Fractions

(i) 3 ÷ 7

A division a ÷ b can be written as the fraction $\frac{a}{b}$.

So, 3 ÷ 7 = $\frac{3}{7}$.

Hence, 3 ÷ 7 = $\frac{3}{7}$.

(ii) 11 ÷ 78

A division a ÷ b can be written as the fraction $\frac{a}{b}$.

So, 11 ÷ 78 = $\frac{11}{78}$.

Hence, 11 ÷ 78 = $\frac{11}{78}$.

(iii) 113 ÷ 128

A division a ÷ b can be written as the fraction $\frac{a}{b}$.

So, 113 ÷ 128 = $\frac{113}{128}$.

Hence, 113 ÷ 128 = $\frac{113}{128}$.

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

Hence, $\frac{2}{7}$ in words is two-seventh.

(ii) $\frac{3}{10}$

Hence, $\frac{3}{10}$ in words is three-tenth.

(iii) $\frac{15}{28}$

Hence, $\frac{15}{28}$ in words is fifteen-twenty eighth.

(i) one-sixth

Hence, one-sixth in number form is $\frac{1}{6}$.

(ii) three-eleventh

Hence, three-eleventh in number form is $\frac{3}{11}$.

(iii) seven-fortieth

Hence, seven-fortieth in number form is $\frac{7}{40}$.

(iv) thirteen-one hundred twenty fifth

Hence, thirteen-one hundred twenty fifth in number form is $\frac{13}{125}$.

(i) Total parts = 7

Shaded parts = 4

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{4}{7}$.

Hence, fraction = $\frac{4}{7}$.

(ii) Total parts = 8

Shaded parts = 3

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{3}{8}$.

Hence, fraction = $\frac{3}{8}$.

(iii) Total parts = 4

Shaded parts = $\frac{1}{2}$

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{\frac{1}{2}}{4} = \frac{1}{2 \times 4} = \frac{1}{8}$.

Hence, fraction = $\frac{1}{8}$.

(iv) Total parts = 4

Shaded parts = 1

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{1}{4}$.

Hence, fraction = $\frac{1}{4}$.

(v) Total parts = 6

Shaded parts = 1

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{1}{6}$.

Hence, fraction = $\frac{1}{6}$.

(vi) Total parts = 10

Shaded parts = 3

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{3}{10}$.

Hence, fraction = $\frac{3}{10}$.

(vii) Total parts = 7

Shaded parts = 3

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{3}{7}$.

Hence, fraction = $\frac{3}{7}$.

(viii) Total parts = 4

Shaded parts = 2

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{2}{4}$.

Hence, fraction = $\frac{2}{4}$.

(ix) Total parts = 9

Shaded parts = 4

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{4}{9}$.

Hence, fraction = $\frac{4}{9}$.

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

The figure (square) is divided into 4 equal parts. Shade 3 parts out of 4.

(ii) $\frac{1}{6}$

The figure (circle) is divided into 6 equal parts. Shade 1 part out of 6.

(iii) $\frac{1}{4}$

The figure (triangle) is divided into 4 equal parts. Shade 1 part out of 4.

(iv) $\frac{4}{9}$

The figure (square) is divided into 9 equal parts. Shade 4 parts out of 9.

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

The figure (hexagon) is divided into 6 equal parts (i.e., 2 parts make $\frac{1}{3}$). Shade 2 parts out of 6.

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

The figure (8 circles) shows 8 equal parts. Shade 5 circles out of 8.

(i) numerator = 5 and denominator = 13

Fraction = $\frac{\text{numerator}}{\text{denominator}} = \frac{5}{13}$.

Hence, the required fraction = $\frac{5}{13}$.

(ii) denominator = 23 and numerator = 17

Fraction = $\frac{\text{numerator}}{\text{denominator}} = \frac{17}{23}$.

Hence, the required fraction = $\frac{17}{23}$.

Total number of dresses to be stitched = 35

Number of dresses stitched so far = 21

Fraction of dresses stitched = $\frac{\text{Dresses stitched}}{\text{Total dresses}}$

= $\frac{21}{35}$

Hence, the required fraction = $\frac{21}{35}$.

We know, 1 day = 24 hours.

Fraction = $\frac{8 \text{ hours}}{24 \text{ hours}}$

= $\frac{8}{24}$

Hence, the required fraction = $\frac{8}{24}$.

We know, 1 hour = 60 minutes.

Fraction = $\frac{45 \text{ minutes}}{60 \text{ minutes}}$

= $\frac{45}{60}$

Hence, the required fraction = $\frac{45}{60}$.

Natural numbers from 87 to 97 are :

87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97 and their number = 11.

Out of these, the prime numbers are 89, 97.

Fraction of prime numbers = $\frac{\text{Number of primes}}{\text{Total numbers}} = \frac{2}{11}$.

Hence, total natural numbers = 11 and fraction of prime numbers = $\frac{2}{11}$.

Draw a straight line and mark a point O on it representing 0. Mark a point A on the right of O at unit length so that OA represents 1.

Now divide the line segment OA into 5 equal parts. Each part represents $\frac{1}{5}$.

Starting from O, mark the points at $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}$ and $\frac{5}{5}$ (= 1).

Hence, the fractions $\frac{2}{5}, \frac{3}{5}, \frac{4}{5}$ and $\frac{5}{5}$ are represented on the number line as shown above.

Draw a straight line and mark a point O on it representing 0. Mark a point A on the right of O at unit length so that OA represents 1.

Now divide the line segment OA into 8 equal parts. Each part represents $\frac{1}{8}$.

Starting from O, mark the points at $\frac{1}{8}, \frac{2}{8}, \frac{3}{8}$ and $\frac{7}{8}$.

Hence, the fractions $\frac{1}{8}, \frac{2}{8}, \frac{3}{8}$ and $\frac{7}{8}$ are represented on the number line as shown above.

Draw a straight line and mark a point O on it representing 0. Mark a point A on the right of O at unit length so that OA represents 1.

Now divide the line segment OA into 10 equal parts. Each part represents $\frac{1}{10}$.

Starting from O, mark the points at $\frac{0}{10}$ (= 0), $\frac{1}{10}, \frac{3}{10}, \frac{5}{10}, \frac{7}{10}$ and $\frac{10}{10}$ (= 1).

Hence, the fractions $\frac{0}{10}, \frac{1}{10}, \frac{3}{10}, \frac{5}{10}, \frac{7}{10}$ and $\frac{10}{10}$ are represented on the number line as shown above.

(i) $\frac{15}{26}$

Here, numerator (15) < denominator (26).

Hence, $\frac{15}{26}$ is a proper fraction.

(ii) $\frac{17}{12}$

Here, numerator (17) > denominator (12).

Hence, $\frac{17}{12}$ is an improper fraction.

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

It consists of a natural number 5 and a proper fraction $\frac{2}{3}$.

Hence, $5\frac{2}{3}$ is a mixed fraction.

(iv) $\frac{6}{8}$

Here, numerator (6) < denominator (8).

Hence, $\frac{6}{8}$ is a proper fraction.

(v) $11\frac{5}{7}$

It consists of a natural number 11 and a proper fraction $\frac{5}{7}$.

Hence, $11\frac{5}{7}$ is a mixed fraction.

(vi) $\frac{117}{8}$

Here, numerator (117) > denominator (8).

Hence, $\frac{117}{8}$ is an improper fraction.

(vii) $\frac{222}{333}$

Here, numerator (222) < denominator (333).

Hence, $\frac{222}{333}$ is a proper fraction.

(viii) $\frac{531}{247}$

Here, numerator (531) > denominator (247).

Hence, $\frac{531}{247}$ is an improper fraction.

(i) $\frac{17}{3}$

Dividing 17 by 3, we get quotient = 5 and remainder = 2.

$\Rightarrow \frac{17}{3} = 5\frac{2}{3}.$

Hence, $\frac{17}{3} = 5\frac{2}{3}$.

(ii) $\frac{119}{15}$

Dividing 119 by 15, we get quotient = 7 and remainder = 14.

$\Rightarrow \frac{119}{15} = 7\frac{14}{15}.$

Hence, $\frac{119}{15} = 7\frac{14}{15}$.

(iii) $\frac{961}{13}$

Dividing 961 by 13, we get quotient = 73 and remainder = 12.

$\Rightarrow \frac{961}{13} = 73\frac{12}{13}.$

Hence, $\frac{961}{13} = 73\frac{12}{13}$.

(iv) $\frac{117}{32}$

Dividing 117 by 32, we get quotient = 3 and remainder = 21.

$\Rightarrow \frac{117}{32} = 3\frac{21}{32}.$

Hence, $\frac{117}{32} = 3\frac{21}{32}$.

We use the rule: improper fraction = $\frac{(\text{natural number} \times \text{denominator}) + \text{numerator}}{\text{denominator}}$.

(i) $7\frac{2}{11}$

$$\begin{aligned}\Rightarrow 7\frac{2}{11} = \frac{7 \times 11 + 2}{11} \\[1em] = \frac{77 + 2}{11} \\[1em] = \frac{79}{11}.\end{aligned}$$

Hence, the required fraction is $7\frac{2}{11} = \frac{79}{11}$.

(ii) $3\frac{5}{48}$

$$\begin{aligned}\Rightarrow 3\frac{5}{48} = \frac{3 \times 48 + 5}{48} \\[1em] = \frac{144 + 5}{48} \\[1em] = \frac{149}{48}.\end{aligned}$$

Hence, the required fraction is $3\frac{5}{48} = \frac{149}{48}$.

(iii) $13\frac{7}{64}$

$$\begin{aligned}\Rightarrow 13\frac{7}{64} = \frac{13 \times 64 + 7}{64} \\[1em] = \frac{832 + 7}{64} \\[1em] = \frac{839}{64}.\end{aligned}$$

Hence, the required fraction is $13\frac{7}{64} = \frac{839}{64}$.

(iv) $7\frac{2}{3}$

$$\begin{aligned}\Rightarrow 7\frac{2}{3} = \frac{7 \times 3 + 2}{3} \\[1em] = \frac{21 + 2}{3} \\[1em] = \frac{23}{3}.\end{aligned}$$

Hence, the required fraction is $7\frac{2}{3} = \frac{23}{3}$.

The fractions representing the shaded regions are:

(i) Total parts = 2

Shaded parts = 1

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{1}{2}$.

Hence, fraction = $\frac{1}{2}$

(ii) Total parts = 6

Shaded parts = 4

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{4}{6}$.

Hence, fraction = $\frac{4}{6}$

(iii) Total parts = 9

Shaded parts = 3

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{3}{9}$.

Hence, fraction = $\frac{3}{9}$

(iv) Total parts = 8

Shaded parts = 2

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{2}{8}$.

Hence, fraction = $\frac{2}{8}$

(v) Total parts = 4

Shaded parts = 3

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{3}{4}$.

Hence, fraction = $\frac{3}{4}$

(a) Total parts = 8

Shaded parts = 2

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{2}{8}$.

Hence, fraction = $\frac{2}{8}$

(b) Total parts = 12

Shaded parts = 8

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{8}{12}$.

Hence, fraction = $\frac{8}{12}$

(c) Total parts = 16

Shaded parts = 12

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{12}{16} = \frac{3}{4}$.

Hence, fraction = $\frac{3}{4}$

(d) Total parts = 8

Shaded parts = 4

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{4}{8}$.

Hence, fraction = $\frac{4}{8}$

(e) Total parts = 9

Shaded parts = 3

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{3}{9}$.

Hence, fraction = $\frac{3}{9}$

Hence, the equivalent pairs are:

(i) ↔ (d), (ii) ↔ (b), (iii) ↔ (e), (iv) ↔ (a), (v) ↔ (c).

(i) Equivalent fraction of $\frac{15}{35}$ with denominator 7

To get 7 from 35, we have to divide 35 by 5.

So, divide both numerator and denominator by 5.

$$\begin{aligned}\Rightarrow \frac{15}{35} = \frac{15 \div 5}{35 \div 5} \\[1em] = \frac{3}{7}.\end{aligned}$$

Hence, the required equivalent fraction = $\frac{3}{7}$.

(ii) Equivalent fraction of $\frac{2}{9}$ with denominator 63

To get 63 from 9, we have to multiply 9 by 7.

So, multiply both numerator and denominator by 7.

$$\begin{aligned}\Rightarrow \frac{2}{9} = \frac{2 \times 7}{9 \times 7} \\[1em] = \frac{14}{63}.\end{aligned}$$

Hence, the required equivalent fraction = $\frac{14}{63}$.

(i) Equivalent fraction of $\frac{3}{5}$ with denominator 30

To get 30 from 5, we have to multiply 5 by 6.

So, multiply both numerator and denominator by 6.

$$\begin{aligned}\Rightarrow \frac{3}{5} = \frac{3 \times 6}{5 \times 6} \\[1em] = \frac{18}{30}.\end{aligned}$$

Hence, the required equivalent fraction = $\frac{18}{30}$.

(ii) Equivalent fraction of $\frac{3}{5}$ with numerator 27

To get 27 from 3, we have to multiply 3 by 9.

So, multiply both numerator and denominator by 9.

$$\begin{aligned}\Rightarrow \frac{3}{5} = \frac{3 \times 9}{5 \times 9} \\[1em] = \frac{27}{45}.\end{aligned}$$

Hence, the required equivalent fraction = $\frac{27}{45}$.

(i) $\frac{2}{3} = \frac{\boxed{\phantom{653}}}{15}$

To get 15 from 3, we have to multiply 3 by 5.

So, multiply both numerator and denominator by 5.

$$\begin{aligned}\Rightarrow \frac{2}{3} = \frac{2 \times 5}{3 \times 5} \\[1em] = \frac{10}{15}.\end{aligned}$$

Hence, replace $\boxed{\phantom{653}}$ by 10.

(ii) $\frac{7}{18} = \frac{42}{\boxed{\phantom{653}}}$

To get 42 from 7, we have to multiply 7 by 6.

So, multiply both numerator and denominator by 6.

$$\begin{aligned}\Rightarrow \frac{7}{18} = \frac{7 \times 6}{18 \times 6} \\[1em] = \frac{42}{108}.\end{aligned}$$

Hence, replace $\boxed{\phantom{653}}$ by 108.

(iii) $\frac{4}{\boxed{\phantom{653}}} = \frac{12}{15}$

To get 4 from 12, we have to divide 12 by 3.

So, divide both numerator and denominator by 3.

$$\begin{aligned}\Rightarrow \frac{12}{15} = \frac{12 \div 3}{15 \div 3} \\[1em] = \frac{4}{5}.\end{aligned}$$

Hence, replace $\boxed{\phantom{653}}$ by 5.

(iv) $\frac{\boxed{\phantom{653}}}{11} = \frac{70}{154}$

To get 11 from 154, we have to divide 154 by 14.

So, divide both numerator and denominator by 14.

$$\begin{aligned}\Rightarrow \frac{70}{154} = \frac{70 \div 14}{154 \div 14} \\[1em] = \frac{5}{11}.\end{aligned}$$

Hence, replace $\boxed{\phantom{653}}$ by 5.

(i) $\frac{3}{10}, \frac{12}{40}$

By cross multiplication,

3 × 40 = 120 and 10 × 12 = 120

Here, the cross products are equal.

Hence, $\frac{3}{10}$ and $\frac{12}{40}$ are equivalent fractions.

(ii) $\frac{5}{8}, \frac{30}{48}$

By cross multiplication,

5 × 48 = 240 and 8 × 30 = 240

Here, the cross products are equal.

Hence, $\frac{5}{8}$ and $\frac{30}{48}$ are equivalent fractions.

(iii) $\frac{4}{6}, \frac{30}{20}$

By cross multiplication,

4 × 20 = 80 and 6 × 30 = 180

Here, the cross products are not equal.

Hence, $\frac{4}{6}$ and $\frac{30}{20}$ are not equivalent fractions.

(iv) $\frac{7}{13}, \frac{5}{11}$

By cross multiplication,

7 × 11 = 77 and 13 × 5 = 65

Here, the cross products are not equal.

Hence, $\frac{7}{13}$ and $\frac{5}{11}$ are not equivalent fractions.

(i) $\frac{12}{27}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{12}{27} = \frac{2 \times 2 \times 3}{3 \times 3 \times 3} \\[1em] = \frac{2 \times 2}{3 \times 3} \\[1em] = \frac{4}{9}.\end{aligned}$$

Hence, $\frac{12}{27}$ in simplest form is $\frac{4}{9}$.

(ii) $\frac{150}{350}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{150}{350} = \frac{2 \times 3 \times 5 \times 5}{2 \times 5 \times 5 \times 7} \\[1em] = \frac{3}{7}.\end{aligned}$$

Hence, $\frac{150}{350}$ in simplest form is $\frac{3}{7}$.

(iii) $\frac{18}{81}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{18}{81} = \frac{2 \times 3 \times 3}{3 \times 3 \times 3 \times 3} \\[1em] = \frac{2}{3 \times 3} \\[1em] = \frac{2}{9}.\end{aligned}$$

Hence, $\frac{18}{81}$ in simplest form is $\frac{2}{9}$.

(iv) $\frac{276}{115}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{276}{115} = \frac{2 \times 2 \times 3 \times 23}{5 \times 23} \\[1em] = \frac{2 \times 2 \times 3}{5} \\[1em] = \frac{12}{5} \\[1em] = 2\frac{2}{5}.\end{aligned}$$

Hence, $\frac{276}{115}$ in simplest form is $\frac{12}{5}$ or $2\frac{2}{5}$.

(i) $\frac{7}{8}, \frac{5}{14}$

L.C.M. of 8 and 14 is :

2	8, 14
2	4, 7
2	2, 7
7	1, 7
	1, 1

LCM of 8 and 14 = 2 x 2 x 2 x 7 = 56

$$\begin{aligned}\Rightarrow \frac{7}{8} = \frac{7 \times 7}{8 \times 7} = \frac{49}{56} \\[1em] \Rightarrow \frac{5}{14} = \frac{5 \times 4}{14 \times 4} = \frac{20}{56}\end{aligned}$$

Hence, the equivalent like fractions are $\frac{49}{56}$ and $\frac{20}{56}$.

(ii) $\frac{5}{6}, \frac{7}{16}$

L.C.M. of 6 and 16 is :

2	6, 16
2	3, 8
2	3, 4
2	3, 2
3	3, 1
	1, 1

LCM of 6 and 16 = 2 x 2 x 2 x 2 x 3 = 48.

$$\begin{aligned}\Rightarrow \frac{5}{6} = \frac{5 \times 8}{6 \times 8} = \frac{40}{48} \\[1em] \Rightarrow \frac{7}{16} = \frac{7 \times 3}{16 \times 3} = \frac{21}{48}\end{aligned}$$

Hence, the equivalent like fractions are $\frac{40}{48}$ and $\frac{21}{48}$.

(iii) $\frac{3}{4}, \frac{5}{6}, \frac{7}{8}$

L.C.M. of 4, 6 and 8 is :

2	4, 6, 8
2	2, 3, 4
2	1, 3, 2
3	1, 3, 1
	1, 1, 1

LCM of 4, 6 and 8 = 2 x 2 x 2 x 3 = 24.

$$\begin{aligned}\Rightarrow \frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24} \\[1em] \Rightarrow \frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24} \\[1em] \Rightarrow \frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24}\end{aligned}$$

Hence, the equivalent like fractions are $\frac{18}{24}, \frac{20}{24}$ and $\frac{21}{24}$.

Each unit is divided into 5 equal parts, so each part represents $\frac{1}{5}$.

The bold line shown in the figure starts from 0 and goes up to a length of 2 parts, so it represents $\frac{2}{5}$.

The remaining boxes are filled as follows (counting parts from 0 to the end of bold line):

The fraction in the first box (after $\frac{1}{5}$) = $\frac{2}{5}$.

The fraction in the second box (after $\frac{3}{5}$) = $\frac{4}{5}$.

The fraction in the third box (after 1) = $\frac{6}{5}$ or $1\frac{1}{5}$.

Hence, the missing fractions are $\frac{2}{5}, \frac{4}{5}$ and $\frac{6}{5}$ (i.e. $1\frac{1}{5}$).

(i) $\frac{3}{4} = \frac{\boxed{\phantom{653}}}{8}$

To get 8 from 4, we have to multiply 4 by 2.

So, multiply numerator and denominator by 2.

$\Rightarrow \frac{3}{4} = \frac{3 \times 2}{4 \times 2} = \frac{6}{8}.$

Hence, the missing number is 6 i.e. 6 glasses of juice shared equally among 8 friends.

(ii) $\frac{4}{3} = \frac{12}{\boxed{\phantom{653}}}$

To get 12 from 4, we have to multiply 4 by 3.

So, multiply numerator and denominator by 3.

$\Rightarrow \frac{4}{3} = \frac{4 \times 3}{3 \times 3} = \frac{12}{9}.$

Hence, the missing number is 9 i.e. 12 kg of sugar divided equally in 9 bags.

(iii) $\frac{7}{5} = \frac{\boxed{\phantom{653}}}{\boxed{\phantom{653}}}$

We can multiply both numerator and denominator by 2.

$\Rightarrow \frac{7}{5} = \frac{7 \times 2}{5 \times 2} = \frac{14}{10}.$

Hence, the missing numbers are 14 and 10 i.e. 14 gulabjamuns divided among 10 children.

Age at which Raghu got the job = 24 years.

Age of retirement = 60 years.

Number of years he was in job = 60 - 24 = 36 years.

Total age till retirement = 60 years.

Fraction of his age in job = $\frac{\text{Years in job}}{\text{Total age}}$

$$\begin{aligned}= \frac{36}{60} \\[1em] = \frac{3}{5}.\end{aligned}$$

Hence, the required fraction = $\frac{3}{5}$.

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

These are like fractions (same denominator 6).

In like fractions, the fraction with greater numerator is greater.

Since 3 < 5, therefore, $\frac{3}{6} \lt \frac{5}{6}$.

Hence, $\frac{3}{6} \lt \frac{5}{6}$.

(ii) $\frac{2}{7}$ and $\frac{2}{5}$

These are unlike fractions with same numerator 2.

In fractions with same numerator, the fraction with smaller denominator is greater.

Since 7 > 5, therefore, $\frac{2}{7} \lt \frac{2}{5}$.

Hence, $\frac{2}{7} \lt \frac{2}{5}$.

(iii) $\frac{3}{5}$ and $\frac{4}{5}$

These are like fractions (same denominator 5).

In like fractions, the fraction with greater numerator is greater.

Since 3 < 4, therefore, $\frac{3}{5} \lt \frac{4}{5}$.

Hence, $\frac{3}{5} \lt \frac{4}{5}$.

(iv) $\frac{4}{7}$ and $\frac{4}{9}$

These are unlike fractions with same numerator 4.

In fractions with same numerator, the fraction with smaller denominator is greater.

Since 7 < 9, therefore, $\frac{4}{7} \gt \frac{4}{9}$.

Hence, $\frac{4}{7} \gt \frac{4}{9}$.

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

These are unlike fractions with same numerator 1.

In fractions with same numerator, the fraction with smaller denominator is greater.

Since 2 < 5, therefore, $\frac{1}{2} \gt \frac{1}{5}$.

Hence, $\frac{1}{2} \gt \frac{1}{5}$.

(ii) $\frac{2}{4}$ and $\frac{3}{6}$

By cross multiplication,

2 × 6 = 12 and 4 × 3 = 12

The cross products are equal.

Hence, $\frac{2}{4} = \frac{3}{6}$.

(iii) $\frac{7}{9}$ and $\frac{3}{9}$

These are like fractions (same denominator 9).

In like fractions, the fraction with greater numerator is greater.

Since 7 > 3, therefore, $\frac{7}{9} \gt \frac{3}{9}$.

Hence, $\frac{7}{9} \gt \frac{3}{9}$.

(iv) $\frac{3}{4}$ and $\frac{2}{8}$

By cross multiplication,

3 × 8 = 24 and 4 × 2 = 8

Since 24 > 8, therefore, $\frac{3}{4} \gt \frac{2}{8}$.

Hence, $\frac{3}{4} \gt \frac{2}{8}$.

(i) $\frac{5}{9}$ and $\frac{4}{5}$

LCM of 9 and 5 = 45.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{5}{9} = \frac{5 \times 5}{9 \times 5} = \frac{25}{45} \\[1em] \Rightarrow \frac{4}{5} = \frac{4 \times 9}{5 \times 9} = \frac{36}{45}\end{aligned}$$

As 25 < 36, $\frac{25}{45} \lt \frac{36}{45} \Rightarrow \frac{5}{9} \lt \frac{4}{5}$.

Hence, $\frac{5}{9} \lt \frac{4}{5}$.

(ii) $\frac{9}{16}$ and $\frac{5}{9}$

LCM of 16 and 9 = 144.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{9}{16} = \frac{9 \times 9}{16 \times 9} = \frac{81}{144} \\[1em] \Rightarrow \frac{5}{9} = \frac{5 \times 16}{9 \times 16} = \frac{80}{144}\end{aligned}$$

As 81 > 80, $\frac{81}{144} \gt \frac{80}{144} \Rightarrow \frac{9}{16} \gt \frac{5}{9}$.

Hence, $\frac{9}{16} \gt \frac{5}{9}$.

(i) $\frac{5}{11}$ and $\frac{3}{7}$

By cross multiplication,

5 × 7 = 35 and 11 × 3 = 33

Since 35 > 33, therefore, $\frac{5}{11} \gt \frac{3}{7}$.

Hence, $\frac{5}{11} \gt \frac{3}{7}$.

(ii) $\frac{8}{15}$ and $\frac{3}{5}$

By cross multiplication,

8 × 5 = 40 and 15 × 3 = 45

Since 40 < 45, therefore, $\frac{8}{15} \lt \frac{3}{5}$.

Hence, $\frac{8}{15} \lt \frac{3}{5}$.

(iii) $\frac{11}{14}$ and $\frac{29}{35}$

LCM of 14 and 35 = 70.

$$\begin{aligned}\Rightarrow \frac{11}{14} = \frac{11 \times 5}{14 \times 5} = \frac{55}{70} \\[1em] \Rightarrow \frac{29}{35} = \frac{29 \times 2}{35 \times 2} = \frac{58}{70}\end{aligned}$$

As 55 < 58,

$\frac{55}{70} \lt \frac{58}{70} \Rightarrow \frac{11}{14} \lt \frac{29}{35}$.

Hence, $\frac{11}{14} \lt \frac{29}{35}$.

(iv) $\frac{13}{27}$ and $\frac{15}{48}$

By cross multiplication,

13 × 48 = 624 and 27 × 15 = 405

Since 624 > 405, therefore, $\frac{13}{27} \gt \frac{15}{48}$.

Hence, $\frac{13}{27} \gt \frac{15}{48}$.

(i) $\frac{5}{17}, \frac{4}{9}, \frac{7}{12}$

LCM of 17, 9 and 12 = 612.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{5}{17} = \frac{5 \times 36}{17 \times 36} = \frac{180}{612} \\[1em] \Rightarrow \frac{4}{9} = \frac{4 \times 68}{9 \times 68} = \frac{272}{612} \\[1em] \Rightarrow \frac{7}{12} = \frac{7 \times 51}{12 \times 51} = \frac{357}{612}\end{aligned}$$

As 357 > 272 > 180,

$\frac{357}{612} \gt \frac{272}{612} \gt \frac{180}{612} \Rightarrow \frac{7}{12} \gt \frac{4}{9} \gt \frac{5}{17}$.

Hence, the given fractions in descending order are $\frac{7}{12}, \frac{4}{9}, \frac{5}{17}$.

(ii) $\frac{7}{12}, \frac{11}{36}, \frac{37}{72}$

LCM of 12, 36 and 72 = 72.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{7}{12} = \frac{7 \times 6}{12 \times 6} = \frac{42}{72} \\[1em] \Rightarrow \frac{11}{36} = \frac{11 \times 2}{36 \times 2} = \frac{22}{72} \\[1em] \Rightarrow \frac{37}{72} = \frac{37}{72}\end{aligned}$$

As 42 > 37 > 22,

$\frac{42}{72} \gt \frac{37}{72} \gt \frac{22}{72} \Rightarrow \frac{7}{12} \gt \frac{37}{72} \gt \frac{11}{36}$.

Hence, the given fractions in descending order are $\frac{7}{12}, \frac{37}{72}, \frac{11}{36}$.

(i) $\frac{7}{8}, \frac{15}{16}, \frac{5}{6}$

LCM of 8, 16 and 6 = 48.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{7}{8} = \frac{7 \times 6}{8 \times 6} = \frac{42}{48} \\[1em] \Rightarrow \frac{15}{16} = \frac{15 \times 3}{16 \times 3} = \frac{45}{48} \\[1em] \Rightarrow \frac{5}{6} = \frac{5 \times 8}{6 \times 8} = \frac{40}{48}\end{aligned}$$

As 40 < 42 < 45,

$\frac{40}{48} \lt \frac{42}{48} \lt \frac{45}{48} \Rightarrow \frac{5}{6} \lt \frac{7}{8} \lt \frac{15}{16}$.

Hence, the given fractions in ascending order are $\frac{5}{6}, \frac{7}{8}, \frac{15}{16}$.

(ii) $\frac{3}{4}, \frac{15}{22}, \frac{26}{33}$

LCM of 4, 22 and 33 = 132.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{3}{4} = \frac{3 \times 33}{4 \times 33} = \frac{99}{132} \\[1em] \Rightarrow \frac{15}{22} = \frac{15 \times 6}{22 \times 6} = \frac{90}{132} \\[1em] \Rightarrow \frac{26}{33} = \frac{26 \times 4}{33 \times 4} = \frac{104}{132}\end{aligned}$$

As 90 < 99 < 104,

$\frac{90}{132} \lt \frac{99}{132} \lt \frac{104}{132} \Rightarrow \frac{15}{22} \lt \frac{3}{4} \lt \frac{26}{33}$.

Hence, the given fractions in ascending order are $\frac{15}{22}, \frac{3}{4}, \frac{26}{33}$.

(iii) $\frac{5}{12}, \frac{1}{4}, \frac{7}{8}, \frac{5}{6}$

LCM of 12, 4, 8 and 6 = 24.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{5}{12} = \frac{5 \times 2}{12 \times 2} = \frac{10}{24} \\[1em] \Rightarrow \frac{1}{4} = \frac{1 \times 6}{4 \times 6} = \frac{6}{24} \\[1em] \Rightarrow \frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24} \\[1em] \Rightarrow \frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24}\end{aligned}$$

As 6 < 10 < 20 < 21,

$\frac{6}{24} \lt \frac{10}{24} \lt \frac{20}{24} \lt \frac{21}{24} \Rightarrow \frac{1}{4} \lt \frac{5}{12} \lt \frac{5}{6} \lt \frac{7}{8}$.

Hence, the given fractions in ascending order are $\frac{1}{4}, \frac{5}{12}, \frac{5}{6}, \frac{7}{8}$.

(i) $\frac{8}{15} + \frac{3}{15}$

$$\begin{aligned}\Rightarrow \frac{8 + 3}{15}\\[1em] \Rightarrow \frac{11}{15}\end{aligned}$$

Hence, $\frac{8}{15} + \frac{3}{15} = \frac{11}{15}$.

(ii) $\frac{12}{15} - \frac{7}{15}$

$$\begin{aligned}\Rightarrow \frac{12 - 7}{15}\\[1em] \Rightarrow \frac{5}{15}\\[1em] \Rightarrow \frac{1}{3}\end{aligned}$$

Hence, $\frac{12}{15} - \frac{7}{15} = \frac{1}{3}$.

(iii) $1 - \frac{2}{3}$

$$\begin{aligned}\Rightarrow \frac{3}{3} - \frac{2}{3}\\[1em] \Rightarrow \frac{3 - 2}{3}\\[1em] \Rightarrow \frac{1}{3}\end{aligned}$$

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

(iv) $\frac{7}{13} + \frac{2}{13} - \frac{5}{13}$

$$\begin{aligned}\Rightarrow \frac{7 + 2 - 5}{13}\\[1em] \Rightarrow \frac{4}{13}\end{aligned}$$

Hence, $\frac{7}{13} + \frac{2}{13} - \frac{5}{13} = \frac{4}{13}$.

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

$$\begin{aligned}\Rightarrow \frac{14}{5} + \frac{18}{5}\\[1em] \Rightarrow \frac{14 + 18}{5}\\[1em] \Rightarrow \frac{32}{5}\\[1em] \Rightarrow 6\frac{2}{5}\end{aligned}$$

Hence, $2\frac{4}{5} + 3\frac{3}{5} = 6\frac{2}{5}$.

(vi) $3\frac{2}{7} - 1\frac{4}{7}$

$$\begin{aligned}\Rightarrow \frac{23}{7} - \frac{11}{7}\\[1em] \Rightarrow \frac{23 - 11}{7}\\[1em] \Rightarrow \frac{12}{7}\\[1em] \Rightarrow 1\frac{5}{7}\end{aligned}$$

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

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

LCM of 3 and 4 = 12.

$$\begin{aligned}\Rightarrow \frac{2 \times 4}{3 \times 4} + \frac{3 \times 3}{4 \times 3}\\[1em] \Rightarrow \frac{8}{12} + \frac{9}{12}\\[1em] \Rightarrow \frac{8 + 9}{12}\\[1em] \Rightarrow \frac{17}{12}\\[1em] \Rightarrow 1\frac{5}{12}\end{aligned}$$

Hence, $\frac{2}{3} + \frac{3}{4} = 1\frac{5}{12}$.

(ii) $\frac{5}{7} - \frac{4}{9}$

LCM of 7 and 9 = 63.

$$\begin{aligned}\Rightarrow \frac{5 \times 9}{7 \times 9} - \frac{4 \times 7}{9 \times 7}\\[1em] \Rightarrow \frac{45}{63} - \frac{28}{63}\\[1em] \Rightarrow \frac{45 - 28}{63}\\[1em] \Rightarrow \frac{17}{63}\end{aligned}$$

Hence, $\frac{5}{7} - \frac{4}{9} = \frac{17}{63}$.

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

LCM of 2 and 5 = 10.

$$\begin{aligned}\Rightarrow \frac{1 \times 5}{2 \times 5} + \frac{3 \times 2}{5 \times 2}\\[1em] \Rightarrow \frac{5}{10} + \frac{6}{10}\\[1em] \Rightarrow \frac{5 + 6}{10}\\[1em] \Rightarrow \frac{11}{10}\\[1em] \Rightarrow 1\frac{1}{10}\end{aligned}$$

Hence, $\frac{1}{2} + \frac{3}{5} = 1\frac{1}{10}$.

(iv) $1\frac{4}{9} + 3\frac{5}{12}$

LCM of 9 and 12 = 36.

$$\begin{aligned}\Rightarrow \frac{13}{9} + \frac{41}{12}\\[1em] \Rightarrow \frac{13 \times 4}{9 \times 4} + \frac{41 \times 3}{12 \times 3}\\[1em] \Rightarrow \frac{52}{36} + \frac{123}{36}\\[1em] \Rightarrow \frac{52 + 123}{36}\\[1em] \Rightarrow \frac{175}{36}\\[1em] \Rightarrow 4\frac{31}{36}\end{aligned}$$

Hence, $1\frac{4}{9} + 3\frac{5}{12} = 4\frac{31}{36}$.

(v) $2\frac{1}{4} - 1\frac{7}{10}$

LCM of 4 and 10 = 20.

$$\begin{aligned}\Rightarrow \frac{9}{4} - \frac{17}{10}\\[1em] \Rightarrow \frac{9 \times 5}{4 \times 5} - \frac{17 \times 2}{10 \times 2}\\[1em] \Rightarrow \frac{45}{20} - \frac{34}{20}\\[1em] \Rightarrow \frac{45 - 34}{20}\\[1em] \Rightarrow \frac{11}{20}\end{aligned}$$

Hence, $2\frac{1}{4} - 1\frac{7}{10} = \frac{11}{20}$.

(vi) $3\frac{5}{6} - 2\frac{7}{15}$

LCM of 6 and 15 = 30.

$$\begin{aligned}\Rightarrow \frac{23}{6} - \frac{37}{15}\\[1em] \Rightarrow \frac{23 \times 5}{6 \times 5} - \frac{37 \times 2}{15 \times 2}\\[1em] \Rightarrow \frac{115}{30} - \frac{74}{30}\\[1em] \Rightarrow \frac{115 - 74}{30}\\[1em] \Rightarrow \frac{41}{30}\\[1em] \Rightarrow 1\frac{11}{30}\end{aligned}$$

Hence, $3\frac{5}{6} - 2\frac{7}{15} = 1\frac{11}{30}$.

(i) $1\frac{2}{3} + 2\frac{1}{2} + \frac{3}{4}$

LCM of 3, 2 and 4 = 12.

$$\begin{aligned}\Rightarrow \frac{5}{3} + \frac{5}{2} + \frac{3}{4}\\[1em] \Rightarrow \frac{5 \times 4}{3 \times 4} + \frac{5 \times 6}{2 \times 6} + \frac{3 \times 3}{4 \times 3}\\[1em] \Rightarrow \frac{20}{12} + \frac{30}{12} + \frac{9}{12}\\[1em] \Rightarrow \frac{20 + 30 + 9}{12}\\[1em] \Rightarrow \frac{59}{12}\\[1em] \Rightarrow 4\frac{11}{12}\end{aligned}$$

Hence, $1\frac{2}{3} + 2\frac{1}{2} + \frac{3}{4} = 4\frac{11}{12}$.

(ii) $3\frac{2}{9} + 2\frac{1}{3} + 2\frac{7}{12}$

LCM of 9, 3 and 12 = 36.

$$\begin{aligned}\Rightarrow \frac{29}{9} + \frac{7}{3} + \frac{31}{12}\\[1em] \Rightarrow \frac{29 \times 4}{9 \times 4} + \frac{7 \times 12}{3 \times 12} + \frac{31 \times 3}{12 \times 3}\\[1em] \Rightarrow \frac{116}{36} + \frac{84}{36} + \frac{93}{36}\\[1em] \Rightarrow \frac{116 + 84 + 93}{36}\\[1em] \Rightarrow \frac{293}{36}\\[1em] \Rightarrow 8\frac{5}{36}\end{aligned}$$

Hence, $3\frac{2}{9} + 2\frac{1}{3} + 2\frac{7}{12} = 8\frac{5}{36}$.

(iii) $\frac{7}{12} + \frac{8}{9} - \frac{5}{6}$

LCM of 12, 9 and 6 = 36.

$$\begin{aligned}\Rightarrow \frac{7 \times 3}{12 \times 3} + \frac{8 \times 4}{9 \times 4} - \frac{5 \times 6}{6 \times 6}\\[1em] \Rightarrow \frac{21}{36} + \frac{32}{36} - \frac{30}{36}\\[1em] \Rightarrow \frac{21 + 32 - 30}{36}\\[1em] \Rightarrow \frac{23}{36}\end{aligned}$$

Hence, $\frac{7}{12} + \frac{8}{9} - \frac{5}{6} = \frac{23}{36}$.

(iv) $1\frac{3}{25} + \frac{7}{20} - \frac{2}{5}$

LCM of 25, 20 and 5 = 100.

$$\begin{aligned}\Rightarrow \frac{28}{25} + \frac{7}{20} - \frac{2}{5}\\[1em] \Rightarrow \frac{28 \times 4}{25 \times 4} + \frac{7 \times 5}{20 \times 5} - \frac{2 \times 20}{5 \times 20}\\[1em] \Rightarrow \frac{112}{100} + \frac{35}{100} - \frac{40}{100}\\[1em] \Rightarrow \frac{112 + 35 - 40}{100}\\[1em] \Rightarrow \frac{107}{100}\\[1em] \Rightarrow 1\frac{7}{100}\end{aligned}$$

Hence, $1\frac{3}{25} + \frac{7}{20} - \frac{2}{5} = 1\frac{7}{100}$.

(v) $1\frac{13}{14} - 2\frac{5}{6} + 1\frac{6}{7}$

LCM of 14, 6 and 7 = 42.

$$\begin{aligned}\Rightarrow \frac{27}{14} - \frac{17}{6} + \frac{13}{7}\\[1em] \Rightarrow \frac{27 \times 3}{14 \times 3} - \frac{17 \times 7}{6 \times 7} + \frac{13 \times 6}{7 \times 6}\\[1em] \Rightarrow \frac{81}{42} - \frac{119}{42} + \frac{78}{42}\\[1em] \Rightarrow \frac{81 - 119 + 78}{42}\\[1em] \Rightarrow \frac{40}{42}\\[1em] \Rightarrow \frac{20}{21}\end{aligned}$$

Hence, $1\frac{13}{14} - 2\frac{5}{6} + 1\frac{6}{7} = \frac{20}{21}$.

(vi) $3 - 1\frac{1}{6} - \frac{7}{15}$

LCM of 1, 6 and 15 = 30.

$$\begin{aligned}\Rightarrow \frac{3}{1} - \frac{7}{6} - \frac{7}{15}\\[1em] \Rightarrow \frac{3 \times 30}{1 \times 30} - \frac{7 \times 5}{6 \times 5} - \frac{7 \times 2}{15 \times 2}\\[1em] \Rightarrow \frac{90}{30} - \frac{35}{30} - \frac{14}{30}\\[1em] \Rightarrow \frac{90 - 35 - 14}{30}\\[1em] \Rightarrow \frac{41}{30}\\[1em] \Rightarrow 1\frac{11}{30}\end{aligned}$$

Hence, $3 - 1\frac{1}{6} - \frac{7}{15} = 1\frac{11}{30}$.

(i) Required number = $2\frac{3}{8} - \frac{5}{12}$

LCM of 8 and 12 = 24.

$$\begin{aligned}\Rightarrow \frac{19}{8} - \frac{5}{12}\\[1em] \Rightarrow \frac{19 \times 3}{8 \times 3} - \frac{5 \times 2}{12 \times 2}\\[1em] \Rightarrow \frac{57}{24} - \frac{10}{24}\\[1em] \Rightarrow \frac{57 - 10}{24}\\[1em] \Rightarrow \frac{47}{24}\\[1em] \Rightarrow 1\frac{23}{24}\end{aligned}$$

Hence, the required number = $1\frac{23}{24}$.

(ii) Required number = $5 - 1\frac{5}{13}$

$$\begin{aligned}\Rightarrow \frac{5}{1} - \frac{18}{13}\\[1em] \Rightarrow \frac{5 \times 13}{1 \times 13} - \frac{18}{13}\\[1em] \Rightarrow \frac{65}{13} - \frac{18}{13}\\[1em] \Rightarrow \frac{65 - 18}{13}\\[1em] \Rightarrow \frac{47}{13}\\[1em] \Rightarrow 3\frac{8}{13}\end{aligned}$$

Hence, the required number = $3\frac{8}{13}$.

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

$$\begin{aligned}\Rightarrow \frac{2 \times 3}{5 \times 7}\\[1em] \Rightarrow \frac{6}{35}\end{aligned}$$

Hence, $\frac{2}{5} \times \frac{3}{7} = \frac{6}{35}$.

(ii) $\frac{3}{5} \times \frac{8}{9}$

$$\begin{aligned}\Rightarrow \frac{3 \times 8}{5 \times 9}\\[1em] \Rightarrow \frac{3}{9} \times \frac{8}{5}\\[1em] \Rightarrow \frac{1}{3} \times \frac{8}{5}\\[1em] \Rightarrow \frac{1 \times 8}{3 \times 5}\\[1em] \Rightarrow \frac{8}{15}\end{aligned}$$

Hence, $\frac{3}{5} \times \frac{8}{9} = \frac{8}{15}$.

(iii) $7 \times 1\frac{2}{3}$

$$\begin{aligned}\Rightarrow 7 \times \frac{5}{3}\\[1em] \Rightarrow \frac{7 \times 5}{3}\\[1em] \Rightarrow \frac{35}{3}\\[1em] \Rightarrow 11\frac{2}{3}\end{aligned}$$

Hence, $7 \times 1\frac{2}{3} = 11\frac{2}{3}$.

(i) $\frac{2}{3} \times 60$

$$\begin{aligned}\Rightarrow \frac{2 \times 60}{3}\\[1em] \Rightarrow 2 \times 20\\[1em] \Rightarrow 40\end{aligned}$$

Hence, $\frac{2}{3} \times 60 = 40$.

(ii) $\frac{4}{7} \times 280$

$$\begin{aligned}\Rightarrow \frac{4 \times 280}{7}\\[1em] \Rightarrow 4 \times 40\\[1em] \Rightarrow 160\end{aligned}$$

Hence, $\frac{4}{7} \times 280 = 160$.

(iii) $\frac{2}{3}$ of $1\frac{4}{9}$

$$\begin{aligned}\Rightarrow \frac{2}{3} \times 1\frac{4}{9}\\[1em] \Rightarrow \frac{2}{3} \times \frac{13}{9}\\[1em] \Rightarrow \frac{2 \times 13}{3 \times 9}\\[1em] \Rightarrow \frac{26}{27}\end{aligned}$$

Hence, $\frac{2}{3}$ of $1\frac{4}{9} = \frac{26}{27}$.

(i) $\frac{9}{13}$

The reciprocal of a fraction $\frac{a}{b}$ is $\frac{b}{a}$.

So, the reciprocal of $\frac{9}{13}$ is $\frac{13}{9}$.

Hence, the reciprocal of $\frac{9}{13}$ is $\frac{13}{9}$.

(ii) $2\frac{3}{8}$

First, convert the mixed fraction into improper fraction.

$\Rightarrow 2\frac{3}{8} = \frac{2 \times 8 + 3}{8} = \frac{19}{8}$

The reciprocal of $\frac{19}{8}$ is $\frac{8}{19}$.

Hence, the reciprocal of $2\frac{3}{8}$ is $\frac{8}{19}$.

(i) $\frac{8}{21} \div 4$

$$\begin{aligned}\Rightarrow \frac{8}{21} \div \frac{4}{1}\\[1em] \Rightarrow \frac{8}{21} \times \frac{1}{4}\\[1em] \Rightarrow \frac{8}{4} \times \frac{1}{21}\\[1em] \Rightarrow 2 \times \frac{1}{21}\\[1em] \Rightarrow \frac{2}{21}\end{aligned}$$

Hence, $\frac{8}{21} \div 4 = \frac{2}{21}$.

(ii) $\frac{4}{15} \div \frac{2}{5}$

$$\begin{aligned}\Rightarrow \frac{4}{15} \times \frac{5}{2}\\[1em] \Rightarrow \frac{4}{2} \times \frac{5}{15}\\[1em] \Rightarrow 2 \times \frac{1}{3}\\[1em] \Rightarrow \frac{2}{3}\end{aligned}$$

Hence, $\frac{4}{15} \div \frac{2}{5} = \frac{2}{3}$.

(iii) $8 \div \frac{5}{6}$

$$\begin{aligned}\Rightarrow \frac{8}{1} \div \frac{5}{6}\\[1em] \Rightarrow \frac{8}{1} \times \frac{6}{5}\\[1em] \Rightarrow \frac{8 \times 6}{1 \times 5}\\[1em] \Rightarrow \frac{48}{5}\\[1em] \Rightarrow 9\frac{3}{5}\end{aligned}$$

Hence, $8 \div \frac{5}{6} = 9\frac{3}{5}$.

(iv) $5\frac{1}{4} \div \frac{7}{8}$

$$\begin{aligned}\Rightarrow \frac{21}{4} \div \frac{7}{8}\\[1em] \Rightarrow \frac{21}{4} \times \frac{8}{7}\\[1em] \Rightarrow \frac{21}{7} \times \frac{8}{4}\\[1em] \Rightarrow 3 \times 2\\[1em] \Rightarrow 6\end{aligned}$$

Hence, $5\frac{1}{4} \div \frac{7}{8} = 6$.

(v) $5\frac{1}{3} \div 1\frac{1}{9}$

$$\begin{aligned}\Rightarrow \frac{16}{3} \div \frac{10}{9}\\[1em] \Rightarrow \frac{16}{3} \times \frac{9}{10}\\[1em] \Rightarrow \frac{16}{10} \times \frac{9}{3}\\[1em] \Rightarrow \frac{8}{5} \times 3\\[1em] \Rightarrow \frac{24}{5}\\[1em] \Rightarrow 4\frac{4}{5}\end{aligned}$$

Hence, $5\frac{1}{3} \div 1\frac{1}{9} = 4\frac{4}{5}$.

(i) $\frac{2}{11}$ of ₹ 715

$$\begin{aligned}\Rightarrow \frac{2}{11} \times 715\\[1em] \Rightarrow \frac{2 \times 715}{11}\\[1em] \Rightarrow 2 \times 65\\[1em] \Rightarrow ₹130\end{aligned}$$

Hence, $\frac{2}{11}$ of ₹ 715 = ₹ 130.

(ii) $\frac{4}{9}$ of 405 kg

$$\begin{aligned}\Rightarrow \frac{4}{9} \times 405\\[1em] \Rightarrow \frac{4 \times 405}{9}\\[1em] \Rightarrow 4 \times 45\\[1em] \Rightarrow 180 \text{ kg}\end{aligned}$$

Hence, $\frac{4}{9}$ of 405 kg = 180 kg.

(iii) $\frac{3}{7}$ of 4 hours 40 minutes

4 hours 40 minutes = (4 × 60 + 40) minutes = (240 + 40) minutes = 280 minutes.

$$\begin{aligned}\Rightarrow \frac{3}{7} \times 280 \text{ minutes}\\[1em] \Rightarrow \frac{3 \times 280}{7} \text{ minutes}\\[1em] \Rightarrow 3 \times 40 \text{ minutes}\\[1em] \Rightarrow 120 \text{ minutes}\\[1em] \Rightarrow 2 \text{ hours}\end{aligned}$$

Hence, $\frac{3}{7}$ of 4 hours 40 minutes = 2 hours.

(i) 9³

$$\begin{aligned}\Rightarrow 9 \times 9 \times 9\\[1em] \Rightarrow 729\end{aligned}$$

Hence, 9³ = 729.

(ii) 8⁴

$$\begin{aligned}\Rightarrow 8 \times 8 \times 8 \times 8\\[1em] \Rightarrow 4096\end{aligned}$$

Hence, 8⁴ = 4096.

(iii) $\left(\frac{3}{5}\right)^3$

$$\begin{aligned}\Rightarrow \frac{3}{5} \times \frac{3}{5} \times \frac{3}{5}\\[1em] \Rightarrow \frac{3 \times 3 \times 3}{5 \times 5 \times 5}\\[1em] \Rightarrow \frac{27}{125}\end{aligned}$$

Hence, $\left(\frac{3}{5}\right)^3 = \frac{27}{125}$.

(iv) 3³ × 5²

$$\begin{aligned}\Rightarrow (3 \times 3 \times 3) \times (5 \times 5)\\[1em] \Rightarrow 27 \times 25\\[1em] \Rightarrow 675\end{aligned}$$

Hence, 3³ × 5² = 675.

Length of ribbon bought by Nayamat = $\frac{2}{5}$ metre.

Length of ribbon bought by Amanat = $\frac{3}{4}$ metre.

Total length of ribbon = $\frac{2}{5} + \frac{3}{4}$

LCM of 5 and 4 = 20.

$$\begin{aligned}\Rightarrow \frac{2 \times 4}{5 \times 4} + \frac{3 \times 5}{4 \times 5}\\[1em] \Rightarrow \frac{8}{20} + \frac{15}{20}\\[1em] \Rightarrow \frac{8 + 15}{20}\\[1em] \Rightarrow \frac{23}{20}\\[1em] \Rightarrow 1\frac{3}{20} \text{ metre}\end{aligned}$$

Hence, the total length of ribbon they bought = $1\frac{3}{20}$ metre.

Total length of bamboo = $2\frac{3}{4}$ metre.

Length of one piece = $\frac{7}{8}$ metre.

Length of other piece = $2\frac{3}{4} - \frac{7}{8}$

LCM of 4 and 8 = 8.

$$\begin{aligned}\Rightarrow \frac{11}{4} - \frac{7}{8}\\[1em] \Rightarrow \frac{11 \times 2}{4 \times 2} - \frac{7}{8}\\[1em] \Rightarrow \frac{22}{8} - \frac{7}{8}\\[1em] \Rightarrow \frac{22 - 7}{8}\\[1em] \Rightarrow \frac{15}{8}\\[1em] \Rightarrow 1\frac{7}{8} \text{ metre}\end{aligned}$$

Hence, the length of the other piece = $1\frac{7}{8}$ metre.

Total distance from house to school = $1\frac{9}{10}$ km.

Distance travelled by bus = $1\frac{1}{2}$ km.

Distance walked = $1\frac{9}{10} - 1\frac{1}{2}$

LCM of 10 and 2 = 10.

$$\begin{aligned}\Rightarrow \frac{19}{10} - \frac{3}{2}\\[1em] \Rightarrow \frac{19}{10} - \frac{3 \times 5}{2 \times 5}\\[1em] \Rightarrow \frac{19}{10} - \frac{15}{10}\\[1em] \Rightarrow \frac{19 - 15}{10}\\[1em] \Rightarrow \frac{4}{10}\\[1em] \Rightarrow \frac{2}{5} \text{ km}\end{aligned}$$

Hence, Nidhi walked $\frac{2}{5}$ km.

Total length of rope = $20\frac{1}{2}$ m.

Length of piece cut off = $3\frac{5}{8}$ m.

Length of remaining rope = $20\frac{1}{2} - 3\frac{5}{8}$

LCM of 2 and 8 = 8.

$$\begin{aligned}\Rightarrow \frac{41}{2} - \frac{29}{8}\\[1em] \Rightarrow \frac{41 \times 4}{2 \times 4} - \frac{29}{8}\\[1em] \Rightarrow \frac{164}{8} - \frac{29}{8}\\[1em] \Rightarrow \frac{164 - 29}{8}\\[1em] \Rightarrow \frac{135}{8}\\[1em] \Rightarrow 16\frac{7}{8} \text{ m}\end{aligned}$$

Hence, the length of the remaining rope = $16\frac{7}{8}$ m.

Weights of three packets = $2\frac{3}{4}$ kg, $3\frac{1}{3}$ kg and $5\frac{2}{5}$ kg.

Total weight = $2\frac{3}{4} + 3\frac{1}{3} + 5\frac{2}{5}$

LCM of 4, 3 and 5 = 60.

$$\begin{aligned}\Rightarrow \frac{11}{4} + \frac{10}{3} + \frac{27}{5}\\[1em] \Rightarrow \frac{11 \times 15}{4 \times 15} + \frac{10 \times 20}{3 \times 20} + \frac{27 \times 12}{5 \times 12}\\[1em] \Rightarrow \frac{165}{60} + \frac{200}{60} + \frac{324}{60}\\[1em] \Rightarrow \frac{165 + 200 + 324}{60}\\[1em] \Rightarrow \frac{689}{60}\\[1em] \Rightarrow 11\frac{29}{60} \text{ kg}\end{aligned}$$

Hence, the total weight of the three packets = $11\frac{29}{60}$ kg.

Total pages of book = 100.

Fraction of book read by Shivani = $\frac{25}{100} = \frac{1}{4}$.

Fraction of book read by Nandini = $\frac{2}{5}$.

Now, compare $\frac{1}{4}$ and $\frac{2}{5}$.

By cross multiplication,

1 × 5 = 5 and 4 × 2 = 8.

Since 5 < 8, therefore, $\frac{1}{4} \lt \frac{2}{5}$.

So, Shivani read less than Nandini.

Hence, Shivani read less.

Time for which Ajay exercised = $\frac{3}{6}$ hour.

Time for which Vijay exercised = $\frac{3}{4}$ hour.

Compare $\frac{3}{6}$ and $\frac{3}{4}$.

These are unlike fractions with same numerator 3.

In fractions with same numerator, the fraction with smaller denominator is greater.

Since 4 < 6, therefore, $\frac{3}{4} \gt \frac{3}{6}$.

So, Vijay exercised for a longer time.

Difference = $\frac{3}{4} - \frac{3}{6}$

LCM of 4 and 6 = 12.

$$\begin{aligned}\Rightarrow \frac{3 \times 3}{4 \times 3} - \frac{3 \times 2}{6 \times 2}\\[1em] \Rightarrow \frac{9}{12} - \frac{6}{12}\\[1em] \Rightarrow \frac{9 - 6}{12}\\[1em] \Rightarrow \frac{3}{12}\\[1em] \Rightarrow \frac{1}{4} \text{ hour}\end{aligned}$$

Hence, Vijay exercised for a longer time by $\frac{1}{4}$ hour.

(i) A fraction is a number which represent a part of whole.

(ii) A proper fraction lies between 0 and 1.

(iii) A mixed fraction can be converted into an improper fraction.

(iv) Fractions having different denominators are called unlike fractions.

(v) Check by cross multiplication: 18 × 540 = 9720 and 135 × 72 = 9720.

Since cross products are equal, the fractions are equivalent.

So, $\frac{18}{135}$ and $\frac{72}{540}$ are proper, unlike and equivalent fractions.

(vi) In two like fractions, the fraction having smaller numerator is smaller.

(vii) $\frac{144}{180}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{144}{180} = \frac{2 \times 2 \times 2 \times 2 \times 3 \times 3}{2 \times 2 \times 3 \times 3 \times 5}\\[1em] = \frac{2 \times 2}{5} \\[1em] = \frac{4}{5}.\end{aligned}$$

So, $\frac{144}{180}$ reduced to simplest form is $\frac{4}{5}$.

(viii) $7\frac{2}{5}$ + .... = 12

Required number = $12 - 7\frac{2}{5} = \frac{12}{1} - \frac{37}{5} = \frac{60}{5} - \frac{37}{5} = \frac{60 - 37}{5} = \frac{23}{5} = 4\frac{3}{5}$.

So, $7\frac{2}{5}$ + $4\frac{3}{5}$ = 12.

(ix) $\frac{42}{56} = \frac{6}{...}$

To get 6 from 42, we have to divide 42 by 7. So, divide 56 by 7.

$\frac{42}{56} = \frac{42 \div 7}{56 \div 7} = \frac{6}{8}$.

So, $\frac{42}{56} = \frac{6}{\textbf{8}}$.

(i) False.

Reason: Two fractions with same denominator are called like fractions, not same numerator.

(ii) True.

Reason: A fraction in which the numerator is greater than (or equal to) its denominator is called an improper fraction.

(iii) True.

Reason: Every improper fraction can be converted into a mixed fraction by dividing the numerator by the denominator.

(iv) True.

Reason: Every fraction has a unique position on the number line.

(v) False.

Reason: In two unlike fractions with same numerator, the fraction having smaller denominator is greater (not greater denominator).

(vi) False.

Reason: $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ have different denominators (2, 3, 4), so they are unlike fractions, not like fractions.

Total parts = 7

Shaded parts = 3

Fraction = $\frac{\text{No. of shaded parts}}{\text{Total parts}} = \frac{3}{7}$.

Hence, option 2 is the correct option.

Total parts = 12

Shaded parts = $1 + \frac{1}{2} + \frac{1}{2} = \frac{5}{2}$

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}} = \frac{\frac{5}{2}}{12} = \frac{5}{2 \times 12} = \frac{5}{24}$.

Hence, option 3 is the correct option.

The fraction $\frac{5}{7}$ is a proper fraction (numerator < denominator).

All proper fractions lie between 0 and 1.

Hence, option 4 is the correct option.

Check each option using cross multiplication:

1. 3 × 20 = 60 and 4 × 15 = 60. Equal, so equivalent.

2. 14 × 6 = 84 and 21 × 4 = 84. Equal, so equivalent.

3. 8 × 15 = 120 and 10 × 12 = 120. Equal, so equivalent.

4. 6 × 25 = 150 and 14 × 10 = 140. Not equal, so not equivalent.

Hence, option 4 is the correct option.

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{45}{81} = \frac{3 \times 3 \times 5}{3 \times 3 \times 3 \times 3}\\[1em] = \frac{5}{3 \times 3}\\[1em] = \frac{5}{9}.\end{aligned}$$

Hence, option 4 is the correct option.

Check each option using cross multiplication with $\frac{4}{5}$:

1. 4 × 50 = 200 and 5 × 40 = 200. Equal, so equivalent.

2. 4 × 15 = 60 and 5 × 9 = 45. Not equal, so not equivalent.

3. 4 × 15 = 60 and 5 × 12 = 60. Equal, so equivalent.

4. 4 × 40 = 160 and 5 × 32 = 160. Equal, so equivalent.

Hence, option 2 is the correct option.

A fraction is in simplest form if HCF of numerator and denominator is 1.

1. HCF of 27 and 28 = 1. So, simplest form.

2. HCF of 13 and 33 = 1. So, simplest form.

3. HCF of 39 and 87 = 3. So, not in simplest form.

$\frac{39}{87} = \frac{39 \div 3}{87 \div 3} = \frac{13}{29}$.

4. HCF of 14 and 9 = 1. So, simplest form.

Hence, option 3 is the correct option.

Like fractions have the same denominator.

In option 2, $\frac{3}{7}$ and $\frac{16}{7}$ have the same denominator 7.

Hence, option 2 is the correct option.

These are unlike fractions with same numerator 5.

In fractions with same numerator, the fraction with smaller denominator is greater.

The smallest denominator is 6.

So, $\frac{5}{6}$ is the greatest.

Hence, option 1 is the correct option.

These are unlike fractions with same numerator 11.

In fractions with same numerator, the fraction with greater denominator is smaller.

The greatest denominator is 10.

So, $\frac{11}{10}$ is the smallest.

Hence, option 3 is the correct option.

Check each option:

1. Cross multiplication: 1 × 14 = 14 and 7 × 3 = 21. Since 14 < 21, $\frac{1}{7} \lt \frac{3}{14}$. True.

2. Cross multiplication: 5 × 24 = 120 and 8 × 15 = 120. Equal, so $\frac{5}{8} = \frac{15}{24}$. True.

3. Cross multiplication: 3 × 16 = 48 and 4 × 6 = 24. Not equal, so $\frac{3}{4} \neq \frac{6}{16}$. False.

4. Cross multiplication: 5 × 6 = 30 and 12 × 2 = 24. Since 30 > 24, $\frac{5}{12} \gt \frac{2}{6}$. True.

Hence, option 3 is the correct option.

LCM of 7 and 14 = 14.

$$\begin{aligned}\Rightarrow \frac{1 \times 2}{7 \times 2} + \frac{4}{14}\\[1em] \Rightarrow \frac{2}{14} + \frac{4}{14}\\[1em] \Rightarrow \frac{2 + 4}{14}\\[1em] \Rightarrow \frac{6}{14}\\[1em] \Rightarrow \frac{3}{7}\end{aligned}$$

Hence, option 4 is the correct option.

LCM of 9 and 18 = 18.

$$\begin{aligned}\Rightarrow \frac{7 \times 2}{9 \times 2} - \frac{5}{18}\\[1em] \Rightarrow \frac{14}{18} - \frac{5}{18}\\[1em] \Rightarrow \frac{14 - 5}{18}\\[1em] \Rightarrow \frac{9}{18}\\[1em] \Rightarrow \frac{1}{2}\end{aligned}$$

Hence, option 3 is the correct option.

Fraction of pizza eaten = $\frac{4}{7}$.

Fraction of pizza left = $1 - \frac{4}{7}$

$$\begin{aligned}\Rightarrow \frac{7}{7} - \frac{4}{7}\\[1em] \Rightarrow \frac{7 - 4}{7}\\[1em] \Rightarrow \frac{3}{7}\end{aligned}$$

Hence, option 1 is the correct option.

Smallest odd prime number = 3.

Smallest composite number = 4.

So, the required fraction = $\frac{3}{4}$.

Hence, option 1 is the correct option.

According to Statement I, $3\frac{4}{7}$ consists of a natural number 3 and a proper fraction $\frac{4}{7}$.

So, $3\frac{4}{7}$ is a mixed fraction.

∴ Statement I is true.

According to Statement II, a natural number added to a proper fraction forms a mixed fraction (which can also be written as an improper fraction).

For example, $3 + \frac{4}{7} = 3\frac{4}{7} = \frac{25}{7}$, which is an improper fraction.

∴ Statement II is true.

Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

According to Statement I,

Total area = 1000 m²

Area used = 200 + 500 = 700 m²

Unused area = 1000 - 700 = 300 m²

Fraction of unused area = $\frac{300}{1000} = \frac{3}{10}$.

∴ Statement I is true.

According to Statement II, every natural number can be written with denominator 1.

For example, $5 = \frac{5}{1}$, $21 = \frac{21}{1}$, etc., which are improper fractions (numerator ≥ denominator).

∴ Statement II is true.

Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

According to Statement I,

Total salary = ₹50,000.

Total spent = 5000 + 6000 = ₹11,000.

Savings = 50,000 - 11,000 = ₹39,000.

Fraction of savings = $\frac{39,000}{50,000} = \frac{39}{50}$.

Reciprocal of $\frac{39}{50}$ = $\frac{50}{39}$.

∴ Statement I is true.

According to Statement II, two numbers are reciprocals of each other if their product is 1.

$\frac{50}{39} \times \frac{39}{50} = \frac{50 \times 39}{39 \times 50} = 1$.

So, $\frac{50}{39}$ and $\frac{39}{50}$ are reciprocals of each other.

∴ Statement II is true.

Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

(i) False.

Reason: $\frac{2}{3}$ is a proper fraction (numerator < denominator), so it lies between 0 and 1, not between 2 and 3.

(ii) False.

Reason: To find an equivalent fraction, we must multiply or divide the numerator and denominator by the same non-zero number, not add or subtract.

Natural numbers between 102 and 112 are:

103, 104, 105, 106, 107, 108, 109, 110, 111 and their number = 9.

Out of these, the prime numbers are:

103, 107, 109 and their number = 3.

Fraction of prime numbers = $\frac{\text{Number of primes}}{\text{Total numbers}} = \frac{3}{9} = \frac{1}{3}$.

Hence, total natural numbers = 9 and fraction of prime numbers = $\frac{1}{3}$.

To find p:

$\frac{5}{8} = \frac{20}{p}$

To get 20 from 5, we have to multiply 5 by 4.

So, multiply both numerator and denominator by 4.

$\Rightarrow \frac{5}{8} = \frac{5 \times 4}{8 \times 4} = \frac{20}{32}.$

So, p = 32.

To find q:

$\frac{5}{8} = \frac{q}{104}$

To get 104 from 8, we have to multiply 8 by 13.

So, multiply both numerator and denominator by 13.

$\Rightarrow \frac{5}{8} = \frac{5 \times 13}{8 \times 13} = \frac{65}{104}.$

So, q = 65.

Hence, p = 32 and q = 65.

Reduce each fraction to its simplest form:

(i) $\frac{250}{400} = \frac{250 \div 50}{400 \div 50} = \frac{5}{8}$. Matches with (d).

(ii) $\frac{180}{200} = \frac{180 \div 20}{200 \div 20} = \frac{9}{10}$. Matches with (e).

(iii) $\frac{660}{990} = \frac{660 \div 330}{990 \div 330} = \frac{2}{3}$. Matches with (a).

(iv) $\frac{180}{360} = \frac{180 \div 180}{360 \div 180} = \frac{1}{2}$. Matches with (c).

(v) $\frac{220}{550} = \frac{220 \div 110}{550 \div 110} = \frac{2}{5}$. Matches with (b).

Hence, the matches are:

(i) ↔ (d), (ii) ↔ (e), (iii) ↔ (a), (iv) ↔ (c), (v) ↔ (b).

(i) $\frac{5}{6}$ and $\frac{13}{15}$

By cross multiplication,

5 × 15 = 75 and 6 × 13 = 78.

Since 75 < 78, therefore, $\frac{5}{6} \lt \frac{13}{15}$.

Hence, $\frac{5}{6} \lt \frac{13}{15}$.

(ii) $\frac{4}{5}$ and $\frac{7}{9}$

By cross multiplication,

4 × 9 = 36 and 5 × 7 = 35.

Since 36 > 35, therefore, $\frac{4}{5} \gt \frac{7}{9}$.

Hence, $\frac{4}{5} \gt \frac{7}{9}$.

(iii) $\frac{11}{12}$ and $\frac{13}{14}$

By cross multiplication,

11 × 14 = 154 and 12 × 13 = 156.

Since 154 < 156, therefore, $\frac{11}{12} \lt \frac{13}{14}$.

Hence, $\frac{11}{12} \lt \frac{13}{14}$.

LCM of 30, 15, 10 and 5 = 30.

Write the given fractions as equivalent like fractions.

$$\begin{aligned}\Rightarrow \frac{7}{30} = \frac{7}{30} \\[1em] \Rightarrow \frac{13}{15} = \frac{13 \times 2}{15 \times 2} = \frac{26}{30} \\[1em] \Rightarrow \frac{9}{10} = \frac{9 \times 3}{10 \times 3} = \frac{27}{30} \\[1em] \Rightarrow \frac{3}{5} = \frac{3 \times 6}{5 \times 6} = \frac{18}{30}\end{aligned}$$

As 27 > 26 > 18 > 7,

$\frac{27}{30} \gt \frac{26}{30} \gt \frac{18}{30} \gt \frac{7}{30} \Rightarrow \frac{9}{10} \gt \frac{13}{15} \gt \frac{3}{5} \gt \frac{7}{30}$.

Hence, the given fractions in descending order are $\frac{9}{10}, \frac{13}{15}, \frac{3}{5}, \frac{7}{30}$.

LCM of 2, 4 and 6 = 12.

$$\begin{aligned}\Rightarrow \frac{5}{2} - \frac{13}{4} + \frac{35}{6}\\[1em] \Rightarrow \frac{5 \times 6}{2 \times 6} - \frac{13 \times 3}{4 \times 3} + \frac{35 \times 2}{6 \times 2}\\[1em] \Rightarrow \frac{30}{12} - \frac{39}{12} + \frac{70}{12}\\[1em] \Rightarrow \frac{30 - 39 + 70}{12}\\[1em] \Rightarrow \frac{61}{12}\\[1em] \Rightarrow 5\frac{1}{12}\end{aligned}$$

Hence, $2\frac{1}{2} - 3\frac{1}{4} + 5\frac{5}{6} = 5\frac{1}{12}$.

(i) $\frac{3}{5} \times 180$

$$\begin{aligned}\Rightarrow \frac{3 \times 180}{5}\\[1em] \Rightarrow 3 \times 36\\[1em] \Rightarrow 108\end{aligned}$$

Hence, $\frac{3}{5} \times 180 = 108$.

(ii) $\frac{3}{7}$ of $5\frac{5}{6}$

$$\begin{aligned}\Rightarrow \frac{3}{7} \times \frac{35}{6}\\[1em] \Rightarrow \frac{3}{6} \times \frac{35}{7}\\[1em] \Rightarrow \frac{1}{2} \times 5\\[1em] \Rightarrow \frac{5}{2}\\[1em] \Rightarrow 2\frac{1}{2}\end{aligned}$$

Hence, $\frac{3}{7}$ of $5\frac{5}{6} = 2\frac{1}{2}$.

(iii) $\frac{5}{18} \div \frac{2}{3}$

$$\begin{aligned}\Rightarrow \frac{5}{18} \times \frac{3}{2}\\[1em] \Rightarrow \frac{5}{2} \times \frac{3}{18}\\[1em] \Rightarrow \frac{5}{2} \times \frac{1}{6}\\[1em] \Rightarrow \frac{5}{12}\end{aligned}$$

Hence, $\frac{5}{18} \div \frac{2}{3} = \frac{5}{12}$.

Asha's shelf = $\frac{5}{6}$ full.

Samuel's shelf = $\frac{3}{5}$ full.

Compare $\frac{5}{6}$ and $\frac{3}{5}$ by cross multiplication:

5 × 5 = 25 and 6 × 3 = 18.

Since 25 > 18, therefore, $\frac{5}{6} \gt \frac{3}{5}$.

So, Asha's bookshelf is more full.

Difference = $\frac{5}{6} - \frac{3}{5}$

LCM of 6 and 5 = 30.

$$\begin{aligned}\Rightarrow \frac{5 \times 5}{6 \times 5} - \frac{3 \times 6}{5 \times 6}\\[1em] \Rightarrow \frac{25}{30} - \frac{18}{30}\\[1em] \Rightarrow \frac{25 - 18}{30}\\[1em] \Rightarrow \frac{7}{30}\end{aligned}$$

Hence, Asha's bookshelf is more full by $\frac{7}{30}$.

Fraction of land used for wheat crop = $\frac{4}{5}$.

Fraction of land used for cereal crop = $\frac{1}{7}$.

Total fraction used = $\frac{4}{5} + \frac{1}{7}$

LCM of 5 and 7 = 35.

$$\begin{aligned}\Rightarrow \frac{4 \times 7}{5 \times 7} + \frac{1 \times 5}{7 \times 5}\\[1em] \Rightarrow \frac{28}{35} + \frac{5}{35}\\[1em] \Rightarrow \frac{28 + 5}{35}\\[1em] \Rightarrow \frac{33}{35}\end{aligned}$$

Fraction of land available for other crops = $1 - \frac{33}{35}$

$$\begin{aligned}\Rightarrow \frac{35}{35} - \frac{33}{35}\\[1em] \Rightarrow \frac{35 - 33}{35}\\[1em] \Rightarrow \frac{2}{35}\end{aligned}$$

Hence, the fraction of land available for other crops = $\frac{2}{35}$.

Let the total number of parts be x.

Given, 16 parts represent the fraction $\frac{2}{5}$.

$$\begin{aligned}\Rightarrow \frac{16}{x} = \frac{2}{5}\\[1em] \Rightarrow 16 \times 5 = 2 \times x\\[1em] \Rightarrow 80 = 2x\\[1em] \Rightarrow x = \frac{80}{2}\\[1em] \Rightarrow x = 40\end{aligned}$$

Hence, the rectangle is divided into 40 equal parts.

(i) 5⁵

$$\begin{aligned}\Rightarrow 5 \times 5 \times 5 \times 5 \times 5\\[1em] \Rightarrow 3125\end{aligned}$$

Hence, 5⁵ = 3125.

(ii) $\left(\frac{3}{4}\right)^4$

$$\begin{aligned}\Rightarrow \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}\\[1em] \Rightarrow \frac{3 \times 3 \times 3 \times 3}{4 \times 4 \times 4 \times 4}\\[1em] \Rightarrow \frac{81}{256}\end{aligned}$$

Hence, $\left(\frac{3}{4}\right)^4 = \frac{81}{256}$.

A proper fraction has numerator less than denominator.

We need pairs of (numerator, denominator) such that numerator + denominator = 12 and numerator < denominator.

Numerator + denominator = 12 with numerator < denominator means numerator < 6.

The possible fractions are:

If numerator = 1, denominator = 11 → $\frac{1}{11}$.

If numerator = 2, denominator = 10 → $\frac{2}{10}$.

If numerator = 3, denominator = 9 → $\frac{3}{9}$.

If numerator = 4, denominator = 8 → $\frac{4}{8}$.

If numerator = 5, denominator = 7 → $\frac{5}{7}$.

Hence, the proper fractions are: $\frac{1}{11}, \frac{2}{10}, \frac{3}{9}, \frac{4}{8}$ and $\frac{5}{7}$.

$$\begin{aligned}\Rightarrow 4\frac{x}{12} + 1\frac{1}{3} = 5\frac{3}{4}\\[1em] \Rightarrow 4\frac{x}{12} = 5\frac{3}{4} - 1\frac{1}{3}\\[1em] \Rightarrow 4\frac{x}{12} = \frac{23}{4} - \frac{4}{3}\end{aligned}$$

LCM of 4 and 3 = 12.

$$\begin{aligned}\Rightarrow 4\frac{x}{12} = \frac{23 \times 3}{4 \times 3} - \frac{4 \times 4}{3 \times 4}\\[1em] \Rightarrow 4\frac{x}{12} = \frac{69}{12} - \frac{16}{12}\\[1em] \Rightarrow 4\frac{x}{12} = \frac{69 - 16}{12}\\[1em] \Rightarrow 4\frac{x}{12} = \frac{53}{12}\\[1em] \Rightarrow 4\frac{x}{12} = 4\frac{5}{12}\end{aligned}$$

Comparing both sides, x = 5.

Hence, x = 5.

Total number of students = 540.

From the figure, the fractions are:

Orange juice = $\frac{1}{4}$, Milk = $\frac{1}{3}$, Tea = $\frac{1}{12}$, Coffee = $\frac{1}{6}$, Mango shake = $\frac{1}{6}$.

(i) Number of students who prefer coffee = $\frac{1}{6} \times 540$

$$\begin{aligned}\Rightarrow \frac{540}{6}\\[1em] \Rightarrow 90\end{aligned}$$

Hence, the number of students who prefer coffee = 90.

(ii) Number of milk drinkers = $\frac{1}{3} \times 540 = \frac{540}{3} = 180$.

Number of orange juice drinkers = $\frac{1}{4} \times 540 = \frac{540}{4} = 135$.

Since 180 > 135, milk drinkers are more than orange juice drinkers.

Difference = 180 - 135 = 45.

Hence, milk drinkers are greater than orange juice drinkers by 45.

(iii) Number of mango shake drinkers = $\frac{1}{6} \times 540 = 90$.

Number of coffee drinkers = 90.

Total = 90 + 90 = 180.

Number of milk drinkers = 180.

Yes, total of mango shake and coffee drinkers (180) is equal to milk drinkers (180).

Hence, the total number of students who drink mango shake or coffee = 180, which is equal to the number of milk drinkers.

(iv) Sum of all fractions = $\frac{1}{4} + \frac{1}{3} + \frac{1}{12} + \frac{1}{6} + \frac{1}{6}$

LCM of 4, 3, 12, 6 = 12.

$$\begin{aligned}\Rightarrow \frac{1 \times 3}{4 \times 3} + \frac{1 \times 4}{3 \times 4} + \frac{1}{12} + \frac{1 \times 2}{6 \times 2} + \frac{1 \times 2}{6 \times 2}\\[1em] \Rightarrow \frac{3}{12} + \frac{4}{12} + \frac{1}{12} + \frac{2}{12} + \frac{2}{12}\\[1em] \Rightarrow \frac{3 + 4 + 1 + 2 + 2}{12}\\[1em] \Rightarrow \frac{12}{12}\\[1em] \Rightarrow 1\end{aligned}$$

Hence, yes, the sum of all fractions in the given figure is equal to 1.