Class 8 Maths Chapter 10 Practice Question Answers - Exponents and Powers

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Q1: Write 3.61492 x 10⁶ in usual form.
Sol: 3.61492 x 10⁶
= 3.61492 x 1000000
= 3614920

Q2: Simplify [25 x t^-4]/[5^-3 x 10 x t^-8]
Sol:
We can write the given expression as;
[5²x t^-4]/[5^-3 x 5 x 2 x t^-8]
= [5² x t^-4+8]/[5^-3+1 x2]
= [5²⁺² x t⁴]/[2]
= [5⁴ x t⁴]/[2]
= [625/2] t⁴

Q3: 5 books and 5 paper sheets are placed in a stack. Find the total thickness of the stack if each book has a thickness of 20 mm and each sheet has a thickness of 0.016 mm.
Sol:
Given,
Thickness of 1 book = 20 mm
And,
Thickness of one paper = 0.016 mm
So, thickness of 5 books = 20 x 5 = 100 mm
And,
Thickness of 5 papers = 0.016 × 5 = 0.08 mm
Now, the total thickness of a stack is:
= 100 + 0.08 = 100.08 mm
= 100.08 10² / 10² mm
= 1.0008 × 10² mm

Q4: Find the value of x for which 2^x ÷ 2^-4 = 4⁵
Sol:
Given,
2^x ÷ 2^-4 = 4⁵
Now, 2^x × (½)^-4 = (2²)⁵
Or, 2^x × (½)^-4 = 2¹⁰
Thus, 2^x+4 = 2¹⁰
⇒ x + 4 = 10
Hence, x + 4 = 10
So, x = 6

Q5: Express 4^-3 as a power with base 2.
Sol:
4^-3 can be written as:
4^-3 = (2²)^-3
Now, by using exponential law i.e. (a^m)ⁿ = a^mn
4^-3 = 2^-6 (which is in base 2 form).

Q6: Simplify the following expression and express the result in positive power notation:
(−4)⁵ ÷ (−4)⁸
Sol:
Using a^m ÷ aⁿ = a^m-n
(−4)⁵ ÷ (−4)⁸ = (-4)⁵/(-4)⁸
⇒ (-4)^5-8 = 1/ (-4)³

Q7: Find the value of (4⁰ + 4^-1) × 2²
Sol:

(4⁰ + 4 ^-1) × 2² = (1 + ¼) × 4
= 5/4 x 4
= 5

Q8: Solve 3^-4 and (½)^-2
Sol:
We know, b^-n = 1/bⁿ
So, 3^-4 = 1/3⁴ = 1/81
And, (½)^-2 = 1^-2/2^-2 
= 2²/1² = 4

Q9: Evaluate a² × a³ × a^-5
Sol:
a² × a³ × a^-5 = a^2+3-5
= a^5-5
= a⁰ 
= 1

Q10: Evaluate (√4)^-3
Sol:
(√4)^-3 = (4^½)^-3
= 4^-3/2 = 1/ 4^3/2
= 1/(4³)^½ = 1/(64)^½
= 1/(8²)^½ = 1/8

Q11: Calculate the missing value of “x” in the following expression: (11/9)³ × (9/11)⁶ = (11/9)2^x-1
Sol:
Given: (11/9)³ × (9/11)⁶ = (11/9)2^x-1
The multiplier of L.H.S of the equation can be written as:
(11/9)³ × (11/9)^-6 = (11/9)2^x-1
⇒ (11/9)^3-6 = (11/9)2^x-1
Therefore, -3 = 2x – 1
2x = -3 + 1
x = -2/2
x = -1

Q12: If a new-born bear weighs 4 kg, calculate how many kilograms a five-year-old bear weigh if its weight increases by the power of 2 in 5 years?
Sol:
Given,
Weight of new-born bear = 4 kg
Rate of weight increase in 5 years = power to 2
Thus, the weight of the 5-year old bear = 42
= 16 kg

Q13: Express 0.00000000837 in standard form.
Sol:
0.00000000837
= 0.00000000837 x 109 / 109
= 8.37 ×10^-9