Keep the units: model lengths are in cm and real lengths are in m.
2. A 21 m model would be almost as long as the real ship, so cm is the sensible unit.

Section C: Short Answer Questions (3 Marks Each)

Question 21. Simplify: (a) \( \left[\left(\frac{1}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1}\right]^{-1} \) (b) \( \left(\frac{5}{8}\right)^{-7} \times \left(\frac{8}{5}\right)^{-4} \)
Answer:
(a) \( \left(\frac{1}{3}\right)^{-1} = 3 \) and \( \left(\frac{1}{4}\right)^{-1} = 4 \).
\( \left[3 - 4\right]^{-1} = (-1)^{-1} = \frac{1}{-1} = -1 \). Answer: −1.
(b) \( \left(\frac{5}{8}\right)^{-7} = \left(\frac{8}{5}\right)^{7} \).
\( \left(\frac{8}{5}\right)^{7} \times \left(\frac{8}{5}\right)^{-4} = \left(\frac{8}{5}\right)^{7 - 4} = \left(\frac{8}{5}\right)^{3} \).
\( = \frac{8^3}{5^3} = \frac{512}{125} \). Answer: \( \frac{512}{125} \).
In simple words: A negative power flips the fraction; then use the rules for powers with the same base.

Teacher's Note:
1. \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n} \).
2. In (b), first make both bases the same (8/5), then add the powers 7 + (−4) = 3.

Teacher's Note:
1. Change all lengths to the same unit (cm) before setting up the proportion.
2. 560 ÷ 320 = 1.75; you can use this ratio as a quick check.

Question 23. (a) 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 the children is reduced by 4?
(b) A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Answer:
(a) Fewer children means more sweets each, so this is inverse variation.
New number of children = 24 − 4 = 20. Let each get x sweets.
24 × 5 = 20 × x, so x = \( \frac{120}{20} \) = 6.
Each child would get 6 sweets.
(b) More animals means the food lasts fewer days, so this is inverse variation.
New number of animals = 20 + 10 = 30. Let the food last y days.
20 × 6 = 30 × y, so y = \( \frac{120}{30} \) = 4.
The food would last 4 days.
In simple words: In both parts the total (120 sweets, 120 animal-days of food) stays the same, so divide it by the new number.

Teacher's Note:
1. For inverse variation, multiply the two quantities and keep the product fixed.
2. Read carefully: "reduced by 4" and "10 more" change the numbers to 20 and 30.

Question 24. Factorise: (i) 121b² − 88bc + 16c² (ii) a⁴ + 2a²b² + b⁴ (iii) (l + m)² − 4lm
Answer:
(i) 121b² − 88bc + 16c² = (11b)² − 2 × 11b × 4c + (4c)².
Using a² − 2ab + b² = (a − b)²: = (11b − 4c)².
(ii) a⁴ + 2a²b² + b⁴ = (a²)² + 2 × a² × b² + (b²)².
Using a² + 2ab + b² = (a + b)²: = (a² + b²)².
(iii) (l + m)² − 4lm = l² + 2lm + m² − 4lm = l² − 2lm + m².
= (l − m)².
In simple words: Each expression is a perfect square; spot the two squares and check the middle term is twice their product.

Teacher's Note:
1. Check the middle term: 2 × 11b × 4c = 88bc.
2. In (iii), expand first and then recognise the identity.

Section D: Long Answer Questions (5 Marks Each)

Question 25. Obtain the volume of rectangular boxes with the following length, breadth and height, respectively.
(i) 5a, 3a², 7a⁴ (ii) 2p, 4q, 8r (iii) xy, 2x²y, 2xy² (iv) a, 2b, 3c (v) xy, yz, zx
Answer:
Volume of a cuboid = length × breadth × height, so multiply the three monomials (numbers first, then variables).
(i) 5a × 3a² × 7a⁴ = (5 × 3 × 7)(a¹⁺²⁺⁴) = 105a⁷.
(ii) 2p × 4q × 8r = (2 × 4 × 8) pqr = 64pqr.
(iii) xy × 2x²y × 2xy² = (1 × 2 × 2)(x¹⁺²⁺¹)(y¹⁺¹⁺²) = 4x⁴y⁴.
(iv) a × 2b × 3c = (1 × 2 × 3) abc = 6abc.
(v) xy × yz × zx = x × x × y × y × z × z = x²y²z².
In simple words: Multiply the numbers together and add the powers of each letter.

Teacher's Note:
1. When the same letter appears more than once, add its powers.
2. A letter with no power written has power 1.

Teacher's Note:
1. Choose scales so that all the points fit and are easy to read.
2. Dashed guide lines from the axes to the graph make the readings clear.

Question 27. Factorise the expressions and divide them as directed. (1 + 2 + 2)
(a) 10y(6y + 21) ÷ 5(2y + 7)
(b) 9x²y²(3z − 24) ÷ 27xy(z − 8)
(c) 96abc(3a − 12)(5b − 30) ÷ 144(a − 4)(b − 6)
Answer:
(a) 6y + 21 = 3(2y + 7).
\( \frac{10y \times 3(2y + 7)}{5(2y + 7)} = \frac{30y}{5} = 6y \). Answer: 6y.
(b) 3z − 24 = 3(z − 8).
\( \frac{9x^2y^2 \times 3(z - 8)}{27xy(z - 8)} = \frac{27x^2y^2}{27xy} = xy \). Answer: xy.
(c) 3a − 12 = 3(a − 4) and 5b − 30 = 5(b − 6).
\( \frac{96abc \times 3(a - 4) \times 5(b - 6)}{144(a - 4)(b - 6)} = \frac{1440abc}{144} = 10abc \). Answer: 10abc.
In simple words: Take out the common number from each bracket so that the same bracket appears above and below, then cancel it.

Teacher's Note:
1. The aim of factorising is to make a common factor that can be cancelled.
2. 96 × 3 × 5 = 1440 and 1440 ÷ 144 = 10.

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

Question 28. A team of scientists is studying the effect of heat on different substances. To analyse the process of phase changes, they conducted an experiment by heating a block of ice and recording its temperature at regular intervals. The following graph shows the change in temperature of a block of ice when heated. Use the graph to answer the following questions:
(a) For how many seconds at the start of heating did the ice block have no change in temperature? (1)
(b) For how long was there a change in temperature? (1)
(c) After how many seconds of heating did the temperature become constant at 100°C? (1)
(d) What will be the temperature after 1.5 minutes? Justify your answer. (1)
Answer:
(a) From 0 to 20 seconds the graph stays on 0°C, so the temperature did not change for the first 20 seconds.
(b) The temperature rises from 20 s to 50 s, so it changed for 50 − 20 = 30 seconds.
(c) The graph becomes flat at 100°C at 50 seconds.
(d) 1.5 minutes = 90 seconds. After 50 s the temperature stays at 100°C (the water is boiling), so at 90 s it will still be 100°C.
In simple words: Flat parts of the graph mean no change in temperature; the sloping part shows the temperature rising.

Teacher's Note:
1. A horizontal segment on a temperature–time graph means the temperature is constant.
2. Convert minutes to seconds before reading or extending the graph.

Question 29. A manufacturing company specialises in producing custom plastic storage boxes for various industrial and household applications. To design an efficient and cost-effective box, they have planned to create a rectangular plastic box with the following dimensions: Length = 1.5 m, Width = 1.25 m, Depth = 65 cm (0.65 m).
The box is open at the top, meaning only the bottom and four side walls require material. The company aims to determine the amount of plastic sheet required to construct the box efficiently while keeping costs minimal.
(i) Find the area of the sheet required for making the box. (2)
(ii) Find the cost of sheet for it, if a sheet measuring 1 m² costs ₹ 20. (2)
Answer:
Length l = 1.5 m, breadth b = 1.25 m, height h = 65 cm = 0.65 m.
(i) The box is open at the top, so it has the base and four walls only.
Area = lb + 2h(l + b) = 1.5 × 1.25 + 2 × 0.65 × (1.5 + 1.25).
= 1.875 + 1.3 × 2.75 = 1.875 + 3.575 = 5.45 m².
Sheet required = 5.45 m².
(ii) Cost = 5.45 × ₹ 20 = ₹ 109.
In simple words: Add the base area to the area of the four walls, then multiply the total by ₹ 20.

Teacher's Note:
1. The four walls together have area 2h(l + b), like the lateral surface of a cuboid.
2. Change 65 cm to 0.65 m so that all lengths are in metres.