CBSE Class 8 Mathematics Annual Exam Sample Paper 05 with Answers (Old NCERT Syllabus)

This Class 8 Maths annual exam sample paper has 29 questions for 60 marks, arranged in sections A to E, and every question comes with a worked, step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, which makes it handy for state-board students and for revising these topics. It checks polynomials and like terms, surface area and volume of cubes, cuboids and cylinders, area of a trapezium and a quadrilateral, laws of exponents, direct and inverse proportion word problems, factorisation using identities, a distance–time graph to draw and read, and a forecast-versus-actual temperature line 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 thing to subtract is the first expression minus the result you want.

Teacher's Note:
1. If A − ? = B, then ? = A − B.
2. Subtracting −1 means adding 1, so the constant is 7 + 1 = 8 (the original answer key wrongly gave 6).

Question 17. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Answer:
Let the side along the road be x m; then the side along the river is 2x m. Height h = 100 m.
Area of trapezium = \( \frac{1}{2} \) × (sum of parallel sides) × h
10500 = \( \frac{1}{2} \) × (x + 2x) × 100 = 150x
x = 10500 ÷ 150 = 70 m
Side along the river = 2x = 140 m
In simple words: Call the road side x, so the river side is 2x; the area formula gives x = 70, so the river side is 140 m.

Teacher's Note:
1. Read carefully which side is twice the other.
2. The question asks for the river side, so double x at the end.

Question 18. Find the value of m for which 5^m ÷ 5⁻³ = 5⁵
Answer:
Using a^m ÷ aⁿ = a^{m − n}: 5^{m − (−3)} = 5⁵
5^{m + 3} = 5⁵
Equating exponents: m + 3 = 5
m = 2
In simple words: Combine the left side into one power of 5 and match the exponent with 5.

Teacher's Note:
1. Minus a negative becomes plus: m − (−3) = m + 3.
2. Check: 5² ÷ 5⁻³ = 5⁵.

Question 19. Factorise: 100 − (x − 5)²
Answer:
Write 100 as 10²: 10² − (x − 5)²
Using a² − b² = (a + b)(a − b): = {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, so it splits into (sum)(difference).

Teacher's Note:
1. Keep (x − 5) in a bracket so the minus sign applies to both terms.
2. 10 − (x − 5) = 15 − x, not 5 − x.

Question 20. If the cost of 93 m of a certain kind of plastic sheet is Rs 1395, then what would it cost to buy 105 m of such plastic sheet?
Answer:
More sheet costs more, so length and cost are in direct proportion.
Let the cost of 105 m be Rs x. Then \( \frac{93}{1395} \) = \( \frac{105}{x} \)
x = \( \frac{105 \times 1395}{93} \) = 105 × 15 = 1575
The cost is Rs 1575.
In simple words: One metre costs Rs 15, so 105 m costs 105 × 15 = Rs 1575.

Teacher's Note:
1. 1395 ÷ 93 = 15 is the cost per metre, which makes the working quick.
2. In direct proportion the ratio of the two quantities stays the same.

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:
\( \frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^{3}}{5^{-7} \times 2^{-5} \times 3^{-5}} \)
Powers of 3: 3^{−5 − (−5)} = 3⁰ = 1; powers of 2: 2^{−5 − (−5)} = 2⁰ = 1
Powers of 5: 5^{−5 + 3 − (−7)} = 5⁵
= 5⁵ = 3125
In simple words: Break every number into prime powers, then cancel or combine the powers of 2, 3 and 5.

Teacher's Note:
1. Any non-zero number to the power 0 is 1.
2. 125 = 5³, so it combines with the other powers of 5.

Question 22. 68 boxes of a certain commodity require a shelf-length of 13.6 m. How many boxes of the same commodity would occupy a shelf length of 20.4 m?
Answer:
More shelf length holds more boxes, so this is direct proportion.
Let x boxes fit in 20.4 m. Then \( \frac{68}{13.6} \) = \( \frac{x}{20.4} \)
x = \( \frac{68 \times 20.4}{13.6} \) = 5 × 20.4 = 102
102 boxes will occupy 20.4 m.
In simple words: Each metre holds 5 boxes, so 20.4 m holds 102 boxes.

Teacher's Note:
1. 68 ÷ 13.6 = 5 boxes per metre simplifies the calculation.
2. Your answer should be a whole number of boxes.

Question 23. 1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted now for 32 days. How many soldiers left the fort?
Answer:
After 4 days, the remaining food is enough for 1200 soldiers for 28 − 4 = 24 days.
Let x soldiers leave. The remaining (1200 − x) soldiers make it last 32 days.
Soldiers and days are in inverse proportion: 1200 × 24 = (1200 − x) × 32
1200 − x = \( \frac{1200 \times 24}{32} \) = 900
x = 1200 − 900 = 300 soldiers left the fort.
In simple words: The leftover food is 24 days' food for 1200 soldiers; for it to last 32 days only 900 soldiers can stay, so 300 left.

Teacher's Note:
1. Use the food left after 4 days (24 days' worth), not the original 28 days.
2. Fewer soldiers make the food last longer: inverse proportion.

Question 24. Factorise: (i) p² + 6p − 16   (ii) x² − 10x + 24
Answer:
(i) Numbers with product −16 and sum 6: 8 and −2.
p² + 8p − 2p − 16 = p(p + 8) − 2(p + 8) = (p + 8)(p − 2)
(ii) Numbers with product 24 and sum −10: −6 and −4.
x² − 6x − 4x + 24 = x(x − 6) − 4(x − 6) = (x − 6)(x − 4)
In simple words: Split the middle term into two parts whose product equals the constant term times 1.

Teacher's Note:
1. A negative constant means the two numbers have opposite signs.
2. A positive constant with a negative middle term means both numbers are negative.

Section D: Long Answer Questions (5 Marks Each)

Teacher's Note:
1. Use dashed guide lines from the axis to the graph when reading a value.
2. A straight-line distance–time graph means the speed is constant.

Question 26. Simplify 7x²(3x − 9) + 3 and find its values for x = 4 and x = 6
Answer:
7x²(3x − 9) + 3 = 21x³ − 63x² + 3 = 21x³ − 63x² + 3
For x = 4: 21 × 64 − 63 × 16 + 3 = 1344 − 1008 + 3 = 339
For x = 6: 21 × 216 − 63 × 36 + 3 = 4536 − 2268 + 3 = 2271
In simple words: Multiply out the bracket first, then put in the given values of x.

Teacher's Note:
1. Check with the original form: at x = 4, 7 × 16 × 3 + 3 = 339.
2. Work out the powers before multiplying by the coefficients.

Question 27. Factorise the following expressions. (1 + 2 + 2 marks)
(i) p² − 10p + 25   (ii) 25m² + 30m + 9   (iii) 49y² + 84yz + 36z²
Answer:
(i) p² − 10p + 25 = p² − 2 × p × 5 + 5² = (p − 5)² [using (a − b)² = a² − 2ab + b²]
(ii) 25m² + 30m + 9 = (5m)² + 2 × 5m × 3 + 3² = (5m + 3)² [using (a + b)² = a² + 2ab + b²]
(iii) 49y² + 84yz + 36z² = (7y)² + 2 × 7y × 6z + (6z)² = (7y + 6z)²
In simple words: Each expression is a perfect square: spot the two square terms and check that the middle term is twice their product.

Teacher's Note:
1. First and last terms that are perfect squares hint at a square identity.
2. The sign of the middle term decides (a + b)² or (a − b)².

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

Teacher's Note:
1. Read each point carefully against the grid; one big square here is 4 °C.
2. A flat part of the actual graph shows the same temperature on two days.

Question 29. A packaging company is designing boxes for shipping different products efficiently. They have two types of boxes to consider:
Cubical Box – Each edge measures 10 cm.
Cuboidal Box – Dimensions are 12.5 cm in length, 10 cm in width, and 8 cm in height.
The company wants to optimize packaging by considering factors such as volume, surface area, and material cost. Additionally, they need to determine which box would be more efficient for shipping and storage, minimizing empty space while ensuring product safety.
(i) Which box has the greater lateral surface area and by how much? (2)
(ii) Which box has the smaller total surface area and by how much? (2)
Answer:
Cube: a = 10 cm. Cuboid: l = 12.5 cm, b = 10 cm, h = 8 cm.
(i) Lateral surface area of cube = 4a² = 4 × 100 = 400 cm²
Lateral surface area of cuboid = 2h(l + b) = 2 × 8 × (12.5 + 10) = 16 × 22.5 = 360 cm²
The cubical box has the greater lateral surface area, by 400 − 360 = 40 cm².
(ii) Total surface area of cube = 6a² = 600 cm²
Total surface area of cuboid = 2(lb + bh + hl) = 2(125 + 80 + 100) = 2 × 305 = 610 cm²
The cubical box has the smaller total surface area, by 610 − 600 = 10 cm².
In simple words: The cube has more area on its four sides but less area overall once the top and bottom are included.

Teacher's Note:
1. Lateral surface area leaves out the top and bottom faces.
2. Both boxes have the same volume (1000 cm³), yet their surface areas differ.