Fractions

(i) Total parts = 8

Shaded parts = 5

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}}$

= $\frac{5}{8}$.

Hence, fraction = $\frac{5}{8}$.

(ii) Total parts = 12

Shaded parts = 7

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}}$

= $\frac{7}{12}$.

Hence, fraction = $\frac{7}{12}$.

(iii) Total parts = 7

Shaded parts = 2

Fraction = $\frac{\text{Shaded parts}}{\text{Total parts}}$

= $\frac{2}{7}$

Hence, fraction = $\frac{2}{7}$.

(i) Total parts = 9

Given parts = 4

Fraction = $\frac{\text{Given parts}}{\text{Total parts}}$

= $\frac{4}{9}$

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

(ii) Total parts = 11

Given parts = 5

Fraction = $\frac{\text{Given parts}}{\text{Total parts}}$

= $\frac{5}{11}$

Hence, fraction = $\frac{5}{11}$.

(iii) Total parts = 5

Given parts = 2

Fraction = $\frac{\text{Given parts}}{\text{Total parts}}$

= $\frac{2}{5}$

Hence, fraction = $\frac{2}{5}$.

(iv) Total parts = 7

Given parts = 4

Fraction = $\frac{\text{Given parts}}{\text{Total parts}}$

= $\frac{4}{7}$

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

(v) Total parts = 10

Given parts = 7

Fraction = $\frac{\text{Given parts}}{\text{Total parts}}$

= $\frac{7}{10}$

Hence, fraction = $\frac{7}{10}$.

(vi) Total parts = 8

Given parts = 6

Fraction = $\frac{\text{Given parts}}{\text{Total parts}}$

= $\frac{6}{8}$

Hence, fraction = $\frac{6}{8}$.

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

Hence, numerator = 5 and denominator = 6.

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

Hence, numerator = 3 and denominator = 7.

(iii) $\frac{9}{14}$

Hence, numerator = 9 and denominator = 14.

(iv) $\frac{1}{20}$

Hence, numerator = 1 and denominator = 20.

(v) $\frac{12}{25}$

Hence, numerator = 12 and denominator = 25.

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

To find equivalent fractions, we multiply the numerator and denominator by a sequence of numbers (e.g., 2, 3, 4, 5, 6).

Multiply by 2: $\frac{2 \times 2}{3 \times 2} = \frac{4}{6}$

Multiply by 3: $\frac{2 \times 3}{3 \times 3} = \frac{6}{9}$

Multiply by 4: $\frac{2 \times 4}{3 \times 4} = \frac{8}{12}$

Multiply by 5: $\frac{2 \times 5}{3 \times 5} = \frac{10}{15}$

Multiply by 6: $\frac{2 \times 6}{3 \times 6} = \frac{12}{18}$

Hence, the equivalent fractions are : $\frac{4}{6}, \frac{6}{9}, \frac{8}{12}, \frac{10}{15}, \frac{12}{18}$.

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

To find equivalent fractions, we multiply the numerator and denominator by a sequence of numbers (e.g., 2, 3, 4, 5, 6).

Multiply by 2: $\frac{3 \times 2}{4 \times 2} = \frac{6}{8}$

Multiply by 3: $\frac{3 \times 3}{4 \times 3} = \frac{9}{12}$

Multiply by 4: $\frac{3 \times 4}{4 \times 4} = \frac{12}{16}$

Multiply by 5: $\frac{3 \times 5}{4 \times 5} = \frac{15}{20}$

Multiply by 6: $\frac{3 \times 6}{4 \times 6} = \frac{18}{24}$

Hence, the equivalent fractions are: $\frac{6}{8}, \frac{9}{12}, \frac{12}{16}, \frac{15}{20}, \frac{18}{24}$.

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

To find equivalent fractions, we multiply the numerator and denominator by a sequence of numbers (e.g., 2, 3, 4, 5, 6).

Multiply by 2: $\frac{5 \times 2}{6 \times 2} = \frac{10}{12}$

Multiply by 3: $\frac{5 \times 3}{6 \times 3} = \frac{15}{18}$

Multiply by 4: $\frac{5 \times 4}{6 \times 4} = \frac{20}{24}$

Multiply by 5: $\frac{5 \times 5}{6 \times 5} = \frac{25}{30}$

Multiply by 6: $\frac{5 \times 6}{6 \times 6} = \frac{30}{36}$

Hence, the equivalent fractions are: $\frac{10}{12}, \frac{15}{18}, \frac{20}{24}, \frac{25}{30}, \frac{30}{36}$.

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

To find equivalent fractions, we multiply the numerator and denominator by a sequence of numbers (e.g., 2, 3, 4, 5, 6).

Multiply by 2: $\frac{7 \times 2}{9 \times 2} = \frac{14}{18}$

Multiply by 3: $\frac{7 \times 3}{9 \times 3} = \frac{21}{27}$

Multiply by 4: $\frac{7 \times 4}{9 \times 4} = \frac{28}{36}$

Multiply by 5: $\frac{7 \times 5}{9 \times 5} = \frac{35}{45}$

Multiply by 6: $\frac{7 \times 6}{9 \times 6} = \frac{42}{54}$

Hence, the equivalent fractions are: $\frac{14}{18}, \frac{21}{27}, \frac{28}{36}, \frac{35}{45}, \frac{42}{54}$.

(v) $\frac{3}{14}$

To find equivalent fractions, we multiply the numerator and denominator by a sequence of numbers (e.g., 2, 3, 4, 5, 6).

Multiply by 2: $\frac{3 \times 2}{14 \times 2} = \frac{6}{28}$

Multiply by 3: $\frac{3 \times 3}{14 \times 3} = \frac{9}{42}$

Multiply by 4: $\frac{3 \times 4}{14 \times 4} = \frac{12}{56}$

Multiply by 5: $\frac{3 \times 5}{14 \times 5} = \frac{15}{70}$

Multiply by 6: $\frac{3 \times 6}{14 \times 6} = \frac{18}{84}$

Hence, the equivalent fractions are: $\frac{6}{28}, \frac{9}{42}, \frac{12}{56}, \frac{15}{70}, \frac{18}{84}$.

(i) $\frac{4}{5}$ and $\frac{24}{30}$

By cross multiplication,

So, 4 x 30 = 120 and 5 x 24 = 120

Here, the cross products are equal.

Hence, $\frac{4}{5} \text{ and } \frac{24}{30}$ are equivalent fractions.

(ii) $\frac{6}{8}$ and $\frac{18}{24}$

By cross multiplication,

So, 6 x 24 = 144 and 8 x 18 = 144

Here, the cross products are equal.

Hence, $\frac{6}{8}\text{ and } \frac{18}{24}$ are equivalent fractions.

(iii) $\frac{9}{16}$ and $\frac{3}{4}$

By cross multiplication,

So, 9 x 4 = 36 and 16 x 3 = 48

Here, the cross products are not equal.

Hence, $\frac{9}{16}\text{ and } \frac{3}{4}$ are not equivalent fractions.

(iv) $\frac{8}{12}$ and $\frac{14}{21}$

By cross multiplication,

So, 8 x 21 = 168 and 12 x 14 = 168

Here, the cross products are equal.

Hence, $\frac{8}{12}\text{ and } \frac{14}{21}$ are equivalent fractions.

(v) $\frac{7}{11}$ and $\frac{21}{32}$

By cross multiplication,

So, 7 x 32 = 224 and 11 x 21 = 231

Here, the cross products are not equal.

Hence, $\frac{7}{11}\text{ and } \frac{21}{32}$ are not equivalent fractions.

(vi) $\frac{5}{12}$ and $\frac{35}{84}$

By cross multiplication,

So, 5 x 84 = 420 and 12 x 35 = 420

Here, the cross products are equal.

Hence, $\frac{5}{12}\text{ and } \frac{35}{84}$ are equivalent fractions.

(i) $\frac{4}{5}$ with numerator 32

To find an equivalent fraction whose numerator is 32, we multiply the fraction $\frac{4}{5}$ with 8,

$\frac{4}{5} = \frac{4 \times 8}{5 \times 8} = \frac{32}{40}$.

Hence, an equivalent fraction of $\frac{4}{5} = \frac{32}{40}$.

(ii) $\frac{7}{9}$ with numerator 42

To find an equivalent fraction whose numerator is 42, we multiply the fraction $\frac{7}{9}$ with 6,

$\frac{7}{9} = \frac{7 \times 6}{9 \times 6} = \frac{42}{54}$.

Hence, an equivalent fraction of $\frac{7}{9} = \frac{42}{54}$.

(iii) $\frac{3}{7}$ with denominator 63

To find an equivalent fraction whose denominator is 63, we multiply the fraction $\frac{3}{7}$ with 9,

$\frac{3}{7} = \frac{3 \times 9}{7 \times 9} = \frac{27}{63}$.

Hence, an equivalent fraction of $\frac{3}{7} = \frac{27}{63}$.

(iv) $\frac{10}{13}$ with denominator 78

To find an equivalent fraction whose denominator is 78, we multiply the fraction $\frac{10}{13}$ with 6,

$\frac{10}{13} = \frac{10 \times 6}{13 \times 6} = \frac{60}{78}$.

Hence, an equivalent fraction of $\frac{10}{13} = \frac{60}{78}$.

(i) $\frac{32}{48}$ with numerator 2

The H.C.F. of 32 and 48 is 16. Thus,

To find an equivalent fraction whose numerator is 2, we divide the numerator and denominator of the given fraction by 16,

$\frac{32}{48} = \frac{32 ÷ 16}{48 ÷ 16} = \frac{2}{3}$.

Hence, an equivalent fraction of $\frac{32}{48} = \frac{2}{3}$.

(ii) $\frac{21}{28}$ with numerator 3

The H.C.F. of 21 and 28 is 7. Thus,

To find an equivalent fraction whose numerator is 3, we divide the numerator and denominator of the given fraction by 7,

$\frac{21}{28} = \frac{21 ÷ 7}{28 ÷ 7} = \frac{3}{4}$.

Hence, an equivalent fraction of $\frac{21}{28} = \frac{3}{4}$.

(iii) $\frac{24}{56}$ with denominator 7

The H.C.F. of 24 and 56 is 8. Thus,

To find an equivalent fraction whose denominator is 7, we divide the numerator and denominator of the given fraction by 8,

$\frac{24}{56} = \frac{24 ÷ 8}{56 ÷ 8} = \frac{3}{7}$.

Hence, an equivalent fraction of $\frac{24}{56} = \frac{3}{7}$.

(iv) $\frac{121}{132}$ with denominator 12

The H.C.F. of 121 and 132 is 11. Thus,

To find an equivalent fraction whose denominator is 12, we divide the numerator and denominator of the given fraction by 11,

$\frac{121}{132} = \frac{121 ÷ 11}{132 ÷ 11} = \frac{11}{12}$.

Hence, an equivalent fraction of $\frac{121}{132} = \frac{11}{12}$.

(i) $\frac{45}{54}$ with numerator 20.

The H.C.F. of 45 and 54 is 9. Thus,

To find an equivalent fraction whose numerator is 20,

We divide the numerator and denominator of the given fraction by 9,

$\frac{45}{54} = \frac{45 ÷ 9}{54 ÷ 9} = \frac{5}{6}$.

Now, to get a fraction with numerator 20, we multiply both the numerator and denominator by 4,

$\frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24}$.

Hence, the equivalent fraction = $\frac{20}{24}$.

(ii) $\frac{75}{90}$ with denominator 42.

The H.C.F. of 75 and 90 is 15. Thus,

To find an equivalent fraction whose denominator is 42,

We divide the numerator and denominator of the given fraction by 15,

$\frac{75}{90} = \frac{75 ÷ 15}{90 ÷ 15} = \frac{5}{6}$.

Now, to get a fraction with denominator 42, we multiply both the numerator and denominator by 7,

$\frac{5}{6} = \frac{5 \times 7}{6 \times 7} = \frac{35}{42}$.

Hence, the equivalent fraction = $\frac{35}{42}$.

(i) $\frac{4}{9}$ = $\frac{\boxed{\phantom{63}}}{63}$

$\frac{4}{9} = \frac{4 \times 7}{9 \times 7} = \frac{28}{63}$

Hence, $\frac{4}{9}$ = $\frac{\boxed{28}}{63}$.

(ii) $\frac{3}{4}$ = $\frac{18}{\boxed{\phantom{63}}}$

$\frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}$

Hence, $\frac{3}{4}$ = $\frac{18}{\boxed{24}}$.

(iii) $\frac{42}{70}$ = $\frac{3}{\boxed{\phantom{63}}}$

Simplifying the fraction,

$\frac{42}{70} = \frac{42 ÷ 14}{70 ÷ 14} = \frac{3}{5}$

Hence, $\frac{42}{70}$ = $\frac{3}{\boxed{5}}$.

(iv) $\frac{52}{78}$ = $\frac{\boxed{\phantom{63}}}{6}$

Simplifying the fraction,

$\frac{52}{78} = \frac{52 ÷ 13}{78 ÷ 13} = \frac{4}{6}$

Hence, $\frac{52}{78}$ = $\frac{\boxed{4}}{6}$.

Given minutes = 25 minutes

Total minutes = 60 minutes

Fraction = $\frac{\text{Given minutes}}{\text{Total minutes}} = \frac{25}{60}$.

Hence, required fraction = $\frac{25}{60}$.

Given hours = 9 hours

Total hours = 24 hours

Fraction = $\frac{\text{Given hours}}{\text{Total hours}} = \frac{9}{24}$.

Hence, required fraction = $\frac{9}{24}$.

A fraction is in its simplest form when the highest common factor (HCF) of its numerator and denominator is 1.

(i) $\frac{21}{40}$

First, we find the H.C.F. of 21 and 40.

$$\begin{array}{l} 21\overline{\smash{\big)}\quad 40\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-21} \\ \phantom{{x^2 =)}} 19 \overline{\smash{\big)}\quad 21\smash{\big(}} 1 \\ \phantom{ac = sc)}\phantom{)}\underline{-19} \\ \phantom{{x^2 )} + 2x =}2\overline{\smash{\big)}\quad 19 \smash{\big(}} 9 \\ \phantom{ac = sc + sa)}\phantom{)}\underline{-18} \\ \phantom{{x^2 )} + 2x = + dca}1\overline{\smash{\big)}\quad 2 \smash{\big(}} 2 \\ \phantom{ac = sc + sa + sa)}\phantom{)}\underline{-2} \\ \phantom{{x^2 + 3x - 54)} + 2x)}\times \\ \end{array}$$

So, HCF of 21 and 40 = 1.

Hence, $\frac{21}{40}$ is in simplest form.

(ii) $\frac{35}{49}$

$$\begin{array}{l} 35\overline{\smash{\big)}\quad 49\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-35} \\ \phantom{{x^2 =)}} 14 \overline{\smash{\big)}\quad 35\smash{\big(}} 2 \\ \phantom{ac = sc)}\phantom{)}\underline{-28} \\ \phantom{{x^2 )} + 2x =}7\overline{\smash{\big)}\quad 14 \smash{\big(}} 2 \\ \phantom{ac = sc + sa)}\phantom{)}\underline{-14} \\ \phantom{{x^2 + 3x - 54)} + }\times \\ \end{array}$$

So, HCF of 35 and 49 = 7.

Hence, $\frac{35}{49}$ is not in simplest form.

(iii) $\frac{42}{54}$

$$\begin{array}{l} 42\overline{\smash{\big)}\quad 54\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-42} \\ \phantom{{x^2 =)}} 12 \overline{\smash{\big)}\quad 42\smash{\big(}} 3 \\ \phantom{ac = sc))}\phantom{)}\underline{-36} \\ \phantom{{x^2 )} + 2x =}6\overline{\smash{\big)}\quad 12 \smash{\big(}} 2 \\ \phantom{ac = sc + sa)}\phantom{)}\underline{-12} \\ \phantom{{x^2 + 3x - 54)} + }\times \\ \end{array}$$

So, HCF of 42 and 54 = 6.

Hence, $\frac{42}{54}$ is not in simplest form.

(iv) $\frac{64}{81}$

$$\begin{array}{l} 64\overline{\smash{\big)}\quad 81\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-64} \\ \phantom{{x^2 =)}} 17 \overline{\smash{\big)}\quad 64\smash{\big(}} 3 \\ \phantom{ac = sc))}\phantom{)}\underline{-51} \\ \phantom{{x^2 )} + 2x =}13\overline{\smash{\big)}\quad 17 \smash{\big(}} 1 \\ \phantom{ac = sc + sa)))}\phantom{)}\underline{-13} \\ \phantom{{x^2 )} + 2x = acdvu}4\overline{\smash{\big)}\quad 13 \smash{\big(}} 3 \\ \phantom{ac = sc + sa+ df)}\phantom{)}\underline{-12} \\ \phantom{{x^2 )} + 2x = acdv + df}1\overline{\smash{\big)}\quad 4 \smash{\big(}} 4 \\ \phantom{ac = sc + sa+ df + dc)}\phantom{)}\underline{-4} \\ \phantom{{x^2 + 3x - 54)} + advnscj }\times \\ \end{array}$$

So, HCF of 64 and 81 = 1.

Hence, $\frac{64}{81}$ is in simplest form.

(v) $\frac{56}{65}$

$$\begin{array}{l} 56\overline{\smash{\big)}\quad 65\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-56} \\ \phantom{{x^2 =)}} 9 \overline{\smash{\big)}\quad 56\smash{\big(}} 6 \\ \phantom{ac = sc)}\phantom{)}\underline{-54} \\ \phantom{{x^2 )} + 2x =}2\overline{\smash{\big)}\quad 9 \smash{\big(}} 4 \\ \phantom{ac = sc + sa))}\phantom{)}\underline{-8} \\ \phantom{{x^2 )} + 2x = acdv}1\overline{\smash{\big)}\quad 2 \smash{\big(}} 2 \\ \phantom{ac = sc + sa+ df)}\phantom{)}\underline{-2} \\ \phantom{{x^2 + 3x - 54)} + aax}\times \\ \end{array}$$

So, HCF of 56 and 65 = 1.

Hence, $\frac{56}{65}$ is in simplest form.

(vi) $\frac{23}{92}$

$$\begin{array}{l} 23\overline{\smash{\big)}\quad 92\smash{\big(}} 4 \\ \phantom{ac)}\phantom{)}\underline{-92} \\ \phantom{{x^2 + 3}}\times \\ \end{array}$$

So, HCF of 23 and 92 = 23.

Hence, $\frac{23}{92}$ is not in simplest form.

(vii) $\frac{102}{119}$

$$\begin{array}{l} 102\overline{\smash{\big)}\quad 119\smash{\big(}} 1 \\ \phantom{ac =)}\phantom{}\underline{-102} \\ \phantom{{x^2 = aa)}} 17 \overline{\smash{\big)}\quad 102\smash{\big(}} 6 \\ \phantom{ac = sc =s)}\phantom{)}\underline{-102} \\ \phantom{{x^2 + 3x - 54)} x}\times \\ \end{array}$$

So, HCF of 102 and 119 = 17.

Hence, $\frac{102}{119}$ is not in simplest form.

(viii) $\frac{91}{114}$

$$\begin{array}{l} 91\overline{\smash{\big)}\quad 114\smash{\big(}} 1 \\ \phantom{ac +)}\phantom{)}\underline{-91} \\ \phantom{{x^2 =)))}} 23 \overline{\smash{\big)}\quad 91\smash{\big(}} 3 \\ \phantom{ac = sc))}\phantom{)}\underline{-69} \\ \phantom{{x^2 )} + 2x =}22\overline{\smash{\big)}\quad 23 \smash{\big(}} 1 \\ \phantom{ac = sc + sa =)}\phantom{)}\underline{-22} \\ \phantom{{x^2 )} + 2x = ac + ad}1\overline{\smash{\big)}\quad 22 \smash{\big(}} 22 \\ \phantom{ac = sc + sa+ df -)}\phantom{)}\underline{-22} \\ \phantom{{x^2 + 3x - 54)} + aax+ s} \times \\ \end{array}$$

So, HCF of 91 and 114 = 1.

Hence, $\frac{91}{114}$ is in simplest form.

(i) Prime factorizing,

$\frac{27}{36} = \frac{3 \times 3 \times 3}{2 \times 2 \times 3 \times 3} = \frac{3}{4}$.

Hence, $\frac{27}{36}$ in lowest terms is $\frac{3}{4}$.

(ii) Prime factorizing,

$\frac{45}{54} = \frac{3 \times 3 \times 5}{3 \times 3 \times 3 \times 2} = \frac{5}{6}$.

Hence, $\frac{45}{54}$ in lowest terms is $\frac{5}{6}$.

(iii) Prime factorizing,

$\frac{38}{95} = \frac{2 \times 19}{5 \times 19} = \frac{2}{5}$.

Hence, $\frac{38}{95}$ in lowest terms is $\frac{2}{5}$.

(iv) Prime factorizing,

$\frac{58}{87} = \frac{2 \times 29}{3 \times 29} = \frac{2}{3}$.

Hence, $\frac{58}{87}$ in lowest terms is $\frac{2}{3}$.

(v) Prime factorizing,

$\frac{85}{153} = \frac{5 \times 17}{3 \times 3 \times 17} = \frac{5}{9}$.

Hence, $\frac{85}{153}$ in lowest terms is $\frac{5}{9}$.

(vi) Prime factorizing,

$\frac{105}{168} = \frac{3 \times 5 \times 7}{2 \times 2 \times 2\times 3 \times 7} = \frac{5}{8}$.

Hence, $\frac{105}{168}$ in lowest terms is $\frac{5}{8}$.

(vii) Prime factorizing,

$\frac{117}{143} = \frac{3 \times 3 \times 13}{11 \times 13} = \frac{9}{11}$.

Hence, $\frac{117}{143}$ in lowest terms is $\frac{9}{11}$.

(viii) Prime factorizing,

$\frac{135}{150} = \frac{3 \times 3 \times 3 \times 5}{2 \times 3 \times 5 \times 5} = \frac{9}{10}$.

Hence, $\frac{135}{150}$ in lowest terms is $\frac{9}{10}$.

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

A proper fraction is a fraction where the numerator (the top number) is less than the denominator (the bottom number).

Here, 7 < 9

Hence, $\frac{7}{9}$ is a proper fraction.

(ii) $\frac{16}{11}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 16 > 11

Hence, $\frac{16}{11}$ is an improper fraction.

(iii) $\frac{18}{25}$

A proper fraction is a fraction where the numerator (the top number) is less than the denominator (the bottom number).

Here, 18 < 25

Hence, $\frac{18}{25}$ is a proper fraction.

(iv) $\frac{10}{10}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 10 = 10

Hence, $\frac{10}{10}$ is an improper fraction.

(v) $\frac{37}{23}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 37 > 23

Hence, $\frac{37}{23}$ is an improper fraction.

(vi) $\frac{21}{8}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 21 > 8

Hence, $\frac{21}{8}$ is an improper fraction.

(vii) $\frac{56}{57}$

A proper fraction is a fraction where the numerator (the top number) is less than the denominator (the bottom number).

Here, 56 < 57

Hence, $\frac{56}{57}$ is a proper fraction.

(viii) $\frac{137}{105}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 137 > 105

Hence, $\frac{137}{105}$ is an improper fraction.

(ix) $\frac{2}{1}$

An improper fraction is a fraction where the numerator is greater than or equal to the denominator.

Here, 2 > 1

Hence, $\frac{2}{1}$ is an improper fraction.

(x) $\frac{100}{101}$

A proper fraction is a fraction where the numerator (the top number) is less than the denominator (the bottom number).

Here, 100 < 101

Hence, $\frac{100}{101}$ is a proper fraction.

(i) $8\frac{3}{4}$

$8\frac{3}{4} = \frac{8 \times 4 + 3}{4} = \frac{35}{4}$

Hence, the improper fraction = $\frac{35}{4}$.

(ii) $5\frac{11}{13}$

$5\frac{11}{13} = \frac{5 \times 13 + 11}{13} = \frac{76}{13}$

Hence, the improper fraction = $\frac{76}{13}$.

(iii) $10\frac{7}{9}$

$10\frac{7}{9} = \frac{10 \times 9 + 7}{9} = \frac{97}{9}$

Hence, the improper fraction = $\frac{97}{9}$.

(iv) $33\frac{1}{3}$

$33\frac{1}{3} = \frac{33 \times 3 + 1}{3} = \frac{100}{3}$

Hence, the improper fraction = $\frac{100}{3}$.

(v) $9\frac{7}{16}$

$9\frac{7}{16} = \frac{9 \times 16 + 7}{16} = \frac{151}{16}$

Hence, the improper fraction = $\frac{151}{16}$.

(i) $\frac{31}{5}$

On dividing 31 by 5, we get quotient = 6 and remainder = 1.

$\therefore \frac{31}{5} = 6 + \frac{1}{5} = 6\frac{1}{5}$.

Hence, the mixed fraction is $6\frac{1}{5}$.

(ii) $\frac{80}{7}$

On dividing 80 by 7, we get quotient = 11 and remainder = 3.

$\therefore \frac{80}{7} = 11 + \frac{3}{7} = 11\frac{3}{7}$.

Hence, the mixed fraction is $11\frac{3}{7}$.

(iii) $\frac{107}{3}$

On dividing 107 by 3, we get quotient = 35 and remainder = 2.

$\therefore \frac{107}{3} = 35 + \frac{2}{3} = 35\frac{2}{3}$.

Hence, the mixed fraction is $35\frac{2}{3}$.

(iv) $\frac{115}{13}$

On dividing 115 by 13, we get quotient = 8 and remainder = 11.

$\therefore \frac{115}{13} = 8 + \frac{11}{13} = 8\frac{11}{13}$.

Hence, the mixed fraction is $8\frac{11}{13}$.

(v) $\frac{200}{9}$

On dividing 200 by 9, we get quotient = 22 and remainder = 2.

$\therefore \frac{200}{9} = 22 + \frac{2}{9} = 22\frac{2}{9}.$

Hence, the mixed fraction is $22\frac{2}{9}$.

(i) $\frac{4}{5}, \frac{7}{10}, \frac{11}{15}, \frac{13}{20}$

Denominators: 5, 10, 15, 20

LCM = 60.

Equivalent fractions:

$$\begin{aligned}\Rightarrow \frac{4 \times 12}{5 \times 12}, \frac{7 \times 6}{10 \times 6}, \frac{11 \times 4}{15 \times 4}, \frac{13 \times 3}{20 \times 3}\\[1em] \Rightarrow \frac{48}{60}, \frac{42}{60}, \frac{44}{60}, \frac{39}{60}\end{aligned}$$

Hence, the like fractions are $\frac{48}{60}, \frac{42}{60}, \frac{44}{60}, \frac{39}{60}$.

(ii) $\frac{2}{3}, \frac{1}{4}, \frac{5}{6}, \frac{7}{8}, \frac{11}{12}$

Denominators: 3, 4, 6, 8 and 12

LCM = 24.

Equivalent fractions:

$$\begin{aligned}\Rightarrow \frac{2 \times 8}{3 \times 8}, \frac{1 \times 6}{4 \times 6}, \frac{5 \times 4}{6 \times 4}, \frac{7 \times 3}{8 \times 3}, \frac{11 \times 2}{12 \times 2}\\[1em] \Rightarrow \frac{16}{24}, \frac{6}{24}, \frac{20}{24}, \frac{21}{24}, \frac{22}{24}\end{aligned}$$

Hence, the like fractions are $\frac{16}{24}, \frac{6}{24}, \frac{20}{24}, \frac{21}{24}, \frac{22}{24}$.

(iii) $\frac{1}{3}, \frac{3}{4}, \frac{5}{12}, \frac{9}{16}, \frac{17}{24}$

Denominators: 3, 4, 12, 16, 24

LCM = 48.

Equivalent fractions:

$$\begin{aligned}\Rightarrow \frac{1 \times 16}{3 \times 16}, \frac{3 \times 12}{4 \times 12}, \frac{5 \times 4}{12 \times 4}, \frac{9 \times 3}{16 \times 3}, \frac{17 \times 2}{24 \times 2}\\[1em] \Rightarrow \frac{16}{48}, \frac{36}{48}, \frac{20}{48}, \frac{27}{48}, \frac{34}{48}\end{aligned}$$

Hence, the like fractions are $\frac{16}{48}, \frac{36}{48}, \frac{20}{48}, \frac{27}{48}, \frac{34}{48}$.

(iv) $\frac{2}{3}, \frac{1}{6}, \frac{5}{9}, \frac{7}{12}, \frac{13}{18}$

Denominators: 3, 6, 9, 12, 18

LCM = 36.

Equivalent fractions:

$$\begin{aligned}\Rightarrow \frac{2 \times 12}{3 \times 12}, \frac{1 \times 6}{6 \times 6}, \frac{5 \times 4}{9 \times 4}, \frac{7 \times 3}{12 \times 3}, \frac{13 \times 2}{18 \times 2}\\[1em] \Rightarrow \frac{24}{36}, \frac{6}{36}, \frac{20}{36}, \frac{21}{36}, \frac{26}{36}\end{aligned}$$

Hence, the like fractions are $\frac{24}{36}, \frac{6}{36}, \frac{20}{36}, \frac{21}{36}, \frac{26}{36}$.

(v) $\frac{1}{2}, \frac{4}{7}, \frac{9}{14}, \frac{11}{21}, \frac{37}{42}$

Denominators: 2, 7, 14, 21, 42

LCM = 42.

Equivalent fractions:

$$\begin{aligned}\Rightarrow \frac{1 \times 21}{2 \times 21}, \frac{4 \times 6}{7 \times 6}, \frac{9 \times 3}{14 \times 3}, \frac{11 \times 2}{21 \times 2}, \frac{37 \times 1}{42 \times 1}\\[1em] \Rightarrow \frac{21}{42}, \frac{24}{42}, \frac{27}{42}, \frac{22}{42}, \frac{37}{42}\end{aligned}$$

Hence, the like fractions are $\frac{21}{42}, \frac{24}{42}, \frac{27}{42}, \frac{22}{42}, \frac{37}{42}$.

(vi) $\frac{2}{7}, \frac{5}{8}, \frac{11}{14}, \frac{9}{16}, \frac{3}{4}$

Denominators: 7, 8, 14, 16, 4

LCM = 112.

Equivalent fractions:

$$\begin{aligned}\Rightarrow \frac{2 \times 16}{7 \times 16}, \frac{5 \times 14}{8 \times 14}, \frac{11 \times 8}{14 \times 8}, \frac{9 \times 7}{16 \times 7}, \frac{3 \times 28}{4 \times 28}\\[1em] \Rightarrow \frac{32}{112}, \frac{70}{112}, \frac{88}{112}, \frac{63}{112}, \frac{84}{112}\end{aligned}$$

Hence, the like fractions are $\frac{32}{112}, \frac{70}{112}, \frac{88}{112}, \frac{63}{112}, \frac{84}{112}$.

(i) $\frac{7}{9} {\boxed{\phantom{63}}} \frac{5}{9}$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

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

(ii) $\frac{9}{13} {\boxed{\phantom{63}}} \frac{11}{13}$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

Hence, $\frac{9}{13} {\boxed \lt} \frac{11}{13}$.

(iii) $\frac{3}{5} {\boxed{\phantom{63}}} \frac{3}{4}$

Among two fractions with same numerator, the one with smaller denominator is greater of the two.

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

(iv) $\frac{7}{9}$ ${\boxed{\phantom{63}}}$ $\frac{7}{11}$

Among two fractions with same numerator, the one with smaller denominator is greater of the two.

Hence, $\frac{7}{9}$ ${\boxed{\gt}}$ $\frac{7}{11}$

(v) $\frac{5}{8} {\boxed{\phantom{63}}} \frac{5}{6}$

Among two fractions with same numerator, the one with smaller denominator is greater of the two.

Hence, $\frac{5}{8} {\boxed{\lt}} \frac{5}{6}$.

(vi) $\frac{7}{9} {\boxed{\phantom{63}}} \frac{6}{11}$

Cross multiply; 7 x 11 and 6 x 9

⇒ 77 and 54

Now, comparing both numbers;

⇒ 77 > 54

Hence, $\frac{7}{9} {\boxed{\gt}} \frac{6}{11}$.

(vii) $\frac{6}{5} {\boxed{\phantom{63}}} \frac{5}{4}$

Cross multiply; 6 x 4 and 5 x 5

⇒ 24 and 25

Now, comparing both numbers;

⇒ 24 < 25

Hence, $\frac{6}{5} {\boxed{\lt}} \frac{5}{4}$.

(viii) $\frac{7}{11} {\boxed{\phantom{63}}} \frac{8}{13}$

Cross multiply; 7 x 13 and 8 x 11

⇒ 91 and 88

Now, comparing both numbers;

⇒ 91 > 88

Hence, $\frac{7}{11} {\boxed{\gt}} \frac{8}{13}$.

(ix) $\frac{10}{13} {\boxed{\phantom{63}}} \frac{13}{16}$

Cross multiply; 10 x 16 and 13 x 13

⇒ 160 and 169

Now, comparing both numbers;

⇒ 160 < 169

Hence, $\frac{10}{13} {\boxed{\lt}} \frac{13}{16}$.

(x) $\frac{2}{9} {\boxed{\phantom{63}}} \frac{3}{14}$

Cross multiply; 2 x 14 and 3 x 9

⇒ 28 and 27

Now, comparing both numbers;

⇒ 28 > 27

Hence, $\frac{2}{9} {\boxed{\gt}} \frac{3}{14}$.

(xi) $\frac{7}{12} {\boxed{\phantom{63}}} \frac{5}{9}$

Cross multiply; 7 x 9 and 12 x 5

⇒ 63 and 60

Now, comparing both numbers;

⇒ 63 > 60

Hence, $\frac{7}{12} {\boxed{\gt}} \frac{5}{9}$.

(xii) $\frac{15}{19} {\boxed{\phantom{63}}} \frac{3}{4}$

Cross multiply; 15 x 4 and 3 x 19

⇒ 60 and 57

Now, comparing both numbers;

⇒ 60 > 57

Hence, $\frac{15}{19} {\boxed{\gt}} \frac{3}{4}$.

(i) $\frac{3}{11}, \frac{9}{11}, \frac{4}{11}, \frac{5}{11}, \frac{1}{11}$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

Hence, $\frac{1}{11} \lt \frac{3}{11} \lt \frac{4}{11} \lt \frac{5}{11} \lt \frac{9}{11}$

(ii) $\frac{2}{13}, \frac{2}{9}, \frac{2}{15}, \frac{2}{7}, \frac{2}{5}$

Among two fractions with same numerator, the one with smaller denominator is greater of the two.

Hence, $\frac{2}{15} \lt \frac{2}{13} \lt \frac{2}{9} \lt \frac{2}{7} \lt \frac{2}{5}$

(iii) $\frac{2}{3}, \frac{5}{6}, \frac{7}{9}, \frac{11}{12}, \frac{13}{18}$

LCM of the denominators (3, 6, 9, 12, 18) = 36

$$\begin{aligned}\Rightarrow \frac{2 \times 12}{3 \times 12}, \frac{5 \times 6}{6 \times 6}, \frac{7 \times 4}{9 \times 4}, \frac{11 \times 3}{12 \times 3}, \frac{13 \times 2}{18 \times 2}\\[1em] \Rightarrow \frac{24}{36}, \frac{30}{36}, \frac{28}{36}, \frac{33}{36}, \frac{26}{36}\end{aligned}$$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

$\frac{24}{36} \lt \frac{26}{36} \lt \frac{28}{36} \lt \frac{30}{36} \lt \frac{33}{36}$

Hence, $\frac{2}{3} \lt \frac{13}{18} \lt \frac{7}{9} \lt \frac{5}{6} \lt \frac{11}{12}$

(iv) $\frac{2}{3}, \frac{1}{4}, \frac{5}{6}, \frac{3}{8}, \frac{7}{12}$

LCM of the denominators (3, 4, 6, 8, 12) = 24

$$\begin{aligned}\Rightarrow \frac{2 \times 8}{3 \times 8}, \frac{1 \times 6}{4 \times 6}, \frac{5 \times 4}{6 \times 4}, \frac{3 \times 3}{8 \times 3}, \frac{7 \times 2}{12 \times 2}\\[1em] \Rightarrow \frac{16}{24}, \frac{6}{24}, \frac{20}{24}, \frac{9}{24}, \frac{14}{24}\end{aligned}$$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

$\frac{6}{24} \lt \frac{9}{24} \lt \frac{14}{24} \lt \frac{16}{24} \lt \frac{20}{24}$

Hence, $\frac{1}{4} \lt \frac{3}{8} \lt \frac{7}{12} \lt \frac{2}{3} \lt \frac{5}{6}$

(i) $\frac{5}{12}, \frac{1}{12}, \frac{7}{12}, \frac{11}{12}, \frac{9}{12}$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

$\frac{11}{12} \gt \frac{9}{12} \gt \frac{7}{12} \gt \frac{5}{12} \gt \frac{1}{12}$

Hence, $\frac{11}{12} \gt \frac{9}{12} \gt \frac{7}{12} \gt \frac{5}{12} \gt \frac{1}{12}$.

(ii) $\frac{4}{7}, \frac{4}{3}, \frac{4}{9}, \frac{4}{5}, \frac{4}{11}$

Among two fractions with same numerator, the one with smaller denominator is greater of the two.

$\frac{4}{3} \gt \frac{4}{5} \gt \frac{4}{7} \gt \frac{4}{9} \gt \frac{4}{11}$

Hence, $\frac{4}{3} \gt \frac{4}{5} \gt \frac{4}{7} \gt \frac{4}{9} \gt \frac{4}{11}$.

(iii) $\frac{2}{3}, \frac{5}{6}, \frac{7}{9}, \frac{3}{4}, \frac{1}{2}$

LCM of the denominators (3, 6, 9, 4, 2) = 36

$$\begin{aligned}\Rightarrow \frac{2 \times 12}{3 \times 12}, \frac{5 \times 6}{6 \times 6}, \frac{7 \times 4}{9 \times 4}, \frac{3 \times 9}{4 \times 9}, \frac{1 \times 18}{2 \times 18}\\[1em] \Rightarrow \frac{24}{36}, \frac{30}{36}, \frac{28}{36}, \frac{27}{36}, \frac{18}{36}\end{aligned}$$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

$\frac{30}{36} \gt \frac{28}{36} \gt \frac{27}{36} \gt \frac{24}{36} \gt \frac{18}{36}$

Hence, $\frac{5}{6} \gt \frac{7}{9} \gt \frac{3}{4} \gt \frac{2}{3} \gt \frac{1}{2}$

(iv) $\frac{2}{3}, \frac{3}{5}, \frac{7}{10}, \frac{8}{15}, \frac{11}{20}$

LCM of the denominators (3, 5, 10, 15, 20) = 60

$$\begin{aligned}\Rightarrow \frac{2 \times 20}{3 \times 20}, \frac{3 \times 12}{5 \times 12}, \frac{7 \times 6}{10 \times 6}, \frac{8 \times 4}{15 \times 4}, \frac{11 \times 3}{20 \times 3}\\[1em] \Rightarrow \frac{40}{60}, \frac{36}{60}, \frac{42}{60}, \frac{32}{60}, \frac{33}{60}\end{aligned}$$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

$\frac{42}{60} \gt \frac{40}{60} \gt \frac{36}{60} \gt \frac{33}{60} \gt \frac{32}{60}$

Hence, $\frac{7}{10} \gt \frac{2}{3} \gt \frac{3}{5} \gt \frac{11}{20} \gt \frac{8}{15}$

(v) $\frac{17}{32}, \frac{7}{12}, \frac{19}{48}, \frac{13}{24}, \frac{9}{16}$

LCM of the denominators (32, 12, 48, 24, 16) = 96

$$\begin{aligned}\Rightarrow \frac{17 \times 3}{32 \times 3}, \frac{7 \times 8}{12 \times 8}, \frac{19 \times 2}{48 \times 2}, \frac{13 \times 4}{24 \times 4}, \frac{9 \times 6}{16 \times 6}\\[1em] \Rightarrow \frac{51}{96}, \frac{56}{96}, \frac{38}{96}, \frac{52}{96}, \frac{54}{96}\end{aligned}$$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

$\frac{56}{96} \gt \frac{54}{96} \gt \frac{52}{96} \gt \frac{51}{96} \gt \frac{38}{96}$

Hence, $\frac{7}{12} \gt \frac{9}{16} \gt \frac{13}{24} \gt \frac{17}{32} \gt \frac{19}{48}$

(vi) $\frac{5}{6}, \frac{7}{9}, \frac{17}{24}, \frac{3}{4}, \frac{23}{36}$

LCM of the denominators (6, 9, 24, 4, 36) = 72

$$\begin{aligned}\Rightarrow \frac{5 \times 12}{6 \times 12}, \frac{7 \times 8}{9 \times 8}, \frac{17 \times 3}{24 \times 3}, \frac{3 \times 18}{4 \times 18}, \frac{23 \times 2}{36 \times 2}\\[1em] \Rightarrow \frac{60}{72}, \frac{56}{72}, \frac{51}{72}, \frac{54}{72}, \frac{46}{72}\end{aligned}$$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

$\frac{60}{72} \gt \frac{56}{72} \gt \frac{54}{72} \gt \frac{51}{72} \gt \frac{46}{72}$

Hence, $\frac{5}{6} \gt \frac{7}{9} \gt \frac{3}{4} \gt \frac{17}{24} \gt \frac{23}{36}$

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

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

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

(ii) $\frac{5}{8} + \frac{1}{8}$

$$\begin{aligned}\Rightarrow \frac{5 + 1}{8}\\[1em] \Rightarrow \frac{6}{8}\\[1em] \Rightarrow \frac{3}{4}\\[1em]\end{aligned}$$

Hence, $\frac{5}{8} + \frac{1}{8} = \frac{3}{4}$.

(iii) $\frac{7}{9} + \frac{4}{9}$

$$\begin{aligned}\Rightarrow \frac{7 + 4}{9}\\[1em] \Rightarrow \frac{11}{9}\\[1em] \Rightarrow 1\frac{2}{9}\\[1em]\end{aligned}$$

Hence, $\frac{7}{9} + \frac{4}{9} = 1\frac{2}{9}$.

(iv) $1\frac{5}{6} + \frac{1}{6}$

$$\begin{aligned}\Rightarrow \frac{11}{6} + \frac{1}{6}\\[1em] \Rightarrow \frac{11 + 1}{6}\\[1em] \Rightarrow \frac{12}{6}\\[1em] \Rightarrow 2\end{aligned}$$

Hence, $1\frac{5}{6} + \frac{1}{6}$ = 2.

(i) $\frac{5}{14} + \frac{3}{14} + \frac{1}{14}$

$$\begin{aligned}\Rightarrow \frac{5 + 3 + 1}{14}\\[1em] \Rightarrow \frac{9}{14}\\[1em]\end{aligned}$$

Hence, $\frac{5}{14} + \frac{3}{14} + \frac{1}{14} = \frac{9}{14}$.

(ii) $\frac{7}{12} + \frac{5}{12} + \frac{11}{12}$

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

Hence, $\frac{7}{12} + \frac{5}{12} + \frac{11}{12} = 1\frac{11}{12}$.

(iii) $1\frac{7}{8} + \frac{3}{8} + 1\frac{5}{8}$

$$\begin{aligned}\Rightarrow \frac{15}{8} + \frac{3}{8} + \frac{13}{8}\\[1em] \Rightarrow \frac{15 + 3 + 13}{8}\\[1em] \Rightarrow \frac{31}{8}\\[1em] \Rightarrow 3\frac{7}{8}\\[1em]\end{aligned}$$

Hence, $1\frac{7}{8} + \frac{3}{8} + 1\frac{5}{8} = 3\frac{7}{8}$.

(i) $\frac{4}{9} + \frac{5}{6}$

LCM of 9 and 6 = 18

$$\begin{aligned}\Rightarrow \frac{4 \times 2}{9 \times 2} + \frac{5 \times 3}{6 \times 3}\\[1em] \Rightarrow \frac{8}{18} + \frac{15}{18}\\[1em] \Rightarrow \frac{8 + 15}{18}\\[1em] \Rightarrow \frac{23}{18}\\[1em] \Rightarrow 1\frac{5}{18}\end{aligned}$$

Hence, $\frac{4}{9} + \frac{5}{6} = 1\frac{5}{18}$

(ii) $\frac{5}{12} + \frac{9}{16}$

LCM of 12 and 16 = 48

$$\begin{aligned}\Rightarrow \frac{5 \times 4}{12 \times 4} + \frac{9 \times 3}{16 \times 3}\\[1em] \Rightarrow \frac{20}{48} + \frac{27}{48}\\[1em] \Rightarrow \frac{20 + 27}{48}\\[1em] \Rightarrow \frac{47}{48}\end{aligned}$$

Hence, $\frac{5}{12} + \frac{9}{16} = \frac{47}{48}$

(iii) $\frac{7}{12} + \frac{13}{18}$

LCM of 12 and 18 = 36

$$\begin{aligned}\Rightarrow \frac{7 \times 3}{12 \times 3} + \frac{13 \times 2}{18 \times 2}\\[1em] \Rightarrow \frac{21}{36} + \frac{26}{36}\\[1em] \Rightarrow \frac{21 + 26}{36}\\[1em] \Rightarrow \frac{47}{36}\\[1em] \Rightarrow 1\frac{11}{36}\end{aligned}$$

Hence, $\frac{7}{12} + \frac{13}{18} = 1\frac{11}{36}$

(i) $\frac{3}{10} + \frac{8}{15} + \frac{7}{20}$

LCM of 10, 15 and 20 = 60

$$\begin{aligned}\Rightarrow \frac{3 \times 6}{10 \times 6} + \frac{8 \times 4}{15 \times 4} + \frac{7 \times 3}{20 \times 3}\\[1em] \Rightarrow \frac{18}{60} + \frac{32}{60} + \frac{21}{60}\\[1em] \Rightarrow \frac{18 + 32 + 21}{60}\\[1em] \Rightarrow \frac{71}{60}\\[1em] \Rightarrow 1\frac{11}{60}\end{aligned}$$

Hence, $\frac{3}{10} + \frac{8}{15} + \frac{7}{20} = 1\frac{11}{60}$

(ii) $\frac{7}{8} + \frac{9}{16} + \frac{17}{24}$

LCM of 8, 16 and 24 = 48

$$\begin{aligned}\Rightarrow \frac{7 \times 6}{8 \times 6} + \frac{9 \times 3}{16 \times 3} + \frac{17 \times 2}{24 \times 2}\\[1em] \Rightarrow \frac{42}{48} + \frac{27}{48} + \frac{34}{48}\\[1em] \Rightarrow \frac{42 + 27 + 34}{48}\\[1em] \Rightarrow \frac{103}{48}\\[1em] \Rightarrow 2\frac{7}{48}\end{aligned}$$

Hence, $\frac{7}{8} + \frac{9}{16} + \frac{17}{24} = 2\frac{7}{48}$

(iii) $\frac{5}{6} + \frac{8}{9} + \frac{11}{18} + \frac{13}{27}$

LCM of 6, 9, 18 and 27 = 54

$$\begin{aligned}\Rightarrow \frac{5 \times 9}{6 \times 9} + \frac{8 \times 6}{9 \times 6} + \frac{11 \times 3}{18 \times 3} + \frac{13 \times 2}{27 \times 2}\\[1em] \Rightarrow \frac{45}{54} + \frac{48}{54} + \frac{33}{54} + \frac{26}{54}\\[1em] \Rightarrow \frac{45 + 48 + 33 + 26}{54}\\[1em] \Rightarrow \frac{152}{54}\\[1em] \Rightarrow \frac{76}{27}\\[1em] \Rightarrow 2\frac{22}{27}\end{aligned}$$

Hence, $\frac{5}{6} + \frac{8}{9} + \frac{11}{18} + \frac{13}{27} = 2\frac{22}{27}$

(i) $4\frac{1}{6} + 2\frac{5}{8} + 3\frac{7}{12}$

LCM of 6, 8 and 12 = 24

$$\begin{aligned}\Rightarrow \frac{25}{6} + \frac{21}{8} + \frac{43}{12}\\[1em] \Rightarrow \frac{25 \times 4}{6 \times 4} + \frac{21 \times 3}{8 \times 3} + \frac{43 \times 2}{12 \times 2}\\[1em] \Rightarrow \frac{100}{24} + \frac{63}{24} + \frac{86}{24}\\[1em] \Rightarrow \frac{100 + 63 + 86}{24}\\[1em] \Rightarrow \frac{249}{24}\\[1em] \Rightarrow \frac{83}{8}\\[1em] \Rightarrow 10\frac{3}{8}\end{aligned}$$

Hence, $4\frac{1}{6} + 2\frac{5}{8} + 3\frac{7}{12} = 10\frac{3}{8}$

(ii) $1\frac{1}{2} + 2\frac{2}{3} + 3\frac{3}{4} + 4\frac{4}{5}$

LCM of 2, 3, 4 and 5 = 60

$$\begin{aligned}\Rightarrow \frac{3}{2} + \frac{8}{3} + \frac{15}{4} + \frac{24}{5}\\[1em] \Rightarrow \frac{3 \times 30}{2 \times 30} + \frac{8 \times 20}{3 \times 20} + \frac{15 \times 15}{4 \times 15} + \frac{24 \times 12}{5 \times 12}\\[1em] \Rightarrow \frac{90}{60} + \frac{160}{60} + \frac{225}{60} + \frac{288}{60}\\[1em] \Rightarrow \frac{90 + 160 + 225 + 288}{60}\\[1em] \Rightarrow \frac{763}{60}\\[1em] \Rightarrow 12\frac{43}{60}\end{aligned}$$

Hence, $1\frac{1}{2} + 2\frac{2}{3} + 3\frac{3}{4} + 4\frac{4}{5} = 12\frac{43}{60}$

(iii) $3\frac{1}{3} + 2\frac{2}{9} + 4\frac{1}{2} + 1\frac{13}{18}$

LCM of 3, 9, 2 and 18 = 18

$$\begin{aligned}\Rightarrow \frac{10}{3} + \frac{20}{9} + \frac{9}{2} + \frac{31}{18}\\[1em] \Rightarrow \frac{10 \times 6}{3 \times 6} + \frac{20 \times 2}{9 \times 2} + \frac{9 \times 9}{2 \times 9} + \frac{31 \times 1}{18 \times 1}\\[1em] \Rightarrow \frac{60}{18} + \frac{40}{18} + \frac{81}{18} + \frac{31}{18}\\[1em] \Rightarrow \frac{60 + 40 + 81 + 31}{18}\\[1em] \Rightarrow \frac{212}{18}\\[1em] \Rightarrow \frac{106}{9}\\[1em] \Rightarrow 11\frac{7}{9}\\[1em]\end{aligned}$$

Hence, $3\frac{1}{3} + 2\frac{2}{9} + 4\frac{1}{2} + 1\frac{13}{18} = 11\frac{7}{9}$

(i) $\frac{9}{11} - \frac{5}{11}$

$$\begin{aligned}\Rightarrow \frac{9 - 5}{11}\\[1em] \Rightarrow \frac{4}{11}\end{aligned}$$

Hence, $\frac{9}{11} - \frac{5}{11} = \frac{4}{11}$.

(ii) $8\frac{5}{7} - \frac{6}{7}$

$$\begin{aligned}\Rightarrow \frac{61}{7} - \frac{6}{7}\\[1em] \Rightarrow \frac{61 - 6}{7}\\[1em] \Rightarrow \frac{55}{7} = 7\frac{6}{7}\end{aligned}$$

Hence, $8\frac{5}{7} - \frac{6}{7} = 7\frac{6}{7}$.

(iii) $3\frac{3}{8} - 1\frac{7}{8}$

$$\begin{aligned}\Rightarrow \frac{27}{8} - \frac{15}{8}\\[1em] \Rightarrow \frac{27 - 15}{8}\\[1em] \Rightarrow \frac{12}{8}\\[1em] \Rightarrow \frac{3}{2} = 1\frac{1}{2}\end{aligned}$$

Hence, $3\frac{3}{8} - 1\frac{7}{8} = 1\frac{1}{2}$.

(iv) $\frac{7}{4} - \frac{3}{4}$

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

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

(i) $\frac{7}{15} - \frac{9}{20}$

LCM of 15 and 20 = 60

$$\begin{aligned}\Rightarrow \frac{7 \times 4}{15 \times 4} - \frac{9 \times 3}{20 \times 3}\\[1em] \Rightarrow \frac{28}{60} - \frac{27}{60}\\[1em] \Rightarrow \frac{28 - 27}{60}\\[1em] \Rightarrow \frac{1}{60}\\[1em]\end{aligned}$$

Hence, $\frac{7}{15} - \frac{9}{20} = \frac{1}{60}$

(ii) $4\frac{1}{7} - 2\frac{1}{14}$

LCM of 7 and 14 = 14

$$\begin{aligned}\Rightarrow \frac{29}{7} - \frac{29}{14}\\[1em] \Rightarrow \frac{29 \times 2}{7 \times 2} - \frac{29 \times 1}{14 \times 1}\\[1em] \Rightarrow \frac{58}{14} - \frac{29}{14}\\[1em] \Rightarrow \frac{58 - 29}{14}\\[1em] \Rightarrow \frac{29}{14}\\[1em] \Rightarrow 2\frac{1}{14}\\[1em]\end{aligned}$$

Hence, $4\frac{1}{7} - 2\frac{1}{14} = 2\frac{1}{14}$

(iii) $1\frac{7}{8} - \frac{5}{12}$

LCM of 8 and 12 = 24

$$\begin{aligned}\Rightarrow \frac{15}{8} - \frac{5}{12}\\[1em] \Rightarrow \frac{15 \times 3}{8 \times 3} - \frac{5 \times 2}{12 \times 2}\\[1em] \Rightarrow \frac{45}{24} - \frac{10}{24}\\[1em] \Rightarrow \frac{45 - 10}{24}\\[1em] \Rightarrow \frac{35}{24}\\[1em] \Rightarrow 1\frac{11}{24}\\[1em]\end{aligned}$$

Hence, $1\frac{7}{8} - \frac{5}{12} = 1\frac{11}{24}$

(iv) $5 - 2\frac{3}{4}$

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

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

(i) $5\frac{1}{8} - 4\frac{1}{12}$

LCM of 8 and 12 = 24

$$\begin{aligned}\Rightarrow \frac{41}{8} - \frac{49}{12}\\[1em] \Rightarrow \frac{41 \times 3}{8 \times 3} - \frac{49 \times 2}{12 \times 2}\\[1em] \Rightarrow \frac{123}{24} - \frac{98}{24}\\[1em] \Rightarrow \frac{123 - 98}{24}\\[1em] \Rightarrow \frac{25}{24}\\[1em] \Rightarrow 1\frac{1}{24}\end{aligned}$$

Hence, $5\frac{1}{8} - 4\frac{1}{12} = 1\frac{1}{24}$

(ii) $6\frac{1}{6} - 3\frac{7}{10}$

LCM of 6 and 10 = 30

$$\begin{aligned}\Rightarrow \frac{37}{6} - \frac{37}{10}\\[1em] \Rightarrow \frac{37 \times 5}{6 \times 5} - \frac{37 \times 3}{10 \times 3}\\[1em] \Rightarrow \frac{185}{30} - \frac{111}{30}\\[1em] \Rightarrow \frac{185 - 111}{30}\\[1em] \Rightarrow \frac{74}{30}\\[1em] \Rightarrow \frac{37}{15}\\[1em] \Rightarrow 2\frac{7}{15}\end{aligned}$$

Hence, $6\frac{1}{6} - 3\frac{7}{10} = 2\frac{7}{15}$

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

$$\begin{aligned}\Rightarrow \frac{12}{1} - \frac{53}{8}\\[1em] \Rightarrow \frac{12 \times 8}{1 \times 8} - \frac{53 \times 1}{8 \times 1}\\[1em] \Rightarrow \frac{96}{8} - \frac{53}{8}\\[1em] \Rightarrow \frac{96 - 53}{8}\\[1em] \Rightarrow \frac{43}{8}\\[1em] \Rightarrow 5\frac{3}{8}\end{aligned}$$

Hence, $12 - 6\frac{5}{8} = 5\frac{3}{8}$

LCM of 9, 12 and 4 = 36

$$\begin{aligned}\Rightarrow \frac{4 \times 4}{9 \times 4} - \frac{5 \times 3}{12 \times 3} + \frac{1 \times 9}{4 \times 9}\\[1em] \Rightarrow \frac{16}{36} - \frac{15}{36} + \frac{9}{36}\\[1em] \Rightarrow \frac{16 - 15 + 9}{36}\\[1em] \Rightarrow \frac{16 - 6}{36}\\[1em] \Rightarrow \frac{10}{36}\\[1em] \Rightarrow \frac{5}{18}\end{aligned}$$

Hence, $\frac{4}{9} - \frac{5}{12} + \frac{1}{4} = \frac{5}{18}$.

Solving,

$$\begin{aligned}\Rightarrow 6\frac{2}{3} + 4\frac{1}{6} - 2\frac{2}{9} \\[1em] \Rightarrow \frac{20}{3} + \frac{25}{6} - \frac{20}{9}\\[1em] \Rightarrow \frac{20 \times 6}{3 \times 6} + \frac{25 \times 3}{6 \times 3} - \frac{20 \times 2}{9 \times 2}\\[1em] \Rightarrow \frac{120}{18} + \frac{75}{18} - \frac{40}{18}\\[1em] \Rightarrow \frac{120 + 75 - 40}{18}\\[1em] \Rightarrow \frac{195 - 40}{18}\\[1em] \Rightarrow \frac{155}{18}\\[1em] \Rightarrow 8\frac{11}{18}.\end{aligned}$$

Hence, $6\frac{2}{3} + 4\frac{1}{6} - 2\frac{2}{9} = 8\frac{11}{18}$.

LCM of 2 and 5 = 10

$$\begin{aligned}\Rightarrow \frac{9}{1} - \frac{9}{2} - \frac{11}{5}\\[1em] \Rightarrow \frac{9 \times 10}{1 \times 10} - \frac{9 \times 5}{2 \times 5} - \frac{11 \times 2}{5 \times 2}\\[1em] \Rightarrow \frac{90}{10} - \frac{45}{10} - \frac{22}{10}\\[1em] \Rightarrow \frac{90 - 45 - 22}{10}\\[1em] \Rightarrow \frac{45 - 22}{10}\\[1em] \Rightarrow \frac{23}{10}\\[1em] \Rightarrow 2\frac{3}{10}\\[1em]\end{aligned}$$

Hence, $9 - 4\frac{1}{2} - 2\frac{1}{5} = 2\frac{3}{10}$.

LCM of 4, 8 and 12 = 24

$$\begin{aligned}\Rightarrow \frac{43}{4} - \frac{41}{8} - \frac{53}{12}\\[1em] \Rightarrow \frac{43 \times 6}{4 \times 6} - \frac{41 \times 3}{8 \times 3} - \frac{53 \times 2}{12 \times 2}\\[1em] \Rightarrow \frac{258}{24} - \frac{123}{24} - \frac{106}{24}\\[1em] \Rightarrow \frac{258 - 123 - 106}{24}\\[1em] \Rightarrow \frac{135 - 106}{24}\\[1em] \Rightarrow \frac{29}{24}\\[1em] \Rightarrow 1\frac{5}{24}\\[1em]\end{aligned}$$

Hence, $10\frac{3}{4} - 5\frac{1}{8} - 4\frac{5}{12} = 1\frac{5}{24}$.

LCM of 5, 15 and 10 = 30

$$\begin{aligned}\Rightarrow \frac{34}{5} - \frac{49}{15} + \frac{43}{10}\\[1em] \Rightarrow \frac{34 \times 6}{5 \times 6} - \frac{49 \times 2}{15 \times 2} + \frac{43 \times 3}{10 \times 3}\\[1em] \Rightarrow \frac{204}{30} - \frac{98}{30} + \frac{129}{30}\\[1em] \Rightarrow \frac{204 - 98 + 129}{30}\\[1em] \Rightarrow \frac{204 + 31}{30}\\[1em] \Rightarrow \frac{235}{30}\\[1em] \Rightarrow \frac{47}{6}\\[1em] \Rightarrow 7\frac{5}{6}\end{aligned}$$

Hence, $6\frac{4}{5} - 3\frac{4}{15} + 4\frac{3}{10} = 7\frac{5}{6}$.

LCM of 12, 18 and 24 = 72

$$\begin{aligned}\Rightarrow \frac{53}{12} + \frac{65}{18} - \frac{55}{24}\\[1em] \Rightarrow \frac{53 \times 6}{12 \times 6} + \frac{65 \times 4}{18 \times 4} - \frac{55 \times 3}{24 \times 3}\\[1em] \Rightarrow \frac{318}{72} + \frac{260}{72} - \frac{165}{72}\\[1em] \Rightarrow \frac{318 + 260 - 165}{72}\\[1em] \Rightarrow \frac{578 - 165}{72}\\[1em] \Rightarrow \frac{413}{72}\\[1em] \Rightarrow 5\frac{53}{72}\end{aligned}$$

Hence, $4\frac{5}{12} + 3\frac{11}{18} - 2\frac{7}{24} = 5\frac{53}{72}$.

Sum of $9\frac{3}{4}$ and $3\frac{5}{6}$.

LCM of 4 and 6 = 12

$$\begin{aligned}\Rightarrow \frac{39}{4} + \frac{23}{6}\\[1em] \Rightarrow \frac{39 \times 3}{4 \times 3} + \frac{23 \times 2}{6 \times 2}\\[1em] \Rightarrow \frac{117}{12} + \frac{46}{12}\\[1em] \Rightarrow \frac{117 + 46}{12}\\[1em] \Rightarrow \frac{163}{12}\\[1em]\end{aligned}$$

Subtract $\frac{163}{12}$ from $15\frac{7}{12}$

$$\begin{aligned}\Rightarrow \frac{187}{12} - \frac{163}{12}\\[1em] \Rightarrow \frac{187 - 163}{12}\\[1em] \Rightarrow \frac{24}{12}\\[1em] \Rightarrow 2\end{aligned}$$

Hence, $15\frac{7}{12} - (9\frac{3}{4} + 3\frac{5}{6}) = 2$.

The sum of $2\frac{5}{12}$ and $3\frac{3}{4}$.

LCM of 12 and 4 = 12

$$\begin{aligned}\Rightarrow \frac{29}{12} + \frac{15}{4}\\[1em] \Rightarrow \frac{29}{12} + \frac{15 \times 3}{4 \times 3}\\[1em] \Rightarrow \frac{29}{12} + \frac{45}{12}\\[1em] \Rightarrow \frac{29 + 45}{12}\\[1em] \Rightarrow \frac{74}{12}\\[1em] \Rightarrow \frac{37}{6}\\[1em]\end{aligned}$$

The sum of $7\frac{1}{3}$ and $5\frac{1}{6}$.

LCM of 3 and 6 = 6

$$\begin{aligned}\Rightarrow \frac{22}{3} + \frac{31}{6}\\[1em] \Rightarrow \frac{22 \times 2}{3 \times 2} + \frac{31}{6}\\[1em] \Rightarrow \frac{44}{6} + \frac{31}{6}\\[1em] \Rightarrow \frac{44 + 31}{6}\\[1em] \Rightarrow \frac{75}{6}\\[1em] \Rightarrow \frac{25}{2}\\[1em]\end{aligned}$$

Subtract $\frac{37}{6}$ from $\frac{25}{2}$

LCM of 6 and 2 = 6

$$\begin{aligned}\Rightarrow \frac{25}{2} - \frac{37}{6}\\[1em] \Rightarrow \frac{25 \times 3}{2 \times 3} - \frac{37}{6} \\[1em] \Rightarrow \frac{75}{6} - \frac{37}{6} \\[1em] \Rightarrow \frac{75 - 37}{6} \\[1em] \Rightarrow \frac{38}{6} \\[1em] \Rightarrow \frac{19}{3}\\[1em] \Rightarrow 6\frac{1}{3}\\[1em]\end{aligned}$$

Hence, final result = $6\frac{1}{3}$.

Let the number that should be added to $9\frac{4}{7}$ be x.

$$\begin{aligned}\Rightarrow 9\frac{4}{7} + x = 16\\[1em] \Rightarrow \frac{67}{7} + x = 16\\[1em] \Rightarrow x = 16 - \frac{67}{7}\\[1em] \Rightarrow x = \frac{16 \times 7}{7} - \frac{67}{7}\\[1em] \Rightarrow x = \frac{112}{7} - \frac{67}{7}\\[1em] \Rightarrow x = \frac{112 - 67}{7} \\[1em] \Rightarrow x = \frac{45}{7} \\[1em] \Rightarrow x = 6\frac{3}{7}\end{aligned}$$

Hence, the number = $6\frac{3}{7}$.

Let the number that can be subtracted be x.

$$\begin{aligned}\Rightarrow 9\frac{1}{6} - x = 6\frac{1}{9}\\[1em] \Rightarrow \frac{55}{6} - x = \frac{55}{9}\\[1em] \Rightarrow x = \frac{55}{6} - \frac{55}{9}\\[1em]\end{aligned}$$

LCM of 6 and 9 = 18

$$\begin{aligned}\Rightarrow x = \frac{55 \times 3}{6 \times 3} - \frac{55 \times 2}{9 \times 2}\\[1em] \Rightarrow x = \frac{165}{18} - \frac{110}{18}\\[1em] \Rightarrow x = \frac{165 - 110}{18}\\[1em] \Rightarrow x = \frac{55}{18}\\[1em] \Rightarrow x = 3\frac{1}{18}\\[1em]\end{aligned}$$

Hence, the number = $3\frac{1}{18}$.

To compare the fractions $\frac{17}{20}$ and $\frac{21}{25}$ ​, we find a common denominator, which is the LCM of 20 and 25. The LCM = 100.

$$\begin{aligned}\Rightarrow \frac{17}{20} - \frac{21}{25}\\[1em] \Rightarrow \frac{17 \times 5}{20 \times 5} - \frac{21 \times 4}{25 \times 4}\\[1em] \Rightarrow \frac{85}{100} - \frac{84}{100}\\[1em] \Rightarrow \frac{85 - 84}{100}\\[1em] \Rightarrow \frac{1}{100}\end{aligned}$$

Hence, $\frac{17}{20}$ is greater by $\frac{1}{100}$.

Given, the sum of two fractions = $14\frac{5}{12}$.

One of them = $7\frac{2}{3}$.

Other number

$$\begin{aligned}\Rightarrow 14\frac{5}{12} - 7\frac{2}{3}\\[1em] \Rightarrow \frac{173}{12} - \frac{23}{3}\\[1em]\end{aligned}$$

LCM of 12 and 3 = 12.

$$\begin{aligned}\Rightarrow \frac{173}{12} - \frac{23\times 4}{3\times 4}\\[1em] \Rightarrow \frac{173}{12} - \frac{92}{12}\\[1em] \Rightarrow \frac{173 - 92}{12}\\[1em] \Rightarrow \frac{81}{12}\\[1em] \Rightarrow \frac{27}{4}\\[1em] \Rightarrow 6\frac{3}{4}\end{aligned}$$

Hence, the other number = $6\frac{3}{4}$.

Given, total length of wire = $12\frac{3}{4}$ m

Piece of wire that has been cut off = $3\frac{5}{6}$ m

Remaining piece

$$\begin{aligned}\Rightarrow 12\frac{3}{4} - 3\frac{5}{6}\\[1em] \Rightarrow \frac{51}{4} - \frac{23}{6}\\[1em]\end{aligned}$$

LCM of 4 and 6 = 12

$$\begin{aligned}\Rightarrow \frac{51 \times 3}{4 \times 3} - \frac{23 \times 2}{6 \times 2}\\[1em] \Rightarrow \frac{153}{12} - \frac{46}{12}\\[1em] \Rightarrow \frac{153 - 46}{12}\\[1em] \Rightarrow \frac{107}{12}\\[1em] \Rightarrow 8\frac{11}{12}\\[1em]\end{aligned}$$

Hence, the length of the remaining piece = $8\frac{11}{12}$ m.

Given, weight of three boxes = $9\frac{1}{2}$ kg, $14\frac{1}{5}$ kg and $18\frac{3}{4}$ kg

LCM of 2, 5 and 4 = 20

Total weight

$$\begin{aligned}\Rightarrow 9\frac{1}{2} + 14\frac{1}{5} + 18\frac{3}{4}\\[1em] \Rightarrow \frac{19}{2} + \frac{71}{5} + \frac{75}{4}\\[1em] \Rightarrow \frac{19 \times 10}{2 \times 10} + \frac{71 \times 4}{5 \times 4} + \frac{75 \times 5}{4 \times 5}\\[1em] \Rightarrow \frac{190}{20} + \frac{284}{20} + \frac{375}{20}\\[1em] \Rightarrow \frac{190 + 284 + 375}{20}\\[1em] \Rightarrow \frac{849}{20}\\[1em] \Rightarrow 42\frac{9}{20}\end{aligned}$$

Hence, the total weight carried by the porter = $42\frac{9}{20}$ kg.

Given, expenses = ₹ $68\frac{1}{2}$ on food, ₹ $20\frac{3}{4}$ on tea and ₹ $16\frac{2}{5}$ on other eatables

Total

$$\begin{aligned}\Rightarrow 68\frac{1}{2} + 20\frac{3}{4} + 16\frac{2}{5}\\[1em] \Rightarrow \frac{137}{2} + \frac{83}{4} + \frac{82}{5}\\[1em]\end{aligned}$$

LCM of 2, 4 and 5 = 20

$$\begin{aligned}\Rightarrow \frac{137 \times 10}{2 \times 10} + \frac{83 \times 5}{4 \times 5} + \frac{82 \times 4}{5 \times 4}\\[1em] \Rightarrow \frac{1370}{20} + \frac{415}{20} + \frac{328}{20}\\[1em] \Rightarrow \frac{1370 + 415 + 328}{20} \\[1em] \Rightarrow \frac{2113}{20} \\[1em] \Rightarrow 105\frac{13}{20}\end{aligned}$$

Savings = Income − Total expenses

$$\begin{aligned}\text{Savings }= 125 - \frac{2113}{20}\\[1em] = \frac{125 \times 20}{20} - \frac{2113}{20}\\[1em] = \frac{2500}{20} - \frac{2113}{20}\\[1em] = \frac{2500 - 2113}{20}\\[1em] = \frac{387}{20}\\[1em] = 19\frac{7}{20}\end{aligned}$$

Hence, savings = $₹19\frac{7}{20}$.

$$\begin{aligned}\Rightarrow \frac{5}{7} \times \frac{2}{3}\\[1em] \Rightarrow \frac{5 \times 2}{7 \times 3}\\[1em] \Rightarrow \frac{10}{21}\end{aligned}$$

Hence, $\frac{5}{7} \times \frac{2}{3} = \frac{10}{21}$.

$$\begin{aligned}\Rightarrow \frac{3}{5} \times \frac{4}{9}\\[1em] \Rightarrow \frac{3 \times 4}{5 \times 9}\\[1em] \Rightarrow \frac{12}{45}\\[1em] \Rightarrow \frac{4}{15}\end{aligned}$$

Hence, $\frac{3}{5} \times \frac{4}{9} = \frac{4}{15}$.

$$\begin{aligned}\Rightarrow \frac{12}{35} \times \frac{14}{27}\\[1em] \Rightarrow \frac{12}{27} \times \frac{14}{35} \\[1em] \Rightarrow \frac{4}{9} \times \frac{2}{5} \\[1em] \Rightarrow \frac{8}{45}.\end{aligned}$$

Hence, $\frac{12}{35} \times \frac{14}{27} = \frac{8}{45}$.

$$\begin{aligned}\Rightarrow \frac{30}{51} \times \frac{17}{25}\\[1em] \Rightarrow \frac{30}{25} \times \frac{17}{51} \\[1em] \Rightarrow \frac{6}{5} \times \frac{1}{3} \\[1em] \Rightarrow \frac{2}{5}.\end{aligned}$$

Hence, $\frac{30}{51} \times \frac{17}{25} = \frac{2}{5}$.

$$\begin{aligned}\Rightarrow \frac{9}{16} \times \frac{14}{15}\\[1em] \Rightarrow \frac{9}{15} \times \frac{14}{16} \\[1em] \Rightarrow \frac{3}{5} \times \frac{7}{8} \\[1em] \Rightarrow \frac{21}{40}.\end{aligned}$$

Hence, $\frac{9}{16} \times \frac{14}{15} = \frac{21}{40}$.

$$\begin{aligned}\Rightarrow 2\frac{11}{27} \times 18\\[1em] \Rightarrow \frac{65}{27} \times 18\\[1em] \Rightarrow 65 \times \frac{18}{27} \\[1em] \Rightarrow 65 \times \frac{2}{3} \\[1em] \Rightarrow \frac{130}{3} \\[1em] \Rightarrow 43\frac{1}{3}.\end{aligned}$$

Hence, $2\frac{11}{27} \times 18 = 43\frac{1}{3}$.

$$\begin{aligned}\Rightarrow 7\frac{1}{12} \times \frac{8}{17}\\[1em] \Rightarrow \frac{85}{12} \times \frac{8}{17}\\[1em] \Rightarrow \frac{85}{17} \times \frac{8}{12} \\[1em] \Rightarrow 5 \times \frac{2}{3} \\[1em] \Rightarrow \frac{10}{3}\\[1em] \Rightarrow 3\frac{1}{3}.\end{aligned}$$

Hence, $7\frac{1}{12} \times \frac{8}{17} = 3\frac{1}{3}$.

$$\begin{aligned}\Rightarrow 8\frac{1}{21} \times 1\frac{1}{13}\\[1em] \Rightarrow \frac{169}{21} \times \frac{14}{13}\\[1em] \Rightarrow \frac{169}{13} \times \frac{14}{21} \\[1em] \Rightarrow 13 \times \frac{2}{3} \\[1em] \Rightarrow \frac{26}{3} \\[1em] \Rightarrow 8\frac{2}{3}.\end{aligned}$$

Hence, $8\frac{1}{21} \times 1\frac{1}{13} = 8\frac{2}{3}$.

$$\begin{aligned}\Rightarrow 7\frac{7}{11} \times 6\frac{3}{16}\\[1em] \Rightarrow \frac{84}{11} \times \frac{99}{16}\\[1em] \Rightarrow \frac{84}{16} \times \frac{99}{11} \\[1em] \Rightarrow \frac{21}{4} \times 9 \\[1em] \Rightarrow \frac{189}{4} \\[1em] \Rightarrow 47\frac{1}{4}.\end{aligned}$$

Hence, $7\frac{7}{11} \times 6\frac{3}{16} = 47\frac{1}{4}$.

$$\begin{aligned}\Rightarrow 11\frac{1}{4} \times 6\frac{7}{9}\\[1em] \Rightarrow \frac{45}{4} \times \frac{61}{9}\\[1em] \Rightarrow \frac{45}{9} \times \frac{61}{4} \\[1em] \Rightarrow 5 \times \frac{61}{4} \\[1em] \Rightarrow \frac{305}{4} \\[1em] \Rightarrow 76\frac{1}{4}.\end{aligned}$$

Hence, $11\frac{1}{4} \times 6\frac{7}{9} = 76\frac{1}{4}$.

$$\begin{aligned}\Rightarrow 1\frac{3}{4} \times 2\frac{1}{7} \times 4\frac{4}{5}\\[1em] \Rightarrow \frac{7}{4} \times \frac{15}{7} \times \frac{24}{5}\\[1em] \Rightarrow \frac{7}{7} \times \frac{15}{5} \times \frac{24}{4} \\[1em] \Rightarrow 1 \times 3\times 6 \\[1em] \Rightarrow 18.\end{aligned}$$

Hence, $1\frac{3}{4} \times 2\frac{1}{7} \times 4\frac{4}{5} = 18$.

$$\begin{aligned}\Rightarrow 3\frac{1}{6} \times 2\frac{3}{4} \times 2\frac{4}{11}\\[1em] \Rightarrow \frac{19}{6} \times \frac{11}{4} \times \frac{26}{11}\\[1em] \Rightarrow \frac{19}{6} \times \frac{11}{11} \times \frac{26}{4} \\[1em] \Rightarrow \frac{19}{6} \times 1 \times \frac{13}{2} \\[1em] \Rightarrow \frac{247}{12} \\[1em] \Rightarrow 20\frac{7}{12}.\end{aligned}$$

Hence, $3\frac{1}{6} \times 2\frac{3}{4} \times 2\frac{4}{11} = 20\frac{7}{12}$.

The reciprocal of a non-zero fraction is a fraction obtained by swapping its numerator and denominator, so that their product is 1.

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

Hence, the reciprocal of $\frac{5}{9} = \frac{9}{5}$.

(ii) $\frac{7}{13}$

Hence, the reciprocal of $\frac{7}{13} = \frac{13}{7}$.

(iii) $8$

Hence, the reciprocal of $8 = \frac{1}{8}$.

(iv) $\frac{1}{5}$

Hence, the reciprocal of $\frac{1}{5} = \frac{5}{1} = 5$.

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

$\Rightarrow \frac{17}{5}$

Hence, the reciprocal of $\frac{17}{5} = \frac{5}{17}$.

(vi) $8\frac{1}{9}$

$\Rightarrow \frac{73}{9}$

Hence, the reciprocal of $\frac{73}{9} = \frac{9}{73}$.

(vii) $5\frac{3}{8}$

$\Rightarrow \frac{43}{8}$

Hence, the reciprocal of $\frac{43}{8} = \frac{8}{43}$.

(viii) 23

Hence, the reciprocal of $23 = \frac{23}{1} = \frac{1}{23}$.

$$\begin{aligned}\Rightarrow \frac{13}{21} ÷ \frac{2}{7}\\[1em] \Rightarrow \frac{13}{21} \times \frac{7}{2}\\[1em] \Rightarrow \frac{13}{2} \times \frac{7}{21} \\[1em] \Rightarrow \frac{13}{2} \times \frac{1}{3} \\[1em] \Rightarrow \frac{13}{6} \\[1em] \Rightarrow 2\frac{1}{6}.\end{aligned}$$

Hence, $\frac{13}{21} ÷ \frac{2}{7} = 2\frac{1}{6}$.

$$\begin{aligned}\Rightarrow 3\frac{1}{8} ÷ 5\\[1em] \Rightarrow \frac{25}{8} \times \frac{1}{5}\\[1em] \Rightarrow \frac{25 \times 1}{8 \times 5}\\[1em] \Rightarrow \frac{25}{40}\\[1em] \Rightarrow \frac{5}{8}.\end{aligned}$$

Hence, $3\frac{1}{8} ÷ 5 = \frac{5}{8}$.

$$\begin{aligned}\Rightarrow 1 ÷ 2\frac{3}{5}\\[1em] \Rightarrow 1 ÷ \frac{13}{5}\\[1em] \Rightarrow 1 \times \frac{5}{13}\\[1em] \Rightarrow \frac{5}{13}.\end{aligned}$$

Hence, $1 ÷ 2\frac{3}{5} = \frac{5}{13}$.

$$\begin{aligned}\Rightarrow 7\frac{5}{9} ÷ 34\\[1em] \Rightarrow \frac{68}{9} ÷ 34\\[1em] \Rightarrow \frac{68}{9} \times \frac{1}{34}\\[1em] \Rightarrow \frac{68}{34} \times \frac{1}{9} \\[1em] \Rightarrow 2 \times \frac{1}{9} \\[1em] \Rightarrow \frac{2}{9}.\end{aligned}$$

Hence, $7\frac{5}{9} ÷ 34 = \frac{2}{9}$.

Solving,

$$\begin{aligned}\Rightarrow 70 ÷ 8\frac{2}{5}\\[1em] \Rightarrow \frac{70}{1} ÷ \frac{42}{5}\\[1em] \Rightarrow \frac{70}{1} \times \frac{5}{42}\\[1em] \Rightarrow \frac{70}{42} \times 5 \\[1em] \Rightarrow \frac{5}{3} \times 5 \\[1em] \Rightarrow \frac{25}{3}\\[1em] \Rightarrow 8\frac{1}{3}.\end{aligned}$$

Hence, $70 ÷ 8\frac{2}{5} = 8\frac{1}{3}$.

Solving,

$$\begin{aligned}\Rightarrow 4\frac{1}{2} ÷ 6\frac{1}{2}\\[1em] \Rightarrow \frac{9}{2} ÷ \frac{13}{2}\\[1em] \Rightarrow \frac{9}{2} \times \frac{2}{13}\\[1em] \Rightarrow \frac{9 \times 2}{2 \times 13}\\[1em] \Rightarrow \frac{9}{13}.\end{aligned}$$

Hence, $4\frac{1}{2} ÷ 6\frac{1}{2} = \frac{9}{13}$.

Solving,

$$\begin{aligned}\Rightarrow 5\frac{7}{10} ÷ 3\frac{1}{6}\\[1em] \Rightarrow \frac{57}{10} ÷ \frac{19}{6}\\[1em] \Rightarrow \frac{57}{10} \times \frac{6}{19}\\[1em] \Rightarrow \frac{57}{19} \times \frac{6}{10} \\[1em] \Rightarrow 3 \times \frac{3}{5}\\[1em] \Rightarrow \frac{9}{5}\\[1em] \Rightarrow 1\frac{4}{5}.\end{aligned}$$

Hence, $5\frac{7}{10} ÷ 3\frac{1}{6} = 1\frac{4}{5}$.

Solving,

$$\begin{aligned}\Rightarrow 10\frac{5}{7} ÷ 1\frac{11}{14}\\[1em] \Rightarrow \frac{75}{7} ÷ \frac{25}{14}\\[1em] \Rightarrow \frac{75}{7} \times \frac{14}{25}\\[1em] \Rightarrow \frac{75}{25} \times \frac{14}{7}\\[1em] \Rightarrow 3 \times 2 \\[1em] \Rightarrow 6.\end{aligned}$$

Hence, $10\frac{5}{7} ÷ 1\frac{11}{14} = 6$.

Solving,

$$\begin{aligned}\Rightarrow 15\frac{8}{9} ÷ 3\frac{2}{3}\\[1em] \Rightarrow \frac{143}{9} ÷ \frac{11}{3}\\[1em] \Rightarrow \frac{143}{9} \times \frac{3}{11}\\[1em] \Rightarrow \frac{143}{11} \times \frac{3}{9} \\[1em] \Rightarrow 13 \times \frac{1}{3} \\[1em] \Rightarrow \frac{13}{3} \\[1em] \Rightarrow 4\frac{1}{3}.\end{aligned}$$

Hence, $15\frac{8}{9} ÷ 3\frac{2}{3} = 4\frac{1}{3}$

Solving,

$$\begin{aligned}\Rightarrow 9\frac{4}{5} ÷ 3\frac{23}{25}\\[1em] \Rightarrow \frac{49}{5} ÷ \frac{98}{25}\\[1em] \Rightarrow \frac{49}{5} \times \frac{25}{98}\\[1em] \Rightarrow \frac{49}{98} \times \frac{25}{5} \\[1em] \Rightarrow \frac{1}{2} \times 5 \\[1em] \Rightarrow \frac{5}{2} \\[1em] \Rightarrow 2\frac{1}{2}.\end{aligned}$$

Hence, $9\frac{4}{5} ÷ 3\frac{23}{25} = 2\frac{1}{2}$.

Solving,

$$\begin{aligned}\Rightarrow 2\frac{17}{38} ÷ 1\frac{12}{19}\\[1em] \Rightarrow \frac{93}{38} ÷ \frac{31}{19}\\[1em] \Rightarrow \frac{93}{38} \times \frac{19}{31}\\[1em] \Rightarrow \frac{93}{31} \times \frac{19}{38}\\[1em] \Rightarrow 3 \times \frac{1}{2}\\[1em] \Rightarrow \frac{3}{2}\\[1em] \Rightarrow 1\frac{1}{2}.\end{aligned}$$

Hence, $2\frac{17}{38} ÷ 1\frac{12}{19} = 1\frac{1}{2}$.

Solving,

$$\begin{aligned}\Rightarrow 8\frac{7}{25} ÷ 3\frac{1}{15}\\[1em] \Rightarrow \frac{207}{25} ÷ \frac{46}{15}\\[1em] \Rightarrow \frac{207}{25} \times \frac{15}{46}\\[1em] \Rightarrow \frac{207}{46} \times \frac{15}{25} \\[1em] \Rightarrow \frac{9}{2} \times \frac{3}{5} \\[1em] \Rightarrow \frac{27}{10}\\[1em] \Rightarrow 2\frac{7}{10}.\end{aligned}$$

Hence, $8\frac{7}{25} ÷ 3\frac{1}{15} = 2\frac{7}{10}$.

Given, the cost of 1 litre of milk = ₹ $42\frac{3}{5}$

The cost of $12\frac{1}{2}$ litres of milk :

$$\begin{aligned}= ₹42\frac{3}{5} \times 12\frac{1}{2}\\[1em] = ₹\frac{213}{5} \times \frac{25}{2}\\[1em] = ₹\frac{213}{2} \times \frac{25}{5}\\[1em] = ₹\frac{213}{2} \times 5 \\[1em] = ₹\frac{1065}{2}\\[1em] = ₹ 532\frac{1}{2}.\end{aligned}$$

Hence, the cost of $12\frac{1}{2}$ litres of milk = ₹ $532\frac{1}{2}$.

Given, the cost of 1 litre of petrol = ₹ $65\frac{3}{4}$

The cost of 36 litres of petrol

$$\begin{aligned}= ₹ 65\frac{3}{4} \times 36\\[1em] = ₹ \frac{263}{4} \times 36\\[1em] = ₹ \frac{263 \times 36}{4} \\[1em] = ₹ \frac{9468}{4} \\[1em] = ₹ 2,367\end{aligned}$$

Hence, the cost of 36 litres of petrol = ₹ 2,367.

Given, the cost of $3\frac{1}{2}$ metres of cloth = ₹ $547\frac{3}{4}$

The cost of 1 m of cloth :

$$\begin{aligned}= 547\frac{3}{4} ÷ 3\frac{1}{2}\\[1em] = \frac{2,191}{4} ÷ \frac{7}{2}\\[1em] = \frac{2,191}{4} \times \frac{2}{7}\\[1em] = \frac{2,191}{7} \times \frac{2}{4}\\[1em] = 313 \times \frac{1}{2}\\[1em] = \frac{313}{2}\\[1em] = 156\frac{1}{2}.\end{aligned}$$

Hence, the cost of 1 metre of cloth = ₹ $156\frac{1}{2}$.

Given, total length of cloth = 54 m

Length of each piece = $3\frac{3}{8}$ m

Let n be the number of pieces.

By formula,

Total length = Number of pieces x length of each piece

$$\begin{aligned}\Rightarrow n \times 3\frac{3}{8} = 54\\[1em] \Rightarrow n \times \frac{27}{8} = 54\\[1em] \Rightarrow n = 54 ÷ \frac{27}{8}\\[1em] \Rightarrow n = 54 \times \frac{8}{27}\\[1em] \Rightarrow n = \frac{54}{27} \times 8 \\[1em] \Rightarrow n = 2 \times 8 \\[1em] \Rightarrow n = 16.\end{aligned}$$

Hence, the number of pieces = 16.

Given, total length of cord = $126\frac{1}{2}$ m

Number of pieces = 46

Let x be the length of each piece.

By formula,

Total length = Number of pieces x length of each piece

$$\begin{aligned}\Rightarrow 46 \times x = 126\frac{1}{2}\\[1em] \Rightarrow 46 \times x = \frac{253}{2} \\[1em] \Rightarrow x = \frac{253}{2} ÷ 46\\[1em] \Rightarrow x = \frac{253}{2} \times \frac{1}{46} \\[1em] \Rightarrow x = \frac{253}{46} \times \frac{1}{2} \\[1em] \Rightarrow x = \frac{253}{92} \\[1em] \Rightarrow x = \frac{23 \times 11}{23 \times 4} \\[1em] \Rightarrow x = \frac{11}{4} \\[1em] \Rightarrow x = 2\frac{3}{4}.\end{aligned}$$

Hence, the length of each piece = $2\frac{3}{4}$ m.

Given, total distance = $283\frac{1}{2}$ km

Total time = $4\frac{2}{3}$ hours

Let x be the distance covered in 1 hour.

By formula,

Distance covered in 1 hour x Total time = Total distance

$$\begin{aligned}\Rightarrow x \times 4\frac{2}{3} = 283\frac{1}{2}\\[1em] \Rightarrow x \times \frac{14}{3} = \frac{567}{2}\\[1em] \Rightarrow x = \frac{567}{2} ÷ \frac{14}{3}\\[1em] \Rightarrow x = \frac{567}{2} \times \frac{3}{14}\\[1em] \Rightarrow x = \frac{567}{14} \times \frac{3}{2} \\[1em] \Rightarrow x = \frac{81}{2} \times \frac{3}{2} \\[1em] \Rightarrow x = \frac{243}{4} \\[1em] \Rightarrow x = 60\frac{3}{4}.\end{aligned}$$

Hence, distance covered in 1 hour = $60\frac{3}{4}$ km.

Given, the area of a rectangular plot of land = $46\frac{2}{5}$ sq. m

Length of rectangular plot = $7\frac{1}{4}$ m

Let b be the breath of the rectangular plot.

By formula, length x breadth = area

$$\begin{aligned}\Rightarrow 7\frac{1}{4} \times b = 46\frac{2}{5}\\[1em] \Rightarrow \frac{29}{4} \times b = \frac{232}{5}\\[1em] \Rightarrow b = \frac{232}{5} ÷ \frac{29}{4}\\[1em] \Rightarrow b = \frac{232}{5} \times \frac{4}{29}\\[1em] \Rightarrow b = \frac{232 \times 4}{5 \times 29}\\[1em] \Rightarrow b = \frac{232}{29} \times \frac{4}{5} \\[1em] \Rightarrow b = 8 \times \frac{4}{5} \\[1em] \Rightarrow b = \frac{32}{5}\\[1em] \Rightarrow b = 6\frac{2}{5}.\end{aligned}$$

Hence, the breadth of the rectangular plot = $6\frac{2}{5}$ m.

Given, the product of two fractions = $15\frac{3}{4}$

One of the fraction = $4\frac{1}{2}$

Let x be the other fraction.

$$\begin{aligned}\Rightarrow x \times 4\frac{1}{2} = 15\frac{3}{4} \\[1em] \Rightarrow x = 15\frac{3}{4} ÷ 4\frac{1}{2}\\[1em] = \frac{63}{4} ÷ \frac{9}{2}\\[1em] = \frac{63}{4} \times \frac{2}{9}\\[1em] = \frac{63 \times 2}{4 \times 9}\\[1em] = \frac{126}{36}\\[1em] = \frac{7}{2}\\[1em] = 3\frac{1}{2}.\end{aligned}$$

Hence, the other fraction = $3\frac{1}{2}$.

(i) $\frac{1}{8}$ of 40

$$\begin{aligned}\Rightarrow \frac{1}{8} \times 40\\[1em] \Rightarrow \frac{1 \times 40}{8}\\[1em] \Rightarrow \frac{40}{8}\\[1em] \Rightarrow 5\end{aligned}$$

Hence, $\frac{1}{8}$ of 40 = 5.

(ii) $\frac{4}{11}$ of $4\frac{2}{5}$

$$\begin{aligned}\Rightarrow \frac{4}{11} \times 4\frac{2}{5}\\[1em] \Rightarrow \frac{4}{11} \times \frac{22}{5}\\[1em] \Rightarrow \frac{4}{5} \times \frac{22}{11}\\[1em] \Rightarrow \frac{4}{5} \times 2 \\[1em] \Rightarrow \frac{8}{5}\\[1em] \Rightarrow 1\frac{3}{5}.\end{aligned}$$

Hence, $\frac{4}{11} \text{ of } 4\frac{2}{5} = 1\frac{3}{5}$.

(iii) $1\frac{3}{5}$ of $6\frac{1}{4}$

$$\begin{aligned}\Rightarrow 1\frac{3}{5} \times 6\frac{1}{4}\\[1em] \Rightarrow \frac{8}{5} \times \frac{25}{4}\\[1em] \Rightarrow \frac{8 \times 25}{5 \times 4}\\[1em] \Rightarrow \frac{200}{20}\\[1em] \Rightarrow 10.\end{aligned}$$

Hence, $1\frac{3}{5} \text{ of } 6\frac{1}{4} = 10$.

(iv) $\frac{3}{4}$ of ₹ 1

$$\begin{aligned}\Rightarrow \frac{3}{4} \times ₹ 1\\[1em] \Rightarrow \frac{3}{4} \times 100 \text{ paise}\\[1em] \Rightarrow \frac{3 \times 100}{4}\text{ paise}\\[1em] \Rightarrow \frac{300}{4}\text{ paise}\\[1em] \Rightarrow 75\text{ paise}\end{aligned}$$

Hence, $\frac{3}{4}$ of ₹ 1 = 75 paise.

(v) $\frac{5}{8}$ of 1 km

$$\begin{aligned}\Rightarrow \frac{5}{8} \times 1\text{ km}\\[1em] \Rightarrow \frac{5}{8} \times 1000\text{ m}\\[1em] \Rightarrow \frac{5 \times 1000}{8}\text{ m}\\[1em] \Rightarrow \frac{5000}{8}\text{ m}\\[1em] \Rightarrow 625\text{ m}\\[1em]\end{aligned}$$

Hence, $\frac{5}{8}$ of 1 km = 625 m.

(vi) $\frac{5}{12}$ of 1 hour

$$\begin{aligned}\Rightarrow \frac{5}{12} \times 1 \text{ hour}\\[1em] \Rightarrow \frac{5}{12} \times 60 \text{ min}\\[1em] \Rightarrow \frac{5 \times 60}{12} \text{ min}\\[1em] \Rightarrow \frac{300}{12} \text{ min}\\[1em] \Rightarrow 25 \text{ min}\\[1em]\end{aligned}$$

Hence, $\frac{5}{12}$ of 1 hour = 25 min.

Solving,

$$\begin{aligned}\Rightarrow 1\frac{1}{6} ÷ 1\frac{5}{9} \times 3\frac{1}{3}\\[1em] \Rightarrow \frac{7}{6} ÷ \frac{14}{9} \times \frac{10}{3}\\[1em] \Rightarrow \frac{7}{6} \times \frac{9}{14} \times \frac{10}{3}\\[1em] \Rightarrow \frac{7}{14} \times \frac{9}{3} \times \frac{10}{6}\\[1em] \Rightarrow \frac{1}{2} \times 3 \times \frac{5}{3}\\[1em] \Rightarrow \frac{5}{2}\\[1em] \Rightarrow 2\frac{1}{2}.\end{aligned}$$

Hence, $1\frac{1}{6} ÷ 1\frac{5}{9} \times 3\frac{1}{3} = 2\frac{1}{2}$.

Solving,

$$\begin{aligned}\Rightarrow 1\frac{1}{3} ÷ \frac{3}{7} \text{ of } 2\frac{5}{8} + 1\frac{1}{9}\\[1em] \Rightarrow \frac{4}{3} ÷ \Big(\frac{3}{7} \times \frac{21}{8}\Big) + \frac{10}{9}\\[1em] \Rightarrow \frac{4}{3} ÷ \frac{3 \times 21}{7 \times 8} + \frac{10}{9}\\[1em] \Rightarrow \frac{4}{3} ÷ \frac{63}{56} + \frac{10}{9}\\[1em] \Rightarrow \frac{4}{3} ÷ \frac{9}{8} + \frac{10}{9}\\[1em] \Rightarrow \frac{4}{3} \times \frac{8}{9} + \frac{10}{9}\\[1em] \Rightarrow \frac{4 \times 8}{3 \times 9} + \frac{10}{9}\\[1em] \Rightarrow \frac{32}{27} + \frac{10}{9}\\[1em] \Rightarrow \frac{32}{27} + \frac{10 \times 3}{9 \times 3}\\[1em] \Rightarrow \frac{32}{27} + \frac{30}{27}\\[1em] \Rightarrow \frac{32 + 30}{27}\\[1em] \Rightarrow \frac{62}{27}\\[1em] \Rightarrow 2\frac{8}{27}\end{aligned}$$

Hence, $1\frac{1}{3} ÷ \frac{3}{7}$ of $2\frac{5}{8} + 1\frac{1}{9} = 2\frac{8}{27}$.

$$\begin{aligned}\Rightarrow 6\frac{1}{5} ÷ 3\frac{1}{10} \text{ of } 2\frac{1}{2} ÷ \frac{1}{4}\\[1em] \Rightarrow 6\frac{1}{5} ÷ \Big(\frac{31}{10} \times \frac{5}{2}\Big) ÷ \frac{1}{4}\\[1em] \Rightarrow 6\frac{1}{5} ÷ \frac{31 \times 5}{10 \times 2} ÷ \frac{1}{4}\\[1em] \Rightarrow 6\frac{1}{5} ÷ \frac{155}{20} ÷ \frac{1}{4}\\[1em] \Rightarrow \Big(6\frac{1}{5} ÷ \frac{31}{4}\Big) ÷ \frac{1}{4}\\[1em] \Rightarrow \Big(\frac{31}{5} ÷ \frac{31}{4}\Big) ÷ \frac{1}{4}\\[1em] \Rightarrow \Big(\frac{31}{5} \times \frac{4}{31}\Big) ÷ \frac{1}{4}\\[1em] \Rightarrow \frac{4}{5} ÷ \frac{1}{4} \\[1em] \Rightarrow \frac{4}{5} \times 4 \\[1em] \Rightarrow \frac{16}{5}\\[1em] \Rightarrow 3\frac{1}{5}.\end{aligned}$$

Hence, $6\frac{1}{5} ÷ 3\frac{1}{10}$ of $2\frac{1}{2} ÷ \frac{1}{4} = 3\frac{1}{5}$.

$$\begin{aligned}\Rightarrow 3\frac{2}{3} - \frac{3}{11} \text{ of } 2\frac{3}{4} ÷ 1\frac{1}{4} \times 1\frac{2}{3} + \frac{1}{3}\\[1em] \Rightarrow 3\frac{2}{3} - \frac{3}{11} \times 2\frac{3}{4} ÷ 1\frac{1}{4} \times 1\frac{2}{3} + \frac{1}{3}\\[1em] \Rightarrow 3\frac{2}{3} - \frac{3}{11} \times \frac{11}{4} ÷ 1\frac{1}{4} \times 1\frac{2}{3} + \frac{1}{3}\\[1em] \Rightarrow 3\frac{2}{3} - \frac{3 \times 11}{11 \times 4} ÷ 1\frac{1}{4} \times 1\frac{2}{3} + \frac{1}{3}\\[1em] \Rightarrow 3\frac{2}{3} - \frac{3 \times 11}{11 \times 4} ÷ 1\frac{1}{4} \times 1\frac{2}{3} + \frac{1}{3}\\[1em] \Rightarrow 3\frac{2}{3} - \frac{3}{4} ÷ 1\frac{1}{4} \times 1\frac{2}{3} + \frac{1}{3}\\[1em] \Rightarrow 3\frac{2}{3} - \frac{3}{4} ÷ \frac{5}{4} \times 1\frac{2}{3} + \frac{1}{3}\\[1em] \Rightarrow 3\frac{2}{3} - \frac{3}{4} \times \frac{4}{5} \times \frac{5}{3} + \frac{1}{3}\\[1em] \Rightarrow 3\frac{2}{3} - \frac{3 \times 4 \times 5}{4 \times 5 \times 3} + \frac{1}{3}\\[1em] \Rightarrow 3\frac{2}{3} - 1 + \frac{1}{3}\\[1em] \Rightarrow \frac{11}{3} + \frac{1}{3} - 1\\[1em] \Rightarrow \frac{11 + 1}{3} - 1\\[1em] \Rightarrow \frac{12}{3} - 1\\[1em] \Rightarrow 4 - 1\\[1em] \Rightarrow 3\end{aligned}$$

Hence, $3\frac{2}{3} - \frac{3}{11}$ of $2\frac{3}{4} ÷ 1\frac{1}{4} \times 1\frac{2}{3} + \frac{1}{3} = 3$.

$$\begin{aligned}\Rightarrow \Big(2\frac{2}{7} \text{ of } 15\frac{3}{4}\Big) \times 2\frac{1}{4} ÷ \Big(\frac{4}{7} \text{ of } 2\frac{5}{8}\Big)\\[1em] \Rightarrow \frac{16}{7} \times \frac{63}{4} \times 2\frac{1}{4} ÷ \frac{4}{7} \times \frac{21}{8}\\[1em] \Rightarrow \frac{16 \times 63}{7 \times 4} \times 2\frac{1}{4} ÷ \frac{4 \times 21}{7 \times 8} \\[1em] \Rightarrow 4 \times 9 \times 2\frac{1}{4} ÷ \frac{3}{2} \\[1em] \Rightarrow 36 \times 2\frac{1}{4} ÷ \frac{3}{2} \\[1em] \Rightarrow 36 \times 2\frac{1}{4} \times \frac{2}{3} \\[1em] \Rightarrow 36 \times \frac{9}{4} \times \frac{2}{3} \\[1em] \Rightarrow 36 \times \frac{3}{2} \\[1em] \Rightarrow 18 \times 3 \\[1em] \Rightarrow 54.\end{aligned}$$

Hence, $2\frac{2}{7}$ of $15\frac{3}{4} \times 2\frac{1}{4} ÷ \frac{4}{7}$ of $2\frac{5}{8} = 54$.

$$\begin{aligned}\Rightarrow 1 ÷ \frac{4}{7} - \Big(\frac{1}{3} \text{ of } 3\frac{3}{4}\Big) + \frac{1}{2} ÷ 3\\[1em] \Rightarrow 1 ÷ \frac{4}{7} - \Big(\frac{1}{3} \times 3\frac{3}{4}\Big) + \frac{1}{2} ÷ 3\\[1em] \Rightarrow 1 ÷ \frac{4}{7} - \frac{1}{3} \times \frac{15}{4} + \frac{1}{2} ÷ 3\\[1em] \Rightarrow 1 ÷ \frac{4}{7} - \frac{1 \times 15}{3 \times 4} + \frac{1}{2} ÷ 3\\[1em] \Rightarrow 1 ÷ \frac{4}{7} - \frac{15}{12} + \frac{1}{2} ÷ 3\\[1em] \Rightarrow 1 ÷ \frac{4}{7} - \frac{5}{4} + \frac{1}{2} ÷ 3\\[1em] \Rightarrow 1 \times \frac{7}{4} - \frac{5}{4} + \frac{1}{2} \times \frac{1}{3}\\[1em] \Rightarrow \frac{7}{4} - \frac{5}{4} + \frac{1 \times 1}{2 \times 3}\\[1em] \Rightarrow \frac{7}{4} - \frac{5}{4} + \frac{1}{6}\\[1em] \Rightarrow \frac{7 \times 3}{4 \times 3} - \frac{5 \times 3}{4 \times 3} + \frac{1 \times 2}{6 \times 2}\\[1em] \Rightarrow \frac{21}{12} - \frac{15}{12} + \frac{2}{12}\\[1em] \Rightarrow \frac{21 - 15 + 2}{12}\\[1em] \Rightarrow \frac{21 - 13}{12}\\[1em] \Rightarrow \frac{8}{12}\\[1em] \Rightarrow \frac{2}{3}\end{aligned}$$

Hence, $1 ÷ \frac{4}{7} - \frac{1}{3}$ of $3\frac{3}{4} + \frac{1}{2} ÷ 3 = \frac{2}{3}$.

$$\begin{aligned}\Rightarrow \Big(\frac{2}{3} + \frac{4}{9}\Big) \text{ of } \frac{3}{5} ÷ 1\frac{2}{3} \times 1\frac{1}{4} - \frac{1}{3}\\[1em] \Rightarrow \Big(\frac{2 \times 3}{3 \times 3} + \frac{4}{9}\Big) \text{ of } \frac{3}{5} ÷ 1\frac{2}{3} \times 1\frac{1}{4} - \frac{1}{3}\\[1em] \Rightarrow \Big(\frac{6}{9} + \frac{4}{9}\Big) \text{ of } \frac{3}{5} ÷ 1\frac{2}{3} \times 1\frac{1}{4} - \frac{1}{3}\\[1em] \Rightarrow \Big(\frac{6 + 4}{9}\Big) \text{ of } \frac{3}{5} ÷ 1\frac{2}{3} \times 1\frac{1}{4} - \frac{1}{3}\\[1em] \Rightarrow \frac{10}{9} \times \frac{3}{5} ÷ 1\frac{2}{3} \times 1\frac{1}{4} - \frac{1}{3}\\[1em] \Rightarrow \frac{10 \times 3}{9 \times 5} ÷ 1\frac{2}{3} \times 1\frac{1}{4} - \frac{1}{3}\\[1em] \Rightarrow \frac{30}{45} ÷ 1\frac{2}{3} \times 1\frac{1}{4} - \frac{1}{3}\\[1em] \Rightarrow \frac{2}{3} ÷ 1\frac{2}{3} \times 1\frac{1}{4} - \frac{1}{3}\\[1em] \Rightarrow \frac{2}{3} ÷ \frac{5}{3} \times \frac{5}{4} - \frac{1}{3}\\[1em] \Rightarrow \frac{2}{3} \times \frac{3}{5} \times \frac{5}{4} - \frac{1}{3}\\[1em] \Rightarrow \frac{2 \times 3 \times 5}{3 \times 5 \times 4} - \frac{1}{3}\\[1em] \Rightarrow \frac{2 \times \cancel{3} \times \cancel{5}}{\cancel{3} \times \cancel{5} \times 4} - \frac{1}{3}\\[1em] \Rightarrow \frac{2}{4} - \frac{1}{3}\\[1em] \Rightarrow \frac{1}{2} - \frac{1}{3}\\[1em] \Rightarrow \frac{1 \times 3}{2 \times 3} - \frac{1 \times 2}{3 \times 2}\\[1em] \Rightarrow \frac{3}{6} - \frac{2}{6}\\[1em] \Rightarrow \frac{3 - 2}{6}\\[1em] \Rightarrow \frac{1}{6}\end{aligned}$$

Hence, $\Big(\frac{2}{3} + \frac{4}{9}\Big)$ of $\frac{3}{5} ÷ 1\frac{2}{3} \times 1\frac{1}{4} - \frac{1}{3} = \frac{1}{6}$

$$\begin{aligned}\Rightarrow \Big(\frac{14}{15} ÷ 1\frac{1}{6} + \frac{7}{10}\Big) \times \frac{3}{4}\\[1em] \Rightarrow \Big(\frac{14}{15} ÷ \frac{7}{6} + \frac{7}{10}\Big) \times \frac{3}{4}\\[1em] \Rightarrow \Big(\frac{14}{15} \times \frac{6}{7} + \frac{7}{10}\Big) \times \frac{3}{4}\\[1em] \Rightarrow \Big(\frac{14 \times 6}{15 \times 7} + \frac{7}{10}\Big) \times \frac{3}{4}\\[1em] \Rightarrow \Big(\frac{84}{105} + \frac{7}{10}\Big) \times \frac{3}{4}\\[1em] \Rightarrow \Big(\frac{4}{5} + \frac{7}{10}\Big) \times \frac{3}{4}\\[1em] \Rightarrow \Big(\frac{4 \times 2}{5 \times 2} + \frac{7}{10}\Big) \times \frac{3}{4}\\[1em] \Rightarrow \Big(\frac{8}{10} + \frac{7}{10}\Big) \times \frac{3}{4}\\[1em] \Rightarrow \Big(\frac{8 + 7}{10}\Big) \times \frac{3}{4}\\[1em] \Rightarrow \frac{15}{10} \times \frac{3}{4}\\[1em] \Rightarrow \frac{15 \times 3}{10 \times 4}\\[1em] \Rightarrow \frac{45}{40}\\[1em] \Rightarrow \frac{9}{8}\\[1em] \Rightarrow 1\frac{1}{8}\end{aligned}$$

Hence, $\Big(\frac{14}{15} ÷ 1\frac{1}{6} + \frac{7}{10}\Big) \times \frac{3}{4} = 1\frac{1}{8}$

$$\begin{aligned}\Rightarrow \frac{1}{3}\Big(2\frac{1}{2} + 3\frac{1}{3}\Big) ÷ \frac{2}{9}\Big(3\frac{1}{8} - 1\frac{1}{12}\Big)\\[1em] \Rightarrow \frac{1}{3}\Big(\frac{5}{2} + \frac{10}{3}\Big) ÷ \frac{2}{9}\Big(\frac{25}{8} - \frac{13}{12}\Big)\\[1em] \Rightarrow \frac{1}{3}\Big(\frac{5 \times 3}{2 \times 3} + \frac{10 \times 2}{3 \times 2}\Big) ÷ \frac{2}{9}\Big(\frac{25 \times 3}{8 \times 3} - \frac{13 \times 2}{12 \times 2}\Big)\\[1em] \Rightarrow \frac{1}{3}\Big(\frac{15}{6} + \frac{20}{6}\Big) ÷ \frac{2}{9}\Big(\frac{75}{24} - \frac{26}{24}\Big)\\[1em] \Rightarrow \frac{1}{3}\Big(\frac{15 + 20}{6}\Big) ÷ \frac{2}{9}\Big(\frac{75 - 26}{24}\Big)\\[1em] \Rightarrow \frac{1}{3}\Big(\frac{35}{6}\Big) ÷ \frac{2}{9}\Big(\frac{49}{24}\Big)\\[1em] \Rightarrow \frac{35 \times 1}{6 \times 3} ÷ \frac{49 \times 2}{24 \times 9}\\[1em] \Rightarrow \frac{35}{18} ÷ \frac{98}{216}\\[1em] \Rightarrow \frac{35}{18} \times \frac{216}{98}\\[1em] \Rightarrow \frac{35}{98} \times \frac{216}{18} \\[1em] \Rightarrow \frac{5}{14} \times 12 \\[1em] \Rightarrow \frac{60}{14}\\[1em] \Rightarrow \frac{30}{7} \\[1em] \Rightarrow 4\frac{2}{7}\end{aligned}$$

Hence, $\frac{1}{3}\Big(2\frac{1}{2} + 3\frac{1}{3}\Big) ÷ \frac{2}{9}\Big(3\frac{1}{8} - 1\frac{1}{12}\Big) = 4\frac{2}{7}$.

$$\begin{aligned}\Rightarrow \Big(\frac{1}{4} - \frac{1}{9}\Big) ÷ \Big(\frac{1}{2} + \frac{1}{4} ÷ \frac{1}{3}\Big)\\[1em] \Rightarrow \Big(\frac{1 \times 9}{4 \times 9} - \frac{1 \times 4}{9 \times 4}\Big) ÷ \Big(\frac{1}{2} + \frac{1}{4} \times \frac{3}{1}\Big)\\[1em] \Rightarrow \Big(\frac{9}{36} - \frac{4}{36}\Big) ÷ \Big(\frac{1}{2} + \frac{1 \times 3}{4 \times 1}\Big)\\[1em] \Rightarrow \Big(\frac{9 - 4}{36}\Big) ÷ \Big(\frac{1}{2} + \frac{3}{4}\Big)\\[1em] \Rightarrow \Big(\frac{5}{36}\Big) ÷ \Big(\frac{1}{2} + \frac{3}{4}\Big)\\[1em] \Rightarrow \Big(\frac{5}{36}\Big) ÷ \Big(\frac{1 \times 2}{2\times 2} + \frac{3}{4}\Big)\\[1em] \Rightarrow \Big(\frac{5}{36}\Big) ÷ \Big(\frac{2}{4} + \frac{3}{4}\Big)\\[1em] \Rightarrow \Big(\frac{5}{36}\Big) ÷ \Big(\frac{2 + 3}{4}\Big)\\[1em] \Rightarrow \Big(\frac{5}{36}\Big) ÷ \Big(\frac{5}{4}\Big)\\[1em] \Rightarrow \frac{5}{36} \times \frac{4}{5}\\[1em] \Rightarrow \frac{5 \times 4}{36 \times 5}\\[1em] \Rightarrow \frac{4}{36}\\[1em] \Rightarrow \frac{1}{9}.\end{aligned}$$

Hence, $\Big(\frac{1}{4} - \frac{1}{9}\Big) ÷ \Big(\frac{1}{2} + \frac{1}{4} ÷ \frac{1}{3}\Big) = \frac{1}{9}$.

$$\begin{aligned}\Rightarrow 3\frac{7}{8} - \left\lbrace1\frac{3}{8} \div \Big(2\frac{4}{5} - 1\frac{7}{10}\Big)\right\rbrace\\[1em] \Rightarrow 3\frac{7}{8} - \left\lbrace1\frac{3}{8} \div \Big(\frac{14}{5} - \frac{17}{10}\Big)\right\rbrace\\[1em] \Rightarrow 3\frac{7}{8} - \left\lbrace1\frac{3}{8} \div \Big(\frac{14 \times 2}{5 \times 2} - \frac{17}{10}\Big)\right\rbrace\\[1em] \Rightarrow 3\frac{7}{8} - \left\lbrace1\frac{3}{8} \div \Big(\frac{28}{10} - \frac{17}{10}\Big)\right\rbrace\\[1em] \Rightarrow 3\frac{7}{8} - \left\lbrace1\frac{3}{8} \div \Big(\frac{28 - 17}{10}\Big)\right\rbrace\\[1em] \Rightarrow 3\frac{7}{8} - \left\lbrace1\frac{3}{8} \div \Big(\frac{11}{10}\Big)\right\rbrace\\[1em] \Rightarrow 3\frac{7}{8} - \left\lbrace\frac{11}{8} \div \frac{11}{10}\right\rbrace\\[1em] \Rightarrow 3\frac{7}{8} - \left\lbrace\frac{11}{8} \times \frac{10}{11}\right\rbrace\\[1em] \Rightarrow 3\frac{7}{8} - \frac{10}{8}\\[1em] \Rightarrow \frac{31}{8} - \frac{10}{8}\\[1em] \Rightarrow \frac{31 - 10}{8}\\[1em] \Rightarrow \frac{21}{8}\\[1em] \Rightarrow 2\frac{5}{8}\end{aligned}$$

Hence, $3\frac{7}{8} - \left\lbrace1\frac{3}{8} \div \Big(2\frac{4}{5} - 1\frac{7}{10}\Big)\right\rbrace = 2\frac{5}{8}$.

$$\begin{aligned}\Rightarrow 3 \div \Big[3\times \left\lbrace3 - \Big(3 - \frac{1}{4}\Big)\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[3\times \left\lbrace3 - \Big(\frac{3}{1} - \frac{1}{4}\Big)\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[3\times \left\lbrace3 - \Big(\frac{3 \times 4}{1 \times 4} - \frac{1}{4}\Big)\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[3\times \left\lbrace3 - \Big(\frac{12}{4} - \frac{1}{4}\Big)\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[3\times \left\lbrace3 - \Big(\frac{12 - 1}{4}\Big)\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[3\times \left\lbrace3 - \Big(\frac{11}{4}\Big)\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[3\times \left\lbrace\frac{3}{1} - \frac{11}{4}\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[3\times \left\lbrace\frac{3 \times 4}{1 \times 4} - \frac{11}{4}\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[3\times \left\lbrace\frac{12}{4} - \frac{11}{4}\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[3\times \left\lbrace\frac{12 - 11}{4}\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[3\times \left\lbrace\frac{1}{4}\right\rbrace \Big]\\[1em] \Rightarrow 3 \div \Big[\frac{3 \times 1}{4} \Big]\\[1em] \Rightarrow 3 \div \frac{3}{4}\\[1em] \Rightarrow 3 \times \frac{4}{3}\\[1em] \Rightarrow \frac{4 \times 3}{3}\\[1em] \Rightarrow 4\end{aligned}$$

Hence, $3 \div \Big[3\times \left\lbrace3 - \Big(3 - \frac{1}{4}\Big)\right\rbrace \Big] = 4$.

$$\begin{aligned}\Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{3}{4} - \frac{1}{2} \times \Big(\frac{7}{10} - \frac{3}{5}\Big)\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{3}{4} - \frac{1}{2} \times \Big(\frac{7}{10} - \frac{3 \times 2}{5 \times 2}\Big)\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{3}{4} - \frac{1}{2} \times \Big(\frac{7}{10} - \frac{6}{10}\Big)\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{3}{4} - \frac{1}{2} \times \Big(\frac{7 - 6}{10}\Big)\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{3}{4} - \frac{1}{2} \times \frac{1}{10}\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{3}{4} - \frac{1 \times 1}{2 \times 10}\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{3}{4} - \frac{1}{20}\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{3 \times 5}{4 \times 5} - \frac{1}{20}\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{15}{20} - \frac{1}{20}\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{15 - 1}{20}\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{14}{20}\right\rbrace\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[\frac{7}{3} \div \frac{14}{20}\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[\frac{7}{3} \times \frac{20}{14}\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[\frac{7 \times 20}{3 \times 14}\Big]\\[1em] \Rightarrow 5\frac{1}{3} - \Big[\frac{140}{42}\Big]\\[1em] \Rightarrow \frac{16}{3} - \frac{10}{3}\\[1em] \Rightarrow \frac{16 - 10}{3}\\[1em] \Rightarrow \frac{6}{3}\\[1em] \Rightarrow 2\end{aligned}$$

Hence, $5\frac{1}{3} - \Big[2\frac{1}{3} \div \left\lbrace\frac{3}{4} - \frac{1}{2} \times \Big(\frac{7}{10} - \frac{3}{5}\Big)\right\rbrace\Big] = 2$.

(i) Given, a man earns = ₹ 18,720 per month

$$\begin{aligned}\text{Expenditure }= \frac{5}{6} \text{ of his income}\\[1em] = \frac{5}{6} \times 18720\\[1em] = \frac{5 \times 18720}{6}\\[1em] = 5 \times 3120\\[1em] = 15,600.\end{aligned}$$

Hence, expenditure = ₹ 15,600.

(ii) Saving = Income - Expenditure

= 18,720 - 15,600

= ₹ 3,120.

Hence, saving = ₹ 3,120.

Given,

Total capacity of tank = $56\frac{1}{4}$ = $\frac{225}{4}$ litres

Fraction full = $\frac{8}{15}$

$$\begin{aligned}\text{Water in the tank }= \frac{8}{15} \times \frac{225}{4}\\[1em] = \frac{8 \times 225}{15 \times 4}\\[1em] = \frac{1800}{60}\\[1em] = 30.\end{aligned}$$

Hence, water in the tank = 30 litres.

Let the total number of pages be x.

Given, number of pages read = $\frac{5}{8}$ x

Pages are left = 168

Number of pages left

$$\begin{aligned}\Rightarrow 1x - \frac{5}{8}x = 168\\[1em] \Rightarrow \frac{8x}{8} - \frac{5x}{8} = 168\\[1em] \Rightarrow \frac{8x - 5x}{8} = 168\\[1em] \Rightarrow \frac{3x}{8} = 168\\[1em] \Rightarrow x = \frac{168 \times 8}{3}\\[1em] \Rightarrow x = \frac{1344}{3}\\[1em] \Rightarrow x = 448\end{aligned}$$

Hence, total number of pages = 448.

(i) Given,

Number of boys in a class = $\frac{2}{3}$ of the students

Number of girls in a class = 17

Let the number of the students in the class be x.

Number of girls = Total students - number of boys

$$\begin{aligned}\Rightarrow x - \frac{2}{3}x = 17\\[1em] \Rightarrow \frac{3}{3}x - \frac{2}{3}x = 17\\[1em] \Rightarrow \frac{3 - 2}{3}x = 17\\[1em] \Rightarrow \frac{1}{3}x = 17\\[1em] \Rightarrow x = 17\times 3\\[1em] \Rightarrow x = 51\end{aligned}$$

Number of boys = $\frac{2}{3} \times 51 = \frac{2 \times 51}{3} = \frac{102}{3} = 34$

Hence, number of boys in the class = 34.

(ii) Hence, number of students in the class = 51.

Given,

Total salary = ₹ 25440

House rent = ​$\frac{1}{4}$ part of the salary

Food & clothes = ​$\frac{3}{8}$ part of the salary

Insurance = $\frac{1}{10}$ part of the salary

Other items = $\frac{1}{5}$​ part of the salary

$$\begin{aligned}\text{Total fraction spent } = \frac{1}{4} + \frac{3}{8} + \frac{1}{10} + \frac{1}{5}\\[1em] = \frac{1 \times 10}{4 \times 10} + \frac{3 \times 5}{8 \times 5} + \frac{1 \times 4}{10 \times 4} + \frac{1 \times 8}{5 \times 8}\\[1em] = \frac{10}{40} + \frac{15}{40} + \frac{4}{40} + \frac{8}{40}\\[1em] = \frac{10 + 15 + 4 + 8}{40}\\[1em] = \frac{37}{40}\\[1em]\end{aligned}$$

Saving fraction = 1 - Total spent fraction

$$\begin{aligned}\Rightarrow \text{Saving fraction } = 1 - \frac{37}{40} \\[1em] = \frac{40}{40} - \frac{37}{40} \\[1em] = \frac{40 - 37}{40} \\[1em] = \frac{3}{40} \\[1em]\end{aligned}$$

Saving amount = Saving fraction × Salary

= $\frac{3}{40} \times 25,440 = \frac{3 \times 25,440}{40} = \frac{76,320}{40} = 1,908$.

Hence, saving per month = ₹ 1,908.

Given,

Total students = 168

Part of students who gave all correct answers = $\frac{5}{6}$

Part of students who made 1 or more mistakes = $1 - \frac{5}{6} = \frac{6 - 5}{6} = \frac{1}{6}$.

Students who made 1 or more mistakes = $\frac{1}{6}$ × Total number of students

= $\frac{1}{6} \times 168$

= 28.

Hence, no. of students who made 1 or more mistakes = 28.

Given,

Part of guava trees = $\frac{1}{3}$

Part of banana trees = $\frac{1}{8}$

Part of mango trees = 1 - (part of guava trees + part of banana trees)

$$\begin{aligned}= 1 - \Big(\frac{1}{3} + \frac{1}{8}\Big) \\[1em] = 1 - \Big(\frac{8 + 3}{24}\Big) \\[1em] = 1 - \frac{11}{24} \\[1em] = \frac{24 - 11}{24} \\[1em] = \frac{13}{24}.\end{aligned}$$

No. of mango trees = Part of mango trees × Total number of trees

$$\begin{aligned}117 = \frac{13}{24} \times \text{Total number of trees} \\[1em] \text{Total number of trees} = 117 \times \frac{24}{13} \\[1em] \text{Total number of trees} = 9 \times 24 = 216.\end{aligned}$$

Hence, total number of trees = 216.

Given,

Total students present on a certain day = 720

Part of students absent on a certain day = $\frac{1}{25}$

Part of students present on a certain day = $1 - \frac{1}{25}$

$$\begin{aligned}= \frac{25 - 1}{25} \\[1em] = \frac{24}{25}.\end{aligned}$$

Number of students present on a certain day = Part of students present on a certain day × Total number of students

$$\begin{aligned}\Rightarrow 720 = \frac{24}{25} \times \text{ Total number of students} \\[1em] \Rightarrow \text{ Total number of students} = 720 \times \frac{25}{24} \\[1em] \Rightarrow \text{ Total number of students} = 30 \times 25 = 750.\end{aligned}$$

Hence, total number of students in the school = 750.

Given, product of three numbers = $7\frac{1}{2}$

Two numbers = $1\frac{1}{7}$ and $3\frac{3}{4}$

Let the other number be x.

$$\begin{aligned}\Rightarrow 1\frac{1}{7} \times 3\frac{3}{4} \times x = 7\frac{1}{2}\\[1em] \Rightarrow \frac{8}{7} \times \frac{15}{4} \times x = \frac{15}{2}\\[1em] \Rightarrow \frac{8 \times 15}{7 \times 4} \times x = \frac{15}{2}\\[1em] \Rightarrow \frac{120}{28} \times x = \frac{15}{2}\\[1em] \Rightarrow x = \frac{15 \times 28}{2 \times 120}\\[1em] \Rightarrow x = \frac{420}{240}\\[1em] \Rightarrow x = \frac{7}{4}\\[1em] \Rightarrow x = 1\frac{3}{4}.\end{aligned}$$

Hence, the third number = $1\frac{3}{4}$.

Draw a line. Take a point O in the middle. From O, set off equal distances on right as well as on the left. Mark them, as shown. Now, distance OA = 1 unit.

Divide OA into 3 equal parts. Take 2 parts out of it to reach the point P.

This point P represents $\frac{2}{3}$ on the number line.

Draw a line. Take a point O in the middle. From O, set off equal distances on right as well as on the left. Mark them, as shown. Now, distance OA = 1 unit.

Divide OA into 8 equal parts. Take 5 parts out of it to reach the point P.

This point P represents $\frac{5}{8}$ on the number line.

Draw a line. Take a point O in the middle. From O, set off equal distances on right as well as on the left. Mark them, as shown. Now, distance OA = 1 unit.

Divide OA into 2 equal parts. Take 1 part out of it to reach the point P.

This point P represents $\frac{1}{2}$ on the number line.

Let O be the origin. Set off equal distances OA, AB, BC, etc., on right hand side of O, denoting 1, 2, 3, etc.

Take 3 full distances and divide CD into 4 equal parts and take 3 parts out of it to reach the point P.

Hence, the point P represents $3\frac{3}{4}$.

Let O be the origin. Set off equal distances OA, AB, BC, etc., on right hand side of O, denoting 1, 2, 3, etc.

Take 1 full distance and divide AB into 7 equal parts. Then, take 4 parts out of it to reach point P.

Hence, the point P represents $1\frac{4}{7}$.

Let O be the origin. Set off equal distances OA, AB, BC, etc., on right hand side of O, denoting 1, 2, 3, etc.

Take 2 full distances and divide BC into 5 equal parts and take 2 parts out of it to reach the point P.

Hence, the point P represents $2\frac{2}{5}$.

We know that the L.C.M. of 2 and 4 is 4.

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

Divide the unit length OA from 0 to 1 into 4 equal parts. Then, each part represents $\frac{1}{4}$.

From 0 move 2 steps to the the right and then 1 step to the right to reach $\frac{3}{4}$.

$\therefore \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$, as shown below.

We know that the L.C.M. of 3 and 4 is 12.

Now, $\frac{1}{3} = \frac{4}{12}$ and $\frac{1}{4} = \frac{3}{12}$

Divide the unit length OA from 0 to 1 into 12 equal parts. Then, each part represents $\frac{1}{12}$.

From 0 move 4 steps to the the right and then 3 steps to the right to reach $\frac{7}{12}$.

$\therefore \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$, as shown below.

Divide the unit length OA from 0 to 1 into 6 equal parts. Then, each part represents $\frac{1}{6}$.

From 0 move 5 steps to the the right and then 1 step to the left to reach $\frac{4}{6}$.

$\therefore \frac{5}{6} - \frac{1}{6} = \frac{4}{6}$, as shown below.

Divide the unit length OA from 0 to 1 into 9 equal parts. Then, each part represents $\frac{1}{9}$.

From 0 move 4 steps to the the right and then 2 steps to the left to reach $\frac{2}{9}$.

$\therefore \frac{4}{9} - \frac{2}{9} = \frac{2}{9}$, as shown below.

By prime factorization,

$$\begin{aligned}\Rightarrow \frac{48}{72} \\[1em] \Rightarrow \frac{2 \times 2 \times 2 \times 2 \times 3}{2 \times 2 \times 2 \times 3 \times 3} \\[1em] \Rightarrow \frac{2}{3}.\end{aligned}$$

Hence, option 3 is the correct option.

By prime factorization,

$$\begin{aligned}\Rightarrow \frac{42}{54} \\[1em] \Rightarrow \frac{2 \times 3 \times 7}{2 \times 3 \times 3 \times 3} \\[1em] \Rightarrow \frac{7}{9}.\end{aligned}$$

Hence, option 4 is the correct option.

HCF of 91 and 114 = 1.

Thus,

$\frac{91}{114}$ in simplest form is $\frac{91}{114}$.

Hence, option 3 is the correct option.

By prime factorization,

$$\begin{aligned}\Rightarrow \frac{117}{143} \\[1em] \Rightarrow \frac{13 \times 9}{13 \times 11} \\[1em] \Rightarrow \frac{9}{11}.\end{aligned}$$

Hence, option 2 is the correct option.

Given, $\frac{3}{4}$ is equivalent to $\frac{x}{20}$

$\frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}$

Comparing,

$\frac{x}{20}$ and $\frac{15}{20}$

Thus, x = 15.

Hence, option 2 is the correct option.

Given, $\frac{3}{x}$ is equivalent to $\frac{45}{60}$

$$\begin{aligned}\Rightarrow \frac{3}{x} = \frac{45}{60} \\[1em] \Rightarrow x = \frac{3 \times 60}{45} \\[1em] \Rightarrow x = \frac{60}{15} = 4.\end{aligned}$$

Hence, option 1 is the correct option.

Like fractions are fractions that have the same denominator.

In option 3,

$\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}$

Denominators = 5, 5, 5 and 5

All denominators are same.

So, they are like fractions.

Hence, option 3 is the correct option.

An improper fraction is a fraction in which the numerator is greater than or equal to the denominator.

In option 3, $\frac{7}{7}$; numerator = denominator = 7

Hence, option 3 is the correct option.

A proper fraction is a fraction in which the numerator is less than the denominator.

In option 4, $\frac{3}{4}$; numerator < denominator

Hence, option 4 is the correct option.

Comparing both fractions; $\frac{2}{3}$ and $\frac{3}{5}$

Cross multiply: 2 x 5 and 3 x 3

⇒ 10 and 9

Now, comparing both numbers;

⇒ 10 > 9

⇒ $\frac{2}{3} \gt \frac{3}{5}$

Hence, option 1 is the correct option.

Given fractions: $\frac{3}{5}, \frac{5}{6}, \frac{7}{10}, \frac{2}{3}$

LCM of the denominators (5, 6, 10, 3) = 30

$$\begin{aligned}\Rightarrow \frac{3 \times 6}{5 \times 6}, \frac{5 \times 5}{6 \times 5}, \frac{7 \times 3}{10 \times 3}, \frac{2 \times 10}{3 \times 10}\\[1em] \Rightarrow \frac{18}{30}, \frac{25}{30}, \frac{21}{30}, \frac{20}{30}\end{aligned}$$

Among fractions with the same denominator, the one with the smaller numerator is smaller.

So, $\frac{18}{30} \lt \frac{20}{30} \lt \frac{21}{30} \lt \frac{25}{30}$

$\frac{3}{5} \lt \frac{2}{3} \lt \frac{7}{10} \lt \frac{5}{6}$

Hence, option 2 is the correct option.

Given fractions: $\frac{5}{6}, \frac{5}{7}, \frac{5}{9}, \frac{5}{11}$

LCM of the denominators (6, 7, 9, 11) = 1,386

$$\begin{aligned}\Rightarrow \frac{5 \times 231}{6 \times 231}, \frac{5 \times 198}{7 \times 198}, \frac{5 \times 154}{9 \times 154}, \frac{5 \times 126}{11 \times 126}\\[1em] \Rightarrow \frac{1155}{1386}, \frac{990}{1386}, \frac{770}{1386}, \frac{630}{1386}\end{aligned}$$

Among fractions with the same denominator, the one with the smaller numerator is smaller.

So, $\frac{630}{1386} \lt \frac{770}{1386} \lt \frac{990}{1386} \lt \frac{1155}{1386}$

$\frac{5}{11} \lt \frac{5}{9} \lt \frac{5}{7} \lt \frac{5}{6}$

Hence, option 4 is the correct option.

Given fractions: $\frac{8}{13}, \frac{9}{13}, \frac{10}{13}, \frac{11}{13}$

Among fractions with same denominator, the one with greater numerator is greater of the two.

So, $\frac{8}{13} \lt \frac{9}{13} \lt \frac{10}{13} \lt \frac{11}{13}$

Hence, option 3 is the correct option.

Given fractions: $\frac{3}{4}, \frac{5}{6}, \frac{7}{12}, \frac{2}{3}$

LCM of the denominators (4, 6, 12, 3) = 12

$$\begin{aligned}\Rightarrow \frac{3 \times 3}{4 \times 3}, \frac{5 \times 2}{6 \times 2}, \frac{7 \times 1}{12 \times 1}, \frac{2 \times 4}{3 \times 4}\\[1em] \Rightarrow \frac{9}{12}, \frac{10}{12}, \frac{7}{12}, \frac{8}{12}\end{aligned}$$

Among two fractions with same denominator, the one with greater numerator is greater of the two.

So, $\frac{7}{12} \lt \frac{8}{12} \lt \frac{9}{12} \lt \frac{10}{12}$

$\frac{7}{12} \lt \frac{2}{3} \lt \frac{3}{4} \lt \frac{5}{6}$

Hence, option 4 is the correct option.

Given; $\frac{37}{5}$

Divide 37 by 5: 37 ÷ 5 = 7 with a remainder of 2.

The quotient is 7 and the remainder 2 becomes the numerator.

$\frac{37}{5} = 7\frac{2}{5}$

Hence, option 2 is the correct option.

$1\frac{5}{7} = \frac{7 \times 1 + 5}{7} = \frac{12}{7}$.

The reciprocal of $\frac{12}{7}$ is $\frac{7}{12}$.

Hence, option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow 1 ÷ \frac{5}{7}\\[1em] \Rightarrow 1 \times \frac{7}{5}\\[1em] \Rightarrow \frac{1 \times 7}{5}\\[1em] \Rightarrow \frac{7}{5}.\end{aligned}$$

Hence, option 4 is the correct option.

Given, $\frac{4}{5}$ of a number is 64.

Let the number be x.

$$\begin{aligned}\Rightarrow \frac{4}{5} \times x = 64\\[1em] \Rightarrow x = 64 ÷ \frac{4}{5} \\[1em] \Rightarrow x = 64 \times \frac{5}{4} \\[1em] \Rightarrow x = \frac{64 \times 5}{4} \\[1em] \Rightarrow x = 16 \times 5 \\[1em] \Rightarrow x = 80.\end{aligned}$$

Half of the number = $\frac{1}{2} \times 80 = \frac{80}{2} = 40$.

Hence, option 2 is the correct option.

(i) $\frac{7}{11}$ ............... $\frac{5}{8}$

Cross multiply; 7 x 8 and 5 x 11

⇒ 56 and 55

Now, comparing both numbers;

⇒ 56 > 55

Hence, $\frac{7}{11}$ ${\boxed{\gt}}$ $\frac{5}{8}$

(ii) $\frac{3}{7}$ ............... $\frac{3}{5}$

Cross multiply; 3 x 5 and 3 x 7

⇒ 15 and 35

Now, comparing both numbers;

⇒ 15 < 35

Hence, $\frac{3}{7}$ ${\boxed{\lt}}$ $\frac{3}{5}$

(iii) $\frac{4}{9}$ ............... $\frac{4}{7}$

Cross multiply; 4 x 7 and 4 x 9

⇒ 28 and 36

Now, comparing both numbers;

⇒ 28 < 36

Hence, $\frac{4}{9}$ ${\boxed{\lt}}$ $\frac{4}{7}$

(iv) $\frac{15}{4} \times$ ............... = $\frac{3}{2}$

Let $\frac{15}{4} \times x = \frac{3}{2}$

Solving,

$$\begin{aligned}\Rightarrow x = \frac{3}{2} ÷ \frac{15}{4}\\[1em] \Rightarrow x = \frac{3}{2} \times \frac{4}{15}\\[1em] \Rightarrow x = \frac{3 \times 4}{2 \times 15}\\[1em] \Rightarrow x = \frac{12}{30}\\[1em] \Rightarrow x = \frac{2}{5}\end{aligned}$$

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

(v) $\frac{14}{15}$ ÷ ............... = $\frac{2}{3}$

Let $\frac{14}{15} \div x = \frac{2}{3}$

Solving,

$$\begin{aligned}\Rightarrow \frac{14}{15} ÷ x = \frac{2}{3}\\[1em] \Rightarrow x = \frac{14}{15} ÷ \frac{2}{3} \\[1em] \Rightarrow x = \frac{14}{15} \times \frac{3}{2} \\[1em] \Rightarrow x = \frac{14 \times 3}{15 \times 2}\\[1em] \Rightarrow x = \frac{42}{30}\\[1em] \Rightarrow x = \frac{7}{5}.\end{aligned}$$

Hence, $\frac{14}{15} ÷ \frac{7}{5} = \frac{2}{3}$

(vi) Reciprocal of $2\frac{3}{4}$ is ...............

Reciprocal of $2\frac{3}{4} = \frac{11}{4}$ is $\frac{4}{11}$

Hence, reciprocal of $2\frac{3}{4}$ is $\frac{4}{11}$.

(i) $\frac{8}{5}$ ÷ 1 = ...............

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

Hence, $\frac{8}{5} ÷ 1 = 1\frac{3}{5}$.

(ii) $1\frac{3}{8}$ of $\frac{4}{5}$ ...............

$$\begin{aligned}\Rightarrow 1\frac{3}{8} \times \frac{4}{5}\\[1em] \Rightarrow \frac{11}{8} \times \frac{4}{5}\\[1em] \Rightarrow \frac{11 \times 4}{8 \times 5}\\[1em] \Rightarrow \frac{44}{40}\\[1em] \Rightarrow \frac{11}{10}\\[1em] \Rightarrow 1\frac{1}{10}\\[1em]\end{aligned}$$

Hence, $1\frac{3}{8}$ of $\frac{4}{5} = \frac{11}{10}$.

(iii) $\Big(\frac{5}{6} \text{ of 1 hour}\Big)$ = ............... minutes

$$\begin{aligned}\Rightarrow \frac{5}{6} \text{ of 1 hour}\\[1em] \Rightarrow \frac{5}{6} \text{ of 60 minutes}\\[1em] \Rightarrow \frac{5}{6} \times 60 \\[1em] \Rightarrow \frac{5 \times 60}{6} \\[1em] \Rightarrow \frac{300}{6} \\[1em] \Rightarrow 50\end{aligned}$$

Hence, $\Big(\frac{5}{6} \text{ of 1 hour}\Big)$ = 50 minutes.

(iv) $\Big(\frac{3}{8} \text{ of 1 km}\Big)$ = ............... metre

$$\begin{aligned}\Rightarrow \frac{3}{8} \text{ of 1 km}\\[1em] \Rightarrow \frac{3}{8} \text{ of 1,000 m}\\[1em] \Rightarrow \frac{3}{8} \times 1,000\\[1em] \Rightarrow \frac{3 \times 1,000}{8}\\[1em] \Rightarrow \frac{3,000}{8}\\[1em] \Rightarrow 375\end{aligned}$$

Hence, $\Big(\frac{3}{8} \text{ of 1 km}\Big)$ = 375 metres.

(v) $\Big(\frac{3}{5} \text{ of ₹ 1}\Big)$ = ............... paise

$$\begin{aligned}\Rightarrow \Big(\frac{3}{5} \text{ of ₹ 1}\Big)\\[1em] \Rightarrow \Big(\frac{3}{5} \text{ of 100 paise}\Big)\\[1em] \Rightarrow \frac{3}{5} \times 100 \\[1em] \Rightarrow \frac{3 \times 100}{5}\\[1em] \Rightarrow \frac{300}{5}\\[1em] \Rightarrow 60\end{aligned}$$

Hence, $\Big(\frac{3}{5} \text{ of ₹ 1}\Big)$ = 60 paise.

True

Reason

Given, 1 ÷ $\frac{3}{2}$ = $\frac{2}{3}$.

Taking L.H.S.

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

Taking R.H.S. = $\frac{2}{3}$.

Since, L.H.S. = R.H.S.

False

Reason

Given, $1\frac{2}{3}$ ÷ $\frac{3}{4}$ = $1\frac{1}{2}$.

Taking L.H.S.

$$\begin{aligned}\Rightarrow 1\frac{2}{3} ÷ \frac{3}{4}\\[1em] \Rightarrow \frac{5}{3} ÷ \frac{3}{4}\\[1em] \Rightarrow \frac{5}{3} \times \frac{4}{3}\\[1em] \Rightarrow \frac{5 \times 4}{3 \times 3}\\[1em] \Rightarrow \frac{20}{9}\\[1em] \Rightarrow 2\frac{2}{9}\\[1em]\end{aligned}$$

Taking R.H.S. = $1\frac{1}{2}$

Since, L.H.S. ≠ R.H.S.

True

Reason

Given, $\frac{3}{5}$ of 1 kg = 600 g.

Taking L.H.S.

$$\begin{aligned}\Rightarrow \frac{3}{5} \text{ of 1 kg}\\[1em] \Rightarrow \frac{3}{5} \text{ of 1000 g}\\[1em] \Rightarrow \frac{3}{5} \times 1000\\[1em] \Rightarrow \frac{3 \times 1000}{5}\\[1em] \Rightarrow \frac{3000}{5}\\[1em] \Rightarrow 600\end{aligned}$$

Taking R.H.S. = 600 g

Since, L.H.S. = R.H.S.

False

Reason

Given, $\frac{3}{4}$ of ₹ 1 = 65 paise.

Taking L.H.S.

$$\begin{aligned}\Rightarrow \frac{3}{4} \text{ of ₹ 1}\\[1em] \Rightarrow \frac{3}{4} \text{ of 100 paise}\\[1em] \Rightarrow \frac{3 \times 100}{4} \\[1em] \Rightarrow \frac{300}{4} \\[1em] \Rightarrow 75\end{aligned}$$

Taking R.H.S. = 65 paise

Since, L.H.S. ≠ R.H.S.

True

Reason

Given, $\frac{7}{10}$ of 1 hour = 42 minutes.

Taking L.H.S.

$$\begin{aligned}\Rightarrow \frac{7}{10} \text{ of 1 hour}\\[1em] \Rightarrow \frac{7}{10} \text{ of 60 minutes}\\[1em] \Rightarrow \frac{7}{10} \times 60 \\[1em] \Rightarrow \frac{7 \times 60}{10}\\[1em] \Rightarrow \frac{420}{10}\\[1em] \Rightarrow 42\end{aligned}$$

Taking R.H.S. = 42 minutes

Since, L.H.S. = R.H.S.

False

Reason

Given, $\frac{5}{6}$ and $\frac{5}{7}$

Since denominator of both fractions are different. So, these are unlike fractions.

True

Reason

Given, $\frac{1}{4}$ and $\frac{1}{5}$

Since denominator of both fractions are different. So, these are unlike fraction.

1. Given,

Part spent on house-rent = $\frac{2}{5}$

Part spent on food = $\frac{3}{10}$

Part spent on conveyance = $\frac{1}{8}$

Savings = ₹ 10,500.

$$\begin{aligned}\Rightarrow \text{Total fraction spent } = \frac{2}{5} + \frac{3}{10} + \frac{1}{8}\\[1em] \Rightarrow \frac{2 \times 8}{5 \times 8} + \frac{3 \times 4}{10 \times 4} + \frac{1 \times 5}{8 \times 5}\\[1em] \Rightarrow \frac{16}{40} + \frac{12}{40} + \frac{5}{40}\\[1em] \Rightarrow \frac{16 + 12 + 5}{40}\\[1em] \Rightarrow \frac{33}{40}\\[1em]\end{aligned}$$

Savings fraction = $1 - \frac{33}{40} = \frac{40 - 33}{40} = \frac{7}{40}$.

Savings = Savings fraction × Total salary

$$\begin{aligned}\Rightarrow 10500 = \frac{7}{40} \times \text{Total salary} \\[1em] \Rightarrow \text{Total salary} = \frac{10500 \times 40}{7} \\[1em] \Rightarrow \text{Total salary} = 1500 \times 40 = ₹60,000.\end{aligned}$$

Hence, option (d) is the correct option.

2. Amount spent on house rent = $\frac{2}{5}$ of the salary

$$\begin{aligned}= \frac{2}{5} \times 60,000 \\[1em] = \frac{2 \times 60,000}{5}\\[1em] = \frac{1,20,000}{5}\\[1em] = ₹24,000.\end{aligned}$$

Hence, option (c) is the correct option.

3. Amount spent on food = $\frac{3}{10}$ of the salary

$$\begin{aligned}= \frac{3}{10} \times 60,000 \\[1em] = \frac{3 \times 60,000}{10}\\[1em] = \frac{1,80,000}{10}\\[1em] = ₹18,000\end{aligned}$$

Hence, option (a) is the correct option.

4. Amount spent on conveyance = $\frac{1}{8}$ of the salary

$$\begin{aligned}= \frac{1}{8} \times 60,000 \\[1em] = \frac{1 \times 60,000}{8}\\[1em] = \frac{60,000}{8}\\[1em] = ₹7,500.\end{aligned}$$

Hence, option (c) is the correct option.

1. Given,

Initial capacity of drum = $\frac{3}{4}$ full

After removing 30 litres: Capacity of drum = $\frac{7}{12}$ full

Let the total capacity of drum be x litres.

$$\begin{aligned}\Rightarrow \frac{3}{4}x - 30 = \frac{7}{12}x\\[1em] \Rightarrow \frac{3}{4}x - \frac{7}{12}x = 30\\[1em] \Rightarrow \frac{3 \times 3}{4 \times 3}x - \frac{7}{12}x = 30\\[1em] \Rightarrow \frac{9}{12}x - \frac{7}{12}x = 30\\[1em] \Rightarrow \frac{9 - 7}{12}x = 30\\[1em] \Rightarrow \frac{2}{12}x = 30\\[1em] \Rightarrow \frac{1}{6}x = 30\\[1em] \Rightarrow x = 30 \times 6\\[1em] \Rightarrow x = 180 \text{ l}.\end{aligned}$$

Hence, option (c) is the correct option.

2. Initial quantity of kerosene = $\frac{3}{4}$ of the capacity of drum

$$\begin{aligned}\Rightarrow \frac{3}{4} \times 180\\[1em] \Rightarrow \frac{3 \times 180}{4}\\[1em] \Rightarrow \frac{540}{4}\\[1em] \Rightarrow 135\end{aligned}$$

Hence, option (a) is the correct option.

3. If 15 l of kerosene is added to the drum.

So, final quantity of kerosene = 135 + 15 = 150

Fraction of drum filled = $\frac{150}{180} = \frac{5}{6}$

Hence, option (c) is the correct option.

4. If 35 l of kerosene is drawn from the drum.

So, final quantity of kerosene = 135 - 35 = 100

Fraction of drum filled = $\frac{100}{180} = \frac{5}{9}$.

Hence, option (a) is the correct option.

According to assertion; the sum of two fractions is always a fraction.

Example - the two fractions are $\frac{1}{2}$ and $\frac{3}{2}$

Now, $\frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$

∴ Assertion (A) is false.

For two fractions $\frac{a}{b} \text{and} \frac{c}{d}$, we have $\frac{a}{b} + \frac{c}{d}$

LCM of b and d = bd

$\frac{a \times d}{b \times d} + \frac{c \times b}{d \times b} = \frac{ad}{bd} + \frac{cb}{bd} = \frac{ad + cb}{bd}$

∴ Reason (R) is true.

Hence, option 4 is the correct option.

According to assertion, half of a number is obtained by dividing it by 2.

half of n is $\frac{n}{2}$, which is exactly dividing by 2.

∴ Assertion (A) is true.

According to reason, the fractions $\frac{x}{y}$ and $\frac{y}{x}$ are multiplicative inverse of each other.

$$\begin{aligned}\Rightarrow \frac{x}{y} \times \frac{y}{x}\\[1em] \Rightarrow \frac{x \times y}{y \times x}\\[1em] \Rightarrow \frac{x \times y}{y \times x}\\[1em] \Rightarrow 1.\end{aligned}$$

∴ Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

According to assertion, the sum of the fractions represented by the shaded and unshaded portion is 1.

Fraction of shaded portion = $\frac{5}{9}$

Fraction of unshaded portion = $\frac{4}{9}$

Sum of fraction of shaded portion + unshaded portion = $\frac{5}{9} + \frac{4}{9} = \frac{5 + 4}{9} = \frac{9}{9} = 1$.

∴ Assertion (A) is true.

According to reason, we can only add like fractions.

We can add any fractions (like or unlike), but if they are unlike fractions we first convert them to like fractions (common denominator) and then add.

∴ Reason (R) is false.

Hence, option 3 is the correct option.