NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3

NCERT Solutions Class 7 Mathematics 
Chapter – 2 (Fractions and Decimals)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 2 Fractions and Decimals Exercise 2.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 2 Fractions and Decimals

Exercise – 2.3

1. Find:

(i) \mathbf{\frac{1}{4}} of (a) \mathbf{\frac{1}{4}} (b) \mathbf{\frac{3}{5}} (c) \mathbf{\frac{4}{3}}

(ii) \mathbf{\frac{1}{7}} of (a) \mathbf{\frac{2}{9}} (b) \mathbf{\frac{6}{5}} (c) \mathbf{\frac{3}{10}}

Solution –

(i) (a) \mathbf{\frac{1}{4}} of \mathbf{\frac{1}{4}}

= \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}

(b) \mathbf{\frac{1}{4}} of \mathbf{\frac{3}{5}}

= \frac{1}{4} × \frac{3}{5} = \frac{3}{20}

(c) \mathbf{\frac{1}{4}} of  \mathbf{\frac{4}{3}}

= \frac{1}{4} × \frac{4}{3} = \frac{1}{3}

(ii) (a) \mathbf{\frac{1}{7}} of  \mathbf{\frac{2}{9}}

= \frac{1}{7} × \frac{2}{9} = \frac{2}{63}

(b) \mathbf{\frac{1}{7}} of  \mathbf{\frac{6}{5}}

= \frac{1}{7} × \frac{6}{5} = \frac{6}{35}

(c) \mathbf{\frac{1}{7}} of  \mathbf{\frac{3}{10}}

= \frac{1}{7} × \frac{3}{10} = \frac{3}{70}

2. Multiply and reduce to lowest form (if possible):

(i) \mathbf{\frac{2}{3} \times 2\frac{2}{3}} 

(ii) \mathbf{\frac{2}{7} \times \frac{7}{9}} 

(iii) \mathbf{\frac{3}{8} \times \frac{6}{4}}

(iv) \mathbf{\frac{9}{5} \times \frac{3}{5}}

(v) \mathbf{\frac{1}{3} \times \frac{15}{8}}

(vi) \mathbf{\frac{11}{2} \times \frac{3}{10}}

(vii) \mathbf{\frac{4}{5} \times \frac{12}{7}}

Solution –

(i) \mathbf{\frac{2}{3} \times 2\frac{2}{3}} 

= \frac{2}{3} \times \frac{8}{3} = \frac{16}{9} = 1\frac{7}{9}

(ii) \mathbf{\frac{2}{7} \times \frac{7}{9}} 

= \frac{2}{9}

(iii) \mathbf{\frac{3}{8} \times \frac{6}{4}}

= \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}

(iv) \mathbf{\frac{9}{5} \times \frac{3}{5}}

= \frac{27}{25} = 1\frac{2}{25}

(v) \mathbf{\frac{1}{3} \times \frac{15}{8}}

= \frac{5}{8}

(vi) \mathbf{\frac{11}{2} \times \frac{3}{10}}

= \frac{33}{20} = 1\frac{13}{20}

(vii) \mathbf{\frac{4}{5} \times \frac{12}{7}}

= \frac{48}{35} = 1\frac{13}{35}

3. Multiply the following fractions:

(i) \mathbf{\frac{2}{5} \times 5\frac{1}{4}}

(ii) \mathbf{6\frac{2}{5} \times \frac{7}{9}}

(iii) \mathbf{\frac{3}{2} \times 5\frac{1}{3}}

(iv) \mathbf{\frac{5}{6} \times 2\frac{3}{7}}

(v) \mathbf{3\frac{2}{5} \times \frac{4}{7}}

(vi) \mathbf{2\frac{3}{5} \times 3}

(vii) \mathbf{3\frac{4}{7} \times \frac{3}{5}}

Solution –

(i) \mathbf{\frac{2}{5} \times 5\frac{1}{4}}

= \frac{2}{5} \times \frac{21}{4} = \frac{21}{10} = 2\frac{1}{10}

(ii) \mathbf{6\frac{2}{5} \times \frac{7}{9}}

= \frac{32}{5} \times \frac{7}{9} = \frac{224}{45} = 4\frac{44}{45}

(iii) \mathbf{\frac{3}{2} \times 5\frac{1}{3}}

= \frac{3}{2} \times \frac{16}{3} = 8

(iv) \mathbf{\frac{5}{6} \times 2\frac{3}{7}}

= \frac{5}{6} \times \frac{17}{7} = \frac{85}{42} = 2\frac{1}{42}

(v) \mathbf{3\frac{2}{5} \times \frac{4}{7}}

= \frac{17}{5} \times \frac{4}{7} = \frac{68}{35} = 1\frac{33}{35}

(vi) \mathbf{2\frac{3}{5} \times 3}

= \frac{13}{5} \times 3 = \frac{39}{5} = 7\frac{4}{5}

(vii) \mathbf{3\frac{4}{7} \times \frac{3}{5}}

= \frac{25}{7} \times \frac{3}{5} = \frac{5}{7} \times \frac{3}{1} = \frac{15}{7} = 2\frac{1}{7}

4. Which is greater?

(i) \mathbf{\frac{2}{7}} of \mathbf{\frac{3}{4}} or \mathbf{\frac{3}{5}} of \mathbf{\frac{5}{8}}

(ii) \mathbf{\frac{1}{2}} of \mathbf{\frac{6}{7}} or \mathbf{\frac{2}{3}} of \mathbf{\frac{3}{7}}

Solution –

(i) \mathbf{\frac{2}{7}} of \mathbf{\frac{3}{4}} or \mathbf{\frac{3}{5}} of \mathbf{\frac{5}{8}}

= \frac{2}{7} × \frac{3}{4} or  \frac{3}{5} × \frac{5}{8}

= \frac{2}{7} × \frac{3}{4} = \frac{1}{7} × \frac{3}{2} = \frac{3}{14}

= \frac{3}{5} × \frac{5}{8} = \frac{3}{8}

LCM of 14 and 8 is 56
\frac{3}{14} = \frac{3\times 4}{14\times 4}= \frac{12}{56}

\frac{3}{8} = \frac{3\times 7}{8\times7} = \frac{21}{56}

Clearly,

\frac{12}{56} < \frac{21}{56}  Hence, \frac{2}{7} of \frac{3}{4} < \frac{3}{5} of \frac{5}{8}

(ii) \mathbf{\frac{1}{2}} of \mathbf{\frac{6}{7}} or \mathbf{\frac{2}{3}} of \mathbf{\frac{3}{7}}

= \frac{1}{2} × \frac{6}{7} or  \frac{2}{3} × \frac{3}{7}

= \frac{1}{2} × \frac{6}{7} = \frac{3}{7} 

= \frac{2}{3} × \frac{3}{7} = \frac{2}{7}

Clearly,

\frac{3}{7}  > \frac{2}{7} Hence, \frac{1}{2} of \frac{6}{7} > \frac{2}{3} of  \frac{3}{7}

5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is ¾ m. Find the distance between the first and the last sapling.
Solution –
Number of saplings = 4
The distance between two adjacent saplings = ¾ m
∴ Distance between the first and the last sapling = ¾ + ¾ + ¾ + ¾
= \frac{9}{4} m = 2\frac{1}{4} m
Hence, the distance between the first and the last saplings is 2\frac{1}{4} m.

6. Lipika reads a book for 1 ¾ hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Solution –
Lipika reads the book for = 1\frac{3}{4} hours every day = \frac{7}{4} hours.
Number of days she took to read the entire book = 6 days
∴ Total number of hours required by her to complete the book =  6 × \frac{7}{4}  =  3 × \frac{7}{2}  = \frac{21}{2}  = 10\frac{1}{2} hours

Hence, the total number of hours required by her to complete the book is 10\frac{1}{2} hours.

7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using \mathbf{2\frac{3}{4}} litres of petrol.
Solution –
The total number of distance travelled by a car in 1 liter of petrol = 16 km
Total quantity of petrol = 2\frac{3}{4} liter = \frac{11}{4} liters
Total number of distance travelled by car in \frac{11}{4} liters of petrol =  16 × \frac{11}{4}

= 4 × 11   = 44 km

∴ Total number of distance travelled by car in \frac{11}{4} liters of petrol is 44 km.

8. (a) (i) provide the number in the box [ ], such that \mathbf{\frac{2}{3}} × [ ] = \mathbf{\frac{10}{30}}.
(ii) The simplest form of the number obtained in [ ] is ______.
(b) (i) provide the number in the box [ ], such that \mathbf{\frac{3}{5}} × [ ] = \mathbf{\frac{24}{75}}
(ii) The simplest form of the number obtained in [ ] is ______. 

Solution –
(a) (i) provide the number in the box [ ], such that \mathbf{\frac{2}{3}} × [ ] = \mathbf{\frac{10}{30}}.
\mathbf{\frac{2}{3}} × [ ] = \mathbf{\frac{10}{30}}

⇒ [] = \frac{10}{30}\div \frac{2}{3}

⇒ [] = \frac{10}{30} \times \frac{3}{2} = \frac{5}{10}

∴ The required number in the box is \frac{5}{10} 

(ii) The simplest form of the number obtained in [ ] is ____.
The number in the box is \frac{5}{10}
Then,
The simplest form of \frac{5}{10} is ½

(b) (i) provide the number in the box [ ], such that \mathbf{\frac{3}{5}} × [ ] = \mathbf{\frac{24}{75}}
\mathbf{\frac{3}{5}} × [ ] = \mathbf{\frac{24}{75}} ⇒ [ ] = \frac{24}{75} \div \frac{3}{5}

⇒ [ ] = \frac{24}{75} \times \frac{5}{3} = \frac{8}{15}

∴ The required number in the box is \frac{8}{15}

(ii) The simplest form of the number obtained in [ ] is ______.
The number in the box is \frac{8}{15}
Then, The simplest form of \frac{8}{15} is \frac{8}{15}.

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