# GSEB Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7

Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7

Question 1.
Find:
(i) 0.4 Ć· 2
(ii) 0.35 Ć· 5
(iii) 2.48 Ć· 4
(iv) 65.4 Ć· 6
(v) 651.2 Ć· 4
(vi) 14.49 Ć· 7
(vii) 3.96 Ć· 4
(viii) 0.80 Ć· 5
Solution:
(i) 0.4 Ć· 2
Since $$\frac { 4 }{ 2 }$$ = 2 and there is one digit in the decimal part of the given decimal number.
ā“ Place the decimal point in the quotient such that there is one digit to its right.
ā“ 0.4 Ć· 2 = 0.2

(ii) 0.35 Ć· 5
Since 35 Ć· 5 = 7 and there are two digits in the decimal part of the given decimal number.
ā“ Place the decimal point in the quotient such that there are two digits to its right.
ā“ 0.35 Ć· 5 = 0.07

(iii) 2.48 Ć· 4
Since 248 Ć· 4 = 62 and there are two digits in the decimal part of the given decimal number.
ā“ Place the decimal point in the quotient such that there are two digits to its right.
ā“ 2.48 Ć· 4 = 0.62

(iv) 65.4 Ć· 6
Since 654 Ć· 6 = 109 and there is one digit in the decimal part of the given decimal number.
ā“ Place the decimal point in the quotient such that there is one digit to its right.
ā“ 65.4 Ć· 6 = 10.9

(v) 651.2 Ć· 4
Since 6512 Ć· 4 = 1628 and there is one digit in the decimal part of the given decimal number.
ā“ Place the decimal point in the quotient such that there is one digit to its right.
ā“ 651.2 Ć· 4 = 162.8

(vi) 14.49 Ć· 7
Since 1449 Ć· 7 = 207 and there are two digits in the decimal part of the given decimal number.
ā“ Place the decimal point in the quotient such that there are two digits to its right.
ā“ 14.49 Ć· 7 = 2.07

(vii) 3.96 Ć· 4
Since 396 Ć· 4 = 99 and there are two digits in the decimal part of the given decimal number.
ā“ Place the decimal point in the quotient such that there are two digits to its right.
ā“ 3.96 Ć· 4 = 0.99

(viii) 0.80 Ć· 5
Since 80 Ć· 5 = 16 and there are two digits in the decimal part of the given decimal number.
ā“ Place the decimal point in the quotient such that there are two digits to its right.
ā“ 0.80 Ć· 5 = 0.16

Question 2.
Find:
(i) 4.8 Ć· 10
(ii) 52.5 Ć· 10
(iii) 0.7 Ć· 10
(iv) 33.1 Ć· 10
(v) 272.23 Ć· 10
(vi) 0.56 Ć· 10
(vii) 3.97 Ć· 10
Solution:
(i) 4.8 Ć· 10
As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.
ā“ 4.8 Ć· 10 = 0.48

(ii) 52.5 Ć· 10
As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.
ā“ 52.5 Ć· 10 = 5.25

(iii) 0.7 Ć· 10
As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.
ā“ 0.7 + 10 = 0.07

(iv) 33.1 Ć· 10
As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.
ā“ 33.1 Ć· 10 = 3.31

(v) 272.23 Ć· 10
As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.
ā“ 272.23 Ć· 10 = 27.223

(vi) 0.56 Ć· 10
As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.
ā“ 0.56 + 10 = 0.056

(vii) 3.97 Ć· 10
As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.
ā“ 3.97 + 10 = 0.397

Question 3.
Find:
(i) 2.7 Ć· 100
(ii) 0.3 Ć· 100
(iii) 0.78 Ć· 100
(iv) 432.6 Ć· 100
(v) 23.6 Ć· 100
(vi) 98.53 Ć· 100
Solution:
(i) 2.7 Ć· 100
āµ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.
ā“ 2.7 Ć· 100 = 0.027

(ii) 0.3 Ć· 100
āµ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.
ā“ 0.3 Ć· 100 = 0.003

(iii) 0.78 Ć· 100
āµ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.
ā“ 0.78 Ć· 100 = 0.0078

(iv) 432.6 Ć· 100
āµ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.
ā“ 432.6 Ć· 100 = 4.326

(v) 23.6 Ć· 100
āµ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.
ā“ 23.6 Ć· 100 = 0.236

(vi) 98.53 Ć· 100
āµ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.
ā“ 98.53 Ć· 100 = 0.9853

Question 4.
Find:
(i) 0 7.9 Ć· 1000
(ii) 26.3 Ć· 1000
(iii) 38.53 Ć· 1000
(iv) 128.9 Ć· 1000
(v) 0.5 4 Ć· 1000
Solution:
(i) 7.9 Ć· 1000
āµ There are three zeros in 1000, therefore, the decimal point in the quotient is shifted to the left by 3 places.
ā“ 7.9 Ć· 1000 = 0.0079

(ii) 26.3 Ć· 1000
āµ There are three zeros in 1000, therefore, the decimal point in the quotient is shifted to the left by 3 places.
ā“ 26.3 Ć· 1000 = 0.0263

(iii) 38.53 Ć· 1000
āµ There are three zeros in 1000, therefore, the decimal point in the quotient is shifted to the left by 3 places.
ā“38.53 Ć· 1000 = 0.03853

(iv) 128.9 Ć· 1000
āµ There are three zeros in 1000, therefore, the decimal point in the quotient is shifted to the left by 3 places.
ā“ 128.9 4 Ć· 1000 = 0.1289

(v) 0.5 Ć· 1000
āµ There are three zeros in 1000, therefore, the decimal point in the quotient is shifted to the left by 3 places.
ā“ 0.5 4 Ć· 1000 = 0.0005

Question 5.
Find:
(i) 7 Ć· 3.5
(ii) 36 Ć· 0.2
(iii) 3.25 Ć· 0.5
(iv) 30.94 Ć· 0.7
(v) 0.5 Ć· 0.25
(vi) 7.75 Ć· 0.25
(vii) 76.5 Ć· 0.15
(viii) 37.8 Ć· 1.4
(ix) 2.73 Ć· 1.3
Solution:

Question 6.
A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?
Solution:
Total distance covered = 43.2 km
Quantity of petrol used = 2.4 litres
ā“ Distance covered in one litre petrol

= $$\frac { 43.2 }{ 2.4 }$$km
= [ $$\frac { 43.2 }{ 2.4 }$$ Ć· $$\frac { 24 }{ 10 }$$ ]km
= $$\frac { 432 }{ 10 }$$ x $$\frac { 10 }{ 24 }$$
= $$\frac { 18Ć1 }{ 1Ć1 }$$
= 18 km