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Here you will find NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers, which is one of the most important study materials for students preparing for the CBSE board examination. Our NCERT Solutions are based on the most recent CBSE syllabus. These solutions, prepared and designed by subject matter experts, are to the point and explained in a point-by-point manner to make it easier for students to learn, study, and revise at the last minute. All of the shortcuts, equations, rules, theorems, and axioms have also been covered. We have explained all of the topics in detail in the NCERT Solution Class 8 that we have provided, which will undoubtedly help students clear all of the required concepts. Important topics covered here include:

  • Exponents
  • Powers with Negative Exponents
  • Laws of exponents
  • Use of Exponents to Express Small Numbers in Standard Form
  • Comparing very large and very small numbers

Our NCERT Solutions Class 8 Maths is the best resource for practicing and learning all of the key concepts. Regular practice of our solutions will assist students in developing subject-related problem-solving skills. We have answers to all types of questions, including exercise and in-text questions.

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NCERT Solutions Class 8 Maths Chapter 12 Exponents and Powers

1. Evaluate:

(i) 3-2

(ii) (-4)-2

(iii) (1/2)-5  

Solution:

2.  Simplify and express the result in power notation with positive exponent:

(i) (-4)÷ (-4)8   

(ii) (1/23)2  

(iii) -(3)× (5/3)4 

(iv) (3-7 ÷ 3-10) × 3-5  

(v) 2-3 × (-7)-3

Solution:

3. Find the value of :

(i) (3+ 4-1) × 22

(ii) (2-1 × 4-1) ÷ 2 – 2

(iii) (1/2)-2 + (1/3)-2 + (1/4)-2

(iv) (3-1 + 4-1 + 5-1)0

(v) {(-2/3)-2}2

Solution:

4. Evaluate

(i) (8-1 × 53)/2-4

(ii) (5-1 × 2-2) × 6-1 

Solution:

5. Find the value of m for which 5m ÷ 5-3 = 55.

Solution:

5m ÷ 5-3 = 55

⇒ 5m-(-3) = 55 [∵ am ÷ an = am-n]

⇒ 5m+3 = 55

Comparing the powers of equal bases, we have

m + 3 = 5

m = 5 – 3 = 2, i.e., m = 2

6. Evaluate:

Solution:

7. Simplify:

Solution:

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8. Express the following numbers in standard form.

(i) 0.0000000000085

(ii) 0.00000000000942

(iii) 6020000000000000

(iv) 0.00000000837

(v) 31860000000

Solution:

(i) 0.0000000000085 = 0.0000000000085×(1012/1012) = 8.5 ×10-12

(ii) 0.00000000000942 = 0.00000000000942×(1012/1012) = 9.42×10-12

(iii) 6020000000000000 = 6020000000000000×(1015/1015) = 6.02×1015

(iv) 0.00000000837 = 0.00000000837×(109/109) = 8.37×10-9

(v) 31860000000 = 31860000000×(1010/1010) = 3.186×1010

9. Express the following numbers in usual form.

(i) 3.02 × 10-6

(ii) 4.5 × 104

(iii) 3× 10-8

(iv) 1.0001 × 109

(v) 5.8 × 1012

(vi) 3.61492 × 106

Solution:

(i) 3.02 × 10-6 = 3.02/106 = 0 .00000302

(ii) 4.5 × 104 = 4.5 × 10000 = 45000

(iii) 3 × 10-8 = 3/108 = 0.00000003

(iv) 1.0001 × 109 = 1000100000

(v) 5.8×1012 = 5.8 × 1000000000000 = 5800000000000

(vi) 3.61492 × 10= 3.61492 × 1000000 = 3614920

10. Express the number appearing in the following statements in standard form.

(i) 1 micron is equal to 1/1000000 m.

(ii) Charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.

(iii) Size of bacteria is 0.0000005 m

(iv)  Size of a plant cell is 0.00001275 m

(v) Thickness of a thick paper is 0.07 mm

Solution:

(i) 1 micron = 1/1000000

= 1/106

= 1×10-6

(ii) Charge of an electron is 0.00000000000000000016 coulombs.

= 0.00000000000000000016 × 1019/1019

= 1.6 × 10-19 coulomb

(iii) Size of bacteria = 0.0000005

=  5/10000000 = 5/107 = 5 × 10-7 m

(iv) Size of a plant cell is 0.00001275 m

= 0.00001275 × 105/105

= 1.275 × 10-5m

(v) Thickness of a thick paper = 0.07 mm

0.07 mm = 7/100 mm = 7/102 = 7 × 10-2 mm

11. In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

Solution:

Thickness of one book = 20 mm

Thickness of 5 books = 20 × 5 = 100 mm

Thickness of one paper = 0.016 mm

Thickness of 5 papers = 0.016 × 5 = 0.08 mm

Total thickness of a stack = 100+0.08 = 100.08 mm

= 100.08 × 102/102 mm

= 1.0008 x 102 mm

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