Unit: 2: Fractions and Decimals

## Exercise: 1 (Multiple Choice Questions and Answers 1-20)

In questions 1 to 20, out of four options, only one is correct. Write the correct answer.

# Question: 1

$\frac{2}{5}×5\frac{1}{5}$ is equal to:

a.       $\frac{26}{25}$

b.       $\frac{52}{25}$

c.       $\frac{2}{5}$

d.       $6$

## Solution

(b)

$5\frac{1}{5}=\frac{\left(5×5\right)+1}{5}$

$=\frac{25+1}{5}=\frac{26}{5}$
Now,

$\frac{2}{5}×5\frac{1}{5}$$=\frac{2}{5}×5\frac{1}{5}=\frac{2}{5}×\frac{26}{5}=\frac{52}{25}$

# Question: 2

$3\frac{3}{4}÷\frac{3}{4}$ is equal to:

a.  $3$

b.  $4$

c.  $5$

d.  $\frac{45}{16}$

## Solution

(c)

Given expression = $3\frac{3}{4}÷\frac{3}{4}$

$\begin{array}{c}3\frac{3}{4}=\frac{\left(3×4\right)+3}{4}\\ =\frac{12+3}{4}\\ =\frac{15}{4}\end{array}$
Now,

$\begin{array}{l}3\frac{3}{4}÷\frac{3}{4}\\ =\frac{15}{4}×\frac{4}{3}\\ =5\end{array}$

# Question: 3

A ribbon of length  is cut into small pieces each of length

Number of pieces will be:

a.  $5$

b.  $6$

c.  $7$

d.  $8$

## Solution

(c)

Number of pieces $=$ (Total length of ribbon) ÷
(Length of $1$ piece)

$\begin{array}{l}=\frac{\left(5\frac{1}{4}\right)}{\frac{3}{4}}=\frac{\frac{\left(5×4\right)+1}{4}}{\frac{3}{4}}\\ =\frac{\left(\frac{21}{4}\right)}{\frac{3}{4}}\\ =\frac{21}{4}×\frac{4}{3}=7\end{array}$

# Question: 4

The ascending arrangement of $\frac{2}{3},\frac{6}{7},\frac{13}{21}$ is

a.    $\frac{6}{7},\frac{2}{3},\frac{13}{21}$

b.   $\frac{13}{21},\frac{2}{3},\frac{6}{7}$

c.    $\frac{6}{7},\frac{13}{21},\frac{2}{3}$

d.   $\frac{2}{3},\frac{6}{7},\frac{13}{21}$

## Solution

(b)

Given fractions are: $\frac{2}{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{6}{7},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{13}{21}$

LCM of denominators

$\begin{array}{l}\frac{2}{3}=\frac{2}{3}×\frac{7}{7}=\frac{14}{21}\\ \frac{6}{7}=\frac{6}{7}×\frac{3}{3}=\frac{18}{21}\end{array}$

And $\frac{13}{21}=\frac{13}{21}$

Now, compare $\frac{14}{21},\frac{18}{21}$ and $\frac{13}{21}$.

So, $\frac{13}{21}<\frac{14}{21}<\frac{18}{21}$

Hence, $\frac{13}{21}<\frac{2}{3}<\frac{6}{7}$

# Question: 5

Reciprocal of the fraction $\frac{2}{3}$ is:

a.    $2$

b.   $3$

c.    $\frac{2}{3}$

d.   $\frac{3}{2}$

## Solution

(d)

The reciprocal of a non-zero fraction is obtained by interchanging its numerator and denominator.

Hence, the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$

# Question: 6

The product of $\frac{11}{13}$ and $4$ is

a.    $3\frac{5}{13}$

b.   $5\frac{3}{13}$

c.    $13\frac{3}{5}$

d.   $13\frac{5}{3}$

## Solution

(a)

We have, $\frac{11}{13}×4$

$\frac{11}{13}×4=\frac{44}{13}=3\frac{5}{13}$

Hence, the product of $\frac{11}{13}$ and $4$ is $3\frac{5}{13}$.

# Question: 7

The product of $3$ and $4\frac{2}{5}$ is

a.    $17\frac{2}{5}$

b.   $\frac{24}{5}$

c.    $13\frac{1}{5}$

d.   $5\frac{1}{3}$

## Solution

(c)

Given: $3×4\frac{2}{5}$

$\begin{array}{l}4\frac{2}{5}=\frac{\left(4×5\right)+2}{5}=\frac{22}{5}\\ 3×4\frac{2}{5}=3×\frac{22}{5}=\frac{66}{5}=13\frac{1}{5}\end{array}$

Hence, the product of $3$ and $4\frac{2}{5}$ is $13\frac{1}{5}$.

# Question: 8

Pictorial representation of $3×\frac{2}{3}$ is

a. b. c. d. ## Solution

(b)

$3×\frac{2}{3}$ means $3$ times the two third part of anything.

# Question: 9

$\frac{1}{5}÷\frac{4}{5}$ is equal to:

a.    $\frac{4}{5}$

b.   $\frac{1}{5}$

c.    $\frac{5}{4}$

d.   $\frac{1}{4}$

## Solution

(d)

$\frac{1}{5}÷\frac{4}{5}=\frac{1}{5}×\frac{5}{4}=\frac{1}{4}$

# Question: 10

The product of $0.03×0.9$ is:

a.    $2.7$

b.   $0.27$

c.    $0.027$

d.   $0.0027$

## Solution

(c)

Given, $0.03×0.9$

Here, $3×9=27$

Sum of the decimal places to the right of the decimal points in $0.03$ and $0.9$ is $3.$

So, $0.03×0.9=0.027$

# Question: 11

$\frac{5}{7}÷6$ is equal to:

a.    $\frac{30}{7}$

b.   $\frac{5}{42}$

c.    $\frac{30}{42}$

d.   $\frac{6}{7}$

## Solution

(b)

$\frac{5}{7}÷6=\frac{5}{7}×\frac{1}{6}$

$=\frac{5}{42}$

# Question: 12

$5\frac{1}{6}÷\frac{9}{2}$ is equal to:

a.    $\frac{31}{6}$

b.   $\frac{1}{27}$

c.    $5\frac{1}{27}$

d.   $\frac{31}{27}$

## Solution

(d)

$\begin{array}{l}5\frac{1}{6}÷\frac{9}{2}=\frac{\left(5×6\right)+1}{6}÷\frac{9}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{30+1}{6}÷\frac{9}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{31}{6}÷\frac{9}{2}\\ 5\frac{1}{6}÷\frac{9}{2}=\frac{31}{6}×\frac{2}{9}=\frac{31}{27}\end{array}$

# Question: 13

Which of the following represents $\frac{1}{3}$ of $\frac{1}{6}$

a.    $\frac{1}{3}+\frac{1}{6}$

b.   $\frac{1}{3}-\frac{1}{6}$

c.    $\frac{1}{3}×\frac{1}{6}$

d.   $\frac{1}{3}÷\frac{1}{6}$

## Solution

(c) $\frac{1}{3}$ of $\frac{1}{6}=\frac{1}{3}×\frac{1}{6}$

# Question: 14

$\frac{3}{7}$ of $\frac{2}{5}$ is equal to:

a.    $\frac{5}{12}$

b.   $\frac{5}{35}$

c.    $\frac{1}{35}$

d.   $\frac{6}{35}$

## Solution

(d)

$\frac{3}{7}$ of $\frac{2}{5}=\frac{3}{7}×\frac{2}{5}=\frac{6}{35}$

# Question: 15

One packet of biscuits requires $2\frac{1}{2}$ cups of flour and $1\frac{2}{3}$ cups of sugar. Estimated total quantity of both ingredients used in $10$ such packets of biscuits will be

a.    less than $30$ cups

b.   between $30$ cups and $40$ cups

c.    between $40$ cups and $50$ cups

d.   above $50$ cups

## Solution

(c)

Total quantity of both ingredients in $1$ packet of biscuits
$=$ quantity of flour $+$ quantity of sugar

$\begin{array}{l}=\frac{\left(2×2\right)+1}{2}+\frac{\left(1×3\right)+2}{3}\\ =\frac{4+1}{2}+\frac{3+2}{3}\\ =\frac{5}{2}+\frac{5}{3}\end{array}$

$\begin{array}{l}=\frac{5×3+2×5}{6}\\ =\frac{15+10}{6}\\ =\frac{25}{6}\end{array}$

Total quantity of both ingredients used in $10$ packets

$=10×$ (Total quantity of ingredients in one packet)

$=10×\frac{25}{6}=\frac{250}{6}$

$\frac{250}{6}$ lies between $40$ and $50.$

# Question: 16

The product of $7$ and $6\frac{3}{4}$ is

a.           $42\frac{1}{4}$

b.          $47\frac{1}{4}$

c.           $42\frac{3}{4}$

d.          $47\frac{3}{4}$

## Solution

(b)

Given expression: $7×6\frac{3}{4}$

$\begin{array}{l}6\frac{3}{4}=\frac{\left(6×4\right)+3}{4}=\frac{24+3}{4}=\frac{27}{4}\\ 7×6\frac{3}{4}=7×\frac{27}{4}=\frac{189}{4}=47\frac{1}{4}\end{array}$

Hence, the product of $7$ $6\frac{3}{4}$ is $47\frac{1}{4}$.

# Question: 17

On dividing $7$ by $\frac{2}{5}$, the result is

a.    $\frac{14}{2}$

b.   $\frac{35}{4}$

c.    $\frac{14}{5}$

d.   $\frac{35}{2}$

## Solution

(d)

Given, $7÷\frac{2}{5}=7×\frac{5}{2}$

$=\frac{35}{2}$

Hence, on dividing $7$ by $\frac{2}{5}\text{,}$ we get $\frac{35}{2}$.

# Question: 18

$2\frac{2}{3}÷5$ is equal to:

a.    $\frac{8}{15}$

b.   $\frac{40}{30}$

c.    $\frac{40}{5}$

d.   $\frac{8}{3}$

## Solution

(a)

Given, $2\frac{2}{3}÷5=\frac{\left(2×3\right)+2}{3}÷5=\frac{6+2}{3}÷5$

$=\frac{8}{3}×\frac{1}{5}=\frac{8}{15}$

Hence, $2\frac{2}{3}÷5$ is equal to $\frac{8}{15}$.

# Question: 19

$\frac{4}{5}$ of $5$ kg apples were used on Monday. The next day $\frac{1}{3}$ of what was left was used. Weight (in kg) of apples left now is

a.    $\frac{2}{7}$

b.   $\frac{1}{14}$

c.    $\frac{2}{3}$

d.   $\frac{14}{21}$

## Solution

(c)

Apples used on Monday $=\frac{4}{5}$ of $5=\frac{4}{5}×5$

$=4$ kg

Remaining apples (in kg) $=5-4=1$ kg

Apples used next day  $=\frac{1}{3}$ of remaining apples

$=\frac{1}{3}×1$ kg $=\frac{1}{3}$ kg

So, weight of apples left now

$=$ Total weight of apples
$=$ Apples used on Monday
$=$ Apples used on next day

$=\left(5-4-\frac{1}{3}\right)$ kg

$=\frac{15-12-1}{3}=\frac{2}{3}$ kg

# Question: 20

The picture interprets

a.    $\frac{1}{4}÷3$

b.   $3×\frac{1}{4}$

c.    $\frac{3}{4}×3$

d.   $3÷\frac{1}{4}$

## Solution

(b)

part of circle:  $=3×\frac{1}{4}$

Also, picture represents $3×\frac{1}{4}$ i.e.,  part of the circle.

In Questions 21 to 44, fill in the blanks to make the statements true.

# Question: 21

Rani ate $\frac{2}{7}$ part of a cake while her brother Ravi ate $\frac{4}{5}$ of the remaining. Part of the cake left is _______

## Solution

Rani ate $\frac{2}{7}$ part of a cake, then

remaining part $=1-\frac{2}{7}$

$\begin{array}{l}=\frac{7-2}{7}\\ =\frac{5}{7}\end{array}$

Part of cake ate by Ravi
= $\frac{4}{5}$ of $\frac{5}{7}=\frac{4}{5}×\frac{5}{7}$$=\frac{4}{7}$

So, remaining part of the cake $=\frac{5}{7}-\frac{4}{7}$

$\begin{array}{l}=\frac{5-4}{7}\\ =\frac{1}{7}\end{array}$

Hence, part of the cake left is $\frac{1}{7}\text{.}$

# Question: 22

The reciprocal of $\frac{3}{7}$ is _______.

## Solution

The reciprocal of $\frac{3}{7}$ is $\frac{7}{3}$.

# Question: 23

$\frac{2}{3}$ of $27$ is _______

## Solution

$\frac{2}{3}$ of $27$

$\begin{array}{l}=\frac{2}{3}×27\\ =18\end{array}$

Hence, $\frac{2}{3}$ of $27$ is $18.$

# Question: 24

$\frac{4}{5}$ of $45$ is _______

## Solution

$\frac{4}{5}$ of $45=\frac{4}{5}×45=36$

Hence, $\frac{4}{5}$ of $45$ is $36.$

# Question: 25

$4×6\frac{1}{3}$ is equal to _______

## Solution

$4×6\frac{1}{3}$

$\begin{array}{l}=4×\frac{\left(6×3\right)+1}{3}\\ =4×\frac{19}{3}\\ =\frac{76}{3}\\ =25\frac{1}{3}\end{array}$

Hence, $4×6\frac{1}{3}=25\frac{1}{3}$

# Question: 26

$\frac{1}{2}$ of $4\frac{2}{7}$ is _______

## Solution

$\frac{1}{2}$ of $4\frac{2}{7}$

$=\frac{1}{2}×\frac{\left(4×7\right)+2}{7}$

$\begin{array}{l}=\frac{1}{2}×\frac{30}{7}\\ =\frac{30}{14}\\ =\frac{15}{7}\end{array}$

Hence, $\frac{1}{2}$ of $4\frac{2}{7}$ is $\frac{15}{7}$.

# Question: 27

$\frac{1}{9}$ of $\frac{6}{5}$ is _______.

## Solution

$\frac{1}{9}$ of $\frac{6}{5}$

$\begin{array}{l}=\frac{1}{9}×\frac{6}{5}\\ =\frac{2}{15}\end{array}$

Hence, $\frac{1}{9}$ of $\frac{6}{5}$ is $\frac{2}{15}$.

# Question: 28

The lowest form of the product $2\frac{3}{7}×\frac{7}{9}$ is _______

## Solution

$2\frac{3}{7}×\frac{7}{9}$

$\begin{array}{l}=\frac{\left(2×7\right)+3}{7}×\frac{7}{9}\\ =\frac{17}{7}×\frac{7}{9}\\ =\frac{17}{9}\\ =1\frac{8}{9}\end{array}$

Hence, the lowest form of the product $2\frac{3}{7}×\frac{7}{9}$ is $1\frac{8}{9}$.

# Question: 29

$\frac{4}{5}÷4$ is equal to _______

## Solution

$\frac{4}{5}÷4=\frac{4}{5}×\frac{1}{4}=\frac{1}{5}$

Hence, $\frac{4}{5}÷4$ is equal to $\frac{1}{5}$.

# Question: 30

$\frac{2}{5}$ of $25$ is _______

## Solution

$\frac{2}{5}$ of $25=\frac{2}{5}×25=10$

Hence, $\frac{2}{5}$ of $25$ is $10$.

# Question: 31

$\frac{1}{5}÷\frac{5}{6}=\frac{1}{5}$ _______ $\frac{6}{5}$

## Solution

$\frac{1}{5}÷\frac{6}{5}$

$\frac{1}{5}÷\frac{5}{6}=\frac{1}{5}×\frac{6}{5}$

# Question: 32

$3.2×10$ is _______

## Solution

$3.2×10=\frac{32}{10}×10=32$

Hence, $3.2×10=32$

# Question: 33

$25.4×1000=$ _______

## Solution

$25.4×1000=\frac{254}{10}×1000=25400$

Hence, $25.4×1000=25400$

# Question: 34

$93.5×100=$ _______

## Solution

$93.5×100=\frac{935}{10}×100=9350$

Hence, $93.5×100=9350$

# Question: 35

$4.7÷10=$ _______

## Solution

$4.7÷10=\frac{47}{10}×\frac{1}{10}$

$\begin{array}{l}=\frac{47}{100}\\ =0.47\end{array}$

Hence, $4.7÷10=0.47$

# Question: 36

$4.7÷100=$ _______

## Solution

Given, $4.7÷100=\frac{47}{10}×\frac{1}{100}$

$\begin{array}{l}=\frac{47}{1000}\\ =0.047\end{array}$

Hence, $4.7÷100=0.047$

# Question: 37

$4.7÷1000=$ _______

## Solution

Given, $4.7÷1000=\frac{47}{10}×\frac{1}{1000}$

$\begin{array}{l}=\frac{47}{10000}\\ =0.0047\end{array}$

Hence, $4.7÷1000=0.0047$

# Question: 38

The product of two proper fractions is _______ than each of the fractions that are multiplied.

## Solution

The product of two proper fractions is less than each of the fractions that are multiplied.
For example, $\frac{1}{2}×\frac{1}{3}=\frac{1}{6}$

$\frac{1}{6}<\frac{1}{2}$ and $\frac{1}{6}<\frac{1}{3}$

# Question: 39

While dividing a fraction by another fraction, we _________ the first fraction by the _______ of the other fraction.

## Solution

While dividing a fraction by another fraction, we multiply the first fraction by the reciprocal of the other fraction.

# Question: 40

$8.4$ _______ $=2.1$

## Solution

Let $x$ be the missing number, then

$\begin{array}{l}8.4÷x=2.1\\ 8.4×\frac{1}{x}=2.1\\ 8.4=2.1x\\ x=\frac{8.4}{2.1}\\ x=\frac{84}{21}×\frac{10}{10}=4\\ x=4\end{array}$

Hence, $8.4÷4=2.1$

# Question: 41

$52.7÷$ _______ $=0.527$

## Solution

Let $x$ be the missing number, then $52.7÷x=0.527$

$\begin{array}{l}\frac{527}{10}×\frac{1}{x}=\frac{527}{1000}\\ \frac{527}{10}×\frac{1000}{527}=x\\ x=100\end{array}$

Hence, $52.7÷100=0.527$

# Question: 42

$0.5$ _______ $0.7=0.35$

## Solution

$0.5=\frac{5}{10}$ and $0.7=\frac{7}{10}$

$0.5×0.7=\frac{5}{10}×\frac{7}{10}=\frac{35}{100}=0.35$

Hence, $0.5×0.7=0.35$

# Question: 43

$2$ _______ $\frac{5}{3}=\frac{10}{3}$

## Solution

On multiplying $2$ with $\frac{5}{3}$, we get $\frac{10}{3}$.

Hence, $2×\frac{5}{3}=\frac{10}{3}$

# Question: 44

$2.001÷0.003=$ _______

## Solution

$2.001÷0.003$

$2.001=\frac{2001}{1000}$

And $0.003=\frac{3}{1000}$

$\begin{array}{c}2.001÷0.003=\frac{2001}{1000}÷\frac{3}{1000}\\ =\frac{2001}{1000}×\frac{1000}{3}=667\end{array}$

Hence, $2.001÷0.003=667$

In each of the Questions 45 to 54, state whether the statement is True or False.

# Question: 45

The reciprocal of a proper fraction is a proper fraction.

## Solution

False

The reciprocal of a proper fraction is always an improper fraction, e.g. $\frac{5}{6}$ is a proper fraction, but its reciprocal is $\frac{6}{5}\text{,}$ which is. an improper fraction.

# Question: 46

The reciprocal of an improper fraction is an improper fraction.

## Solution

False

The reciprocal of an improper fraction is a proper fraction. E.g. $\frac{6}{5}\text{,}$ is an improper fraction, but its reciprocal is $\frac{5}{6}$ which is a proper fraction.

# Question: 47

Product of two fractions

## Solution

False

Two fractions are multiplied by multiplying their numerators and denominators separately and writing the product as,

Product of two fractions

# Question: 48

The product of two improper fractions is less than both the fractions.

## Solution

False

The product of two improper fractions are greater than both the fractions.

For example: $\frac{3}{2}×\frac{7}{4}=\frac{21}{8}$

Hence, $\frac{21}{8}$ is greater than both $\frac{3}{2}$ and $\frac{7}{4}$.

# Question: 49

A reciprocal of a fraction is obtained by inverting it upside down.

## Solution

True

Let $\frac{a}{b}$ be the fraction. Then, for obtaining its reciprocal, numerator and denominator are interchanged.

$\therefore$ Reciprocal of $\frac{a}{b}=\frac{b}{a}$

# Question: 50

To multiply a decimal number by $1000\text{,}$ we move the decimal point in the number to the right by three places.

## Solution

True

$2.732×1000=2732$ (moving the decimal to the right by three places)

# Question: 51

To divide a decimal number by $100\text{,}$ we move the decimal point in the number to the left by two places.

## Solution

True

Example: $\frac{273.2}{100}=2.732$ (moving decimal to the left by two places.)

# Question: 52

$1$ is the only number which is its own reciprocal.

## Solution

True

For obtaining the reciprocal of a number, we simply interchange the numerator and denominator. Hence, reciprocal of $1$ will be $\frac{1}{1}\text{,}$ i.e. $1$.

# Question: 53

$\frac{2}{3}$ of $8$ is same as $\frac{2}{3}÷8.$

## Solution

False

$\frac{2}{3}$ of $8=\frac{2}{3}×8=\frac{16}{3}$

$\frac{2}{3}÷8=\frac{2}{3}×\frac{1}{8}=\frac{2}{24}=\frac{1}{12}$

Therefore, $\frac{2}{3}$ of $8\ne \frac{2}{3}÷8$

Hence, $\frac{2}{3}$ of $8$ is not same as $\frac{2}{3}÷8.$

# Question: 54

The reciprocal of $\frac{4}{7}$ is $\frac{4}{7}\text{.}$

## Solution

False

Reciprocal of $\frac{4}{7}$ is $\frac{7}{4}$.

# Question: 55

If $5$ is added to both the numerator and the denominator of the fraction $\frac{5}{9}\text{,}$ will the value of the fraction be changed? If so, will the value increase or decrease?

## Solution

Given fraction $=\frac{5}{9}$

Add $5$ to the numerator and denominator. $\frac{5+5}{9+5}=\frac{10}{14}=\frac{5}{7}$

$\frac{5}{7}>\frac{5}{9}$

So, the value will increase.

# Question: 56

What happens to the value of a fraction if the denominator of the fraction is decreased while numerator is kept unchanged?

## Solution

The value of fraction would increase if the denominator of the fraction is decreased while numerator is kept unchanged.

For example, consider a fraction $\frac{2}{3}$.

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

# Question: 57

Which letter comes $\frac{2}{5}$ of the way among A and J?

## Solution

The total number of letters from A to j is $10$.

Determine the letter that would fall $\frac{2}{5}$ of the way from A to J:

$\frac{2}{5}$ of $10=\frac{2}{5}×10$

$=\frac{20}{5}=4$

Two-fifth along the way is $4$ th letter which is ‘d’.

# Question: 58

If $\frac{2}{3}$ of a number is $10\text{,}$ then what is $1.75$ times of that number?

## Solution

Let the number be $x$.

According to the question, $\frac{2}{3}$ of $x=10$

$\frac{2}{3}×x=10$

On multiplying both sides by $\frac{3}{2}\text{,}$ we get

$\begin{array}{l}\frac{2}{3}×x×\frac{3}{2}=10×\frac{3}{2}\\ x=5×3=15\end{array}$

$1.75$ times of $15=1.75×15=\frac{175}{100}×15=\frac{2625}{100}=26.25$

# Question: 59

In a class of 40 students, $\frac{1}{5}$ of the total number of students like to eat rice only, $\frac{2}{5}$ of the total number of students like to eat chapatti only and the remaining students like to eat both. What fraction of the total number of students like to eat both?

## Solution

Total number of students $=40$

Students who eat rice only $=\frac{1}{5}$ of total students $\frac{1}{5}×40=8$

Students who eat chapatti only $=\frac{2}{5}$ of total students

$=\frac{2}{5}×40=16$

Students who eat both chapatti and rice $=$ total no. of students $-$ (students who eat rice only $+$ students who eat chapatti only)

$\begin{array}{l}=40-\left(8+16\right)\\ =40-24=16\end{array}$

Fraction of students who eat both chapatti & rice

# Question: 60

Renu completed $\frac{2}{3}$ part of her home work in $2$ hours. How much part of her home work had she completed in $1\frac{1}{4}$ hours?

## Solution

The part of the work finished by Renu in

So, the part of the work finished by Renu in

The part of the work finished by Renu in

Hence,  of Renu’s home-work is completed by her in

# Question: 61

Reemu read $\frac{1}{5}\text{th}$ pages of a book. If she reads further $40$ pages, she would have read $\frac{7}{10}\text{th}$ pages of the book. How many pages are left to be read?

## Solution

Let total pages of the book be $x$.

According to question, $\frac{1}{5}x+40=\frac{7}{10}x$

$\begin{array}{l}40=\frac{7}{10}x-\frac{1}{5}x=\frac{7x-2x}{10}\\ 40=\frac{5x}{10}\\ x=\frac{400}{5}=80\end{array}$

Total pages of book $=80$

Hence,

$=$ total pages of book $-\left(\frac{7}{10}\right)x$

# Question: 62

Write the number in the box $\overline{)}$ such that

$\frac{3}{7}×\overline{)}=\frac{15}{98}$

## Solution

Let the missing number be $x$.

Then, $\frac{3}{7}×x=\frac{15}{98}$

$\begin{array}{l}x=\frac{15}{98}÷\frac{3}{7}=\frac{15}{98}×\frac{7}{3}\\ x=\frac{5}{14}\end{array}$

Hence, $\frac{3}{7}×\frac{5}{14}=\frac{15}{98}$

# Question: 63

Will the quotient $7\frac{1}{6}÷3\frac{2}{3}$ be a fraction greater than $1.5$ or less than $1.5?$ Explain.

## Solution

$7\frac{1}{6}÷3\frac{2}{3}=\frac{\left(7×6\right)+1}{6}÷\frac{\left(3×3\right)+2}{3}$

$\begin{array}{l}=\frac{42+1}{6}÷\frac{9+2}{3}\\ =\frac{43}{6}÷\frac{11}{3}\\ =\frac{43}{6}×\frac{3}{11}=\frac{43}{22}=1.95\\ 1.95>1.5\end{array}$

Hence, $7\frac{1}{6}÷3\frac{2}{3}>1.5$

# Question: 64

Describe two methods to compare $\frac{13}{17}$ and $0.82.$ Which do you think is easier and why?

## Solution

Method $1$

Convert both numbers into decimal form.

$\begin{array}{l}\frac{13}{17}=0.76\\ 0.76<0.82\\ \frac{13}{17}<0.76\end{array}$

Method $2$

Convert both numbers into fraction form.

$0.82=\frac{82}{100}=\frac{41}{50}$

Now, compare $\frac{13}{17}$ and $\frac{41}{50}$

To compare these fractions, we have to make the denominators same.

$\begin{array}{l}\frac{13}{17}=\frac{13}{17}×\frac{50}{50}=\frac{650}{850}\\ \frac{41}{50}=\frac{41}{50}×\frac{17}{17}=\frac{697}{850}\\ \frac{697}{850}>\frac{650}{850}\end{array}$

Conclusion, method 1 is easier in this case.

# Question: 65

Health: The directions for a pain reliever recommend that an adult of  and over, take $4$ tablets every $4$ hours as needed, and an adult who weighs between $40$ and $50$ kg take only $2\frac{1}{2}$ tablets every $4$ hours as needed. Each tablet weighs $\frac{4}{25}$ gram.

a.    If a  adult takes $4$ tablets, how many grams of pain reliever is he or she receiving?

b.   How many grams of pain reliever is the recommended dose for an adult weighing

## Solution

a.    Given, $4$ tablets for a  adult & each tablet weighs
Total weight of pain reliever, he/she is receiving

b.   Given, adult weighing  takes $2\frac{1}{2}$ tablets & each tablet weighs .
Total weight of pain reliever, he/she is receiving

# Question: 66

Animals: The label on a bottle of pet vitamins lists dosage guidelines. What dosage would you give to each of these animals?

b.   a  cat

c.    a  pregnant dog

## Solution

a.    Dosage prescribed for an adult dog is  per  body weight.

dosage $=\frac{\left(\frac{1}{2}\right)}{9}×18$

b.   For a  cat,
dosage $=\frac{\left(\frac{1}{4}\right)}{9}×6$

c.    For a  pregnant dog,
dosage $=\frac{\left(\frac{1}{2}\right)}{4.5}×18$

# Question: 67

How many  boxes of chocolates can be made with  chocolates?

## Solution

Total chocolates

Number of boxes of chocolates

$=\frac{3}{2}×16=3×8=24$

# Question: 68

Anvi is making bookmarker like the one shown in  How many bookmarker can she make from a  long ribbon? ## Solution

Length of $1$ bookmarker

Length of ribbon

Number of bookmarkers $=$ Length of ribbon/Length of $1$ bookmarker $=\frac{1500}{\frac{21}{2}}$

$\begin{array}{l}=\frac{1500}{21}×2\\ =142.85\\ =142\end{array}$

Hence, $142$ bookmarkers can be made from a  long ribbon.

# Question: 69

A rule for finding the approximate length of diagonal of a square is to multiply the length of a side of the square by $1.414.$ Find the length of the diagonal when:

a.    The length of a side of the square is

b.   The length of a side of the square is exactly

## Solution

a.    Side of a square

Length of diagonal
$=$ Length of side of the square $×1.414$

b.   Side of a square

Length of diagonal

$=$ Length of side of the square $×1.414$

# Question: 70

The largest square that can be drawn in a circle has a side whose length is $0.707$ times the diameter of the circle. By this rule, find the length of the side of such a square when the diameter of the circle is

a.

b.

## Solution

Given,

Side of square

a.    Diameter of circle

Side of square $=0.707×14.35$

b.   Diameter of circle

Side of square $=0.707×8.63$

# Question: 71

To find the distance around a circular disc, multiply the diameter of the disc by $3.14.$ What is the distance around the disc when :

a.    the diameter is

## Solution

Given,

a.    Diameter of  a circular disc

Distance around the circular disc $=18.7×3.14$

Diameter of disc

Distance around the circular disc $=12.9×3.14$

# Question: 72

What is the cost of  of cloth at $\text{\hspace{0.17em}}53.50$ per metre?

## Solution

Cost of  cloth $=\text{\hspace{0.17em}}\text{\hspace{0.17em}}53.50$

Cost of  cloth $=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(53.50×27.5\right)$

$\begin{array}{l}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\frac{5350}{100}×\frac{275}{10}\right)\\ =\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\frac{1471250}{1000}\right)\\ =\text{\hspace{0.17em}}\text{\hspace{0.17em}}1471.25\end{array}$

# Question: 73

In a hurdle race, Nidhi is over hurdle B and $\frac{2}{6}$ of the way through the race, as shown in a.    Where will Nidhi be, when she is $\frac{4}{6}$ of the way through the race?

b.   Where will Nidhi be when she is $\frac{5}{6}$ of the way through the race?

c.    Give two fractions to tell what part of the race Nidhi has finished when she is over hurdle C.

## Solution

As per the given information and figure, 5 hurdles are placed at equal distances between starting and finishing point.

These 5 hurdles divide the whole distance in 6 equal parts.

So, hurdle A is placed after $\frac{1}{6}$ of the way from starting point.

So, if Nidhi is at A, she will cover $\frac{1}{6}$ way.

If Nidhi is at B, then $\frac{2}{6}$ of way is completed. a.    When she is $\frac{4}{6}$ of the way, she will be at $\frac{\frac{4}{6}}{\frac{1}{6}}$ position

b.   When she is $\frac{5}{6}$ of the way, she will be at $\frac{\frac{5}{6}}{\frac{1}{6}}$ position

c.    When she is over hurdle C, she has completed half race.

Hence, she will be at $\frac{3}{6}\text{\hspace{0.17em}}\text{way}=\frac{3}{6}=\frac{1}{2}\text{\hspace{0.17em}}\text{way}$

# Question: 74

Diameter of Earth is  In $1996\text{,}$ a new planet was discovered whose diameter is $\frac{5}{86}$ of the diameter of Earth. Find the diameter of this planet in km.

## Solution

Give, diameter of Earth

According to question,

Diameter of new planet $=\frac{5}{86}$ of diameter of earth

# Question: 75

What is the product of $\frac{5}{129}$ and its reciprocal?

## Solution

Reciprocal of $\frac{5}{129}=\frac{129}{5}$

Product of $\frac{5}{129}$ and its reciprocal

$=\frac{5}{129}×\frac{129}{5}=1$

# Question: 76

Simplify: $\frac{2\frac{1}{2}+\frac{1}{5}}{2\frac{1}{2}÷\frac{1}{5}}$

## Solution

Given, $\frac{2\frac{1}{2}+\frac{1}{5}}{2\frac{1}{2}÷\frac{1}{5}}=\frac{\frac{\left(2×2\right)+1}{2}+\frac{1}{5}}{\frac{\left(2×2\right)+1}{2}÷\frac{1}{5}}$

$\begin{array}{l}=\frac{\frac{5}{2}+\frac{1}{5}}{\frac{5}{2}÷\frac{1}{5}}=\frac{\frac{25+2}{10}}{\frac{5}{2}×5}\\ =\frac{\frac{27}{10}}{\frac{25}{2}}=\frac{27}{10}×\frac{2}{25}\\ =\frac{27}{125}\end{array}$

# Question: 77

Simplify: $\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{3}{8}×\frac{3}{5}}$

## Solution

Given, $\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{3}{8}×\frac{3}{5}}=\frac{\frac{5+4}{20}}{1-\frac{9}{40}}$

$\begin{array}{l}=\frac{\frac{9}{20}}{\frac{31}{40}}=\frac{9}{20}×\frac{40}{31}\\ =\frac{18}{31}\end{array}$

# Question: 78

Divide $\frac{3}{10}$ by

## Solution

$\begin{array}{l}=\frac{3}{10}÷\left(\frac{3}{20}\right)\\ =\frac{3}{10}×\frac{20}{3}=2\end{array}$

# Question: 79

$\frac{1}{8}$ of a number equals $\frac{2}{5}÷\frac{1}{20}\text{.}$ What is the number?

## Solution

Let the number be $x$

Then, $\frac{1}{8}$ of a number $=\frac{2}{5}÷\frac{1}{20}$

$\begin{array}{l}\frac{1}{8}×x=\frac{2}{5}×20\\ \frac{x}{8}=8\\ x=8×8=64\end{array}$

# Question: 80

Heena’s father paid an electric bill of $\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}385.70$ out of a $500$ rupee note. How much change should he have received?

## Solution

Given, total rupees $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}500$ and money paid $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}385.70$

$\therefore$ Change he received $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(500-385.70\right)=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}114.30$

# Question: 81

The normal body temperature is $98.6°\text{F}\text{.}$ When Savitri was ill her temperature rose to $103.1°\text{F}\text{.}$ How many degrees above normal was that?

## Solution

Given, normal body temperature $=98.6°\text{F}$ and temperature rise to $=103.1°\text{F}$

$\therefore$ Rise in temperature $=\left(103.1-98.6\right)°\text{F}=4.5°\text{F}$

# Question: 82

Meteorology: One measure of average global temperature shows how each year varies from a base measure. The table shows results for several years.

 Year $1958$ $1964$ $1965$ $1978$ $2002$ Difference from Base $0.10°\text{C}$ $-0.17°\text{C}$ $-0.10°\text{C}$ $\left(\frac{1}{50}\right)°\text{C}$ $0.54°\text{C}$

See the table and answer the following:

a.    Order the five years from coldest to warmest.

b.   In $1946\text{,}$ the average temperature varied by $-0.03°\text{C}$ from the base measure. Between which two years should $1946$ fall when the years are ordered from coldest to warmest?

## Solution

In year $1978\text{,}$ temperature is $\left(\frac{1}{50}\right)°\text{C}=0.02°\text{C}$

(a) By observing the given temperatures, coldest to warmest in ascending order is:

$-0.17°\text{C}<-0.10°\text{C}<0.02°\text{C}<0.10°\text{C}<0.54°\text{C}$

Order of years is:

$1964<1965<1978<1958<2002$

(b) In $1946\text{,}$ the average temperature varied by $-0.03°\text{C}$ from the base measure. So, temperature from coldest to warmest in ascending order is:

$-0.17°\text{C}<-0.10°\text{C}<-0.03°\text{C}<0.02°\text{C}<0.10°\text{C}<0.54°\text{C}$

Therefore, order of years is:

$1964<1965<1946<1978<1958<2002$

Thus, year $1946$ fall between years 1965 and 1978 when the years are ordered from coldest to warmest.

# Question: 83

In her science class, Jyoti learned that the atomic weight of Helium is $4.0030\text{;}$ of Hydrogen is $1.0080\text{;}$ and of Oxygen is $16.0000.$ Find the difference between the atomic weights of:

a.    Oxygen and Hydrogen

b.   Oxygen and Helium

c.    Helium and Hydrogen

## Solution

Given, atomic weights of Helium $=4.0030\text{,}$ Hydrogen $=1.0080$ and Oxygen $=16.0000$

a.    Difference between atomic weights of Oxygen & Hydrogen $=16.0000-1.0080=14.9920$

b.   Difference between atomic weights of Oxygen & Helium $=16.0000-04.0030=11.9970$

c.    Difference between atomic weights of Helium & Hydrogen $=4.0030-1.0080=2.9950$

# Question: 84

Measurement made in science lab must be as accurate as possible. Ravi measured the length of an iron rod and said it was  long; Kamal said  and Tabish said  The correct length was  How much of error was made by each of the boys?

## Solution

The actual length of an iron rod

Length measured by Ravi

Error $=$ Measured value $-$ Actual value

Length measured by Kamal

Error

Length measured by Tabish

Error

# Question: 85

When $0.02964$ is divided by $0.004\text{,}$ what will be the quotient?

## Solution

Given, $0.02964÷0.004=\frac{2964}{100000}÷\frac{4}{1000}$

$\begin{array}{l}=\frac{2964}{100000}×\frac{1000}{4}\\ =\frac{741}{100}\\ =7.41\end{array}$

# Question: 86

What number divided by $520$ gives the same quotient as $85$ divided by $0.625\text{?}$

## Solution

Let the number be $x$.

According to the question,

$\begin{array}{l}\frac{x}{520}=\frac{85}{0.625}\\ x=\frac{85×520×1000}{625}\\ x=\frac{44200000}{625}\\ x=70720\end{array}$

Hence, the number is $70720.$

# Question: 87

A floor is  long and  wide. A  square tile costs $\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}23.25.$ What will be the cost to cover the floor with these tiles?

## Solution

Area of floor

Convert

Cost of  tile is $\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}23.25$.

Area of each square tile $=6×6\text{\hspace{0.17em}}{\text{cm}}^{2}$

So, the cost for  floor

$=\frac{\left(162000×23.25\right)}{6×6}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}104625$

# Question: 88

Sunita and Rehana want to make dresses for their dolls. Sunita has  of cloth, and she gave $\frac{1}{3}$ of it to Rehana. How much did Rehana have?

## Solution

Given that, Sunita has  of cloth.

She gave one-third of cloth to Rehana.

Hence, Rehana has  of cloth.

# Question: 89

A flower garden is  long. Sheela wants to make a border along one side using bricks that are  long. How many bricks will be needed?

## Solution

Length of flower garden

Length of $1$ brick

Number of bricks used in $1$ side
$=$ length of flower garden/length of $1$ brick

$=\frac{22.50}{0.25}=\frac{2250}{25}=90$

Hence, $90$ bricks will be needed.

# Question: 90

How much cloth will be used in making $6$ shirts, if each required  of cloth, allowing  for waste in cutting and finishing in each shirt?

## Solution

Cloth required in making $1$ shirt $=\left(2\frac{1}{4}+\frac{1}{8}\right)=\frac{\left(2×4\right)+1}{4}+\frac{1}{8}$

Total cloth required in making such $6$ shirts $=6×\frac{19}{8}$

Hence,  cloth will be used in making $6$ shirts.

# Question: 91

A picture hall has seats for $820$ persons. At a recent film show, one usher guessed it was $\frac{3}{4}$ full, another that it was $\frac{2}{3}$ full. The ticket office reported $648$ sales. Which usher (first or second) made the better guess?

## Solution

Given, picture hall has seats $=820$

$1$ usher guessed, picture hall was $\frac{3}{4}$ full.

Another usher guessed, picture hall was $\frac{2}{3}$ full.

Since, $648$ tickets are sold that is near to $615.$

So, first usher made  better guess.

# Question: 92

For the celebrating children’s students of Class VII bought sweets for $\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}740.25$ and cold drink for $\text{Rs}\text{\hspace{0.17em}}70.$ If $35$ students contributed equally what amount was contributed by each student?

## Solution

Cost of sweets $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}740.25$

Cost of cold drink $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}70$

Total cost $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(740.25+70\right)$

$=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}810.25$

Given, $35$ students are contributing equally.

$\begin{array}{l}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{810.25}{35}\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{81025}{35×100}\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{2315}{100}\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}23.15\end{array}$

# Question: 93

The time taken by Rohan in five different races to run a distance of  was $3.20$ minutes, $3.37$ minutes, $3.29$ minutes, $3.17$ minutes and $3.32$ minutes. Find the average time taken by him in the races.

## Solution

Total time taken by Rohan $5$ races $=\left(3.20+3.37+3.29+3.17+3.32\right)$

Average time taken by Rohan $=$ total time taken/total number of observation

$=$ total time taken/ $5$

# Question: 94

A public sewer line is being installed along  of road. The supervisor says that the labourers will be able to complete  in one day. How long will the project take to complete? ## Solution

Total sewer line to be installed

In one day, labourers can complete

Number of days to complete the project $=$ total sewer line to be installed/ $1$ day work

Hence, the number of days to complete the project will be

# Question: 95

The weight of an object on moon is $\frac{1}{6}$ its weight on Earth. If an object weighs  on Earth, how much would it weigh on the moon?

## Solution

Weight of an object on the moon is $\frac{1}{6}$ of its weight on Earth.

Object weights on Earth

Weight of the object on Moon

Hence, weight of the object on the Moon is .

# Question: 96

In a survey, $200$ students were asked what influenced them most to buy their latest CD. The results are shown in the circle graph.

a.    How many students said radio influenced them most?

b.   How many more students were influenced by radio than by a music video channel?

c.    How many said a friend or relative influenced them or they heard the CD in a shop? ## Solution

(a)       Fraction of radio (in figure) $=\frac{9}{20}$

Total number of students $=200$

Number of students influenced by radio the most

$\begin{array}{l}=\frac{9}{20}×200\\ =90\end{array}$

(b)       We have to find,

(Students influenced by radio) $-$ (Students influenced by music video channel) $=$ (fraction of radio $×$ total number of students) $-$ (fraction of music video channel $×$ total number of students)

$\begin{array}{l}=\frac{9}{20}×200-\frac{2}{25}×200\\ =90-16\\ =74\end{array}$

(c)
Number of students influenced by friend or relative or CD in a shop
= Number of students influenced by friend or relative + Number of students who heard a CD in a shop

$\begin{array}{l}=\frac{3}{20}×200+\frac{1}{10}×200\\ =3×10+20\\ =30+20=50\end{array}$

# Question: 97

In the morning, a milkman filled  of milk in his can. He sold to Renu, Kamla and Renuka  each; to Shadma he sold  and to Jassi he gave  How much milk is left in the can?

## Solution

Given, milk in can

If  sold to Renu, Kamla and Renuka

Then, total milk sold $=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}$

Milk sold to Jassi

Total milk sold $=\frac{9}{4}+\frac{7}{8}+\frac{3}{2}$

Total mink left in can $=\frac{11}{2}-\left(\frac{37}{8}\right)$

Hence,  milk is left in the can.

# Question: 98

Anuradha can do a piece of work in $6$ hours. What part of the work can she do in $1$ hour, in $5$ hours, in $6$ hours?

## Solution

It is given that, Anuradha can do a piece of work in

In  Anuradha can do $=$ $1$ complete  work

# Question: 99

What portion of a ‘saree’ can Rehana paint in $1$ hour if it requires $5$ hours to paint the whole saree? In $4\frac{3}{5}$ hours? In $3\frac{1}{2}$ hours?

## Solution

In  Rehana paints $=$ 1 whole saree

In  she paints

In  she paints $=\frac{1}{5}×4\frac{3}{5}$

$\begin{array}{l}=\frac{1}{5}×\frac{\left(5×4\right)+3}{5}\\ =\frac{1}{5}×\frac{23}{5}\end{array}$

In  she paints $=\frac{1}{5}×3\frac{1}{2}$

$\begin{array}{l}=\frac{1}{5}×\frac{\left(3×2\right)+1}{2}\\ =\frac{1}{5}×\frac{7}{2}\end{array}$

# Question: 100

Rama has  of cotton wool for making pillows. If one pillow takes  how many pillows can she make?

## Solution

Give, Rama has  of cotton for making pillows

i.e.,

Where, $1$ pillow can be made from

i.e.,

Number of pillows

$\begin{array}{l}=\frac{\left(\frac{25}{4}\right)}{\frac{5}{4}}\\ =\frac{25}{4}×\frac{4}{5}\\ =\frac{25}{5}\\ =5\end{array}$

# Question: 101

It takes  of cloth to make a shirt. How many shirts can Radhika make from a piece of cloth  long?

## Solution

Radhika takes  cloth to make a shirt

i.e.,

i.e.,

Then, number of shirts that can be made

$\begin{array}{l}=\frac{\frac{28}{3}}{\frac{7}{3}}\\ =\frac{28}{3}×\frac{3}{7}\\ =\frac{28}{7}\\ =4\end{array}$

# Question: 102

Ravi can walk  in one hour. How long will it take him to walk to his office which is  from his home?

## Solution

Given, Ravi can walk  in

Ravi’s speed

Distance between home & office

Time

# Question: 103

Raj travels  on three fifths of his petrol tank. How far would he travel at the same rate with a full tank of petrol?

## Solution

Raj travels  on $\frac{3}{5}$ of his petrol tank.

Total distance travelled with full tank of petrol
$=$

Hence, total distance travelled by Raj with the available petrol tank is

# Question: 104

Kajol has $\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}75.$ This is $\frac{3}{8}$ of the amount she earned. How much did she earn?

## Solution

Kajol has $\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}75$.

According to question,

Amount earned $=\frac{75}{3}×8=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}200$

# Question: 105

It takes $17$ full specific type of trees to make one tonne of paper. If there are $221$ such trees in a forest, then

i.      What fraction of forest will be used to make;

a.    $5$ tonnes of paper.

b.   $10$ tonnes of paper.

ii.            To save $\frac{7}{13}$ part of the forest how much of paper we have to save.

## Solution

(i)

a.    $1$ tonne of paper require $=17$ trees

$5$ tonne of paper require $=17×5$ trees

$=85$ trees

Now, there are $221$ trees on the forest.

So, $85$ trees covers $=\frac{85}{221}$ fraction of forest

$=\frac{5}{13}$ fraction of forest

b.   Similarly

$10$ tonne of paper require $=17×10$ trees

$=170$ trees

So, $170$ trees covers $=\frac{170}{221}$ fraction of forest

$=\frac{10}{13}$ fraction of forest

(ii)          $\frac{7}{13}$ part of forest

Number of tonnes of paper which can be made by

# Question: 106

Simplify and write the result in decimal form :

$\left(1÷\frac{2}{9}\right)+\left(1÷3\frac{1}{5}\right)+\left(1÷2\frac{2}{3}\right)$

## Solution

$\left(1÷\frac{2}{9}\right)+\left(1÷3\frac{1}{5}\right)+\left(1÷2\frac{2}{3}\right)$

$\begin{array}{l}=\left(1÷\frac{2}{9}\right)+\left(1÷\frac{\left(5×3\right)+1}{5}\right)+\left(1÷\frac{\left(2×3\right)+2}{3}\right)\\ =\left(1×\frac{9}{2}\right)+\left(1÷\frac{16}{5}\right)+\left(1÷\frac{8}{3}\right)\\ =\left(1×\frac{9}{2}\right)+\left(1×\frac{5}{16}\right)+\left(1×\frac{3}{8}\right)\\ =\frac{9}{2}+\frac{5}{16}+\frac{3}{8}\\ =\frac{72+5+6}{16}\\ =\frac{83}{16}\\ =5.1875\end{array}$

# Question: 107

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

1.   $2×\frac{1}{4}$

2.   $2×\frac{3}{7}$

3.   $2×\frac{1}{3}$

4.   $\frac{1}{4}×4$

5.   $3×\frac{2}{9}$

6.   $\frac{1}{4}×3$

a. b. c. d. e. f. ## Solution

1.   d. $\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$

2.   f. $\frac{3}{7}+\frac{3}{7}=\frac{6}{7}$

3.   c. $\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

4.   b. $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{4}{4}$

5.   a. $\frac{2}{9}+\frac{2}{9}+\frac{2}{9}=\frac{6}{9}$

6.   e. $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$

# Question: 108

Evaluate: $\left(0.3\right)×\left(0.3\right)-\left(0.2\right)×\left(0.2\right)$

## Solution

Given, $\left(0.3\right)×\left(0.3\right)-\left(0.2\right)×\left(0.2\right)$

# Question: 109

Evaluate: $\frac{0.6}{0.3}+\frac{0.16}{0.4}$

## Solution

Given, $\frac{0.6}{0.3}+\frac{0.16}{0.4}$

$\begin{array}{l}\frac{0.6}{0.3}+\frac{0.16}{0.4}\\ =\frac{\frac{6}{10}}{\frac{3}{10}}+\frac{\frac{16}{100}}{\frac{4}{10}}\\ \left(\frac{6}{10}×\frac{10}{3}\right)+\left(\frac{16}{100}×\frac{10}{4}\right)\\ =\frac{60}{30}+\frac{160}{400}\\ =\frac{6}{3}+\frac{16}{40}\\ =\frac{2}{1}+\frac{4}{10}\\ =\frac{20+4}{10}\\ =\frac{24}{10}\\ =\frac{12}{5}\\ =2.4\end{array}$

# Question: 110

Find the value: $\frac{\left(0.2×0.14\right)+\left(0.5×0.91\right)}{\left(0.1×0.2\right)}$

## Solution

Given, $\frac{\left(0.2×0.14\right)+\left(0.5×0.91\right)}{\left(0.1×0.2\right)}$

$\begin{array}{l}=\frac{\frac{483}{1000}}{\frac{2}{100}}\\ =\frac{483}{1000}×\frac{100}{2}\\ =\frac{483}{10×2}\\ =\frac{241.5}{10}\\ =24.15\end{array}$

# Question: 111

A square and an equilateral triangle have a side in common. If side of triangle is  long, find the perimeter of figure formed (Fig. 2.8). Fig. 2.8

## Solution

The given square & equilateral triangle both have a common side  BC.

So, all the side of square & triangle will be equal and of measure .

Perimeter of the figure $=\text{AB}+\text{BD}+\text{DE}+\text{EC}+\text{AC}$

# Question: 112

Rita has bought a carpet of size  But her room size is  What fraction of area should be cut off to fit wall to wall carpet into the room?

## Solution

Carpet size

$\begin{array}{l}=4×\frac{\left(6×3\right)+2}{3}\\ =4×\frac{18+2}{3}\\ =4×\frac{20}{3}\end{array}$

Room size

Difference between the area of carpet & room

In fraction,

Area that will be cut off

$\begin{array}{l}=\frac{80}{9}×\frac{3}{80}\\ =\frac{1}{3}\end{array}$

Hence, $\frac{1}{3}$ of area should be cut off.

# Question: 113

Family photograph has length  and breadth  It has border of uniform width  Find the area of framed photograph.

## Solution

Length of family photograph

New length including border (from both sides)

New width including border (from both sides)

Area of framed photograph $=$ length $×$ breadth

$=\frac{98}{5}×\frac{78}{5}$

Hence, the area of framed photograph is

# Question: 114

Cost of a burger is $\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}20\frac{3}{4}$ and of Macpuff is $\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}15\frac{1}{2}\text{.}$ Find the cost of $4$ burgers and $14$ macpuffs.

## Solution

Cost of $1$ burger $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}20\frac{3}{4}$

$\begin{array}{l}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\left(20×4\right)+3}{4}\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{83}{4}\end{array}$

Cost of $4$ burger $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4×\frac{83}{4}$

$=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}83$

Cost of $1$ macpuff $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}15\frac{1}{2}$

$=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{31}{2}$

Cost of $14$ macpuff $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}14×\frac{31}{2}$

$=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}217$

Total cost of $4$ burgers & $14$ macpuffs $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(83+217\right)$

$=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}300$

# Question: 115

A hill,  in height, has $\frac{1}{4}\text{th}$ of its height under water. What is the height of the hill visible above the water?

## Solution

Given, height of the hill

Height of the hill under water $=\frac{1}{4}$ of the height of the hill

Height of the hill above the water
$=$ height of the hill – height of the hill under water

# Question: 116

Sports: Reaction time measures how quickly a runner reacts to the starter pistol. In the  dash at the $2004$ Olympic Games, Lauryn Williams had a reaction time of $0.214$ second. Her total race time, including reaction time, was $11.03$ seconds. How long did it take her to run the actual distance?

## Solution

Time taken to run the actual distance $=$ Total race time $-$ reaction time

# Question: 117

State whether the answer is greater than $1$ or less than $1.$ Put a ‘mark in appropriate box.

 Questions Greater than $1$ Less than $1$ $\frac{2}{3}÷\frac{1}{2}$ $\frac{2}{3}÷\frac{2}{1}$ $6÷\frac{1}{4}$ $\frac{1}{5}÷\frac{1}{2}$ $4\frac{1}{3}÷3\frac{1}{2}$ $\frac{2}{3}×8\frac{1}{2}$

## Solution

i.       $\frac{2}{3}÷\frac{1}{2}=\frac{2}{3}×\frac{2}{1}=\frac{4}{3}=1.33\left(>1\right)$

ii.       $\frac{2}{3}÷\frac{2}{1}=\frac{2}{3}×\frac{1}{2}=\frac{1}{3}=0.33\left(<1\right)$

iii.       $6÷\frac{1}{4}=6×\frac{4}{1}=24\left(>1\right)$

iv.       $\frac{1}{5}÷\frac{1}{2}=\frac{1}{5}×\frac{2}{1}=\frac{2}{5}=0.4\left(<1\right)$

v.       $4\frac{1}{3}÷3\frac{1}{2}=\frac{13}{3}÷\frac{7}{2}=\frac{13}{3}×\frac{2}{7}=\frac{26}{21}=1.24\left(>1\right)$

vi.       $\frac{2}{3}×8\frac{1}{2}=\frac{2}{3}×\frac{17}{2}=\frac{17}{3}=5.67\left(>1\right)$

 Questions Greater than $1$ Less than $1$ $\frac{2}{3}÷\frac{1}{2}$ √ $\frac{2}{3}÷\frac{2}{1}$ √ $6÷\frac{1}{4}$ √ $\frac{1}{5}÷\frac{1}{2}$ √ $4\frac{1}{3}÷3\frac{1}{2}$ √ $\frac{2}{3}×8\frac{1}{2}$ √

# Question: 118

There are four containers that are arranged in the ascending order of their heights. If the height of the smallest container given in the figure is expressed as  Find the height of the largest container. ## Solution

It is given that, the smallest cylinder is  high.

Height of the smallest cylinder in terms of $x$ is $\frac{7}{25}x\text{,}$ where $x$ is height of largest cylinder.

Then, $\frac{7}{25}x=10.5$

Hence, height of the largest container is

In Questions 119 to 122, replace ‘?’ with appropriate fraction.

# Question: 119 ## Solution

Given sequence is

We observe that each fraction is divided by $3$ to get next fraction

So, $\text{?}=\frac{7}{216}÷3$

$\begin{array}{l}=\frac{7}{216}×\frac{1}{3}\\ =\frac{7}{648}\end{array}$

Question: 120 ## Solution

Given sequence is

Each fraction is multiplied by $2$ to get the next fraction

So, $\text{?}=\frac{3}{4}×2$

$=\frac{3}{2}$

# Question: 121 ## Solution

Given sequence is

We observe that, each fraction is multiplied by $10$ to get the next number.

$\text{?}=50×10$

$=500$

# Question: 122 ## Solution

Given sequence is

We observe that, each fraction is divided by $10$ to get the next number

$\text{?}=\frac{0.0001}{10}$

$=0.00001$

What is the Error in each of question 123 to 125?

# Question: 123

A student compared $-\frac{1}{4}$ and $-0.3.$ He changed $-\frac{1}{4}$ to the decimal $-0.25$ and wrote, “Since $0.3$ is greater than $0.25\text{,}-0.3$ is greater than $-0.25\text{"}\text{.}$ What was the student’s error?

## Solution

If the numbers are negative, then the numbers whose absolute value is greater, will be smaller.

Hence, $-0.25$ is greater than $-0.3$

So, the student made the error that $\left(-0.3\right)>-\left(0.25\right)$.

# Question: 124

A student multiplied two mixed fractions in the following manner: $2\frac{4}{7}×3\frac{1}{4}=6\frac{1}{7}\text{.}$ What error the student has done?

## Solution

To multiply two mixed fractions, first convert them into improper fraction.

So, $2\frac{4}{7}×3\frac{1}{4}=\frac{2×7+4}{7}×\frac{3×4+1}{4}$

$\begin{array}{l}=\frac{18}{7}×\frac{13}{4}\\ =\frac{9×2}{7}×\frac{13}{2×2}\\ =\frac{117}{14}\\ =\frac{14×8+5}{14}\\ =\frac{14×8}{14}+\frac{5}{14}\\ =8\frac{5}{14}\end{array}$

# Question: 125

In the pattern $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots ..$ which fraction makes the sum greater than $1$ (first time)? Explain.

## Solution

$\begin{array}{l}\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\\ =\frac{20+15+12}{60}\\ =\frac{47}{60}<1\end{array}$

According to the pattern, next number will be $\frac{1}{6}$

$\begin{array}{l}\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\\ =\frac{40+30+24+20}{120}\\ =\frac{114}{120}<1\end{array}$

Now, according to the pattern, next number after

$\begin{array}{l}\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\\ =\frac{280+210+168+140+120}{840}\\ =\frac{918}{840}>1\end{array}$

Hence, first time, $\frac{1}{7}$ makes the sum greater than $1$ .