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Subjects /Class 7 / Mathematics.7 / Fractions and Decimals

INTRODUCTION
27 Sep 2021

• The study of fractions includes proper, improper and mixed fractions as well as their addition and subtraction.
• We also studied comparison of fractions, equivalent fractions, representation of fractions on the number line and ordering of fractions.
• Our study of decimals included, their comparison, their representation on the number line and their addition and subtraction.
• Now we shall next level multiplication and division of fractions as well as of decimals.

Numerator and Denominator

\frac{2}{6}-\frac{\text { Numerator }}{\text { Denominator }}

Numerator:

• The upper part of the fraction is called Numerator.

• It tells the number of parts we have.

Denominator

• The lower part of the fraction is called Denominator.
• It tells the total parts in a whole.

Representation of Fractions on Number Line

For Example:

Mark $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}$ and $\frac{1}{8}$ on the different lines.

#### Types of Fractions

Proper Fractions:

• If the numerator is less than the denominator then it is called proper fraction. If we represent a proper fraction on the number line than it will always lie between 0 and 1.

Example: $\frac{1}{4}, \frac{4}{7}, \frac{1}{9}$ etc.

Improper or Mixed Fractions:

When the numerator is greater than the denominator then it is called Improper fraction.

• Example: $\frac{5}{4}$

The above fraction is made by adding one whole part and one-fourth part.

$1+\frac{1}{4}=1 \frac{1}{4}=\frac{5}{4}$

The fraction made by the combination of whole and a part is called Mixed fraction.

Like fractions:

Fractions which have same denominators are known as Like fractions.

Example: $\frac{5}{13}, \frac{3}{13}, \frac{1}{13} \text { etc. }$

Unlike fractions:

Fractions which have different denominators are known as unlike fractions.

Example: $\frac{2}{3}, \frac{5}{9}, \frac{13}{17} \text { etc. }$

Fraction as an operator 'of'

'of' represent multiplication

So, $\frac{1}{2}$ of $4=\frac{1}{2} \times 4=2$

#### Multiple of a Fraction by a Whole Number

Multiplication of a Fraction by a Whole Number:

To multiply a whole number with a proper or an improper fraction, we multiply the whole number with the numerator of the fraction, keeping the denominator same.

$2 \times \frac{4}{3}=\frac{2 \times 4}{3}=\frac{8}{3}$

To multiply a mixed fraction to whole numbers, first convert the mixed fraction to an improper fraction and then multiple.

$3 \times 2 \frac{5}{7}=3 \times \frac{19}{7}=\frac{57}{7}=8 \frac{1}{7}$

Multiplication of two fractions $=\frac{\text { Product of Numerators }}{\text { Production of Denominators }}$

Example: $=\frac{3}{5} * \frac{1}{7}=\frac{3 * 1}{5 * 7}=\frac{3}{35}$

Value of Product of fractions:

• The value of the product of two proper fractions is smaller than each of the two fractions.

• The value of the product of two improper fractions is more than each of the two fractions.

#### Reciprocal of A Number

Reciprocal of any number $\mathrm{n}$ is written as $\frac{1}{\mathrm{n}}$

Reciprocal of a fraction is obtained by interchanging the numerator and denominator.

Reciprocal of a fraction is obtained by interchanging the numerator and Reciprocal of $2 / 5$ is $5 / 2$

Example $=$ Reciprocal of $\frac{3}{5}$ is $\frac{5}{3}$

Although zero divided by any number means zero itself, we cannot find reciprocals as a number divided by 0 is undefined.

Example $=$ Reciprocal of $\frac{0}{8}$ is $\frac{8}{0}$ (undefined)

#### Division of a Fraction

Division of a whole number by a fraction:

To divide a whole number by any fraction, multiply that whole number by the reciprocal of that fraction.

$\frac{5}{\frac{2}{3}}=5 \times \frac{3}{2}=\frac{15}{2}$

Division of a fraction by a whole number:

We multiply the fraction with the reciprocal of the whole number.

$\frac{\frac{8}{11}}{4}=\frac{8}{11} \times \frac{1}{4}=\frac{2}{11}$

Division of a fraction by another fraction:

To divide two fractions, multiply the first fraction by the reciprocal of the second fraction.

$\frac{\frac{2}{3}}{\frac{4}{6}}=\frac{2}{3} \times \frac{6}{4}=\frac{12}{12}=1$

#### Decimals

Whenever we use dots to write singe or many numbers then that dot is the decimal point.

Example: 0.5, 1.125, 5.0

How to show Decimals on Number Line?

We have to divide the gap of each number into 10 equal parts as the decimal shows the tenth part of the number.

Example

Q. Show 0.3, 0.5 and 0.8 on the number line.

Solution

All the three numbers are greater than 0 and less than 1. So we have to make a number line with 0 and 1 and divide the gap into 10 equal parts.

Then mark as shown below.

Place Value Chart

It show the place value of each digit in the decimal number. It makes it easy to write numbers in decimal form.

A fraction can be written as a decimal:

53 is a part of 100 , so the fraction will be $\frac{53}{100}$

In the decimal form we will write it as $0.53$

#### Multiplication and Division of Decimals

Multiplication of decimal numbers with whole numbers:

Multiply both the numbers as whole Numbers.

Now Count the number of digits to the right of the decimal point in the one decimal.

Put the decimal point in the product by counting the digits from its rightmost place.

Example = 1.3 × 5 = 6.5

Multiplication of decimals with powers of 10:

Example = 53.12 × 10 = 531.20

Multiplication of decimals with decimals:

Example: 1.53 × 2.63 = 4.0239

Division of decimal numbers with whole numbers:

We first divide them as whole numbers.

Then place the decimal point in the quotient as in the decimal number.

Example: $\frac{8.4}{4}=2.1$

Division of decimals with powers of 10:

The digit in the division remains as same as in the decimal number.

Just the decimal point is shifted towards left by the same places as there are zeroes over one.

$\frac{523.1}{100}=5.231$

Division of decimals with decimals:

First shift the decimal points to the right by equal number of places in both, to convert the divisor to a natural number and then divide.

Example: $\frac{1.44}{1.2}=\frac{14.4}{12}=1.2$

#### NCERT SOLUTIONS

EXERCISE 2.1

1. Solve:

(i) $2-\frac{v}{5}$

(ii) $4+\frac{7}{8}$

(iii) $\frac{3}{5}+\frac{2}{7}$

(iv) $\frac{9}{11}-\frac{4}{15}$

(v) $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$

(vi) $2 \frac{2}{3}+3 \frac{1}{2}$

(vii) $8 \frac{1}{2}-3 \frac{5}{8}$

Solution:

(i) $2-\frac{3}{5}=\frac{2}{1}-\frac{3}{5}=\frac{2 \times 5-3 \times 1}{1 \times 5}$

$=\frac{10-3}{5}=\frac{7}{5}=1 \frac{2}{5}$

Hence, $2-\frac{3}{5}=1 \frac{2}{5}$

(ii) $4+\frac{7}{8}=\frac{4}{1}+\frac{7}{8}=\frac{4 \times 8+1 \times 7}{1 \times 8}$

$=\frac{32+7}{8}=\frac{39}{8}=4 \frac{7}{8}$

Hence, $4+\frac{7}{8}=4 \frac{7}{8}$

(iii) $\frac{3}{5}+\frac{2}{7}=\frac{3 \times 7+2 \times 5}{5 \times 7}=\frac{21+10}{35}=\frac{31}{35}$

Hence, $\frac{3}{5}+\frac{2}{7}=\frac{31}{35}$

(iv) $\frac{9}{11}-\frac{4}{15}=\frac{9 \times 15-4 \times 11}{11 \times 15}=\frac{135-44}{165}=\frac{91}{165}$

Hence, $\frac{9}{11}-\frac{4}{15}=\frac{91}{165}$

(v) $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}$

[LCM of 10,5 and $2=10$ ]

$=\frac{26}{10}=\frac{26 \div}{10 \div 2}=\frac{13}{5}=2 \frac{3}{5}$

Hence, $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}=2 \frac{3}{5}$

$\begin{array}{l} (v i) 2 \frac{2}{3}+3 \frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{8 \times 2+3 \times 7}{3 \times 2} \\ =\frac{16+21}{6}=\frac{37}{6}=6 \frac{1}{6} \end{array}$

Hence, $2 \frac{2}{3}+3 \frac{1}{2}=6 \frac{1}{6}$

$(v i i) 8 \frac{1}{2}-3 \frac{5}{8}=\frac{17}{2}-\frac{29}{8}$

$[\mathrm{LCM}$ of 2 and $8=8]$

$\begin{array}{l} =\frac{17 \times 4-29 \times 1}{8} \\ =\frac{68-29}{8}=\frac{39}{8}=4 \frac{7}{8} \end{array}$

$\text { Hence, } 8 \frac{1}{2}-3 \frac{5}{8}=4 \frac{7}{8}$

2. Arrange the following in descending order:

(i) $\frac{2}{9}, \frac{2}{3}, \frac{8}{21}$

(ii) $\frac{1}{5}, \frac{3}{7}, \frac{7}{10}$

Solution:

(i) $\frac{2}{9}, \frac{2}{3}, \frac{8}{21}$

LCM of 9,3 and $21=3 \times 3 \times 7=63$

Making the denominator same, we have

$\frac{2}{9} \times \frac{7}{7}=\frac{14}{63}, \frac{2}{3} \times \frac{21}{21}=\frac{42}{63}$

$\text { and } \frac{8}{21} \times \frac{3}{3}=\frac{24}{63} \mid \\ \text { Since } 42>24>14 \\ \text { Thus } \frac{42}{63}>\frac{24}{63}>\frac{14}{63} \\ \text { Hence, } \frac{2}{3}>\frac{8}{21}>\frac{2}{9}$

(ii) $\frac{1}{5}, \frac{3}{7}, \frac{7}{10}$

LCM of 5,7 and 10

\begin{tabular}{c|c}

2 & $5,7,10$ \\

\hline 5 & $5,7,5$ \\

\hline 7 & $1,7,1$ \\

\hline & $1,1,1$

\end{tabular}

LCM of 5,7 and 10

$=2 \times 5 \times 7$

$=70$

Making the denominator same, we have

$\frac{1}{5} \times \frac{14}{14}=\frac{14}{70} & {[\because 70 \div 5=14]} \\ \frac{3}{7} \times \frac{10}{10}=\frac{30}{70} & {[\because 70 \div 7=10]} \\ \frac{7}{10} \times \frac{7}{7}=\frac{49}{70} & {[\because 70 \div 10=7]} \\ \text { Since } 49>30>14 & \\ \text { Thus } \frac{49}{70}>\frac{30}{70}>\frac{14}{70} \\ \text { Hence, } \frac{7}{10}>\frac{3}{7}>\frac{1}{5}$

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{tabular}{|c|c|c|}

\hline$\frac{4}{11}$ & $\frac{9}{11}$ & $\frac{2}{11}$ \\

\hline$\frac{3}{11}$ & $\frac{5}{11}$ & $\frac{7}{11}$ \\

\hline$\frac{8}{11}$ & $\frac{1}{11}$ & $\frac{6}{11}$ \\

\hline

\end{tabular}

Solution:

Along first row, $\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}$

Along second row, $\frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{15}{11}$

Along third row, $\frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{15}{11}$

Along first column, $\frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{15}{11}$

Along second column, $\frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{15}{11}$

Along third column, $\frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{15}{11}$

Along first diagonal, $\frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{15}{11}$

Along second diagonal, $\frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{15}{11}$

Since the sum of all the fraction row wise, column wise and the diagonal

$\frac{15}{11}$

Hence, it is a magic square.

4. A rectangular sheet of paper is $12 \frac{1}{2}$ cm long and $10 \frac{2}{3}$cm wide. Find its perimeter.

Solution:

Length of sheet $=12 \frac{1}{2} \mathrm{~cm}=\frac{25}{2} \mathrm{~cm}$

Breadth of the sheet $=10 \frac{2}{3}=\frac{32}{3} \mathrm{~cm}$

Perimeter $=2 \times$ [length $+$ breadth $]$

$=2 \times\left[\frac{25}{2}+\frac{32}{3}\right] \mathrm{cm}$

$=2 \times\left[\frac{25 \times 3+32 \times 2}{2 \times 3}\right] \mathrm{cm} \=2 \times\left[\frac{75+64}{6}\right] \mathrm{cm}$

[LCM of 2 and $3=6]$

$= 2 \times \frac{139}{6_{3}}=\frac{139}{3} \\=46 \frac{1}{3} \mathrm{~cm} \\ \text { Hence, the required perimeter }=46 \frac{1}{3} \mathrm{~cm}.$

NOTE:

How to divide ->

5. Find the perimeters of (i) Triangle ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Solution:

(i) Perimeter of Triangle ABE

$=\mathrm{AB}+\mathrm{BE}+\mathrm{AE}$ $=\frac{5}{2} \mathrm{~cm}+2 \frac{3}{4} \mathrm{~cm}+3 \frac{3}{5} \mathrm{~cm}$ $=\left(\frac{5}{2}+\frac{11}{4}+\frac{18}{5}\right) \mathrm{cm}$ $=\left(\frac{5 \times 10+11 \times 5+18 \times 4}{20}\right) \mathrm{cm}$

$[$ LCM of $2,4,5=20]$

\begin{tabular}{r} $2 0 \longdiv { 1 7 7 } 8$ \\ $\frac{160}{17}$ \\ \hline \end{tabular}

$=\left(\frac{50+55+72}{20}\right) \mathrm{cm}$

$=\frac{177}{20} \mathrm{~cm}=8 \frac{17}{20} \mathrm{~cm}$

Hence, the perimeter of $\triangle \mathrm{ABE}=8 \frac{17}{20} \mathrm{~cm}$.

(ii) Perimeter of rectangle BCDE
$=2 \times[$ Length $+$ Breadth $]$
$=2 \times\left[2 \frac{3}{4}+\frac{7}{6}\right] \mathrm{cm} \\ =2 \times\left[\frac{11}{4}+\frac{7}{6}\right] \mathrm{cm} \\ =2 \times\left[\frac{11 \times 3+7 \times 2}{12}\right] \mathrm{cm}$
[LCM of 4 and $6=12]$
$=2 \times\left[\frac{33+14}{12}\right] \mathrm{cm}= 2 \times \frac{47}{12} =\frac{47}{6} \mathrm{~cm}=7 \frac{5}{6} \mathrm{~cm}$
Hence, the required perimeter
$7 \frac{5}{6} \mathrm{~cm}$

$\text { Since } 8 \frac{17}{20}>7 \frac{5}{6}$
Thus perimeter of $\triangle \mathrm{ABE}$ is greater than the perimeter of the rectangle BCDE.

6. Salil wants to put a picture in a frame. The picture is $7 \frac{5}{5} \mathrm{~cm}$ wide. To fit in the frame the picture cannot be more than $7 \frac{3}{10} \mathrm{~cm}$ wide. How much should the picture be trimmed?
Solution:

The width of the picture

$=7 \frac{3}{5} \mathrm{~cm}=\frac{38}{5} \mathrm{~cm}$

The required width of the frame

$=7 \frac{3}{10} \mathrm{~cm}=\frac{73}{10} \mathrm{~cm}$

$\therefore$ The width of the picture to be trimmed of

$=\frac{38}{5} \mathrm{~cm}-\frac{73}{10} \mathrm{~cm}=\left(\frac{38}{5}-\frac{73}{10}\right) \mathrm{cm} \\ =\left(\frac{2 \times 38-73 \times 1}{10}\right) \mathrm{cm}$

$[\mathrm{LCM}$ of 5 and $10=10]$

$=\left(\frac{76-73}{10}\right) \mathrm{cm}=\frac{3}{10} \mathrm{~cm}$

Hence, the required width to be trimmed $=\frac{3}{10} \mathrm{~cm}$.

7. Ritu ate (3/5) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?
Solution:

Let the whole part of the apple be 1.

Part of the apple eaten by Ritu $=\frac{3}{5}$

$\therefore$ Part of the apple eaten by her brother Somu

$=1-\frac{3}{5}=\frac{1}{1}-\frac{3}{5}=\frac{1 \times 5-3 \times 1}{5} \\ =\frac{5-3}{5}=\frac{2}{5} \\ \text { Since } \frac{3}{5}>\frac{2}{5}$

Thus, the share of Ritu was larger.

Difference between the two parts

$=\frac{3}{5}-\frac{2}{5}=\frac{1}{5} \text { part. }$

8. Michael finished colouring a picture in (7/12) hour. Vaibhav finished colouring the same picture in (3/4) hour. Who worked longer? By what fraction was it longer?
Solution:

Time taken by Michael $=\frac{7}{12}$ hour

Time taken by Vaibhav $=\frac{3}{4}$ hour

Comparing $\frac{7}{12}$ and $\frac{3}{4}$, we have

$\frac{7 \times 1}{12}$ and $\frac{3 \times 3}{12} \Rightarrow \frac{7}{12}$ and $\frac{9}{12}$

$[\mathrm{LCM}$ of 12 and $4=12]$

Since $\frac{9}{12}>\frac{7}{12} \Rightarrow \frac{3}{4}$ hour $>\frac{7}{12}$ hour

Hence, time taken by Vaibhav was longer.

Difference $=\frac{3}{4}-\frac{7}{12}=\frac{3 \times 3}{4 \times 3}-\frac{7 \times 1}{12}$ $=\frac{9}{12}-\frac{7}{12}=\frac{2}{12}=\frac{1}{6}$ hour longer.

#### Exercise 3

1. Find:

(i) $\frac{1}{4}$ of
(a) $\frac{1}{4}$
(b) $3 / 5$
(c) $4 / 3$
(ii) $1 / 7$ of
(a) $2 / 9$
(b) $6 / 5$
(c) $3 / 10$
Solution:
(i) (a) $\frac{1}{4}$ of $\frac{1}{4}=\frac{1}{4} \times \frac{1}{4}=\frac{1 \times 1}{4 \times 4}=\frac{1}{16}$
(b) $\frac{1}{4}$ of $\frac{3}{5}=\frac{1}{4} \times \frac{3}{5}=\frac{1 \times 3}{4 \times 5}=\frac{3}{20}$
(c) $\frac{1}{4}$ of $\frac{4}{3}=\frac{1}{4} \times \frac{4}{3}=\frac{1}{3}$

(ii) (a) $\frac{1}{7}$ of $\frac{2}{9}=\frac{1}{7} \times \frac{2}{9}=\frac{1 \times 2}{7 \times 9}=\frac{2}{63}$

(b) $\frac{1}{7}$ of $\frac{6}{5}=\frac{1}{7} \times \frac{6}{5}=\frac{1 \times 6}{7 \times 5}=\frac{6}{35}$
(c) $\frac{1}{7}$ of $\frac{3}{10}=\frac{1}{7} \times \frac{3}{10}=\frac{1 \times 3}{7 \times 10}=\frac{3}{70}$

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

(i) $\frac{2}{3} \times 2 \frac{2}{3}$
(ii) $\frac{2}{7} \times \frac{7}{9}$
(iii) $\frac{3}{8} \times \frac{6}{4}$
$(i v) \frac{9}{5} \times \frac{3}{5}$
(v) $\frac{1}{3} \times \frac{15}{8}$
(vi) $\frac{11}{2} \times \frac{3}{10}$
(vii) $\frac{4}{5} \times \frac{12}{7}$
Solution:
(i) $\frac{2}{3} \times 2 \frac{2}{3}=\frac{2}{3} \times \frac{8}{3}=\frac{2 \times 8}{3 \times 3}$

$=\frac{16}{9}=1 \frac{7}{9}$
(ii) $\frac{2}{7} \times \frac{7}{9}=\frac{2 \times 7}{7 \times 9}=\frac{14}{63}=\frac{14 \div 7}{63 \div 7}=\frac{2}{9}$
(iii) $\frac{3}{8} \times \frac{6}{4}=\frac{3 \times 6}{8 \times 4}=\frac{18}{32}=\frac{18 \div 2}{32 \div 2}=\frac{9}{16}$
(iv) $\frac{9}{5} \times \frac{3}{5}=\frac{9 \times 3}{5 \times 5}=\frac{27}{25}=1 \frac{2}{25}$

(v) $\frac{1}{3} \times \frac{15}{8}=\frac{1 \times 15}{3 \times 8}=\frac{15}{24}=\frac{15 \div 3}{24 \div 3}=\frac{5}{8}$

(vi) $\frac{11}{2} \times \frac{3}{10}=\frac{11 \times 3}{2 \times 10}=\frac{33}{20}=1 \frac{13}{20}$

(vii) $\frac{4}{5} \times \frac{12}{7}=\frac{4 \times 12}{5 \times 7}=\frac{48}{35}=1 \frac{13}{35}$

3. Multiply the following fractions:

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

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

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

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

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

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

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

Solution:

(i) $\frac{2}{5} \times 5 \frac{1}{4}=\frac{2}{5} \times \frac{21}{A_{2}}=\frac{1 \times 21}{5 \times 2}$

$=\frac{21}{10}=2 \frac{1}{10}$

(ii) $6 \frac{2}{5} \times \frac{7}{9}=\frac{32}{5} \times \frac{7}{9}=\frac{32 \times 7}{5 \times 9}$

$=\frac{224}{45}=4 \frac{44}{45}$
(iii) $\frac{3}{2} \times 5 \frac{1}{3}=\frac{3}{2} \times \frac{16}{3}=8$
(iv)$\frac{5}{6} \times 2 \frac{3}{7}=\frac{5}{6} \times \frac{17}{7}=\frac{85}{42}=2 \frac{1}{42}$

(v) $3 \frac{2}{5} \times \frac{4}{7}=\frac{17}{5} \times \frac{4}{7}=\frac{68}{35}=1 \frac{33}{35}$

(vi) $2 \frac{3}{5} \times 3=\frac{13}{5} \times 3=\frac{39}{5}=7 \frac{4}{5}$

(vii) $3 \frac{4}{7} \times \frac{3}{5}=\frac{25}{7} \times \frac{3}{5}=\frac{5 \times 3}{7}$

$=\frac{15}{7}=2 \frac{1}{7}$

4. Which is greater:

(i) $\frac{2}{7}$ of $\frac{3}{4}$ or $\frac{3}{5}$ of $\frac{5}{8}$
(ii) $\frac{1}{2}$ of $\frac{6}{7}$ or $\frac{2}{3}$ of $\frac{3}{7}$
Solution:
(i) $\frac{2}{7}$ of $\frac{3}{4}$ $={\frac{2}{7} \times \frac{3}{4}} = \frac{1 \times 3}{7 \times 2}=\frac{3}{14}$
$\frac{3}{5}$ of $\frac{5}{8}=\frac{3}{5} \times \frac{5}{8}=\frac{3}{8}$
Since in $\frac{3}{14}$ and $\frac{3}{8}$, their numerators are
same and $14>8$
$\therefore \frac{3}{14}<\frac{3}{8}$ or $\frac{3}{8}>\frac{3}{14}$
Hence, $\frac{3}{5}$ of $\frac{5}{8}>\frac{2}{7}$ of $\frac{3}{4}$
(ii) $\frac{1}{2}$ of $\frac{6}{7}$ or $\frac{2}{3}$ of $\frac{3}{7}$
$\frac{1}{2} \text { of } \frac{6}{7}=\frac{1}{2} \times \frac{6}{7}=\frac{1 \times 6}{2 \times 7}=\frac{6}{14}=\frac{3}{7}$
$\frac{2}{3} \text { of } \frac{3}{7}=\frac{2}{3} \times \frac{3}{7}=\frac{2}{7}$
Here, denominators are same.
$\therefore \frac{2}{7}<\frac{3}{7} \text { or } \frac{3}{7}>\frac{2}{7}$
$\text { Hence, } \frac{1}{2} \text { of } \frac{6}{7}>\frac{2}{3} \text { of } \frac{3}{7}$

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 $=4$
Distance between two adjacent saplings $=\frac{3}{4} \mathrm{~m}$

$=\frac{3}{4} \mathrm{~m}+\frac{3}{4} \mathrm{~m}+\frac{3}{4} \mathrm{~m}$
$=3 \times \frac{3}{4} \mathrm{~m}$
$=\frac{9}{4} \mathrm{~m}=2 \frac{1}{4} \mathrm{~m}$

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:
In 1 day Lipika needs $1 \frac{3}{4}$ hours
In 6 days Lipika will need $6 \times 1 \frac{3}{4}$ hours
$= 6 \times \frac{7}{4} \text { hours }=\frac{3 \times 7}{2} \text { hours } \\ =\frac{21}{2} \text { hours }=10 \frac{1}{2} \text { hours }$
$\text { Hence the required hours }=10 \frac{1}{2}$

7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 ¾ litres of petrol.

Solution:
In 1 litre of petrol, the car covers $16 \mathrm{~km}$ distance in $2 \frac{3}{4}$ litres o petrol, the car will cover 2 $\frac{3}{4} \times 16 \mathrm{~km}$ distance
$=2 \frac{3}{4} \times 16 \mathrm{~km}$
$=\frac{11}{4} \times 16 \mathrm{~km}$
$=11 \times 4 \mathrm{~km}=44 \mathrm{~km}$
Hence, the required distance $=44 \mathrm{~km}$.

8. Fill in the blanks

(a) (i) provide the number in the box [], such that $(2 / 3) \times[]=(10 / 30)$.
(ii) The simplest form of the number obtained in [] is.......
(b) (i) Provide the number in the box [], such that $3 / 5 \times[]=(24 / 75)$.
(ii) The simplest form of the number obtained in [] is

Solution:

(a)

$(i) \frac{2}{3} \times \square =\frac{10}{30} \Rightarrow \frac{2}{3} \times \frac{5}{10}=\frac{10}{30}$

Hence, the required number in $\square \text {is} \frac{5}{10}$

(ii) The simplest form of the number obtained $\operatorname{in} \square \mathrm{is} \frac {5}{10}=\frac{1}{2}$

(b)

$(i) \frac{3}{5} \times \square=\frac{24}{75} \Rightarrow \frac{3}{5} \times \frac{8}{15}=\frac{24}{75}$

Hence, the required number in the box $\square$ is $\frac{24}{75}$

Simplest form of $\frac{\not 24_{8}}{\not 5_{25}}=\frac{8}{25}$

(ii) The simplest form of the number obtained in $\square$ is $\frac{8}{25}$

#### Exercise 2

EXERCISE 2.2

1. Which of the drawings (a) to (d) show:

(i) $2 \times \frac{1}{5}$

(ii) $2 \times \frac{1}{2}$

(iii) $3 \times \frac{2}{3}$

(iv) $3 \times \frac{1}{4}$

(a)

(b)

(c)

(d)

Solution:

(i) $2 \times \frac{1}{5}$ represents drawing (d)

(ii) $2 \times \frac{1}{2}$ represents drawing (b)

(iii) $3 \times \frac{2}{3}$ represents drawing (a)

(iv) $3 \times \frac{1}{4}$ represents drawing (c)

2. Some pictures (a) to (c) are given below. Tell which of them show:

(i) $3 \times \frac{1}{5}=\frac{3}{5}$
(ii) $2 \times \frac{1}{3}=\frac{2}{3}$
(iii) $3 \times \frac{3}{4}=2 \frac{1}{4}$

(a)

(b)

(c)

Solution:

(i) $3 \times \frac{1}{5}=\frac{3}{5}$ represents figure (c)
(ii) $2 \times \frac{1}{3}=\frac{2}{3}$ represents figure (a)
(iii) $3 \times \frac{3}{4}=2 \frac{1}{4}$ represents figure (b)

3. Multiply and reduce to lowest from and convert into a mixed fraction:

(i) $7 \times \frac{3}{5}$
(ii) $4 \times \frac{1}{3}$
(iii) $2 \times \frac{6}{7}$
(iv) $5 \times \frac{2}{9}$
(v) $\frac{2}{3} \times 4$
(vi) $\frac{5}{2} \times 6$
(vii) $11 \times \frac{4}{7}$
(viii) $20 \times \frac{4}{5}$
(ix) $13 \times \frac{1}{3}$
(x) $15 \times \frac{3}{5}$
Solution:
(i) $7 \times \frac{3}{5}=\frac{21}{5}=4 \frac{1}{5}$

(ii) $4 \times \frac{1}{3}=\frac{4}{3}=1 \frac{1}{3}$

(iii) $2 \times \frac{6}{7}=\frac{12}{7}=1 \frac{5}{7}$

(iv) $5 \times \frac{2}{9}=\frac{10}{9}=1 \frac{1}{9}$

(v) $\frac{2}{3} \times 4=\frac{8}{3}=2 \frac{2}{3}$

(vi)$\frac{5}{2} \times 6=15$ ()

(vii) $11 \times \frac{4}{7}=\frac{44}{7}=6 \frac{2}{7}$

(viii) $20 \times \frac{4}{5}=16$

(i x) $13 \times \frac{1}{3}=\frac{13}{3}=4 \frac{1}{3}$

(x) $15 \times \frac{3}{5}=9$

(i) $\frac{1}{2}$ of the circles in box (a)

(ii) $\frac{2}{3}$ of the circles in box (b)

(iii) $\frac{3}{5}$ of the circles in box (c)

(a)

(b)

(c)

Solution:

(i) $\frac{1}{2}$ of the circles

$=\frac{1}{2} \times 12 = 6$

= 6 triangles are to be shaded

(ii) $\frac{2}{3}$ of the triangles $=\frac{2}{3} \times 9$

= 6 triangles are to be shaded

(iii) $\frac{3}{5}$ of the squares $=\frac{3}{5} \times 15 = 9$

$=9$ squares are to be shaded

5. Find:

(a) $\frac{1}{2}$ of

(i) 24 (ii) 46

(b) $\frac{2}{3}$ of

(i) 18 (ii) 27

(c) $\frac{3}{4}$ of

(i) 16 (ii) 36

(d) $\frac{4}{5}$ of

(i) 20 (ii) 35

Solution:

(a)

(i)$\frac{1}{2}$ of $24=\frac{1}{2} \times 24 =12$

(ii) $\frac{1}{2}$ of $46=\frac{1}{2} \times 46 =23$

(b)

(i)$\frac{2}{3}$ of $18=\frac{2}{3} \times 18 =12$

(ii) $\frac{2}{3}$ of $27=\frac{2}{3} \times 27 = 18$

(c)

(i)$\frac{3}{4}$ of $16=\frac{3}{4} \times 16 =12$

(i i) $\frac{3}{4}$ of $36=\frac{3}{4} \times 36=27$

(d)

(i) $\frac{4}{5}$ of $20=\frac{3}{4} \times 36 = 16$

(ii) $\frac{4}{5}$ of $35=\frac{4}{5} \times 37=28$

6. Multiply and express as a mixed fraction:

(a) $3 \times 5 \frac{1}{5}$

(b) $5 \times 6 \frac{3}{4}$

(c) $7 \times 2 \frac{1}{4}$

(d) $4 \times 6 \frac{1}{3}$

(e) $3 \frac{1}{4} \times 6$

(f) $3 \frac{2}{5} \times 8$

Solution:

(a) $3 \times 5 \frac{1}{5}=3 \times \frac{26}{5} =\frac{78}{5}\ = 15\frac{3}{5}$

(b) $5 \times 6 \frac{3}{4}=5 \times \frac{27}{4}$

$=\frac{135}{4}$

$= 33\frac{3}{4}$

(c) $7 \times 2 \frac{1}{4}=7 \times \frac{9}{4}$

$=\frac{63}{4}$

$=15 \frac{3}{4}$

(d) $4 \times 6 \frac{1}{3}=4 \times \frac{19}{3}$

$=\frac{76}{3}$

$=25 \frac{1}{3}$

(e) $3 \frac{1}{4} \times 6=\frac{13}{4} \times 6$

$=\frac{13 \times 3}{2}$

$=\frac{39}{2}$

$=19 \frac{1}{2}$

(f)$3 \frac{2}{5} \times 8=\frac{17}{5} \times 8$

=$\frac{136}{5}$

$3 \frac{27}{5}$

7. Find:

Options

(a) $\frac{1}{2}$ of

(i) $2 \frac{3}{4}$

(ii) $4 \frac{2}{9}$

(b) $\frac{5}{8}$ of

(i) $3 \frac{5}{6}$

(ii) $9 \frac{2}{3}$

Solution:

(a)

(i) $\frac{1}{2}$ of $2 \frac{3}{4}=\frac{1}{2} \times \frac{11}{4}=\frac{11}{8}=1 \frac{3}{8}$

(ii) $\frac{1}{2}$ of $4 \frac{2}{9}=\frac{1}{2} \times \frac{38}{9}=\frac{19}{9}=2 \frac{1}{9}$

(b)

(i) $\frac{5}{8} \times 3 \frac{5}{6}=\frac{5}{8} \times \frac{23}{6}=\frac{115}{48}=2 \frac{19}{48}$

(ii) $\frac{5}{8} \times 9 \frac{2}{3}=\frac{5}{8} \times \frac{29}{3}=\frac{145}{24}=6 \frac{1}{24}$

8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 liters water. Vidya consumed 2/5 of the water. Pratap consumed the remaining water.

(i) How much water did Vidya drink?

(ii) What fraction of the total quantity of water did Pratap drink?

Solution:

(i) Water consumed by Vidya $=\frac{2}{5}$ of 5 litres

$\frac{2}{5} \times 5$ litres $=2$ litres

Water consumed by Pratap

$=5$ litres $-2$ litres $=3$ litres

(ii) Fraction of water consumed by Pratap

$=\frac{3}{5}$ litres

#### Exercise 4

EXERCISE 2.4

1. Find

(i) $12 \div \frac{3}{4}$

(ii) $14 \div \frac{5}{6}$

(iii) $8 \div \frac{7}{3}$

(iv) $4 \div \frac{8}{3}$

(v) $3 \div 2 \frac{1}{3}$

(vi) $5 \div 3 \frac{4}{7}$

Solution:

(i) $12 \div \frac{3}{4}={12} \times \frac{4}{3}=4 \times 4=16$

(ii) $14 \div \frac{5}{6}=14 \times \frac{6}{5}$

(iii) $8 \div \frac{7}{3}=8 \times \frac{3}{7}=\frac{24}{7}=3 \frac{3}{7}$

(iv) $4 \div \frac{8}{3}=\quad 4 \times \frac{3}{8_{2}}=\frac{3}{2}=1 \frac{1}{2}$

(v) $3 \div 2 \frac{1}{3}=3 \div \frac{7}{3}=3 \times \frac{3}{7}$

$=\frac{9}{7}=1 \frac{2}{7}$

(vi) $5 \div 3 \frac{4}{7}=5 \div \frac{25}{7}=\ 5 \times \frac{7}{25}$

$=\frac{7}{5}=1 \frac{2}{5}$

2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

(i) $\frac{3}{7}$

(ii) $\frac{5}{8}$

(iii) $\frac{9}{7}$

$(i v) \frac{6}{5}$

$(v) \frac{12}{7}$

$(v i) \frac{1}{8}$

(vii) $\frac{1}{11}$

Solution:

(i) Reciprocal of $\frac{3}{7}=\frac{7}{3}$, which is improper fraction.

(ii) Reciprocal of $\frac{5}{8}=\frac{8}{5}$, which is improper fraction.

(iii) Reciprocal of $\frac{9}{7}=\frac{7}{9}$, which is proper fraction.

(iv) Reciprocal of $\frac{6}{5}=\frac{5}{6}$, which is proper fraction.

(v) Reciprocal of $\frac{12}{7}=\frac{7}{12}$, which is proper fraction.

(vi) Reciprocal of $\frac{1}{8}=8$, which is whole number.

(vii) Reciprocal of $\frac{1}{11}=11$, which is whole number.

3. Find:

(i) $\frac{7}{3} \div 2$

(ii) $\frac{4}{9} \div 5$ (iii) $\frac{6}{13} \div 7$

(iii) $\frac{6}{13} \div 7$ (iv) $4 \frac{1}{3} \div 3$

(v) $3 \frac{1}{2} \div 4$

(vi) $4 \frac{3}{7} \div 7$

Solution:

(i) $\frac{7}{3} \div 2=\frac{7}{3} \times \frac{1}{2}=\frac{7}{6}=1 \frac{1}{6}$

(ii) $\frac{4}{9} \div 5=\frac{4}{9} \times \frac{1}{5}=\frac{4}{45}$

(iii) $\frac{6}{13} \div 7=\frac{6}{13} \times \frac{1}{7}=\frac{6}{91}$

(iv) $4 \frac{1}{3} \div 3=\frac{13}{3} \div 3=\frac{13}{3} \times \frac{1}{3}$

$=\frac{13}{9}=1 \frac{4}{9}$

(v) $3 \frac{1}{2} \div 4=\frac{7}{2} \div 4=\frac{7}{2} \times \frac{1}{4}=\frac{7}{8}$

$($ vi $) 4 \frac{3}{7} \div 7=\frac{31}{7} \div 7=\frac{31}{7} \times \frac{1}{7}=\frac{31}{49}$

4. Find:

(i) $\frac{2}{5} \div \frac{1}{2}$

(ii) $\frac{4}{9} \div \frac{2}{3}$

(iii) $\frac{3}{7} \div \frac{8}{7}$

(iv) $2 \frac{1}{3} \div \frac{3}{5}$

(v) $3 \frac{1}{2} \div \frac{8}{3}$

(vi) $\frac{2}{5} \div 1 \frac{1}{2}$

(vii) $3 \frac{1}{5} \div 1 \frac{2}{3}$

(viii) $2 \frac{1}{5} \div 1 \frac{1}{5}$

Solution:

(i) $\frac{2}{5} \div \frac{1}{2}=\frac{2}{5} \times \frac{2}{1}=\frac{4}{5}$

(ii) $\frac{4}{9} \div \frac{2}{3}= {\frac{4}{9} \times \frac{3}{2}}=\frac{2}{3}$

(iii) $\frac{3}{7} \div \frac{8}{7}=\frac{3}{7} \times \frac{7}{8}=\frac{3}{8}$

(iv) $1 \frac{1}{3} \div \frac{3}{5}=\frac{7}{3} \div \frac{3}{5}=\frac{7}{3} \times \frac{5}{3}$

${=\frac{35}{9}=3 \frac{8}{9}}$

(v) $3 \frac{1}{2} \div \frac{8}{3}=\frac{7}{2} \div \frac{8}{3}=\frac{7}{2} \times \frac{3}{8}$

$=\frac{21}{16}=1 \frac{5}{16}$

(vi) $\frac{2}{5} \div 1 \frac{1}{2}=\frac{2}{5} \div \frac{3}{2}=\frac{2}{5} \times \frac{2}{3}=\frac{4}{15}$

(vii) $3 \frac{1}{5} \div 1 \frac{2}{3}=\frac{16}{5} \div \frac{5}{3}=\frac{16}{5} \times \frac{3}{5}$

$=\frac{48}{25}=1 \frac{23}{25}$

(viii) $2 \frac{1}{5} \div 1 \frac{1}{5}=\frac{11}{5} \div \frac{6}{5}=\frac{11}{5} \times \frac{5}{6}$

$=\frac{11}{6}=1 \frac{5}{6}$

#### Exercise 5

EXERCISE 2.5

1. Which is greater?

(i) $0.5$ or $0.05$

(ii) $0.7$ or $0.5$

(iii) 7 or $0.7$

Solution:

(i) $0.5$ or $0.05$

Comparing the tenths place, we get $5>0$

$\therefore 0.5>0.05$

(ii) $0.7$ or $0.5$

Comparing the tenths place, we get $7>5$

$\therefore 0.7>0.5$

(iii) 7 or $0.7$

Comparing the one's place, we get $7>0$

$\therefore 7>0.7$

(iv) $1.37$ or $1.49$

Comparing the tenths place, we get $3>4$

$\therefore 1.37<1.49$ or $1.49>1.37$

(v) $2.03$ or $2.30$

Comparing the tenths place, we get $0>3$

$\therefore 2.03<2.30$ or $2.30>2.03$

(vi) $0.8$ or $0.88 \Rightarrow 0.80$ or $0.88 \mid$

Since tenths place is same.

Comparing the hundredth place, we get $0>8$

$\therefore 0.80<0.88$ or $0.88>0.80$

2. Express as rupees as decimals:

(i) 7 paise

(ii) 7 rupees 77 paise

(iii) 77 rupees 77 paise

(iv) 50 paise

(v) 235 paise

Solution:

(i) Since 1 rupee $=100$ paise and 1 paise $=\frac{1}{100}$ rupees

7 paise $=\frac{7}{100}$ rupees $=0.07$ rupees

(ii) 7 rupees 7 paise $=7$ rupees $+\frac{7}{100}$ rupees

$=7.07$ rupees

iii) 77 rupees 77 paise $=77$ rupees $+\frac{77}{100}$ rupees

$=77.77$ rupees

iv) 50 paise $=\frac{50}{100}$ rupees $=0.50$ rupees

(v) 235 paise $=\frac{235}{100}$ rupees $=2.35$ rupees

3. Do as directed below:

(i) Express $5 \mathrm{~cm}$ in meter and kilometer

ii) Express $35 \mathrm{~mm}$ in $\mathrm{cm}, \mathrm{m}$ and $\mathrm{km}$.

Solution:

(i) 1 metre $=100 \mathrm{~cm}$ 1 kilometre $=1000$ metre $=100 \times 1000 \mathrm{~cm}$ $=100000 \mathrm{~cm}$

$\therefore 5 \mathrm{~cm}=\frac{5}{100}$ metre $=0.05$ metre

$5 \mathrm{~cm}=\frac{5}{100000} \mathrm{~km}=0.00005 \mathrm{~km}$

Hence, $5 \mathrm{~cm}=0.05 \mathrm{~m}$ and $0.00005 \mathrm{~km}$

(ii) $1 \mathrm{~cm}=10 \mathrm{~mm} \text { and } 1 \mathrm{~km}=100000 \mathrm{~cm}$

$\therefore 35 \mathrm{~mm}=\frac{35}{10} \mathrm{~cm}=3.5 \mathrm{~cm}$

$35 \mathrm{~mm}=\frac{35}{1000} \mathrm{~m}=0.035 \mathrm{~m}$

$35 \mathrm{~mm}=\frac{35}{1000000} \mathrm{~km}=0.000035 \mathrm{~km}$

4. Express in $\mathrm{kg}$

(i) 200 g

(ii) 3470 g

(iii) 4 kg 8 g

Solution:

(i) $200 \mathrm{~g}=\frac{200}{10000}=0.02 \mathrm{~kg}$

$\because 1 \mathrm{~kg}=1000 \mathrm{~g}$

(ii) $3470 \mathrm{~g}=\frac{3470}{10000}=3.47 \mathrm{~kg}$

$\because 1 \mathrm{~kg}=1000 \mathrm{~g}$

(iii) $4 \mathrm{~kg} 8 \mathrm{~g}=4 \mathrm{~kg}+\frac{8}{10000} \mathrm{~kg}$

$\because 1 \mathrm{~kg}=1000 \mathrm{~g}$

$=4 \mathrm{~kg}+0.008 \mathrm{~kg}=4.008 \mathrm{~kg}$

5. Write the following decimal numbers in the expanded form:

(i) $20.03$

(ii) $2.03$

(iii) $200.03$

(iv) $2.034$

Solution:

(i) $20.03=2 \times 10+0 \times 1+0 \times \frac{1}{100}+3 \times \frac{1}{100}$

(i) $2.03=2 \times 1+0 \times \frac{1}{10}+3 \times \frac{1}{100}$

(iii) $200.03=2 \times 100+0 \times 10+0 \times 1+0 \times \frac{1}{10}+3 \times \frac{1}{100}$

(iv) $2.034=2 \times 1+0 \times \frac{1}{10}+3 \times \frac{1}{10}+4 \times \frac{1}{1000}$

6. Write the place value of 2 in the following decimal numbers:

(i) $2.56$

(ii) $21.37$

(iii) $10.25$

(iv) $9.42$

(v) $63.352$

Solution:

(i) Place value of 2 in $2.56=2 \times 1=2$ is, ones

(ii) Place value of 2 in $21.37=2 \times 10=20$ i.e. tens

(iii) Place value of 2 in $10.25=\frac{2}{10}=0.2$ i.e. tenths

(iv) Place value of 2 in $9.42=\frac{2}{100}=0.02$ i.e. hundredths

(v) Place value of 2 in $63.352=\frac{2}{1000}=0.002$ i.e. thousandths.

7. Dinesh went from place A to place B and from there to place C. A is $7.5 \mathrm{~km}$ from $B$ and $B$ is $12.7 \mathrm{~km}$ from C. Ayub went from place A to place $\mathrm{D}$ and from there to place C. D is $9.3 \mathrm{~km}$ from $\mathrm{A}$ and $\mathrm{C}$ is $11.8 \mathrm{~km}$ from D. Who travelled more and by how much?

Solution:

Distance travelled by Dinesh from $\mathrm{A}$ to $\mathrm{C}$

$=\mathrm{AB}+\mathrm{BC}$

$=7.5 \mathrm{~km}+12.7 \mathrm{~km}$

$=20.2 \mathrm{~km}$

Distance travelled by Ayub from $\mathrm{A}$ to $\mathrm{C}$

$=\mathrm{AD}+\mathrm{DC}$

$=9.3 \mathrm{~km}+11.8 \mathrm{~km}$

$=21.1 \mathrm{~km}$

Since $21.1 \mathrm{~km}>20.2 \mathrm{~km}$.

Hence, Ayub travelled more distance.

8. Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought $4 k g 800$ g oranges and $4 \mathrm{~kg} 150 \mathrm{~g}$ bananas. Who bought more fruits?

Solution:

Fruits bought by Shyama

$=5 \mathrm{~kg} 300 \mathrm{~g}$ apples $+3 \mathrm{~kg} 250 \mathrm{~g}$ mangoes

$=5.300 \mathrm{~kg}$ apples $+3.250 \mathrm{~kg}$ mangoes

$=8.550 \mathrm{~kg}$ of fruits

Fruits bought by Sarala

$=4 \mathrm{~kg} 800$ g oranges $+4 \mathrm{~kg} 150 \mathrm{~g}$ bananas

$=4.800 \mathrm{~kg}$ oranges $+4.150 \mathrm{~kg}$ bananas

$=8.950 \mathrm{~kg}$ of fruits

Since $8.950 \mathrm{~kg}>8.550 \mathrm{~kg}$

Hence, Sarala bought more fruits.

9. How much less is $28 \mathrm{~km}$ than $42.6 \mathrm{~km}$ ?

Solution:

Since $28 \mathrm{~km}<42.6 \mathrm{~km}$

$42.6 \mathrm{~km}$ $-28.0 \mathrm{~km}$

$14.6 \mathrm{~km}$

Hence, $28 \mathrm{~km}$ is less than $42.6 \mathrm{~km}$ by $14.6 \mathrm{~km}$.

## Did You Know

• The word fraction is derived from Latin word ‘fractio’ which means to break.
• The Egyptian civilization was the first one to use the fractions.
• Decimal fractions were first used by Chinese in the end of 4th BCE.