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

This Class 8 Maths practice paper on Mensuration and Exponents and Powers carries 40 marks and has 17 questions in sections A to E, with a worked step-by-step answer for every question. It follows the earlier (old) NCERT Class 8 Maths syllabus, which makes it useful for state-board students and for revising both chapters. It tests surface areas and volumes of cuboids and cylinders, the laws of exponents with negative powers, and writing very small and very large numbers in standard form.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. A cuboidal vessel is 10 cm long and 5 cm wide. How high it must be made to hold 300 cm³ of a liquid?
Answer:
Volume = l × b × h, so 300 = 10 × 5 × h = 50h.
h = 300 ÷ 50 = 6 cm.
So the vessel must be 6 cm high.
In simple words: The base is 50 cm², so to hold 300 cm³ it must be 300 ÷ 50 = 6 cm high.

Teacher's Note:
1. Height = volume ÷ base area.
2. Check: 10 × 5 × 6 = 300.

Question 8. The curved surface area of a cylindrical rod is 132 cm². Find its length if the radius is 0.35 cm.
Answer:
Curved surface area = 2πrh.
132 = 2 × \( \frac{22}{7} \) × 0.35 × h = 2.2h (since 2 × 22 × 0.05 = 2.2).
h = 132 ÷ 2.2 = 60 cm.
So the length of the rod is 60 cm.
In simple words: Going once round the rod covers 2.2 cm, so 132 cm² needs a rod 60 cm long.

Teacher's Note:
1. The length of a rod is the height of the cylinder.
2. 0.35 ÷ 7 = 0.05 simplifies the working.

Question 9. 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:
Same base, so add the exponents: \( \left(\frac{5}{7}\right)^{-3 + (-11)} = \left(\frac{5}{7}\right)^{-14} \).
So \( \left(\frac{5}{7}\right)^{-14} = \left(\frac{5}{7}\right)^{7x} \).
Equate the exponents: 7x = −14, so x = −2.
So x = −2.
In simple words: Adding −3 and −11 gives −14, and 7x = −14 means x = −2.

Teacher's Note:
1. Use \( a^{m} \times a^{n} = a^{m+n} \).
2. Watch the signs when adding negative exponents.

Section C: Short Answer Questions (3 Marks Each)

Question 10. A tea-packet measures 10 cm × 6 cm × 4 cm. How many such tea-packets can be placed in a cardboard box of dimensions 50 cm × 30 cm × 0.2 m?
Answer:
0.2 m = 20 cm, so the box is 50 cm × 30 cm × 20 cm.
Volume of the box = 50 × 30 × 20 = 30000 cm³.
Volume of a tea-packet = 10 × 6 × 4 = 240 cm³.
Number of packets = 30000 ÷ 240 = 125.
They fit exactly: 50 ÷ 10 = 5, 30 ÷ 6 = 5 and 20 ÷ 4 = 5 packets along each edge, and 5 × 5 × 5 = 125.
So 125 tea-packets can be placed in the box.
In simple words: Five packets fit along each edge of the box, so 5 × 5 × 5 = 125 packets.

Teacher's Note:
1. Convert 0.2 m into cm before multiplying.
2. Dividing volumes is valid here because the packets fit along each edge exactly.

Question 11. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area.
Answer:
2πr = 88, so r = \( \frac{88 × 7}{44} \) = 14 cm.
Curved surface area = 2πrh = 88 × 15 = 1320 cm².
Total surface area = 2πr(h + r) = 88 × (15 + 14) = 88 × 29 = 2552 cm².
So the curved surface area is 1320 cm² and the total surface area is 2552 cm².
In simple words: The radius is 14 cm; the side area is 1320 cm², and adding the two circular ends gives 2552 cm².

Teacher's Note:
1. Since 2πr is given as 88, use it directly: CSA = 88 × h.
2. Each circular end is πr² = 616 cm², and 1320 + 2 × 616 = 2552.

Question 12. By what number should \( \left(\frac{5}{4}\right)^{-3} \) be divided so that the quotient may be \( \left(\frac{15}{16}\right)^{-2} \)?
Answer:
Let the number be x. Then \( \left(\frac{5}{4}\right)^{-3} \div x = \left(\frac{15}{16}\right)^{-2} \).
So x = \( \left(\frac{5}{4}\right)^{-3} \div \left(\frac{15}{16}\right)^{-2} = \left(\frac{4}{5}\right)^{3} \div \left(\frac{16}{15}\right)^{2} \).
= \( \frac{64}{125} \) ÷ \( \frac{256}{225} \) = \( \frac{64}{125} \) × \( \frac{225}{256} \) = \( \frac{9}{20} \).
So the required number is \( \frac{9}{20} \).
In simple words: Turn the negative powers into positive ones by flipping, then divide: the number is 9/20.

Teacher's Note:
1. Dividend ÷ divisor = quotient, so divisor = dividend ÷ quotient.
2. Cancel before multiplying: 64/256 = 1/4 and 225/125 = 9/5.

Question 13. In a stack, there are 5 books, each having a thickness of 20 mm and 5 paper sheets, each having a thickness of 0.016 mm. What is the total thickness of the stack?
Answer:
Thickness of 5 books = 5 × 20 = 100 mm.
Thickness of 5 sheets = 5 × 0.016 = 0.08 mm.
Total thickness = 100 + 0.08 = 100.08 mm.
In standard form, 100.08 mm = 1.0008 × 10² mm.
So the stack is 1.0008 × 10² mm thick.
In simple words: Books give 100 mm and sheets add 0.08 mm, so the stack is 100.08 mm thick.

Teacher's Note:
1. Add numbers in usual form first, then convert the result to standard form.
2. In standard form there is exactly one non-zero digit before the decimal point.

Section D: Long Answer Questions (4 Marks Each)

Question 14. What will happen to the volume of a cuboid if its:
(i) Length is doubled, height is same and breadth is halved?
(ii) Length is doubled, height is doubled and breadth is same?
Answer:
Let the original length, breadth and height be l, b and h, so the volume V = lbh.
(i) New volume = 2l × \( \frac{b}{2} \) × h = lbh = V. The volume remains the same.
(ii) New volume = 2l × b × 2h = 4lbh = 4V. The volume becomes four times the original.
In simple words: Doubling one side and halving another cancels out; doubling two sides multiplies the volume by 2 × 2 = 4.

Teacher's Note:
1. Multiply the scale factors of the three edges to get the change in volume.
2. (i): 2 × ½ × 1 = 1; (ii): 2 × 1 × 2 = 4.

Question 15. Simplify: (i) \( \left[\left(\frac{2}{5}\right)^{-3}\right]^{4} \)   (ii) \( \left[\left(\frac{-6}{11}\right)^{-5}\right]^{-3} \)   (iii) \( \left(-\frac{2}{3}\right)^{-4} \times \left(\frac{1}{8}\right)^{-4} \)   (iv) \( \left(\frac{5}{7}\right)^{-1} \times \left(\frac{7}{3}\right)^{-1} \)
Answer:
(i) \( \left[\left(\frac{2}{5}\right)^{-3}\right]^{4} = \left(\frac{2}{5}\right)^{-12} = \left(\frac{5}{2}\right)^{12} \), so the answer is \( \left(\frac{5}{2}\right)^{12} \).
(ii) \( \left[\left(\frac{-6}{11}\right)^{-5}\right]^{-3} = \left(\frac{-6}{11}\right)^{15} \), so the answer is \( \left(\frac{-6}{11}\right)^{15} \) (a negative number, since 15 is odd).
(iii) Same power, so multiply the bases: \( \left(-\frac{2}{3} \times \frac{1}{8}\right)^{-4} = \left(-\frac{1}{12}\right)^{-4} = (-12)^{4} = 12^{4} \) = 20736.
(iv) \( \left(\frac{5}{7} \times \frac{7}{3}\right)^{-1} = \left(\frac{5}{3}\right)^{-1} \) = \( \frac{3}{5} \).
In simple words: Multiply powers of powers, add or combine powers with the same exponent, and flip fractions to remove negative exponents.

Teacher's Note:
1. Laws used: \( (a^{m})^{n} = a^{mn} \), \( a^{m} \times b^{m} = (ab)^{m} \) and \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n} \).
2. An even power of a negative number is positive; an odd power stays negative.

Question 16. Express the following numbers in standard form.
(i) 0.0000000000085   (ii) 0.00000000000942
(iii) 6020000000000000   (iv) 0.00000000837
Answer:
(i) The decimal point moves 12 places to the right to give 8.5, so 0.0000000000085 = 8.5 × 10⁻¹².
(ii) The point moves 12 places to the right to give 9.42, so 0.00000000000942 = 9.42 × 10⁻¹².
(iii) The point moves 15 places to the left to give 6.02, so 6020000000000000 = 6.02 × 10¹⁵.
(iv) The point moves 9 places to the right to give 8.37, so 0.00000000837 = 8.37 × 10⁻⁹.
In simple words: Move the decimal point until one non-zero digit is in front of it; the number of places moved is the power of 10 (negative for small numbers).

Teacher's Note:
1. Small numbers (less than 1) get negative powers of 10; large numbers get positive powers.
2. Count digits carefully in groups of three to avoid slips.

Section E: Case Study Based Question (4 Marks)

Question 17. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. There were 35 competitors to participate in this competition. The cost of cardboard is Rs. 20 per 10 square cm.

Based on the above situation, answer the following questions:
(a) Find the lateral surface area of a pen holder. (1)
(b) Find the total surface area of a penholder. (1)
(c) Find the area of cardboard sheet used by 35 competitors. (1)
(d) What is the cost of cardboard? (1)
Answer:
(a) Lateral (curved) surface area = 2πrh = 2 × \( \frac{22}{7} \) × 3 × 10.5 = 44 × 3 × 1.5 = 198 cm².
(b) The penholder is open at the top and has one base, so its surface area = 2πrh + πr² = 198 + \( \frac{22}{7} \) × 9 = 198 + \( \frac{198}{7} \) = \( \frac{1584}{7} \) cm² (about 226.29 cm²).
(c) Cardboard for 35 penholders = 35 × \( \frac{1584}{7} \) = 5 × 1584 = 7920 cm².
(d) Cost = \( \frac{7920}{10} \) × Rs 20 = 792 × 20 = Rs 15840.
In simple words: Each penholder needs 1584/7 cm² of card; 35 of them need 7920 cm², which at Rs 20 per 10 cm² costs Rs 15840.

Teacher's Note:
1. A penholder has only one circular end, so add πr² once, not twice.
2. The rate is per 10 cm², so first find how many 10 cm² pieces are needed.