CBSE Class 8 Mathematics Annual Exam Sample Paper 03 with Answers (Old NCERT Syllabus)

Class 8 Maths Annual Exam Sample Paper 03 is a 60-mark, 29-question paper arranged in Sections A to E, with step-by-step answers for every question. It follows the earlier (old) NCERT Class 8 Maths syllabus, which makes it useful for state-board students and for revising these topics. The questions cover multiplying monomials and binomials, factorisation, laws of exponents and standard form, direct and inverse variation, area of a rhombus and trapezium, surface area of cylinders and cuboids, and drawing and reading time-distance graphs.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 16. Subtract: 3a(a + b + c) − 2b(a − b + c) from 4c(−a + b + c)
Answer:
4c(−a + b + c) = −4ac + 4bc + 4c²
3a(a + b + c) − 2b(a − b + c) = 3a² + 3ab + 3ac − 2ab + 2b² − 2bc = 3a² + ab + 3ac + 2b² − 2bc
Required = (−4ac + 4bc + 4c²) − (3a² + ab + 3ac + 2b² − 2bc)
= −4ac + 4bc + 4c² − 3a² − ab − 3ac − 2b² + 2bc
= −3a² − 2b² + 4c² − ab + 6bc − 7ac
In simple words: Expand both expressions, then subtract the first from the second by changing its signs.

Teacher's Note:
1. “Subtract X from Y” means Y − X.
2. Simplify each expression fully before subtracting.

Question 17. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Answer:
A closed tank needs its total surface area: TSA = 2πr(h + r).
= 2 × \( \frac{22}{7} \) × 7 × (3 + 7)
= 44 × 10 = 440
Metal sheet required = 440 m².
In simple words: The sheet must cover the curved side and both circular ends.

Teacher's Note:
1. “Closed” means include both top and bottom.
2. 2πr(h + r) is the same as 2πrh + 2πr².

Question 18. Find the value of x for which \( \left(\frac{5}{7}\right)^{-3} \times \left(\frac{5}{7}\right)^{-11} = \left(\frac{5}{7}\right)^{7x} \)
Answer:
Using \( a^{m} \times a^{n} = a^{m+n} \): \( \left(\frac{5}{7}\right)^{-3 + (-11)} = \left(\frac{5}{7}\right)^{7x} \)
\( \left(\frac{5}{7}\right)^{-14} = \left(\frac{5}{7}\right)^{7x} \)
Bases are equal, so −14 = 7x.
x = −2
In simple words: Add the exponents on the left and match them with the exponent on the right.

Teacher's Note:
1. The bases must be the same before you compare exponents.
2. Keep track of the signs: −3 + (−11) = −14.

Question 19. Factorise: p² + 6p + 8
Answer:
Numbers with product 8 and sum 6 are 2 and 4.
p² + 6p + 8 = p² + 2p + 4p + 8
= p(p + 2) + 4(p + 2)
= (p + 2)(p + 4)
In simple words: Split 6p into 2p + 4p and take out common factors.

Teacher's Note:
1. Check: (p + 2)(p + 4) = p² + 6p + 8.
2. Both numbers are positive because the constant and middle term are positive.

Teacher's Note:
1. Animals × days stays constant for the same stock of food.
2. Remember to add the extra animals to the original number.

Section C: Short Answer Questions (3 Marks Each)

Question 21. If y varies inversely as x, and if y = 4 when x = 72, find the following:
(a) the constant of variation
(b) the value of y when x = 9.
Answer:
(a) In inverse variation, xy = k.
k = 72 × 4 = 288. The constant of variation is 288.
(b) y = \( \frac{k}{x} \) = \( \frac{288}{9} \) = 32
In simple words: x times y always equals 288, so when x is 9, y must be 32.

Teacher's Note:
1. Inverse variation: y = k/x, so the product xy is constant.
2. Check: 9 × 32 = 288.

Question 22. It is given that s varies directly as q.
(a) Write an equation that specifies the variation and find the constant of variation if s is 4 when q is 6.
(b) Find s when q is 15.
(c) Find q when s is 12.
Answer:
(a) Direct variation: s = kq, where k is the constant of variation.
4 = k × 6 ⇒ k = \( \frac{4}{6} \) = \( \frac{2}{3} \). So the equation is s = \( \frac{2}{3} \)q and k = \( \frac{2}{3} \).
(b) s = \( \frac{2}{3} \) × 15 = 10
(c) 12 = \( \frac{2}{3} \)q ⇒ q = 12 × \( \frac{3}{2} \) = 18
In simple words: s is always two-thirds of q.

Teacher's Note:
1. In direct variation, s ÷ q is constant.
2. Check: 10 ÷ 15 = 12 ÷ 18 = 2/3.

Question 23. Find the value of \( a^{6} - (b + 1)^{3} \) when \( a = \frac{1}{2}, b = \frac{-3}{4} \).
Answer:
\( a^{6} = \left(\frac{1}{2}\right)^{6} = \frac{1}{64} \)
\( b + 1 = \frac{-3}{4} + 1 = \frac{1}{4} \), so \( (b + 1)^{3} = \left(\frac{1}{4}\right)^{3} = \frac{1}{64} \)
\( a^{6} - (b + 1)^{3} = \frac{1}{64} - \frac{1}{64} \) = 0
In simple words: Both parts work out to 1/64, so their difference is zero.

Teacher's Note:
1. Simplify b + 1 first, then raise it to the power.
2. 2⁶ = 64 and 4³ = 64.

Question 24. Factorise the expression m² − 14m − 32 and divide it by (m + 2) as directed.
Answer:
Numbers with product −32 and sum −14 are −16 and 2.
m² − 14m − 32 = m² + 2m − 16m − 32 = m(m + 2) − 16(m + 2) = (m − 16)(m + 2)
(m² − 14m − 32) ÷ (m + 2) = \( \frac{(m - 16)(m + 2)}{m + 2} \) = m − 16
In simple words: Factorise first, then cancel the common factor (m + 2).

Teacher's Note:
1. Dividing by a factor is easiest after factorising.
2. Check: (m − 16)(m + 2) = m² − 14m − 32.

Section D: Long Answer Questions (5 Marks Each)

Teacher's Note:
1. Constant speed gives a straight-line time-distance graph.
2. Use dotted guide lines to read values from the graph.
3. Always check the reading with a quick calculation.

Question 26. Multiply the binomials.
(i) (2pq + 3q²) and (3pq − 2q²)
(ii) (\( \frac{3}{4} \)a² + 3b²) and 4(a² − \( \frac{2}{3} \)b²)
Answer:
(i) (2pq + 3q²)(3pq − 2q²)
= 2pq × 3pq − 2pq × 2q² + 3q² × 3pq − 3q² × 2q²
= 6p²q² − 4pq³ + 9pq³ − 6q⁴
= 6p²q² + 5pq³ − 6q⁴
(ii) 4(a² − \( \frac{2}{3} \)b²) = 4a² − \( \frac{8}{3} \)b²
(\( \frac{3}{4} \)a² + 3b²)(4a² − \( \frac{8}{3} \)b²)
= \( \frac{3}{4} \)a² × 4a² − \( \frac{3}{4} \)a² × \( \frac{8}{3} \)b² + 3b² × 4a² − 3b² × \( \frac{8}{3} \)b²
= 3a⁴ − 2a²b² + 12a²b² − 8b⁴
= 3a⁴ + 10a²b² − 8b⁴
In simple words: Multiply every term of the first bracket by every term of the second, then add like terms.

Teacher's Note:
1. In (ii), first multiply the 4 into its bracket to keep the work neat.
2. pq³ terms are like terms and can be combined.

Question 27. Factorise: (i) x⁴ − (x − z)⁴ (ii) a⁴ − 2a²b² + b⁴
Answer:
(i) x⁴ − (x − z)⁴ = (x²)² − [(x − z)²]²
= [x² − (x − z)²][x² + (x − z)²]
= [x − (x − z)][x + (x − z)][x² + x² − 2xz + z²]
= z(2x − z)(2x² − 2xz + z²)
(ii) a⁴ − 2a²b² + b⁴ = (a²)² − 2(a²)(b²) + (b²)² = (a² − b²)²
= [(a − b)(a + b)]² = (a − b)²(a + b)²
In simple words: Part (i) uses the difference of squares twice; part (ii) is a perfect square of a difference of squares.

Teacher's Note:
1. Treat (x − z)² as a single block in part (i).
2. Recognise a² − 2ab + b² = (a − b)² with a² and b² in place of a and b.

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

Question 28. A courier-person starts his journey from a town to a neighbouring suburban area to deliver a parcel to a merchant. His movement is tracked, and the distance from the town at different times is represented in the given graph.
The graph depicts his journey, including stops, speed variations, and the time taken for delivery. Observe the graph carefully and answer the following questions:
(a) How much time did the person take for the travel? (1)
(b) How far is the place of the merchant from the town? (1)
(c) Did the person stop on his way? Explain. (1)
(d) During which period did he ride fastest? (1)
Answer:
Each small square on the time axis is 15 minutes and on the distance axis is 2 km.
(a) He started at 8 a.m. and reached at 11.30 a.m., so he took \( 3\frac{1}{2} \) hours.
(b) The graph ends at 22 km, so the merchant is 22 km from the town.
(c) Yes. The graph is horizontal from 10 a.m. to 10.30 a.m. (distance stays 16 km), so he stopped for half an hour.
(d) 8 a.m. to 9 a.m.: 10 km in 1 hour; 9 a.m. to 10 a.m.: 6 km; 10.30 a.m. to 11.30 a.m.: 6 km. He rode fastest between 8 a.m. and 9 a.m. (the steepest part).
In simple words: Flat parts of the graph mean he stopped; the steepest part is where he went fastest.

Teacher's Note:
1. First work out the value of one small square on each axis.
2. Compare distance covered in equal time intervals to judge speed.

Question 29. Shanti Sweets and Snacks Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. The cost of the cardboard is Rs 4 for 1000 cm².
Based on the above situation, answer the following questions:
(a) Find the total surface area of the bigger box and the smaller box. (1)
(b) Find the total cardboard sheet required for 250 boxes of each kind considering all overlaps. (2)
(c) Find the cost of cardboard required for supplying 250 boxes of each kind. (1)
Answer:
(a) TSA of a cuboid = 2(lb + bh + hl).
Bigger box: 2(25 × 20 + 20 × 5 + 25 × 5) = 2(500 + 100 + 125) = 1450 cm²
Smaller box: 2(15 × 12 + 12 × 5 + 15 × 5) = 2(180 + 60 + 75) = 630 cm²
(b) With 5% extra for overlaps:
Bigger box: 1450 + 5% of 1450 = 1450 + 72.5 = 1522.5 cm²; for 250 boxes: 1522.5 × 250 = 380625 cm²
Smaller box: 630 + 5% of 630 = 630 + 31.5 = 661.5 cm²; for 250 boxes: 661.5 × 250 = 165375 cm²
Total cardboard = 380625 + 165375 = 546000 cm²
(c) Cost = \( \frac{546000}{1000} \) × Rs 4 = 546 × 4 = Rs 2184
In simple words: Find each box's surface, add 5% for overlaps, multiply by 250 for each kind, then convert area to cost.

Teacher's Note:
1. 5% extra means multiply the area by 1.05.
2. The rate is per 1000 cm², so divide the total area by 1000 first.
3. Work with one box size at a time to avoid mistakes.