Important Questions Class 8 Maths Chapter 12 Exponents And Powers

Important questions for class 8 maths chapter 12 – exponents and powers, with solutions are provided here by our subject experts. Practice these questions to score better marks in the exam as it covers the latest CBSE syllabus (2022-2023) and is as per NCERT book/curriculum.

The chapter exponents and powers will explain to represent very large numbers or very small numbers in a standard form. Students can reach us at Important Questions Class 8 Math to get important questions and solutions for all the chapters of Maths 8th Class.

Also Check:

• Important 2 Marks Questions for CBSE 8th Maths
• Important 3 Marks Questions for CBSE 8th Maths
• Important 4 Marks Questions for CBSE 8th Maths

Exponents And Powers Important Questions For Class 8 (Chapter 12)

Some important short answer type questions and long answer type questions from exponents and powers are given below.

Short Answer Type Questions

1. Find the value of (4⁰ + 4^-1) × 2²

Solution:

(4⁰ + 4 ^-1) × 2² = (1 + ¼) × 4

= 5/4 x 4

= 5

2. Solve 3^-4 and (½)^-2

Solution:

We know, b^-n = 1/bⁿ

So, 3^-4 = 1/3⁴ = 1/81

And, (½)^-2 = 1^-2/2^-2 = 2²/1² = 4

3. Simplify the following expression and express the result in positive power notation:

(−4)⁵ ÷ (−4)⁸

Solution:

Using a^m ÷ aⁿ = a^m-n

(−4)⁵ ÷ (−4)⁸ = (-4)⁵/(-4)⁸

⇒ (-4)^5-8 = 1/ (-4)³

4. Evaluate a² × a³ × a^-5

Solution:

a² × a³ × a^-5 = a^2+3-5

= a^5-5

= a⁰ = 1

5. Express 4^-3 as a power with base 2.

Solution:

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).

Long Answer Type Questions

6. Evaluate (√4)^-3

Solution:

(√4)^-3 = (4^½)^-3

= 4^-3/2 = 1/ 4^3/2

= 1/(4³)^½ = 1/(64)^½

= 1/(8²)^½ = 1/8

7. Find the value of x for which 2^x ÷ 2^-4 = 4⁵

Solution:

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

8. Calculate the missing value of “x” in the following expression: (11/9)³ × (9/11)⁶ = (11/9)^2x-1

Solution:

Given: (11/9)³ × (9/11)⁶ = (11/9)^2x-1

The multiplier of L.H.S of the equation can be written as:

(11/9)³ × (11/9)^-6 = (11/9)^2x-1

⇒ (11/9)^3-6 = (11/9)^2x-1

Therefore, -3 = 2x – 1
2x = -3 + 1
x = -2/2
x = -1

9. 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.

Solution:

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

10. 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?

Solution:

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

Q.11: Simplify [25 x t^-4]/[5^-3 x 10 x t^-8]

Solution:

We can write the given expression as;

= [5² x t^-4+8]/[5^-3+1 x2]

= [5²⁺² x t⁴]/[2]

= [5⁴ x t⁴]/[2]

= [625/2] t⁴

Q.12: Express 0.00000000837 in standard form.

Solution:

0.00000000837

= 0.00000000837 x 10⁹ / 10⁹

= 8.37 ×10^-9

Q.13: Write 3.61492 x 10⁶ in usual form.

Solution: 3.61492 x 10⁶

= 3.61492 x 1000000

= 3614920

Maths Class 8 Chapter 12 Extra Questions

1. Evaluate: (-4)^-3
2. Simplify: (𝟑^-7÷ 𝟑^-9) × 𝟑^-4
3. Find the value of (3⁷ + 4^-3 + 5³)⁰
4. Evaluate: [{1/2}^-1+{1/3}^-1]^-1
5. Express 31860000000 in standard form.
6. Find x so that (-5)^x+1 × (-5)⁵ = (-5)⁷
7. Solve the following: (81)^-4 ÷ (729)^2-x = 9^4x

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