CBSE Class 8 Mathematics Mensuration, Exponents and Proportions Practice Paper 01 with Answers (Old NCERT Syllabus)

This Class 8 Maths practice paper combines Mensuration, Exponents and Powers, and Direct and Inverse Proportions. It carries 40 marks and has 17 questions in sections A to E, each with a step-by-step solution. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it is useful for state-board students and for revising these chapters. It tests inverse-proportion word problems, laws of exponents and usual form, and surface area and volume of cubes, cuboids and cylinders.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface areas.
Answer:
Let the radii be 2r and 3r and the heights 5h and 3h.
CSA of the first = 2π × 2r × 5h = 20πrh.
CSA of the second = 2π × 3r × 3h = 18πrh.
Ratio = 20πrh : 18πrh = 10 : 9.
So the ratio of their curved surface areas is 10 : 9.
In simple words: Multiply the radius ratio by the height ratio: (2 × 5) : (3 × 3) = 10 : 9.

Teacher's Note:
1. Curved surface area is proportional to r × h.
2. Common factors 2π, r and h cancel in the ratio.

Question 8. The curved surface area of a cylinder is 1320 cm² and its base has diameter 21 cm. Find the height of the cylinder.
Answer:
Radius r = 21 ÷ 2 = 10.5 cm.
CSA = 2πrh, so 1320 = 2 × \( \frac{22}{7} \) × 10.5 × h = 66h.
h = 1320 ÷ 66 = 20 cm.
So the height is 20 cm.
In simple words: The base circle has circumference 66 cm, so the height is 1320 ÷ 66 = 20 cm.

Teacher's Note:
1. 2πr is the circumference; with d = 21 cm it is 66 cm.
2. Check: 66 × 20 = 1320.

Question 9. Express each of the following rational numbers with a positive exponent:
(i) \( \left\{\left(\frac{3}{2}\right)^{4}\right\}^{-2} \)   (ii) \( 4^{3} \times 4^{-9} \)
Answer:
(i) \( \left\{\left(\frac{3}{2}\right)^{4}\right\}^{-2} = \left(\frac{3}{2}\right)^{4 \times (-2)} = \left(\frac{3}{2}\right)^{-8} = \left(\frac{2}{3}\right)^{8} \), so the answer is \( \left(\frac{2}{3}\right)^{8} \).
(ii) \( 4^{3} \times 4^{-9} = 4^{3-9} = 4^{-6} = \left(\frac{1}{4}\right)^{6} \), so the answer is \( \left(\frac{1}{4}\right)^{6} \).
In simple words: Combine the powers first, then flip the base to make the negative exponent positive.

Teacher's Note:
1. Laws used: \( (a^{m})^{n} = a^{mn} \) and \( a^{m} \times a^{n} = a^{m+n} \).
2. \( a^{-n} = \left(\frac{1}{a}\right)^{n} \) turns a negative exponent into a positive one.

Section C: Short Answer Questions (3 Marks Each)

Question 10. The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m², find the circumference of its base.
Answer:
Total surface area = 2πr(r + h) and r + h = 37 m.
1628 = 2 × \( \frac{22}{7} \) × r × 37.
r = \( \frac{1628 × 7}{2 × 22 × 37} \) = \( \frac{11396}{1628} \) = 7 m.
Circumference = 2πr = 2 × \( \frac{22}{7} \) × 7 = 44 m.
So the circumference of the base is 44 m.
In simple words: Since r + h is given, the surface-area formula gives r = 7 m directly, and the circumference is 44 m.

Teacher's Note:
1. Look for r + h inside 2πr(r + h) and substitute it straight away.
2. Actually 2πr × 37 = 1628 gives 2πr = 44 at once.

Question 11. Simplify: \( \left\{\left(\frac{2}{3}\right)^{2}\right\}^{3} \times \left(\frac{1}{3}\right)^{-4} \times 3^{-1} \times 6^{-1} \)
Answer:
\( \left\{\left(\frac{2}{3}\right)^{2}\right\}^{3} = \left(\frac{2}{3}\right)^{6} = \frac{64}{729} \).
\( \left(\frac{1}{3}\right)^{-4} = 3^{4} = 81 \), and \( 3^{-1} \times 6^{-1} = \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} \).
Product = \( \frac{64}{729} \) × 81 × \( \frac{1}{18} \) = \( \frac{64}{9} \) × \( \frac{1}{18} \) = \( \frac{64}{162} \) = \( \frac{32}{81} \).
So the value is \( \frac{32}{81} \).
In simple words: Work out each power separately, then multiply and cancel: the answer is 32/81.

Teacher's Note:
1. 729 ÷ 81 = 9, so cancel 81 into 729 early.
2. Negative exponents mean reciprocals.

Teacher's Note:
1. Inverse variation: product constant; direct variation: ratio constant.
2. Simplify fractions: 108/39 cancels by 3 to 36/13.

Question 13. A car can finish a certain journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance?
Answer:
Speed and time vary inversely for a fixed distance.
Distance = 48 × 10 = 480 km.
Speed needed for 8 hours = 480 ÷ 8 = 60 km/h.
Increase = 60 − 48 = 12 km/h.
So the speed should be increased by 12 km/h.
In simple words: The trip is 480 km; doing it in 8 hours needs 60 km/h, which is 12 km/h faster.

Teacher's Note:
1. The question asks for the increase, not the new speed.
2. Check: 60 × 8 = 480 = 48 × 10.

Section D: Long Answer Questions (4 Marks Each)

Question 14. (a) In a hostel of 50 girls, there are food provisions for 40 days. If 30 more girls join the hostel, how long will these provisions last?
(b) If 12 pumps can empty a reservoir in 20 hours, then find the time required by 45 such pumps to empty the same reservoir.
Answer:
(a) Number of girls becomes 50 + 30 = 80. Girls and days vary inversely.
50 × 40 = 80 × x, so x = 2000 ÷ 80 = 25. The provisions last 25 days.
(b) Pumps and time vary inversely: 12 × 20 = 45 × t.
t = 240 ÷ 45 = \( \frac{16}{3} \) hours = 5 hours 20 minutes.
So 45 pumps need 5 hours 20 minutes.
In simple words: More girls finish the food faster (25 days); more pumps empty the tank faster (5 h 20 min).

Teacher's Note:
1. In both parts the product of the two quantities stays the same.
2. \( \frac{1}{3} \) hour = 20 minutes.

Question 15. How many planks each of which is 3 m long, 15 cm broad and 5 cm thick can be prepared from a wooden block 6 m long, 75 cm broad and 45 cm thick?
Answer:
Use centimetres: block = 600 cm × 75 cm × 45 cm; plank = 300 cm × 15 cm × 5 cm.
Volume of the block = 600 × 75 × 45 = 2025000 cm³.
Volume of a plank = 300 × 15 × 5 = 22500 cm³.
Number of planks = 2025000 ÷ 22500 = 90.
The planks fit exactly: 600 ÷ 300 = 2, 75 ÷ 15 = 5 and 45 ÷ 5 = 9, and 2 × 5 × 9 = 90.
So 90 planks can be made.
In simple words: Two planks fit along the length, five across and nine up, so 2 × 5 × 9 = 90 planks.

Teacher's Note:
1. Convert metres to centimetres before calculating volumes.
2. Checking each edge confirms that no wood is wasted.

Question 16. Write the following numbers in the usual form:
(i) 4.83 × 10⁷   (ii) 3.02 × 10⁻⁶   (iii) 4.5 × 10⁴   (iv) 3 × 10⁻⁸
Answer:
(i) Move the decimal point 7 places to the right: 48300000.
(ii) Move the point 6 places to the left: 0.00000302.
(iii) Move the point 4 places to the right: 45000.
(iv) Move the point 8 places to the left: 0.00000003.
In simple words: Positive powers of 10 make the number bigger (point moves right); negative powers make it smaller (point moves left).

Teacher's Note:
1. For \( 10^{-n} \), the first non-zero digit ends up in the n-th decimal place.
2. In the Indian system (i) is written 4,83,00,000.

Section E: Case Study Based Question (4 Marks)

Question 17. 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 bigger box and 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) Bigger box: 2(25 × 20 + 25 × 5 + 20 × 5) = 2(500 + 125 + 100) = 1450 cm².
Smaller box: 2(15 × 12 + 15 × 5 + 12 × 5) = 2(180 + 75 + 60) = 630 cm².
(b) With 5% extra for overlaps: bigger box = 1450 + 72.5 = 1522.5 cm²; smaller box = 630 + 31.5 = 661.5 cm².
For 250 of each: 250 × 1522.5 = 380625 cm² and 250 × 661.5 = 165375 cm².
Total cardboard = 380625 + 165375 = 546000 cm².
(c) Cost = \( \frac{546000}{1000} \) × Rs 4 = 546 × 4 = Rs 2184.
In simple words: One big and one small box together need 2184 cm² of card with overlaps; 250 pairs need 546000 cm², costing Rs 2184.

Teacher's Note:
1. 5% extra means multiply the area by 1.05.
2. Shortcut: (1450 + 630) × 1.05 × 250 = 546000 cm².