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
āµ 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
āµ 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
āµ 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
āµ 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
āµ 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
āµ 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
āµ 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
āµ 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
āµ 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
āµ 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
āµ 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 \(\frac { 10 }{ 24 }\)
= \(\frac { 18Ć1 }{ 1Ć1 }\)
= 18 km