# Frank Learning Maths Class 8 Solutions Chapter 2

## Frank Learning Maths Class 8 Solutions Chapter 2 Exponents

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 1 Exponents. Here students can easily find step by step solutions of all the problems for Exponents. Here students will find solutions for Exercise 2.1 and 2.2. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.

Exponents Exercise 2.1 Solution

Question no – (1)

Solution :

(a) {(1/3)-3 – (1/3)-3} ÷ (1/4)-3

= {1/3-3 – 1/2-3} ÷ (1/4-3)

= {33 – 23} ÷ 43

= (27 – 8) × 1/43

= 19 × 1/64

= 19/64…(Simplified)

(b) {(1/3)-1 × (-9)-1}-1

= {(1/3-1) × (-9-1)}-1

= {31 × 1/-91}-1

= (- 1/3)-1

= -3…(Simplified)

(c) (52 – 32) × (2/3)2

= (25 – 9) × (4/9)

= 16 ×  4/9

= 64/9…(Simplified)

(d) {(2/3)2}3 × (1/3)-4 × (3)-2 × (2)-1

= (4/9)3 × (1/3-4) × 1/32 × 1/21

= 4 × 4 × 4 /9 × 9 × 9 ×  34 × 1/32 × 1/2

= 32/9…(Simplified)

(e) (4-1 ÷ 6-1)3

= (1/4 ÷ 1/6)3

= (1/4 × 6)3

= (3/2)3

= 33/23

= 27/8…(Simplified)

(f) {6-1 – 5-1} ÷ 3-1

= (1/6 – 1/5) × 2/9 ÷ 1/3

= 5 – 6/30 ÷ 1/3

= 1/30 × 3

= 1/10…(Simplified)

Question no – (2)

Solution :

(a) (1/3)-2 + (1/2)-2 + (1/4)-2

= 1/3-2 + 1/2-2 + 1/4-2

= 32 + 22 + 42

= 9 + 4 + 16

= 29

(b) (1/5)45 × (1/5)-60 – (1/5)+28 × (1/5)-43

= 1/545 × 560 – 1/528 × 543

= (5)60 – 45 – (5)43 – 28

= 515 – 515

= 0

(c) {(2+ 3-1) × 92}

= (1 + 1/3) × 92

= 4/3 × 9 ×

= 108

(d) 3-5 × 10-5 × 125/5-7 × 6-5

= 3-5 × 2-5 × 5+5 × 125/5-7 × 2-5 × 3-5

= 125/(5-2)

= 125 × 52

= 3125

(e) (4/3)-2 × (32/4)(2)

= (4/3) -2 × (32/4) -2

= 4 -2/3 -2 × 3 -4/4 -2

= 3 -2 × 4

= 1/9

(f) (13 + 23 + 33)-5/2

= (1 + 8 + 27)-5/2

= (36)-5/2

= (62)-5/2

= 6-5

= 1/65

= 1/7776

(g) (-7)-3 × 492 × 11-4 × 1212

= – 1/73 × 73 × 1/114 × 114

= 1

(h) 3-5 × 10-4 × 625/5-3 × 6-7

= 3-5 × 2-4 × 5-4 × 625/5-3 × 2-7 × 3-7

= 32 × 23 × 625/5

= 9000

(i) 16 × 102 × 64/24 × 42

= 42 × 22 × 52 × 22 × 16/24 × 42

= 25 × 16

= 400

(j) (3-22 × (52-3 × (t-32/(3-25 × (53-2 × (t-43

= 3-4 × 5-6 × t-6/3-10 × 5-6 × t-12

= 36 × t6

= 729t6

Question no – (3)

Solution :

(a) (- 1/7)-5 ÷ (- 1/7)-7 = (-7)x

= 75 ÷ 77 = (-7) x

= 75/77 = (-7) x

= 7-2 = (-7) x

= (-7) -2 = (-7) x

x = -2

Hence, the x will be -2

(b) (2/5) 2x + 6 × (2/5) 3 = (2/5) x + 2

= 2x + 6 + 3 = x + 2

= x = 2 – 9

x = -7

Thus, the x will be -7

(c) 2x + 2x + 2x = 192

= 3 × 2x = 192

= 2x = 192/3

= 2x = (2) 6

x = 6

Therefore, the x will be 6

(d) (-6/7) x – 7 = 1

= (- 6/7) x – 7 = (- 6/7) 0

= x – 7 = 0

x = 7

Hence, the x will be 7

(e) 23x = 82x + 1

= 8x = 82x + 1

= x = 2x + 1

= -x = 1

x = -1

Thus, the x will be -1

(f) 5x + 5x – 1 = 750

= 5x (1 + 1/5) = 750

= 5x 6/5 = 750

= 5x = 125 × 5

= 5x = (5) 4

x = 4

Therefore, the x will be 4

Question no – (4)

Solution :

Let the multiplied number be x

∴ (-5)-1 × x = (11)-1

X = 11-1/-5-1

= – 1/11 × 5

= – 5/11

Hence, -5/11 should be multiplied.

Question no – (5)

Solution :

Let the number should be x

(-1/2)-1 × x = (- 5/9)-1

X = (- 5/9)-1 × (2)-1

= – 9/5 × (1/2)

= – 9/10

Therefore, -9/10 should be multiplied.

Question no – (6)

Solution :

Let the number should be x

(-12)-1/x = (-4)-1

x = -(12)-1/(-4)-1

= -3-1

= 1/3

Thus, 1/3 should be divided.

Question no – (7)

Solution :

Let the number should be x

∴ (2/7)-2 × x = (5/7)-1

= (7/2)2 × x = (7/5)

X = 7/5 × (/2/7)2

= 4/35

Therefore, 4/35 should be multiplied.

Exponents Exercise 2.2 Solution

Question no – (1)

Solution :

(a) 2, 43, 00, 00, 00, 000 in standard form,

= 243 × 109

= 2.43 × 1011

(b) 89, 00, 00, 00, 000 in standard,

= 8.9 × 1010

(c) 16, 70, 000 in standard,

= 1.67 × 106

(d) 0.0000000029 in standard,

= 2.9 × 10-9

(e) 0.00000000007 in standard,

= 7 × 10-10

Question no – (2)

Solution :

(a) 1.732 × 10^8

In usual form = 173200000

(b) 9.7137 × 10^13

In usual form = 97137000000000

(c) 4.01853 × 10^7

In usual form = 40185300

(d) 1.31 × 10^-9

In usual form = 0.0000000131

(e) 5.3 × 10^-13

In usual form = 0.00000000000053

(f) 8.37 × 10^-6

In usual form = 0.0000837

Question no – (3)

Solution :

(a) Size of a bacteria is 0.0000005 m.

In standard form = 5 × 10^-6 m

(b) Diameter of the write on a computer chip is 0.000003 m.

In standard form = 3 × 10^-5 m

(c) Speed of light is 300, 000, 000 m/s.

In standard form = 3 × 10^8 m/s

(d) The distance of the moon from the earth is 384, 460, 000 m (approx.)

In standard form = 3.8446 × 10^8 (approx.)

(e) Size of a plant cell is 0.000 000 012 75 km.

In standard form = 1.275 × 10^-10 km.

(f) Distance of the sun from the earth is 149, 600, 000 km.

In standard form = 1.496 × 10^8 km.

Question no – (4)

Solution :

As per the given question,

The size of an animal cell is = 0.0000068 m

The size of a plant cell is = 0.000015 m

0.000015/0.0000068

= 15/68 × 10-10

= 150/68

= 75 : 34

Hence, the ratio of the size of the plant cell to that of the animal cell will be 75 : 34

Question no – (5)

Solution :

According to the question,

Diameter of the sun = 1.4 × 109 m

Diameter of the earth = 1.275 × 107m

1.4 × 109/1.275 × 107

= 14 × 10/1275 × 104

= 14 × 104/1275

= 5600 : 51

Therefore, the ratios of their diameters will be 5600 : 51

Question no – (6)

Solution :

Total pages,

=  500 × 7

= 3500 pages

Thickness,

= 0.000075m

= 75 × 10-4 m.

Thickness of the stack,

= 3500 × 75 × 10-4 m

= 35 × 75 × 10-2 m

= 2625 × 10-2 m

= 26.25 m

Therefore, the total thickness of the stack will be 26.25 m.

Exponents – Chapter Check-up Solution :

Question no – (1)

Solution :

(a) The reciprocal of (2/5)^-1 is – (2/5)

(b) (- 2)^2×3-1 = (- 2)^6-1 = (- 2)^5 = – 3^2

(c) (- 3/4)^5 × (5/3)^5 is equal to – 3^5/4^5 × 5^5/3^5 = (- 5/4)^5

(d) [2^-1 + 3^-1 + 4^-1]^0 = 1

(e) (6^0 – 7^0) (6^0 + 7^0) = (1 – 1) (1 + 1) = 0

(f) The standard form of 0.000064 is 6.4 × 10^-5

(g) The standard form of 234500000 is 2.345 × 10^8

(h) The usual form of 2.03 × 10^-5 is = 0.000203

(i) Cube of – 1/2 is (- 1/2)^3 = – 1/8

(j) The value of [3^-1 × 4^-1]^2 is 1/3 × 1/4 = 1/12

Question no – (2)

Solution :

(a) (1/4)-2 + (1/2)-2 + (1/3) -2

= (4) 2 + (2) 2 + (3)2

= 16 + 4 + 9

= 29…(Simplified)

(b) (-2/3) -2)3 × (1/3)-4 × 3-1 × 1/6

= ((3/2) 23 × (3) 4 × 1/3 × 1/3×2

= (3 × 2/2 × 2) 3 × 32 × 32 × 1/32 × 2

= 33 × 33/23 × 23 × 32/2

= 38/28…(Simplified)

(c) 49 × z-3/7-3 × 10 × z-5 (z ≠ 0)

= 72 × z2/7-3 × 10

= 75 × z2 × 10-1…(Simplified)

(d) (25 ÷ 28) × 2-7

= 25/28 × 2-7

= 2-7/2-3

= 1/24…(Simplified)

Question no – (3)

Solution :

(a) (5/3) -2 × (5/3) -14 = (5/3) 8x

= (5/3) -16 = (5/3) 8x

= 8x = -16

= x = -2

So, the value of x will be -2

(b) (- 2) 3 × (- 2) -6 = (- 2) 2x – 1

= (- 2) -3 = (- 2) 2x – 1

= 2x -1 = -3

= 2x = -2

= x = -1

Hence, the value of x will be -1

(c) (2-1 + 4-1 + 6-1 + 8-1x = 1

= (2-1 + 4-1 + 6-1 + 8-1x

= (2-1 + 4-1 + 6-1 + 8-10

= x = 0

Therefore, the value of x will be 0.

Question no – (4)

Solution :

(a) The mass of a proton in gram is 1673/1000000000000000000000000000

= 1.673 × 10-24 gram

(b) A Helium atom has a diameter of 0.000000022 cm.

= 2.2 × 10-8 cm

(c) Mass of a molecule of hydrogen gas is about 0.00000000000000000000334 tons.

= 3.34 × 10-22 tons.

(d) Human body has 1 trillion of cells which vary in shapes and sizes.

= 1 × 1012

(e) Express 56 km in m.

= 5.6 × 104 m

(f) Express 5 tons in g.

= 5 × 105 gm

(g) Express 2 years in seconds.

= 6.3072 × 107 s

(h) Express 5 hectares in cm2 (1 hectare = 10000 m2)

= 5 × 108 cm2

Question no – (5)

Solution :

(a) 6n/6-2 = 63

= (6) n+2 = 63

= x + 2 = 3

= x = 1

Hence, the value of n will be 1.

(b) 2× 26/2-3 = 218

= 26+n+3 = 218

= n + 9 = 18

= n = 9

Thus, the value of n will be 9.

Question no – (6)

Solution :

As per the given question,

The diameter of the sun = 1.4 × 109 m

The diameter of the earth = 1.2756 × 107 m

1.4 × 109 m

= 1.2756 × 107 m

= 1.3 × 107 m (Approximately)

Therefore, diameter of the sun is 100 time of the Earth diameter.

Question no – (7)

Solution :

According to the question,

Mass of mars is 6.42 × 1029 kg

Mass of the sun is 1.99 × 1030 kg.

6.42 × 1029 kg + 1.99 × 1030

= 6.42 × 1029 + 19.9 × 1029

= 26.32 × 1029

= 2.632 × 1030 kg

Hence, the total mass will be 2.632 × 1030 kg.

Question no – (8)

Solution :

As per the question,

Distance between sun and earth is = 1.496 × 108 km

Distance between earth and moon is = 3.84 × 108 m

Distance between moon and sun,

= – 1.496 × 108/2

= 7.48 × 107

Therefore, the distance between moon and sun will be 7.48 × 107

Question no – (11)

Solution :

According to the question,

The cells of a bacteria double itself every hour.

How many cells will there be after 8 hours,

The answer in power will be 28

Question no – (12)

Solution :

Given, (2/9)3 × (2/9)-6 = (2/9)2x-1

= – 3 = 2x – 1

= 2x = -2

= x = -1

Therefore, the x will be -1

Question no – (13)

Solution :

Let, we should divided the number by x

∴ (- 3/2) -3/x = (4/27) -2

= -(2/3) 3 × 1/x = (27/4) 2

= – (2/3) 3 × (4/27) 2 = x

= – 23/33 × 24/36 = x

= (- 2) 7/39 = x

Therefore, (-2)7/39 should be divided.

Previous Chapter Solution :

Updated: June 5, 2023 — 5:48 am