# Exercise 2.3 Fractions and Decimals - NCERT Solutions Class 7

## Chapter 2 Ex.2.3 Question 1

Find:

(i) \begin{align} \frac{1}{4}\end{align} of

(a) \begin{align} \frac{1}{4}\end{align}

(b) \begin{align} \frac{3}{5}\end{align}

(c) \begin{align} \frac{4}{3}\end{align}

(ii) \begin{align} \frac{1}{7} \end{align}  of

(a) \begin{align} \frac{2}{9}\end{align}

(b) \begin{align} \frac{6}{5}\end{align}

(c) \begin{align} \frac{3}{10}\end{align}

### Solution

What is known?

Fractions

What is unknown?

Product of the given fractions.

Reasoning:

Find the product by multiplying numerator with numerator and denominator with denominator.

Steps:

\begin{align} \text{ (i) }\left( \text{a} \right)\text{ }\frac{1}{4}\text{ of }\frac{1}{4}&=\frac{1}{4}\times \frac{1}{4}\\&=\frac{1}{16}\\ \\ \text{ }\,\,\,\,\,\,\,\,\text{(b) }\frac{1}{4}\text{ of }\frac{3}{5}&=\frac{1}{4}\times \frac{3}{5}\\&=\frac{3}{20} \\\\ \text{ (c) }\frac{1}{4}\text{ of }\frac{4}{3}&=\frac{1}{4}\times \frac{4}{3}\\&=\frac{4}{12}\\&=\frac{1}{3} \\\end{align}

\begin{align} \text{ (ii) (a) }\frac{1}{7}\text{ of }\frac{2}{9}&=\frac{1}{7}\times \frac{2}{9}\\&=\frac{1\times 2}{7\times 9}\\&=\frac{2}{63} \\\\ \text{ }\,\text{ (b) }\frac{1}{7}\text{ of }\frac{6}{5}&=\frac{1}{7}\times \frac{6}{5}\\&=\frac{1\times 6}{7\times 5}\\&=\frac{6}{35} \\\\\ \text{ (c) }\frac{1}{7}\text{ of }\frac{3}{10}&=\frac{1}{7}\times \frac{6}{5}\\&=\frac{1\times 3}{7\times 10}\\&=\frac{3}{70} \\\end{align}

## Chapter 2 Ex.2.3 Question 2

Multiply and reduce to lowest form (if possible):

(i)    \begin{align} \frac{2}{3} \times 2\frac{2}{3}\end{align}

(ii)   \begin{align} \frac{2}{7} \times \frac{7}{9}\end{align}

(iii)  \begin{align} \frac{3}{8} \times \frac{6}{4}\end{align}

(iv)  \begin{align} \frac{9}{5} \times \frac{3}{5}\end{align}

(v)   \begin{align} \frac{1}{3} \times \frac{{15}}{8}\end{align}

(vi)  \begin{align} \frac{{11}}{2} \times \frac{3}{{10}}\end{align}

(vii) \begin{align} \frac{4}{5} \times \frac{{12}}{7}\end{align}

### Solution

What is known?

Expression

What is unknown?

Product of the given expression.

Reasoning:

Find the product by multiplying numerator with numerator and denominator with denominator.

Steps:

(i)

\begin{align}\frac{2}{3} \times 2\frac{2}{3}&= \frac{2}{3} \times \frac{8}{3}\\{}&= \frac{{2 \times 8}}{{3 \times 3}}\\&= {\frac{{16}}{9}}\end{align}

(improper fraction)

Converting \begin{align} \frac{{16}}{9}\end{align} into mixed fraction,

we get,  \begin{align} \frac{{16}}{9}\end{align} \begin{align} = 1\frac{7}{9}\end{align}

(ii)

\begin{align} \frac{2}{7} \times \frac{7}{9} &= \frac{{2 \times 7}}{{7 \times 9}}\\ &= \frac{{14}}{{63}}\end{align}

Reducing \begin{align} \frac{{14}}{{63}}\end{align}to the lowest form,

we get,  \begin{align} \frac{14}{63}= \frac{2}{9}\end{align}

(iii)

\begin{align}\frac{3}{8} \times \frac{6}{4} &= \frac{{3 \times 6}}{{8 \times 4}}\\&= \frac{{18}}{{32}}\end{align}
Reducing\begin{align} \frac{{18}}{{32}}\end{align} to the lowest form,

we get, \begin{align} \frac{18}{32}= \frac{9}{{16}}\end{align}

(iv)

\begin{align}\frac{9}{3} \times \frac{3}{5}&= \frac{{9 \times 3}}{{5 \times 5}}\\&= \frac{{27}}{{25}}\end{align}

(improper fraction)

Converting \begin{align} \frac{{27}}{{25}}\end{align}into mixed fraction,

we get,  \begin{align}\frac{27}{25} = 1\frac{2}{{25}}\end{align}

(v)

\begin{align}\frac{1}{3} \times \frac{15}{8} &= \frac{1 \times 15}{3 \times 8}\\[0.5mm]&= \frac{15}{24}\end{align}

Reducing \begin{align} \frac{{15}}{{24}}\end{align} to the lowest form,

we get,  \begin{align}\frac{15}{24} = \frac{5}{8}\end{align}

(vi)

\begin{align} \frac{11}{2} \times \frac{3}{10} &= \frac{11 \times 3}{2 \times 10}\\[0.5mm] &= \frac{33}{20}\end{align}

(improper fraction)

Converting \begin{align} \frac{{33}}{{20}}\end{align} into mixed fraction,

we get,  \begin{align} \frac{33}{20}= 1\frac{{13}}{{20}}\end{align}

(vii)

\begin{align}\frac{4}{5} \times \frac{12}{7} &= \frac{4 \times 12}{5 \times 7}\\[0.5mm]&= \frac{48}{35} \end{align}

(improper fraction)

Converting \begin{align} \frac{{48}}{{35}}\end{align}into mixed fraction,

we get,\begin{align} \frac{48}{35}= 1\frac{{13}}{{35}}\end{align}

## Chapter 2 Ex.2.3 Question 3

Multiply the following fractions:

i) \begin{align} \frac{2}{5} \times 5\frac{1}{4}\end{align}

ii) \begin{align} 6\frac{2}{5} \times \frac{7}{9} \end{align}

iii) \begin{align} \frac{3}{2} \times 5\frac{1}{3}\end{align}

iv) \begin{align} \frac{5}{6} \times 2\frac{3}{7}\end{align}

v) \begin{align} 3\frac{2}{5} \times \frac{4}{7}\end{align}

vi) \begin{align} 2\frac{3}{5} \times 3\end{align}

vii) \begin{align} 3\frac{4}{7} \times \frac{3}{5}\end{align}

### Solution

What is known?

Expression

What is unknown?

Product of the given expression.

Reasoning:

Covert mixed fraction into improper fraction then find the product.

Steps:

i) \begin{align} \frac{2}{5} \times 5\frac{1}{4}\end{align}

\begin{align} =\frac{2}{5}\times \frac{21}{4}\end{align}

Reducing to the lowest form, we get

\begin{align}&= \frac{1}{5} \times \frac{{21}}{2}\\&= \frac{{1 \times 21}}{{5 \times 2}}\\&= \frac{{21}}{{20}}\end{align}

(improper fraction)

Converting into mixed fraction, we get \begin{align}= 2\frac{1}{{10}}\end{align}

ii) \begin{align} 6\frac{2}{5} \times \frac{7}{9} \end{align}

\begin{align}&= \frac{{32}}{5} \times \frac{7}{9}\\&= \frac{{224}}{{45}}\end{align}

(improper fraction)

Converting \begin{align} \frac{{224}}{{45}}\end{align} into mixed fraction, we get \begin{align} = 4\frac{{44}}{{45}} \end{align}

iii) \begin{align} \frac{3}{2} \times 5\frac{1}{3}\end{align}

\begin{align}&= \frac{3}{2} \times \frac{{16}}{3}\\&= \frac{{3 \times 16}}{{2 \times 3}}\end{align}

Reducing to the lowest form, we get

\begin{align} &= \frac{{3 \times 16}}{{2 \times 3}} \\&= 8\end{align}

(This is a whole number)

iv) \begin{align} \frac{5}{6} \times 2\frac{3}{7}\end{align}

\begin{align}&= \frac{5}{6} \times \frac{{17}}{7}\\&= \frac{{85}}{{42}}\end{align}

Reducing \begin{align} \frac{{85}}{{42}}\end{align} to the lowest form, we get \begin{align} = 2\frac{1}{{42}} \end{align}

v) \begin{align} 3\frac{2}{5} \times \frac{4}{7}\end{align}

\begin{align}&= \frac{{17}}{5} \times \frac{4}{7}\\&= \frac{{68}}{{35}}\end{align}

(improper fraction)

Converting \begin{align} \frac{{68}}{{35}}\end{align} into mixed fraction, we get \begin{align} = 1\frac{{33}}{{35}}\end{align}

vi) \begin{align} 2\frac{3}{5} \times 3\end{align}

\begin{align}&= \frac{{13}}{5} \times \frac{3}{1}\\&= \frac{{13 \times 3}}{{5 \times 1}}\\&= \frac{{39}}{5}\end{align}

(improper fraction)

Converting \begin{align} \frac{{39}}{5}\end{align} into mixed fraction, we get \begin{align} = 7\frac{4}{5} \end{align}

(vii) \begin{align} 3\frac{4}{7} \times \frac{3}{5}\end{align}

\begin{align}&=\frac{25}{7}\times \frac{3}{5} \\& =\frac{25\times 3}{7\times 5} \\&\\\end{align}

Reducing it to the lowest form, we get

$= \,\frac{15}{7}$

(improper fraction)

Converting \begin{align} \frac{{15}}{7}\end{align} into mixed fraction, we get \begin{align}=2\frac{1}{7} \end{align}

## Chapter 2 Ex.2.3 Question 4

Which is greater?

(i) \begin{align} \frac{2}{7}{\text{ of }}\frac{3}{4}{\text{ or }}\frac{3}{5}{\text{ of }}\frac{5}{8}\end{align}

ii) \begin{align} \frac{1}{2}{\text{ of }}\frac{6}{7}{\text{ or }}\frac{2}{3}{\text{ of }}\frac{3}{7}\end{align}

### Solution

What is known?

Fractions.

What is unknown?

Which is greater.

Reasoning:

Convert the fractions into like fractions then compare by numerator.

Steps:

(i) \begin{align} \frac{2}{7}{\text{ of }}\frac{3}{4}{\text{ or }}\frac{3}{5}{\text{ of }}\frac{5}{8}\end{align}

\begin{align}&= \frac{2}{7} \times \frac{3}{4} = \frac{3}{{14}}\\&= \frac{3}{5} \times \frac{5}{8} = \frac{3}{8}\end{align}

Converting these fractions into like fraction, we get

\begin{align}\frac{3}{{14}} &= \frac{{3 \times 4}}{{14 \times 4}} = \frac{{12}}{{56}}\\\frac{3}{8} &= \frac{{3 \times 7}}{{8 \times 7}} = \frac{{21}}{{56}}\end{align}

Since,

\begin{align}\frac{{21}}{{56}} > \frac{{12}}{{56}}\\\frac{3}{8} > \frac{3}{{14}}\end{align}

Thus \begin{align} \frac{3}{5}{\rm{ of }}\frac{5}{8}\end{align} is greater.

ii) \begin{align} \frac{1}{2}{\text{ of }}\frac{6}{7}{\text{ or }}\frac{2}{3}{\text{ of }}\frac{3}{7}\end{align}

\begin{align}&= \frac{1}{2} \times \frac{6}{7} = \frac{3}{7}\\&= \frac{2}{3} \times \frac{3}{7} = \frac{2}{7} \end{align}

On comparing, we get

$\frac{3}{7} > \frac{2}{7}$

Thus, \begin{align} \frac{1}{2}{\rm{ of }}\frac{6}{7}\end{align} is greater.

## Chapter 2 Ex.2.3 Question 5

Saili plants $$4$$ saplings, in a row, in her garden. The distance between two adjacent saplings is \begin{align} \frac{3}{4}\rm\, m\end{align}. Find the distance between the first and the last sapling.

### Solution

What is known?

Number of saplings and distance between two adjacent saplings.

What is unknown?

Distance between the first and the last sapling.

Reasoning:

Add the distance of first to second, second to third and third to fourth. Since all the sapling are plant equidistance to each other then we can simply multiply \begin{align}\frac{3}{4}\end{align}by $$3$$.

Steps:

Total number of saplings $$\rm= 4$$

Distance between two adjacent saplings\begin{align} = \frac{3}{4}{\rm{m}}\end{align}

Distance between the first and the last sapling \begin{align}= 3 \times \frac{3}{4} = \frac{9}{4}{\rm{m}} = 2\frac{1}{4}{\rm{m}}\end{align}

Thus, the distance between the first and last sapling is \begin{align} 2\frac{1}{4}{\rm{m}}\end{align}

## Chapter 2 Ex.2.3 Question 6

Lipika reads a book for \begin{align} 1\frac{3}{4}\end{align} hour every day. She reads the entire book in $$6$$ days. How many hours in all were required by her to read the book?

### Solution

What is known?

Lipika reads for \begin{align}1\frac{3}{4}\end{align}hours every day and read the entire book in $$6$$ days.

What is unknown?

Total hours required by her to complete the book.

Reasoning:

In a day she reads for \begin{align}1\frac{3}{4}\end{align} hours so for calculating $$6$$ days we can simply multiply \begin{align}1\frac{3}{4}\end{align} hours by $$6$$.

Steps:

\begin{align} \begin{Bmatrix}\text{No. of hours taken by Lipika }\\ \text{to read a book everyday}\end{Bmatrix} \end{align}

\begin{align} =1\frac{3}{4}{\rm{ hour }} = \frac{7}{4}\end{align}

No. of days taken to read the entire book $$= 6 \rm \,days$$

$$\begin{Bmatrix} \text{Total no. of hours required } \\ \text{ by her to read the book} \end{Bmatrix}$$

\begin{align} &= \frac{7}{4} \times 6 \\[0.5mm] &= \frac{{42}}{4}{\rm\,{ hours }} \\ &= 10\frac{1}{2}\end{align}
Thus, \begin{align} 10\frac{1}{2}\end{align} hours in all were required by Lipika to read the book.

## Chapter 2 Ex.2.3 Question 7

A car runs $$16 \rm \,km$$ using $$1$$ liter of petrol. How much distance will it cover using \begin{align} 2\frac{3}{4}\end{align} liters of petrol?

### Solution

What is known?

A car runs $$16 \rm{km}$$ using $$1$$ liter of petrol.

What is unknown?

How much distance car can in \begin{align}2\frac{3}{4}\end{align}  liters of petrol.

Reasoning:

By using Unitary Method we can simply multiply  \begin{align}2\frac{3}{4}\end{align} by $$16$$ to get how much distance will car cover in  \begin{align}2\frac{3}{4}\end{align} liters.

Steps:

Distance covered by the car using $$1$$ liter of petrol $$= 16 \rm \,km$$

Distance covered by using \begin{align} \left( {2\frac{3}{4} = \frac{{11}}{4}{\rm\,{ liter }}} \right)\end{align} of petrol

\begin{align} &=\frac{{16}}{1} \times \frac{{11}}{4}\\&= 44\rm \, km\end{align}

Thus, $$44\rm \, km$$ distance covered by car using \begin{align} 2\frac{3}{4}\end{align} liters of petrol.

## Chapter 2 Ex.2.3 Question 8

a) i) Provide the number in the box $$□$$, such that \begin{align} \frac{2}{3}\times \square =\frac{10}{30}\end{align}

ii) The simplest form of the number obtained in $$□$$ is ―.

b) i) Provide the number in the box $$□$$, such that  \begin{align} \frac{3}{5}\times \,\square =\frac{24}{75}\end{align}

ii) The simplest form of the number obtained in □ is ―.

### Solution

Steps:

(a)

What is known?

Equations.

What is unknown?

Value of the box.

Reasoning:

To make L.H.S. $$=$$ R.H.S. we have to multiple numerator by $$5$$ and denominator by $$10$$.

i)\begin{align} \frac{2}{3}\times \square =\frac{10}{30}\end{align}

$=\frac{2}{3}\times \frac{5}{10}=\frac{10}{30}$

Therefore, the number in the box □, such that \begin{align} \frac{2}{3} \times \square = \frac{{10}}{{30}}{\text{ is }}\frac{5}{{10}}\end{align}

ii) The simplest form of the number obtained in \begin{align} \frac{5}{{10}}{\text{ is }}\frac{1}{2}\end{align} .

(b)

What is known?

Equations.

What is unknown?

Value of the box.

Reasoning:

To make L.H.S. $$=$$ R.H.S. we have to multiple numerator by $$8$$ and denominator by $$15$$.

i)  \begin{align} \frac{3}{5}\times \,\square =\frac{24}{75}\end{align}

\begin{align}=\frac{3}{5}\times \frac{8}{15}=\frac{24}{75} \\\end{align}

Therefore, the number in the box $$□$$, such that \begin{align} \frac{3}{5} × □ = \frac{{24}}{{75}} \rm \,is\, \frac{8}{{15}}\end{align}

ii) As \begin{align} \frac{8}{{15}}\end{align} can’t be simplified further. Therefore, its simplest form is \begin{align} \frac{8}{{15}}\end{align} .

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