# Exercise 2.1 Fractions and Decimals - NCERT Solutions Class 7

Go back to  'Fractions and Decimals'

## Question 1

Solve: -

a) \begin{align}\;2 - \frac{3}{5} \end{align}

b) \begin{align}\;4 + \frac{7}{8}\end{align}

c) \begin{align}\;\frac{3}{5} + \frac{2}{7}\end{align}

d) \begin{align}\;\frac{9}{{11}} - \frac{4}{{15}}\end{align}

e) \begin{align}\;\frac{7}{{10}} + \frac{2}{5} + \frac{3}{2}\end{align}

f) \begin{align}\;2\frac{2}{3} + 3\frac{1}{2}\end{align}

g) \begin{align}\;8\frac{1}{2} - 3\frac{5}{8}\end{align}

### Solution

What is known?

Expression in unlike fraction.

What is unknown?

Sum/difference.

Reasoning:

Convert following fractions into like fraction (fractions with same denominator) then find sum or difference by adding or subtracting numerator.

Steps:

a)

\begin{align}2 - \frac{3}{5} &= \frac{2}{1} - \frac{3}{5} \\&= \frac{{10 - 3}}{5} \\&= \frac{7}{5}\\&( \text{improper fraction})\end{align}

Converting into mixed fraction, we get

\begin{align} = 1\frac{2}{5}\end{align}

b)

\begin{align}4 + \frac{7}{8}& = \frac{4}{1} + \frac{7}{8} \\&= \frac{{32 + 7}}{8} \\&= \frac{{39}}{8} \\& (\text{improper fraction})\end{align}

Converting into mixed fraction, we get

\begin{align} = 4\frac{7}{8}\end{align}

c)

\begin{align}\frac{3}{5} + \frac{2}{7} &= \frac{{21 + 10}}{{35}} \\&= \frac{{31}}{{35}}\end{align}

d)

\begin{align}\frac{9}{{11}} - \frac{4}{{15}} &= \frac{{135 - 44}}{{165}} \\&= \frac{{91}}{{165}}\end{align}

e)

\begin{align}\frac{7}{{10}} + \frac{2}{5} + \frac{3}{2} &= \frac{{7 + 4 + 15}}{{10}} \\&= \frac{{26}}{{10}} \\&= \frac{{13}}{5}\\&({\text{improper fraction}})\end{align}

Converting into mixed fraction, we get

\begin{align} = 2\frac{3}{5}\end{align}

f)

\begin{align}2\frac{2}{3} + 3\frac{1}{2} &= \frac{8}{3} + \frac{7}{2} \\&= \frac{{16 + 21}}{6} \\&= \frac{{37}}{6}\\&({\text{improper fraction}})\end{align}

Converting into mixed fraction, we get

\begin{align} = 6\frac{1}{6}\end{align}

g)

\begin{align}8\frac{1}{2} - 3\frac{5}{8} &= \frac{{17}}{2} - \frac{{29}}{8} \\&= \frac{{68 - 29}}{8} \\&= \frac{{39}}{8}\\&(\text{improper fraction})\end{align}

Converting into mixed fraction, we get

\begin{align}= 4\frac{7}{8}\end{align}

## Question 2

Arrange the following in descending order: -

(i)\begin{align}\,\,\frac{2}{9},\frac{2}{3},\frac{8}{21}\end{align}

(ii)\begin{align}\,\,\frac{1}{5},\frac{3}{7},\frac{7}{10}\end{align}

### Solution

What is Known?

Unlike fractions

What is unknown?

Descending order of the given fractions.

Reasoning:

Convert the given fractions into like fractions so the all of them have the same denominator. Now the fractions can be easily arranged into descending order by comparing their numerators.

Steps:

i)\begin{align}\,\,\frac{2}{9},\frac{2}{3},\frac{8}{21}\end{align}

Converting them into like fractions, we get,

\begin{align} &\frac{2\times 7}{9\times 7},\frac{2\times 21}{3\times 21},\frac{8\times 3}{21\times 3} \\& \Rightarrow \frac{14}{63},\frac{42}{63},\frac{24}{63} \\ \end{align}

Arranging in descending order,

\begin{align}\Rightarrow \frac{42}{63}>\frac{24}{63}>\frac{14}{63}\end{align}

Therefore,\begin{align}\frac{2}{3}>\frac{8}{21}>\frac{2}{9}\end{align}

ii)\begin{align}\,\,\frac{1}{5},\frac{3}{7},\frac{7}{10}\end{align}

Converting them into like fractions, we get,

\begin{align} &\frac{1\times 14}{5\times 14},\frac{3\times 10}{7\times 10},\frac{7\times 7}{10\times 7} \\& \Rightarrow \frac{14}{70},\frac{30}{70},\frac{49}{70} \\ \end{align}

Arranging in descending order,

\begin{align}\Rightarrow \frac{49}{70}>\frac{30}{70}>\frac{14}{70}\end{align}

Therefore,

\begin{align}\frac{2}{3}>\frac{8}{21}>\frac{2}{9}\end{align}

## Question 3

In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?

\begin{align} \frac{4}{11} &\frac{9}{11} \frac{2}{11} \\ \frac{3}{11} & \frac{5}{11} \frac{7}{11} \\ \frac{8}{11} & \frac{1}{11} \frac{6}{11} \\ \end{align}

### Solution

What is Known?

A square with three rows and three columns

What is unknown?

If the square is a magic square or not.

Reasoning:

As stated in the question, in a magic square, the sum of the numbers in each row, in each column and along the diagonals is the same. We can add the fractions in all rows, columns and diagonals to see if sum is the same or not.

Steps:

Sum of rows:

Sum of first row

\begin{align}& =\frac{4}{11}+\frac{9}{11}+\frac{2}{11} \\ {} & =\frac{4+9+2}{11}=\frac{15}{11} \\ \end{align}

Sum of second row

\begin{align} & =\frac{3}{11}+\frac{5}{11}+\frac{7}{11} \\ {} & =\frac{3+5+7}{11}=\frac{15}{11} \\ \end{align}

Sum of third row

\begin{align} & =\frac{8}{11}+\frac{1}{11}+\frac{6}{11} \\ {} & =\frac{8+1+6}{11}=\frac{15}{11} \\ \end{align}

Sum of columns:

Sum of first column

\begin{align} & =\frac{4}{11}+\frac{3}{11}+\frac{8}{11} \\ {} & =\frac{4+3+8}{11}=\frac{15}{11} \\ \end{align}

Sum of second column

\begin{align} & =\frac{9}{11}+\frac{5}{11}+\frac{1}{11} \\ {} & =\frac{9+5+1}{11}=\frac{15}{11} \\ \end{align}

Sum of third column

\begin{align} & =\frac{2}{11}+\frac{7}{11}+\frac{6}{11} \\ & =\frac{2+7+6}{11}=\frac{15}{11} \\ \end{align}

Sum of diagonals:

Sum of first diagonal

\begin{align} & =\frac{4}{11}+\frac{5}{11}+\frac{6}{11} \\ & =\frac{4+5+6}{11}=\frac{15}{11} \\ \end{align}

Sum of second diagonal

\begin{align} & =\frac{2}{11}+\frac{5}{11}+\frac{8}{11} \\ & =\frac{2+5+8}{11}=\frac{15}{11} \\ \end{align}

Since, the sum of fractions in each row, column, and along the diagonal is the same, therefore, the given square is a magic square.

## Question 4

A rectangular sheet of paper is \begin{align}12\frac{1}{2} \,\rm{cm}\end{align} long and \begin{align}10\frac{2}{3}\,\rm{cm}\end{align} wide. Find its perimeter.

### Solution

What is Known?

Length and breadth of a rectangular sheet of paper.

What is unknown?

Perimeter of the sheet.

Reasoning:

Perimeter of a rectangle can be calculated easily by adding lengths of four sides of the rectangle.

Steps:

Length of rectangular sheet  \begin{align}=12\frac{1}{2}\,\text{cm}=\frac{25}{2}\,\text{cm}\end{align}

Breadth of rectangular sheet  \begin{align}=10\frac{2}{3}\,\text{cm}=\frac{32}{3}\,\text{cm}\end{align}

We know that,

Perimeter of Rectangle

\begin{align} &=2 (\rm{Length}\,+ \, \rm{Breadth})\\& =2\left( \frac{25}{2}+\frac{32}{3} \right) \\ & =2\left( \frac{25\times 3+32\times 2}{2\times 3} \right) \\ & =2\left( \frac{75+64}{6} \right) \\ & =2\left( \frac{139}{6} \right) \\ & =2\times \left( \frac{139}{6} \right) \\ & =46\frac{1}{3}\text{cm} \end{align}

Therefore, perimeter of rectangular sheet of the paper is \begin{align}46\frac{1}{3}{\rm{cm}}\end{align}

## Question 5

Find the perimeter of

i)$$∆ABE$$

ii) The rectangle $$BCDE$$ in this figure. Whose perimeter is greater?

### Solution

What is Known?

A rectangle and a triangle

What is unknown?

Perimeter of the given rectangle and triangle.

Reasoning:

Perimeters of the rectangle and triangle can be calculated by adding lengths of four sides of the rectangle and the three sides of the triangle. Perimeters can be compared to decide whose perimeter is greater.

Steps:

i) Perimeter of $$∆ABE$$ $$=$$ sum of all sides $$=$$ $$AB + BE + AE$$

\begin{align} & =\frac{5}{2}\text{cm}+2\frac{3}{4}\text{cm}+3\frac{3}{5}\text{cm} \\ & =\left( \frac{5}{2}+2\frac{3}{4}+3\frac{3}{5} \right)\text{cm} \\ & =\left( \frac{5}{2}+\frac{11}{4}+\frac{18}{5} \right)\text{cm} \\ & =\left( \frac{50+55+72}{20} \right)\text{cm}\\&=\frac{177}{20}\text{cm}\\& =8\frac{17}{20}\text{cm}\end{align}

Therefore, perimeter of $$∆ABE$$  is \begin{align}8\frac{{17}}{{20}}{\rm{cm}}\end{align}

ii) In rectangle, $$BCDE$$

\begin{align}{\rm{BE}} = 2\frac{3}{4}{\rm{cm}},{\rm{ED}} = \frac{7}{6}{\rm{cm}}\end {align}

Perimeter of rectangle $$BCDE = 2 \text{(Length + Breadth)}$$

\begin{align} & ={ 2\left( {2\frac{3}{4} + \frac{7}{6}} \right)}\\ & ={ 2\left( {\frac{{11}}{4} + \frac{7}{6}} \right)}\\ &= {2\left( {\frac{{66 + 28}}{{24}}} \right)}\\&= {2\left( {\frac{{94}}{{24}}} \right)}\\&= {\frac{{47}}{6}}\\&={ 7\frac{5}{6}{\rm{cm}}}\end{align}

Therefore, perimeter of rectangle $$BCDE$$  is \begin{align}7\frac{5}{6}{\rm{cm}}.\end{align}

On comparing,

\begin{align}8\frac{{17}}{{20}}{\rm{cm}}\end{align}> \begin{align}7\frac{5}{6}{\rm{cm}}\end{align}

Therefore, perimeter of $$∆ABE$$ is greater than the perimeter of rectangle $$BCDE$$.

## Question 6

Salil wants to put a picture in a frame. The picture is \begin{align}7\frac{3}{5}{\rm{cm}}\end{align}wide. To fit in the frame the picture cannot be more than \begin{align}7\frac{3}{{10}}{\rm{cm}}\end{align} wide. How much the picture should be trimmed?

### Solution

What is Known?

Width of the picture and frame.

What is unknown?

The frame width is smaller than that the width of the picture. How much the picture should be trimmed so that it fits into the frame.

Reasoning:

Since width of the picture is more, it should be trimmed to make its width equal to the width of the frame. How much picture should be trimmed can be obtained by subtracting width of the frame from the width of the picture.

Steps:

The width of the picture

\begin{align} = 7\frac{3}{5}{\rm{cm}} = \frac{{38}}{5}{\rm{cm}}\end{align}

Width of the picture frame

\begin{align}=7\frac{3}{{10}}{\rm{cm}}=\frac{{73}}{{10}}{\rm{cm}}\end{align}

Therefore,

The picture should be trimmed

\begin{align} &{ = 7\frac{3}{5}{\rm{cm}} - 7\frac{3}{{10}}{\rm{cm}}}\\ {}&{ = \frac{{38}}{5}{\rm{cm}} - \frac{{73}}{{10}}{\rm{cm}}}\\ {}&{ = \left( {\frac{{76 - 73}}{{10}}} \right){\rm{cm}}}\\&= {\frac{3}{{10}}{\rm{cm}}} \end{align}

Thus, the picture should be trimmed \begin{align}\frac{3}{{10}} \,\rm{cm}\end{align}  to fit in the picture.

## Question 7

Ritu ate \begin{align}\frac{3}{5}\end{align} part of an apple and the remaining part was eaten by his brother Somu. What part of the apple did Somu eat? Who has the larger share? By how much?

### Solution

What is Known?

Parts (fraction) of the apple which Ritu ate.

What is unknown?

Who ate the larger part of the apple and by how much.

Reasoning:

Compare the fractions and decide which one is larger.

Steps:

Part (fraction) of apple eaten by Ritu $$=$$ \begin{align}\frac{3}{5}\end{align}

Part of apple eaten by Somu $$=$$ $$1-$$ Part of apple eaten by Ritu

\begin{align} = 1 - \frac{3}{5}=\frac{2}{5}\end{align}

Since\begin{align}\frac{3}{5}\end{align} > \begin{align}\frac{2}{5}\end{align} , Ritu ate larger share.

Difference between their shares

\begin{align}=\frac{3}{5} - \frac{2}{5} = \frac{1}{5}\end{align}

Thus, Ritu’ s part is \begin{align}\frac{1}{5}\end{align} more than Somu’s part.

## Question 8

Michael finished coloring picture in \begin{align}\frac{7}{{12}}\end{align}hours. Vaibhav finished coloring the same picture in \begin{align}\frac{3}{4}\end{align}hours. Who worked longer? By what fraction was it longer?

### Solution

What is Known?

Time taken by Michael and Vaibhav to finish coloring the picture.

What is unknown?

Who took longer to finish colouring the picture and how much longer.

Reasoning:

Time is given in unlike fractions. The first step is to change the fractions to like fractions and compare them to decide which one is bigger.

Steps:

Time taken by Michael \begin{align} = \frac{7}{{12}}\end{align} hour

Time taken by Vaibhav \begin{align}=\frac{3}{4}\end{align} hour

\begin{align} &= \frac{3 \times 3}{4 \times 3}\\&\;\;\;\;(\text{converting into like fractions})\\ & = \frac{9}{12}\text{ hour} \end{align}

Since,\begin{align}\frac{9}{{12}} > \frac{7}{{12}}\end{align}therefore, Vaibhav worked longer.

Difference\begin{align} = \frac{9}{{12}} - \frac{7}{{12}} = \frac{2}{{12}} = \frac{1}{6}\end{align} hour

Thus, Vaibhav took\begin{align}\frac{1}{6}\end{align} hour more than Michael.

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