CBSE Class 8 Mathematics Annual Exam Sample Paper 02 with Answers (Old NCERT Syllabus)

Class 8 Maths Annual Exam Sample Paper 02 is a full 60-mark paper with 29 questions in five sections (A to E), and every question comes with a step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it suits state-board students and anyone revising these chapters. The paper tests reading distance-time and temperature graphs, adding, subtracting and multiplying algebraic expressions, factorisation using identities, laws of exponents, direct and inverse proportion, and surface area and volume of cuboids, cubes and cylinders.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 16. Subtract 4p²q − 3pq + 5pq² − 8p + 7q − 10 from 18 − 3p − 11q + 5pq − 2pq² + 5p²q.
Answer:
Required = (18 − 3p − 11q + 5pq − 2pq² + 5p²q) − (4p²q − 3pq + 5pq² − 8p + 7q − 10)
= 18 − 3p − 11q + 5pq − 2pq² + 5p²q − 4p²q + 3pq − 5pq² + 8p − 7q + 10
Group like terms: (18 + 10) + (−3p + 8p) + (−11q − 7q) + (5pq + 3pq) + (−2pq² − 5pq²) + (5p²q − 4p²q)
= 28 + 5p − 18q + 8pq − 7pq² + p²q
In simple words: Change the sign of every term being subtracted, then add like terms.

Teacher's Note:
1. “Subtract A from B” means B − A, so write B first.
2. pq² and p²q are unlike terms; keep them apart.

Question 17. A carpenter makes a box which has a volume of 13,400 cm³. The base has an area of 670 cm². What is the height of the box?
Answer:
Volume of the box = area of base × height.
13400 = 670 × h
h = 13400 ÷ 670 = 20
Height of the box = 20 cm.
In simple words: Divide the volume by the base area to get the height.

Teacher's Note:
1. For any cuboid, volume = base area × height.
2. Check: 670 × 20 = 13400.

Question 18. Find the value of m for which \( 5^{m} \div 5^{-3} = 5^{5} \).
Answer:
Using \( a^{m} \div a^{n} = a^{m-n} \): \( 5^{m} \div 5^{-3} = 5^{m-(-3)} = 5^{m+3} \)
So \( 5^{m+3} = 5^{5} \).
Bases are equal, so the exponents are equal: m + 3 = 5.
m = 2
In simple words: Dividing powers of 5 means subtracting exponents; then match the exponents.

Teacher's Note:
1. Subtracting a negative exponent is the same as adding it.
2. Check: 5² ÷ 5⁻³ = 5⁵.

Question 19. Factorise: 25a² − 4b² + 28bc − 49c²
Answer:
25a² − 4b² + 28bc − 49c² = 25a² − (4b² − 28bc + 49c²)
= (5a)² − [(2b)² − 2(2b)(7c) + (7c)²]
= (5a)² − (2b − 7c)²
Using x² − y² = (x + y)(x − y):
= (5a + 2b − 7c)(5a − 2b + 7c)
In simple words: The last three terms form a perfect square, so the whole thing is a difference of two squares.

Teacher's Note:
1. Spot the hidden identity (a − b)² inside the bracket.
2. When removing the bracket after the minus sign, change each sign.

Question 20. A batch of bottles was packed in 25 boxes, with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?
Answer:
More bottles per box means fewer boxes, so the quantities are in inverse proportion.
Let the number of boxes be x. Then 12 × 25 = 20 × x.
x = 300 ÷ 20 = 15
15 boxes would be filled.
In simple words: The batch has 300 bottles; packing 20 in a box needs 15 boxes.

Teacher's Note:
1. In inverse proportion, the product of the two quantities stays the same.
2. Total bottles = 12 × 25 = 300 in both cases.

Section C: Short Answer Questions (3 Marks Each)

Question 21. Simplify: \( \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}} \)
Answer:
Write 10 = 2 × 5, 6 = 2 × 3 and 125 = 5³:
\( = \frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^{3}}{5^{-7} \times 2^{-5} \times 3^{-5}} \)
\( = 3^{-5+5} \times 2^{-5+5} \times 5^{-5+3+7} \)
\( = 3^{0} \times 2^{0} \times 5^{5} = 1 \times 1 \times 3125 \)
= 3125
In simple words: Break every number into prime factors; the powers of 2 and 3 cancel, leaving 5⁵ = 3125.

Teacher's Note:
1. Always split composite bases into primes before using exponent laws.
2. Any non-zero number to the power 0 is 1.

Teacher's Note:
1. Choose a scale so that all the data fit on the graph paper.
2. A straight line through the origin shows direct proportion.

Teacher's Note:
1. Persons × days stays constant for the same job.
2. Make a table first so you place the values correctly.

Question 24. Factorise: (i) p² + 6p − 16 (ii) x² − 10x + 24
Answer:
(i) Numbers with product −16 and sum 6 are 8 and −2.
p² + 6p − 16 = p² + 8p − 2p − 16 = p(p + 8) − 2(p + 8) = (p + 8)(p − 2)
(ii) Numbers with product 24 and sum −10 are −6 and −4.
x² − 10x + 24 = x² − 6x − 4x + 24 = x(x − 6) − 4(x − 6) = (x − 6)(x − 4)
In simple words: Split the middle term using two numbers that multiply to the constant and add to the middle coefficient.

Teacher's Note:
1. If the constant is positive and the middle term negative, both numbers are negative.
2. Expand your answer to check it.

Section D: Long Answer Questions (5 Marks Each)

Question 25. The following graph shows the temperature of a patient in a hospital, recorded every hour.
(a) What was the patient’s temperature at 1 p.m.?
(b) When was the patient’s temperature 38.5 °C?
(c) The patient’s temperature was the same two times during the period given. What were these two times?
(d) What was the temperature at 1.30 p.m.? How did you arrive at your answer?
(e) During which periods did the patient’s temperature show an upward trend?
Answer:
(a) At 1 p.m. the temperature was 36.5 °C.
(b) The temperature was 38.5 °C at 12 noon.
(c) The temperature was the same (36.5 °C) at 1 p.m. and 2 p.m.
(d) Between 1 p.m. and 2 p.m. the graph is a horizontal line at 36.5 °C. 1.30 p.m. is midway between 1 p.m. and 2 p.m. on the time axis, and the graph there is still at 36.5 °C. So the temperature at 1.30 p.m. was 36.5 °C.
(e) The graph rises from 9 a.m. to 11 a.m. and from 2 p.m. to 3 p.m., so the upward trend was during 9 a.m. to 11 a.m. and 2 p.m. to 3 p.m.
In simple words: Read each value by going up from the time to the line and across to the temperature scale.

Teacher's Note:
1. Each small square on the temperature axis is 0.5 °C here.
2. A rising line shows an upward trend; a flat line shows no change.
3. The zig-zag on the y-axis shows that the scale does not start from 0.

Question 26. Simplify: (i) (a + b)(c − d) + (a − b)(c + d) + 2(ac + bd)
(ii) (x + y)(2x + y) + (x + 2y)(x − y)
Answer:
(i) (a + b)(c − d) = ac − ad + bc − bd
(a − b)(c + d) = ac + ad − bc − bd
2(ac + bd) = 2ac + 2bd
Sum = (ac + ac + 2ac) + (−ad + ad) + (bc − bc) + (−bd − bd + 2bd) = 4ac
(ii) (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, then collect like terms; many terms cancel in part (i).

Teacher's Note:
1. Expand one product at a time and write it in a bracket.
2. Watch the signs: −bd − bd + 2bd = 0.

Question 27. Factorise: (i) a⁴ − b⁴ (ii) p⁴ − 81
Answer:
(i) a⁴ − b⁴ = (a²)² − (b²)² = (a² − b²)(a² + b²)
= (a − b)(a + b)(a² + b²)
(ii) p⁴ − 81 = (p²)² − 9² = (p² − 9)(p² + 9)
= (p² − 3²)(p² + 9) = (p − 3)(p + 3)(p² + 9)
In simple words: Use the difference of squares twice.

Teacher's Note:
1. After the first step, check if a factor is again a difference of squares.
2. A sum of squares like p² + 9 cannot be factorised further here.

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

Question 28. One day a Maths teacher was explaining direct and inverse proportion with the help of many examples. He gave the example of the relation between the length of a shadow and the height of an object such as a water tank. A water tank casts a shadow 21 m long. A tree of height 9.6 m casts a shadow 8 m long at the same time.
Answer the following questions based on the above information:
(a) Find the height of the water tank in the above situation.
(b) A water tank casts a shadow 27 m long. A tree of height 10.5 m casts a shadow 9 m long at the same time. The lengths of the shadows are directly proportional to their heights. Find the height of the tank.
(c) A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
(d) A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time, a tower casts a shadow of length 40 m on the ground. Determine the height of the tower.
Answer:
At the same time, height and shadow length are in direct proportion, so height ÷ shadow is the same for both objects.
(a) \( \frac{x}{21} \) = \( \frac{9.6}{8} \) ⇒ x = \( \frac{21 \times 9.6}{8} \) = \( \frac{201.6}{8} \) = 25.2 m
(b) \( \frac{x}{27} \) = \( \frac{10.5}{9} \) ⇒ x = \( \frac{27 \times 10.5}{9} \) = \( \frac{283.5}{9} \) = 31.5 m
(c) \( \frac{x}{28} \) = \( \frac{6}{4} \) ⇒ x = \( \frac{28 \times 6}{4} \) = 42 m
(d) \( \frac{x}{40} \) = \( \frac{12}{8} \) ⇒ x = \( \frac{40 \times 12}{8} \) = 60 m
In simple words: At one time of day, a taller object casts a longer shadow in the same ratio.

Teacher's Note:
1. Keep the same order in both ratios: height over shadow.
2. The tree (or pole) gives the ratio; use it to find the unknown height.
3. Check: 25.2 ÷ 21 = 1.2 = 9.6 ÷ 8.

Question 29. Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope (see the figure). She wanted to make a kaleidoscope of length 25 cm with a 7 cm diameter. You may take π = \( \frac{22}{7} \).
Answer the following questions based on the above information:
(i) Find the radius of the kaleidoscope. (1)
(ii) Find the area of chart paper required. (1)
(iii) Find the cost of chart paper if the rate of chart paper is Rs. 2 per 100 cm². (1)
(iv) Find the volume of the kaleidoscope. (1)
Answer:
(i) Radius = diameter ÷ 2 = 7 ÷ 2 = 3.5 cm
(ii) Chart paper covers the curved surface: CSA = 2πrh = 2 × \( \frac{22}{7} \) × 3.5 × 25 = 550 cm²
(iii) Cost = \( \frac{550}{100} \) × Rs. 2 = Rs. 11
(iv) Volume = πr²h = \( \frac{22}{7} \) × 3.5 × 3.5 × 25 = 962.5 cm³
In simple words: The paper wraps around the tube, so we need its curved surface area; the space inside is its volume.

Teacher's Note:
1. Only the curved surface is covered, so do not add the two circular ends.
2. Length of the kaleidoscope is the height h of the cylinder.
3. Area is in cm² and volume in cm³.