Fractions and Decimals

CBSE Class –VII Mathematics
NCERT Solutions
Chapter 2 Fractions and Decimals (Ex. 2.7)

Question 1. Find:

(i) 0.4

\div

2
(ii) 0.35

\div

5
(iii) 2.48

\div

4
(iv) 65.4

\div

6
(v) 651.2

\div

4
(v) 14.49

\div

7
(vii) 3.96

\div

4
(viii) 0.80

\div

Answer: (i) 0.4

\div

2 =

\frac{4}{10} \times \frac{1}{2} = \frac{2}{10}

= 0.2

(ii) 0.35

\div

5 =

\frac{35}{100} \times \frac{1}{5} = \frac{7}{100}

= 0.07

(iii)2.48

\div

4 =

\frac{248}{100} \times \frac{1}{4} = \frac{62}{100}

= 0.62

(iv) 65.4

\div

6 =

\frac{654}{10} \times \frac{1}{6} = \frac{109}{10}

= 10.9

(v) 651.2

\div

4 =

\frac{6512}{10} \times \frac{1}{4} = \frac{1628}{10}

= 162.8

(vi)14.49

\div

7 =

\frac{1449}{100} \times \frac{1}{7} = \frac{207}{100}

= 2.07

(vii)3.96

\div

4 =

\frac{396}{100} \times \frac{1}{4} = \frac{99}{100}

= 0.99

(viii)0.80

\div

5 =

\frac{80}{100} \times \frac{1}{5} = \frac{16}{100}

= 0.16

Question 2. Find:

(i) 4.8

\div

(ii) 52.5

\div

(iii) 0.7

\div

(iv) 33.1

\div

(v) 272.23

\div

(vi) 0.56

\div

(vii) 3.97

\div

Answer: (i) 4.8

\div

10 =

\frac{4.8}{10}

= 0.48

(ii) 52.5

\div

10 =

\frac{52.5}{10}

= 5.25

(iii) 0.7

\div

10 =

\frac{0.7}{10}

= 0.07

(iv) 33.1

\div

10 =

\frac{33.1}{10}

= 3.31

(v) 272.23

\div

10 =

\frac{272.23}{10}

= 27.223

(vi) 0.56

\div

10 =

\frac{0.56}{10}

= 0.056

(vii) 3.97

\div

10 =

\frac{3.97}{10}

= 0.397

Question 3. Find:

(i) 2.7

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100

(ii) 0.3

\div

100

(iii) 0.78

\div

100

(iv) 432.6

\div

100

(v) 23.6

\div

100

(vi) 98.53

\div

100

Answer: (i) 2.7

\div

100 =

\frac{27}{10} \times \frac{1}{100} = \frac{27}{1000}

= 0.027

(ii) 0.3

\div

100 =

\frac{3}{10} \times \frac{1}{100} = \frac{3}{1000}

= 0.003

(iii) 0.78

\div

100 =

\frac{78}{100} \times \frac{1}{100} = \frac{78}{10000}

= 0.0078

(iv) 432.6

\div

100 =

\frac{4326}{10} \times \frac{1}{100} = \frac{4326}{1000}

= 4.326

(v) 23.6

\div

100 =

\frac{236}{10} \times \frac{1}{100} = \frac{236}{1000}

= 0.236

(vi) 98.53

\div

100 =

\frac{9853}{100} \times \frac{1}{100} = \frac{9853}{10000}

0.9853

Question 4. Find:

(i) 7.9

\div

1000

(ii) 26.3

\div

1000

(iii) 38.53

\div

1000

(iv) 128.9

\div

1000

(v) 0.5

\div

1000

Answer: (i) 7.9

\div

1000 =

\frac{79}{10} \times \frac{1}{1000} = \frac{79}{10000}

= 0.0079

(ii) 26.3

\div

1000 =

\frac{263}{10} \times \frac{1}{1000} = \frac{263}{10000}

= 0.0263

(iii) 38.53

\div

1000 =

\frac{3853}{100} \times \frac{1}{1000} = \frac{3853}{100000}

= 0.03853

(iv) 128.9

\div

1000 =

\frac{1289}{10} \times \frac{1}{1000} = \frac{1289}{10000}

= 0.1289

(v) 0.5

\div

1000 =

\frac{5}{10} \times \frac{1}{1000} = \frac{5}{10000}

= 0.0005

Question 5. Find:

(i) 7

\div

3.5

(ii) 36

\div

0.2

(iii) 3.25

\div

0.5

(iv) 30.94

\div

0.7

(v) 0.5

\div

0.25

(vi) 7.75

\div

0.25

(vii) 76.5

\div

0.15

(viii) 37.8

\div

1.4

(ix) 2.73

\div

1.3

Answer: (i) 7

\div

3.5 =

7 \div \frac{35}{10} = 7 \times \frac{10}{35} = \frac{10}{5}

= 2

(ii) 36

\div

0.2 =

36 \div \frac{2}{10} = 36 \times \frac{10}{2} =

18 x 10 = 180

(iii) 3.25

\div

0.5 =

\frac{325}{100} \div \frac{5}{10} = \frac{325}{100} \times \frac{10}{5} = \frac{65}{10}

= 6.5

(iv) 30.94

\div

0.7 =

\frac{3094}{100} \div \frac{7}{10} = \frac{3094}{100} \times \frac{10}{7} = \frac{442}{10}

= 44.2

(v) 0.5

\div

0.25 =

\frac{5}{10} \div \frac{25}{100} = \frac{5}{10} \times \frac{100}{25} = \frac{10}{5}

= 2

(vi) 7.75

\div

0.25 =

\frac{775}{100} \div \frac{25}{100} = \frac{775}{100} \times \frac{100}{25}

= 31

(vii) 76.5

\div

0.15 =

\frac{765}{10} \div \frac{15}{100} = \frac{765}{10} \times \frac{100}{15}

= 51 x 10 = 510

(viii) 37.8

\div

1.4 =

\frac{378}{10} \div \frac{14}{10} = \frac{378}{10} \times \frac{10}{14}

= 27

(ix) 2.73

\div

1.3 =

\frac{273}{100} \div \frac{13}{10} = \frac{273}{100} \times \frac{10}{13} = \frac{21}{10}

= 2.1

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 petrol?

Answer:

∵

In 2.4 litres of petrol, distance covered by the vehicle = 43.2 km

∴

In 1 litre of petrol, distance covered by the vehicle = 43.2

\div

2.4

\frac{432}{10} \div \frac{24}{10}

\frac{432}{10} \times \frac{10}{24}

= 18 km

Thus, it covered 18 km distance in one litre of petrol.

Fractions and Decimals - Solutions 7