Exercise 2.3 Fractions and Decimals - NCERT Solutions Class 7

Go back to  'Fractions and Decimals'

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

Video 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:

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

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

 

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

Video 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,\(\begin{align} \frac{48}{35}= 1\frac{{13}}{{35}}\end{align} \)

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

Video 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 \(\begin{align}=2\frac{1}{7} \end{align}\)

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

Video 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.

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

Video 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 \(\begin{align} 2\frac{1}{4}{\rm{m}}\end{align} \)

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

Video 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.

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

Video 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.

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

Video 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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