CBSE Class 8 Mathematics Annual Exam Sample Paper 01 with Answers (Old NCERT Syllabus)

This Class 8 Maths annual exam sample paper is a full 60-mark paper with 29 questions in five sections (A to E), and every question has a 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 the second-half chapters: Mensuration, Algebraic Expressions and Identities, Exponents and Powers, Direct and Inverse Proportions, Factorisation and Introduction to Graphs. It tests reading and drawing line graphs, surface area and volume, laws of exponents, factorising quadratics and differences of squares, and direct and inverse variation through two case studies.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 16. Factorise: 100 − (x − 5)²
Answer:
Write 100 as 10²: 100 − (x − 5)² = 10² − (x − 5)².
Use a² − b² = (a + b)(a − b) with a = 10 and b = x − 5:
= [10 + (x − 5)][10 − (x − 5)] = (10 + x − 5)(10 − x + 5)
= (x + 5)(15 − x).
In simple words: It is a difference of two squares, 10² − (x − 5)², so it splits into a sum and a difference.

Teacher's Note:
1. Keep (x − 5) in brackets so the minus sign is applied to both terms.
2. Check with x = 5: 100 − 0 = 100 and 10 × 10 = 100.

Question 17. Subtract the following polynomials.
3xy + 5yz − 7xz + 1 from −4xy + 2yz − 2xz + 5xyz + 1
Answer:
(−4xy + 2yz − 2xz + 5xyz + 1) − (3xy + 5yz − 7xz + 1)
= −4xy + 2yz − 2xz + 5xyz + 1 − 3xy − 5yz + 7xz − 1
= (−4xy − 3xy) + (2yz − 5yz) + (−2xz + 7xz) + 5xyz + (1 − 1)
= −7xy − 3yz + 5xz + 5xyz.
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 the second polynomial comes first.
2. xyz is a different term from xy, yz and xz.

Question 18. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³.
Answer:
Volume of a cuboid = base area × height.
900 = 180 × h, so h = 900 ÷ 180 = 5.
So the height is 5 cm.
In simple words: Height = volume ÷ base area = 900 ÷ 180 = 5 cm.

Teacher's Note:
1. Base area already includes length × breadth.
2. Check the units: cm³ ÷ cm² = cm.

Question 19. Find m so that (−3)^{m + 1} × (−3)⁵ = (−3)⁷.
Answer:
Using a^m × aⁿ = a^m+n: (−3)^{m + 1 + 5} = (−3)⁷.
So (−3)^{m + 6} = (−3)⁷.
The base −3 is not 0, 1 or −1, so the exponents must be equal: m + 6 = 7.
So m = 1.
In simple words: Add the powers on the left, then make them equal to 7: m + 6 = 7.

Teacher's Note:
1. You may compare exponents only when the bases are the same and are not 0, 1 or −1.
2. Check: (−3)² × (−3)⁵ = (−3)⁷.

Question 20. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Answer:
At constant speed, distance varies directly with time.
5 hours = 5 × 60 = 300 minutes. Let the distance be x km.
\( \frac{14}{25} \) = \( \frac{x}{300} \), so x = \( \frac{14 \times 300}{25} \) = 14 × 12 = 168.
So the truck can travel 168 km.
In simple words: 300 minutes is 12 times 25 minutes, so the truck goes 12 × 14 = 168 km.

Teacher's Note:
1. Convert hours to minutes so both times are in the same unit.
2. In direct variation, the ratio distance/time stays the same.

Section C: Short Answer Questions (3 Marks Each)

Question 21. Simplify: \( \frac{12^{4} \times 9^{3} \times 4}{6^{3} \times 8^{2} \times 27} \)
Answer:
Write each number as powers of primes: 12 = 2² × 3, 9 = 3², 4 = 2², 6 = 2 × 3, 8 = 2³, 27 = 3³.
\( \frac{(2^{2} \times 3)^{4} \times (3^{2})^{3} \times 2^{2}}{(2 \times 3)^{3} \times (2^{3})^{2} \times 3^{3}} = \frac{2^{8} \times 3^{4} \times 3^{6} \times 2^{2}}{2^{3} \times 3^{3} \times 2^{6} \times 3^{3}} \)
\( = 2^{8+2-3-6} \times 3^{4+6-3-3} = 2^{1} \times 3^{4} \)
= 2 × 81 = 162.
In simple words: Break every number into 2s and 3s, then add powers on top, subtract powers at the bottom.

Teacher's Note:
1. (2² × 3)⁴ = 2⁸ × 3⁴: raise each factor to the power.
2. Treat the powers of 2 and the powers of 3 separately.

Teacher's Note:
1. Choose a scale that fits the largest value (24) comfortably on the y-axis.
2. Extended, this line would pass through the origin, because a square of side 0 has perimeter 0.

Question 23. Factorise the following expressions.
(i) p² − 10p + 25 (ii) 63a² − 112b²
Answer:
(i) p² − 10p + 25 = p² − 2 × p × 5 + 5².
Using a² − 2ab + b² = (a − b)²: p² − 10p + 25 = (p − 5)² = (p − 5)(p − 5).
(ii) 63a² − 112b² = 7(9a² − 16b²) (taking out the common factor 7).
9a² − 16b² = (3a)² − (4b)² = (3a + 4b)(3a − 4b).
So 63a² − 112b² = 7(3a + 4b)(3a − 4b).
In simple words: (i) is a perfect square; in (ii) take out 7 first, then use the difference of squares.

Teacher's Note:
1. Always look for a common factor first: 63 = 7 × 9 and 112 = 7 × 16.
2. Check (i): (p − 5)² = p² − 10p + 25.

Teacher's Note:
1. Find the constant k from the complete column first.
2. The source answer key left out the value of q; it is 36/13.

Section D: Long Answer Questions (5 Marks Each)

Teacher's Note:
1. Check the readings by calculation: 10% of 250 = 25 and 70 is 10% of 700.
2. Draw the dashed read-off lines clearly; examiners look for them.

Question 26. Simplify: (i) (1.5x − 4y)(1.5x + 4y + 3) − 4.5x + 12y
(ii) (a + b + c)(a + b − c)
Answer:
(i) (1.5x − 4y)(1.5x + 4y + 3) = 1.5x(1.5x + 4y + 3) − 4y(1.5x + 4y + 3)
= 2.25x² + 6xy + 4.5x − 6xy − 16y² − 12y.
Now subtract 4.5x and add 12y: 2.25x² + 6xy + 4.5x − 6xy − 16y² − 12y − 4.5x + 12y.
The xy, x and y terms all cancel, leaving 2.25x² − 16y².
(ii) (a + b + c)(a + b − c) = a(a + b − c) + b(a + b − c) + c(a + b − c)
= a² + ab − ac + ab + b² − bc + ac + bc − c²
= a² + 2ab + b² − c².
In simple words: Multiply out fully; most terms cancel and only the squares (and 2ab in part ii) remain.

Teacher's Note:
1. 1.5 × 1.5 = 2.25 and 1.5 × 4 = 6: work carefully with decimals.
2. Part (ii) can also be done as (a + b)² − c².

Question 27. Factorise the expressions and divide them as directed.
(i) (y² + 7y + 10) ÷ (y + 5) (ii) (m² − 14m − 32) ÷ (m + 2)
Answer:
(i) y² + 7y + 10: two numbers with product 10 and sum 7 are 2 and 5.
y² + 2y + 5y + 10 = y(y + 2) + 5(y + 2) = (y + 2)(y + 5).
(y + 2)(y + 5) ÷ (y + 5) = y + 2.
(ii) m² − 14m − 32: two numbers with product −32 and sum −14 are −16 and 2.
m² − 16m + 2m − 32 = m(m − 16) + 2(m − 16) = (m − 16)(m + 2).
(m − 16)(m + 2) ÷ (m + 2) = m − 16.
In simple words: Factorise each quadratic, then cancel the common bracket with the divisor.

Teacher's Note:
1. When the constant term is negative, the two numbers have opposite signs.
2. Check (ii): (m − 16)(m + 2) = m² − 14m − 32.

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

Teacher's Note:
1. In direct proportion, y ÷ x is the same for every column (here 42).
2. Check (iii): 10 m is twice 5 m, so it costs twice 210 = 420.

Question 29. Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured paper with picture of Santa Claus on it (see the figure). She must know the exact quantity of paper to buy for this purpose. The box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively and she wants to use square sheets of paper of side 40 cm for the decoration.
Answer the following questions based on the above information:
(i) Find the total surface area of the box. [1]
(ii) Find the area of each sheet of the paper. [1]
(iii) Find the number of sheets required. [2]
Answer:
(i) Total surface area = 2(lb + bh + hl) = 2(80 × 40 + 40 × 20 + 20 × 80)
= 2(3200 + 800 + 1600) = 2 × 5600 = 11200 cm².
(ii) Area of one sheet = 40² = 1600 cm².
(iii) Number of sheets = 11200 ÷ 1600 = 7.
In simple words: Total paper needed is 11200 cm²; each sheet covers 1600 cm², so 7 sheets are needed.

Teacher's Note:
1. Total surface area covers all six faces of the box.
2. The answer assumes the sheets can be cut and joined with no overlap or waste.