Gujarat Board 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 ÷ 2

(ii) 0.35 ÷ 5

(iii) 2.48 ÷ 4

(iv) 65.4 ÷ 6

(v) 651.2 ÷ 4

(vi) 14.49 ÷ 7

(vii) 3.96 ÷ 4

(viii) 0.80 ÷ 5

Solution:

(i) 0.4 ÷ 2

Since \(\frac { 4 }{ 2 }\) = 2 and there is one digit in the decimal part of the given decimal number.

∴ Place the decimal point in the quotient such that there is one digit to its right.

∴ 0.4 ÷ 2 = 0.2

(ii) 0.35 ÷ 5

Since 35 ÷ 5 = 7 and there are two digits in the decimal part of the given decimal number.

∴ Place the decimal point in the quotient such that there are two digits to its right.

∴ 0.35 ÷ 5 = 0.07

(iii) 2.48 ÷ 4

Since 248 ÷ 4 = 62 and there are two digits in the decimal part of the given decimal number.

∴ Place the decimal point in the quotient such that there are two digits to its right.

∴ 2.48 ÷ 4 = 0.62

(iv) 65.4 ÷ 6

Since 654 ÷ 6 = 109 and there is one digit in the decimal part of the given decimal number.

∴ Place the decimal point in the quotient such that there is one digit to its right.

∴ 65.4 ÷ 6 = 10.9

(v) 651.2 ÷ 4

Since 6512 ÷ 4 = 1628 and there is one digit in the decimal part of the given decimal number.

∴ Place the decimal point in the quotient such that there is one digit to its right.

∴ 651.2 ÷ 4 = 162.8

(vi) 14.49 ÷ 7

Since 1449 ÷ 7 = 207 and there are two digits in the decimal part of the given decimal number.

∴ Place the decimal point in the quotient such that there are two digits to its right.

∴ 14.49 ÷ 7 = 2.07

(vii) 3.96 ÷ 4

Since 396 ÷ 4 = 99 and there are two digits in the decimal part of the given decimal number.

∴ Place the decimal point in the quotient such that there are two digits to its right.

∴ 3.96 ÷ 4 = 0.99

(viii) 0.80 ÷ 5

Since 80 ÷ 5 = 16 and there are two digits in the decimal part of the given decimal number.

∴ Place the decimal point in the quotient such that there are two digits to its right.

∴ 0.80 ÷ 5 = 0.16

Question 2.

Find:

(i) 4.8 ÷ 10

(ii) 52.5 ÷ 10

(iii) 0.7 ÷ 10

(iv) 33.1 ÷ 10

(v) 272.23 ÷ 10

(vi) 0.56 ÷ 10

(vii) 3.97 ÷ 10

Solution:

(i) 4.8 ÷ 10

As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.

∴ 4.8 ÷ 10 = 0.48

(ii) 52.5 ÷ 10

As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.

∴ 52.5 ÷ 10 = 5.25

(iii) 0.7 ÷ 10

As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.

∴ 0.7 + 10 = 0.07

(iv) 33.1 ÷ 10

As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.

∴ 33.1 ÷ 10 = 3.31

(v) 272.23 ÷ 10

As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.

∴ 272.23 ÷ 10 = 27.223

(vi) 0.56 ÷ 10

As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.

∴ 0.56 + 10 = 0.056

(vii) 3.97 ÷ 10

As there is one zero in 10, the decimal point in the quotient is shifted to the left by one place.

∴ 3.97 + 10 = 0.397

Question 3.

Find:

(i) 2.7 ÷ 100

(ii) 0.3 ÷ 100

(iii) 0.78 ÷ 100

(iv) 432.6 ÷ 100

(v) 23.6 ÷ 100

(vi) 98.53 ÷ 100

Solution:

(i) 2.7 ÷ 100

∵ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.

∴ 2.7 ÷ 100 = 0.027

(ii) 0.3 ÷ 100

∵ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.

∴ 0.3 ÷ 100 = 0.003

(iii) 0.78 ÷ 100

∵ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.

∴ 0.78 ÷ 100 = 0.0078

(iv) 432.6 ÷ 100

∵ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.

∴ 432.6 ÷ 100 = 4.326

(v) 23.6 ÷ 100

∵ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.

∴ 23.6 ÷ 100 = 0.236

(vi) 98.53 ÷ 100

∵ There are two zeros in 100, therefore, the decimal point in the quotient is shifted to the left by two places.

∴ 98.53 ÷ 100 = 0.9853

Question 4.

Find:

(i) 0 7.9 ÷ 1000

(ii) 26.3 ÷ 1000

(iii) 38.53 ÷ 1000

(iv) 128.9 ÷ 1000

(v) 0.5 4 ÷ 1000

Solution:

(i) 7.9 ÷ 1000

∵ There are three zeros in 1000, therefore, the decimal point in the quotient is shifted to the left by 3 places.

∴ 7.9 ÷ 1000 = 0.0079

(ii) 26.3 ÷ 1000

∵ There are three zeros in 1000, therefore, the decimal point in the quotient is shifted to the left by 3 places.

∴ 26.3 ÷ 1000 = 0.0263

(iii) 38.53 ÷ 1000

∵ There are three zeros in 1000, therefore, the decimal point in the quotient is shifted to the left by 3 places.

∴38.53 ÷ 1000 = 0.03853

(iv) 128.9 ÷ 1000

∵ There are three zeros in 1000, therefore, the decimal point in the quotient is shifted to the left by 3 places.

∴ 128.9 4 ÷ 1000 = 0.1289

(v) 0.5 ÷ 1000

∵ There are three zeros in 1000, therefore, the decimal point in the quotient is shifted to the left by 3 places.

∴ 0.5 4 ÷ 1000 = 0.0005

Question 5.

Find:

(i) 7 ÷ 3.5

(ii) 36 ÷ 0.2

(iii) 3.25 ÷ 0.5

(iv) 30.94 ÷ 0.7

(v) 0.5 ÷ 0.25

(vi) 7.75 ÷ 0.25

(vii) 76.5 ÷ 0.15

(viii) 37.8 ÷ 1.4

(ix) 2.73 ÷ 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

∴ Distance covered in one litre petrol

= \(\frac { 43.2 }{ 2.4 }\)km

= [ \(\frac { 43.2 }{ 2.4 }\) ÷ \(\frac { 24 }{ 10 }\) ]km

= \(\frac { 432 }{ 10 }\) x \(\frac { 10 }{ 24 }\)

= \(\frac { 18×1 }{ 1×1 }\)

= 18 km