Fractions

As we know, 'out of' means division, so it is written as a fraction $\frac{\text{part}}{\text{whole}}$

(i) 2 out of 7

⇒ 2 out of 7 = $\frac{2}{7}$.

Hence, 2 out of 7 = $\frac{2}{7}$.

(ii) 5 out of 17

⇒ 5 out of 17 = $\frac{5}{17}$.

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

(iii) three-fifths

⇒ three-fifths = $\frac{3}{5}$.

Hence, three-fifths = $\frac{3}{5}$.

In $\frac{5}{8}$, numerator (5) < denominator (8).

Hence, $\frac{5}{8}$ is a proper fraction.

In $\frac{8}{5}$, numerator (8) > denominator (5).

Hence, $\frac{8}{5}$ is an improper fraction.

In $\frac{15}{15}$, numerator (15) = denominator (15), i.e. numerator is not less than denominator.

Hence, $\frac{15}{15}$ is an improper fraction.

$\frac{23}{23} = 23 \div 23 = 1$.

Hence, the value of $\frac{23}{23}$ is 1.

$\frac{5}{5} = 5 \div 5 = 1$.

Hence, the value of $\frac{5}{5}$ is 1.

$3\frac{3}{10}$ consists of a natural number 3 and a proper fraction $\frac{3}{10}$.

Hence, $3\frac{3}{10}$ is a mixed fraction.

$\frac{2}{15} \text{ and } \frac{7}{15}$ have the same denominator 15.

Hence, $\frac{2}{15} \text{ and } \frac{7}{15}$ are like fractions.

$\frac{23}{12} \text{ and } \frac{23}{15}$ have different denominators (12 and 15).

Hence, $\frac{23}{12} \text{ and } \frac{23}{15}$ are unlike fractions.

Reducing to lowest terms, $\frac{6}{15} = \frac{2}{5} \text{ and } \frac{28}{70} = \frac{2}{5}$.

Both are equal to $\frac{2}{5}$.

Hence, $\frac{6}{15} \text{ and } \frac{28}{70}$ are equal (equivalent) fractions.

$\frac{8}{24} \text{ and } \frac{8}{32}$ have different denominators (24 and 32).

Hence, $\frac{8}{24} \text{ and } \frac{8}{32}$ are not like fractions.

$3\frac{2}{13}$

$$\begin{aligned}\Rightarrow 3\frac{2}{13} = \frac{3 \times 13 + 2}{13} \\[1em] = \frac{39 + 2}{13} \\[1em] = \frac{41}{13}.\end{aligned}$$

Hence, $3\frac{2}{13} = \frac{3 \times 13 + 2}{13} = \frac{41}{13}$.

$4\frac{3}{5}$

$$\begin{aligned}\Rightarrow 4\frac{3}{5} = \frac{4 \times 5 + 3}{5} \\[1em] = \frac{20 + 3}{5} \\[1em] = \frac{23}{5}.\end{aligned}$$

Hence, $4\frac{3}{5} = \frac{4 \times 5 + 3}{5} = \frac{23}{5}$.

A fraction is proper if numerator < denominator, and improper if numerator ≥ denominator.

(i) Proper fractions (numerator < denominator) are:

$\frac{2}{9}, \frac{7}{15}, \frac{11}{20}, \frac{18}{23} \text{ and } \frac{27}{35}$.

(ii) Improper fractions (numerator > denominator) are:

$\frac{4}{3} \text{ and } \frac{20}{11}$.

Hence, proper fractions are $\frac{2}{9}, \frac{7}{15}, \frac{11}{20}, \frac{18}{23}, \frac{27}{35}$ and improper fractions are $\frac{4}{3}, \frac{20}{11}$.

$2\frac{1}{5}$

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

Hence, $2\frac{1}{5} = \frac{11}{5}$.

$3\frac{1}{4}$

$$\begin{aligned}\Rightarrow 3\frac{1}{4} = \frac{3 \times 4 + 1}{4} \\[1em] = \frac{12 + 1}{4} \\[1em] = \frac{13}{4}.\end{aligned}$$

Hence, $3\frac{1}{4} = \frac{13}{4}$.

$7\frac{1}{8}$

$$\begin{aligned}\Rightarrow 7\frac{1}{8} = \frac{7 \times 8 + 1}{8} \\[1em] = \frac{56 + 1}{8} \\[1em] = \frac{57}{8}.\end{aligned}$$

Hence, $7\frac{1}{8} = \frac{57}{8}$.

$2\frac{1}{11}$

$$\begin{aligned}\Rightarrow 2\frac{1}{11} = \frac{2 \times 11 + 1}{11} \\[1em] = \frac{22 + 1}{11} \\[1em] = \frac{23}{11}.\end{aligned}$$

Hence, $2\frac{1}{11} = \frac{23}{11}$.

$\frac{100}{17}$

Dividing 100 by 17, we get quotient = 5 and remainder = 15.

$\Rightarrow \frac{100}{17} = 5\frac{15}{17}.$

Hence, $\frac{100}{17} = 5\frac{15}{17}$.

$\frac{81}{11}$

Dividing 81 by 11, we get quotient = 7 and remainder = 4.

$\Rightarrow \frac{81}{11} = 7\frac{4}{11}.$

Hence, $\frac{81}{11} = 7\frac{4}{11}$.

$\frac{209}{7}$

Dividing 209 by 7, we get quotient = 29 and remainder = 6.

$\Rightarrow \frac{209}{7} = 29\frac{6}{7}.$

Hence, $\frac{209}{7} = 29\frac{6}{7}$.

$\frac{113}{15}$

Dividing 113 by 15, we get quotient = 7 and remainder = 8.

$\Rightarrow \frac{113}{15} = 7\frac{8}{15}.$

Hence, $\frac{113}{15} = 7\frac{8}{15}$.

$\frac{1}{3}, \frac{2}{5}, \frac{3}{4}, \frac{1}{6}$

LCM of 3, 5, 4 and 6 = 60.

$$\begin{aligned}\Rightarrow \frac{1}{3} = \frac{1 \times 20}{3 \times 20} = \frac{20}{60} \\[1em] \Rightarrow \frac{2}{5} = \frac{2 \times 12}{5 \times 12} = \frac{24}{60} \\[1em] \Rightarrow \frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60} \\[1em] \Rightarrow \frac{1}{6} = \frac{1 \times 10}{6 \times 10} = \frac{10}{60}\end{aligned}$$

Hence, the required like fractions are $\frac{20}{60}, \frac{24}{60}, \frac{45}{60} \text{ and } \frac{10}{60}$.

$\frac{5}{6}, \frac{7}{8}, \frac{11}{12}, \frac{3}{10}$

LCM of 6, 8, 12 and 10 = 120.

$$\begin{aligned}\Rightarrow \frac{5}{6} = \frac{5 \times 20}{6 \times 20} = \frac{100}{120} \\[1em] \Rightarrow \frac{7}{8} = \frac{7 \times 15}{8 \times 15} = \frac{105}{120} \\[1em] \Rightarrow \frac{11}{12} = \frac{11 \times 10}{12 \times 10} = \frac{110}{120} \\[1em] \Rightarrow \frac{3}{10} = \frac{3 \times 12}{10 \times 12} = \frac{36}{120}\end{aligned}$$

Hence, the required like fractions are $\frac{100}{120}, \frac{105}{120}, \frac{110}{120} \text{ and } \frac{36}{120}$.

$\frac{2}{7}, \frac{7}{8}, \frac{5}{14}, \frac{9}{16}$

LCM of 7, 8, 14 and 16 = 112.

$$\begin{aligned}\Rightarrow \frac{2}{7} = \frac{2 \times 16}{7 \times 16} = \frac{32}{112} \\[1em] \Rightarrow \frac{7}{8} = \frac{7 \times 14}{8 \times 14} = \frac{98}{112} \\[1em] \Rightarrow \frac{5}{14} = \frac{5 \times 8}{14 \times 8} = \frac{40}{112} \\[1em] \Rightarrow \frac{9}{16} = \frac{9 \times 7}{16 \times 7} = \frac{63}{112}\end{aligned}$$

Hence, the required like fractions are $\frac{32}{112}, \frac{98}{112}, \frac{40}{112} \text{ and } \frac{63}{112}$.

$\frac{8}{10}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{8}{10} = \frac{2 \times 2 \times 2}{2 \times 5} \\[1em] = \frac{2 \times 2}{5} \\[1em] = \frac{4}{5}.\end{aligned}$$

Hence, $\frac{8}{10}$ in lowest terms is $\frac{4}{5}$.

$\frac{50}{75}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{50}{75} = \frac{2 \times 5 \times 5}{3 \times 5 \times 5} \\[1em] = \frac{2}{3}.\end{aligned}$$

Hence, $\frac{50}{75}$ in lowest terms is $\frac{2}{3}$.

$\frac{18}{81}$

By prime factorisation,

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

Hence, $\frac{18}{81}$ in lowest terms is $\frac{2}{9}$.

$\frac{40}{120}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{40}{120} = \frac{2 \times 2 \times 2 \times 5}{2 \times 2 \times 2 \times 3 \times 5} \\[1em] = \frac{1}{3}.\end{aligned}$$

Hence, $\frac{40}{120}$ in lowest terms is $\frac{1}{3}$.

$\frac{105}{70}$

By prime factorisation,

$$\begin{aligned}\Rightarrow \frac{105}{70} = \frac{3 \times 5 \times 7}{2 \times 5 \times 7} \\[1em] = \frac{3}{2} \\[1em] = 1\frac{1}{2}.\end{aligned}$$

Hence, $\frac{105}{70}$ in lowest terms is $\frac{3}{2} \text{ or } 1\frac{1}{2}$.

$\frac{2}{5} = \frac{10}{15}$

By cross multiplication, 2 × 15 = 30 and 5 × 10 = 50.

Since 30 ≠ 50, the cross products are not equal.

Hence, the statement is False.

$\frac{35}{42} = \frac{5}{6}$

By cross multiplication, 35 × 6 = 210 and 42 × 5 = 210.

Since the cross products are equal, the fractions are equal.

Hence, the statement is True.

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

By cross multiplication, 5 × 5 = 25 and 4 × 4 = 16.

Since 25 ≠ 16, the cross products are not equal.

Hence, the statement is False.

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

$\frac{7}{9}$ is a proper fraction (numerator < denominator), while $1\frac{1}{7} = \frac{8}{7}$ is an improper fraction (denominator < numerator).

Hence, the statement is False.

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

Converting, improper fraction into mixed fraction:

$\frac{9}{7} = 1\frac{2}{7}$, which is not equal to $1\frac{1}{7}$.

Hence, the statement is False.

$\frac{3}{5} \text{ or } \frac{2}{3}$

LCM of 5 and 3 = 15.

$$\begin{aligned}\Rightarrow \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15} \\[1em] \Rightarrow \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}\end{aligned}$$

Since 9 < 10, $\frac{9}{15} \lt \frac{10}{15}$.

Hence, $\frac{2}{3}$ is greater.

$\frac{5}{9} \text{ or } \frac{3}{4}$

LCM of 9 and 4 = 36.

$$\begin{aligned}\Rightarrow \frac{5}{9} = \frac{5 \times 4}{9 \times 4} = \frac{20}{36} \\[1em] \Rightarrow \frac{3}{4} = \frac{3 \times 9}{4 \times 9} = \frac{27}{36}\end{aligned}$$

Since 20 < 27, $\frac{20}{36} \lt \frac{27}{36}$.

Hence, $\frac{3}{4}$ is greater.

$\frac{11}{14} \text{ or } \frac{26}{35}$

LCM of 14 and 35 = 70.

$$\begin{aligned}\Rightarrow \frac{11}{14} = \frac{11 \times 5}{14 \times 5} = \frac{55}{70} \\[1em] \Rightarrow \frac{26}{35} = \frac{26 \times 2}{35 \times 2} = \frac{52}{70}\end{aligned}$$

Since 55 > 52, $\frac{55}{70} \gt \frac{52}{70}$.

Hence, $\frac{11}{14}$ is greater.

$\frac{3}{8} \text{ or } \frac{4}{5}$

LCM of 8 and 5 = 40.

$$\begin{aligned}\Rightarrow \frac{3}{8} = \frac{3 \times 5}{8 \times 5} = \frac{15}{40} \\[1em] \Rightarrow \frac{4}{5} = \frac{4 \times 8}{5 \times 8} = \frac{32}{40}\end{aligned}$$

Since 15 < 32, $\frac{15}{40} \lt \frac{32}{40}$.

Hence, $\frac{3}{8}$ is smaller.

$\frac{8}{15} \text{ or } \frac{4}{7}$

LCM of 15 and 7 = 105.

$$\begin{aligned}\Rightarrow \frac{8}{15} = \frac{8 \times 7}{15 \times 7} = \frac{56}{105} \\[1em] \Rightarrow \frac{4}{7} = \frac{4 \times 15}{7 \times 15} = \frac{60}{105}\end{aligned}$$

Since 56 < 60, $\frac{56}{105} \lt \frac{60}{105}$.

Hence, $\frac{8}{15}$ is smaller.

$\frac{7}{26} \text{ or } \frac{10}{39}$

LCM of 26 and 39 = 78.

$$\begin{aligned}\Rightarrow \frac{7}{26} = \frac{7 \times 3}{26 \times 3} = \frac{21}{78} \\[1em] \Rightarrow \frac{10}{39} = \frac{10 \times 2}{39 \times 2} = \frac{20}{78}\end{aligned}$$

Since 20 < 21, $\frac{20}{78} \lt \frac{21}{78}$.

Hence, $\frac{10}{39}$ is smaller.

$\frac{5}{16}, \frac{13}{24} \text{ and } \frac{7}{8}$

LCM of 16, 24 and 8 = 48.

$$\begin{aligned}\Rightarrow \frac{5}{16} = \frac{5 \times 3}{16 \times 3} = \frac{15}{48} \\[1em] \Rightarrow \frac{13}{24} = \frac{13 \times 2}{24 \times 2} = \frac{26}{48} \\[1em] \Rightarrow \frac{7}{8} = \frac{7 \times 6}{8 \times 6} = \frac{42}{48}\end{aligned}$$

Since 42 > 26 > 15, we have $\frac{42}{48} \gt \frac{26}{48} \gt \frac{15}{48}$.

Hence, the descending order is $\frac{7}{8} \gt \frac{13}{24} \gt \frac{5}{16}$.

$\frac{4}{5}, \frac{7}{15}, \frac{11}{20} \text{ and } \frac{3}{4}$

LCM of 5, 15, 20 and 4 = 60.

$$\begin{aligned}\Rightarrow \frac{4}{5} = \frac{4 \times 12}{5 \times 12} = \frac{48}{60} \\[1em] \Rightarrow \frac{7}{15} = \frac{7 \times 4}{15 \times 4} = \frac{28}{60} \\[1em] \Rightarrow \frac{11}{20} = \frac{11 \times 3}{20 \times 3} = \frac{33}{60} \\[1em] \Rightarrow \frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60}\end{aligned}$$

Since 48 > 45 > 33 > 28, we have $\frac{48}{60} \gt \frac{45}{60} \gt \frac{33}{60} \gt \frac{28}{60}$.

Hence, the descending order is $\frac{4}{5} \gt \frac{3}{4} \gt \frac{11}{20} \gt \frac{7}{15}$.

$\frac{5}{7}, \frac{3}{8} \text{ and } \frac{9}{11}$

LCM of 7, 8 and 11 = 616.

$$\begin{aligned}\Rightarrow \frac{5}{7} = \frac{5 \times 88}{7 \times 88} = \frac{440}{616} \\[1em] \Rightarrow \frac{3}{8} = \frac{3 \times 77}{8 \times 77} = \frac{231}{616} \\[1em] \Rightarrow \frac{9}{11} = \frac{9 \times 56}{11 \times 56} = \frac{504}{616}\end{aligned}$$

Since 504 > 440 > 231, we have $\frac{504}{616} \gt \frac{440}{616} \gt \frac{231}{616}$.

Hence, the descending order is $\frac{9}{11} \gt \frac{5}{7} \gt \frac{3}{8}$.

$\frac{9}{16}, \frac{7}{12} \text{ and } \frac{1}{4}$

LCM of 16, 12 and 4 = 48.

$$\begin{aligned}\Rightarrow \frac{9}{16} = \frac{9 \times 3}{16 \times 3} = \frac{27}{48} \\[1em] \Rightarrow \frac{7}{12} = \frac{7 \times 4}{12 \times 4} = \frac{28}{48} \\[1em] \Rightarrow \frac{1}{4} = \frac{1 \times 12}{4 \times 12} = \frac{12}{48}\end{aligned}$$

Since 12 < 27 < 28, we have $\frac{12}{48} \lt \frac{27}{48} \lt \frac{28}{48}$.

Hence, the ascending order is $\frac{1}{4} \lt \frac{9}{16} \lt \frac{7}{12}$.

$\frac{5}{6}, \frac{2}{7}, \frac{8}{9} \text{ and } \frac{1}{3}$

LCM of 6, 7, 9 and 3 = 126.

$$\begin{aligned}\Rightarrow \frac{5}{6} = \frac{5 \times 21}{6 \times 21} = \frac{105}{126} \\[1em] \Rightarrow \frac{2}{7} = \frac{2 \times 18}{7 \times 18} = \frac{36}{126} \\[1em] \Rightarrow \frac{8}{9} = \frac{8 \times 14}{9 \times 14} = \frac{112}{126} \\[1em] \Rightarrow \frac{1}{3} = \frac{1 \times 42}{3 \times 42} = \frac{42}{126}\end{aligned}$$

Since 36 < 42 < 105 < 112, we have $\frac{36}{126} \lt \frac{42}{126} \lt \frac{105}{126} \lt \frac{112}{126}$.

Hence, the ascending order is $\frac{2}{7} \lt \frac{1}{3} \lt \frac{5}{6} \lt \frac{8}{9}$.

$\frac{2}{3}, \frac{5}{9}, \frac{5}{6} \text{ and } \frac{3}{8}$

LCM of 3, 9, 6 and 8 = 72.

$$\begin{aligned}\Rightarrow \frac{2}{3} = \frac{2 \times 24}{3 \times 24} = \frac{48}{72} \\[1em] \Rightarrow \frac{5}{9} = \frac{5 \times 8}{9 \times 8} = \frac{40}{72} \\[1em] \Rightarrow \frac{5}{6} = \frac{5 \times 12}{6 \times 12} = \frac{60}{72} \\[1em] \Rightarrow \frac{3}{8} = \frac{3 \times 9}{8 \times 9} = \frac{27}{72}\end{aligned}$$

Since 27 < 40 < 48 < 60, we have $\frac{27}{72} \lt \frac{40}{72} \lt \frac{48}{72} \lt \frac{60}{72}$.

Hence, the ascending order is $\frac{3}{8} \lt \frac{5}{9} \lt \frac{2}{3} \lt \frac{5}{6}$.

$\frac{6}{11} \text{ and } \frac{5}{9}$

By cross multiplication, 6 × 9 = 54 and 11 × 5 = 55.

Since 54 < 55, therefore $\frac{6}{11} \lt \frac{5}{9}$.

Hence, $\frac{6}{11} \lt \frac{5}{9}$.

$\frac{3}{7} \text{ and } \frac{9}{13}$

By cross multiplication, 3 × 13 = 39 and 7 × 9 = 63.

Since 39 < 63, therefore $\frac{3}{7} \lt \frac{9}{13}$.

Hence, $\frac{3}{7} \lt \frac{9}{13}$.

$\frac{56}{64} \text{ and } \frac{7}{8}$

Reducing $\frac{56}{64}$ to lowest terms, $\frac{56}{64} = \frac{7}{8}$.

Hence, $\frac{56}{64} = \frac{7}{8}$.

$\frac{5}{12} \text{ and } \frac{8}{33}$

By cross multiplication, 5 × 33 = 165 and 12 × 8 = 96.

Since 165 > 96, therefore $\frac{5}{12} \gt \frac{8}{33}$.

Hence, $\frac{5}{12} \gt \frac{8}{33}$.

$1\frac{3}{4} \text{ and } \frac{3}{8}$

$$\begin{aligned}\Rightarrow 1\frac{3}{4} + \frac{3}{8} = \frac{7}{4} + \frac{3}{8} \\[1em] \text{ L.C.M. of 4 and 8 = 8 }\\[1em] = \frac{7 \times 2}{4 \times 2} + \frac{3}{8} \\[1em] =\frac{14}{8} + \frac{3}{8} \\[1em] = \frac{17}{8} \\[1em] = 2\frac{1}{8}.\end{aligned}$$

Hence, the required sum = $2\frac{1}{8}$.

$\frac{2}{5}, 2\frac{3}{15} \text{ and } \frac{7}{10}$

LCM of 5, 15 and 10 = 30.

$$\begin{aligned}\Rightarrow \frac{2}{5} + 2\frac{3}{15} + \frac{7}{10} = \frac{2}{5} + \frac{33}{15} + \frac{7}{10} \\[1em] = \frac{12}{30} + \frac{66}{30} + \frac{21}{30} \\[1em] = \frac{99}{30}\\[1em] = \frac{33}{10}\\[1em] = 3\frac{3}{10}.\end{aligned}$$

Hence, the required sum = $3\frac{3}{10}$.

$1\frac{7}{8}, 1\frac{1}{2} \text{ and } 1\frac{3}{4}$

LCM of 8, 2 and 4 = 8.

$$\begin{aligned}\Rightarrow 1\frac{7}{8} + 1\frac{1}{2} + 1\frac{3}{4} = \frac{15}{8} + \frac{3}{2} + \frac{7}{4} \\[1em] = \frac{15}{8} + \frac{12}{8} + \frac{14}{8} \\[1em] = \frac{41}{8} \\[1em] = 5\frac{1}{8}.\end{aligned}$$

Hence, the required sum = $5\frac{1}{8}$.

$3\frac{3}{4}, 2\frac{1}{6} \text{ and } 1\frac{5}{8}$

LCM of 4, 6 and 8 = 24.

$$\begin{aligned}\Rightarrow 3\frac{3}{4} + 2\frac{1}{6} + 1\frac{5}{8} = \frac{15}{4} + \frac{13}{6} + \frac{13}{8} \\[1em] = \frac{90}{24} + \frac{52}{24} + \frac{39}{24} \\[1em] = \frac{181}{24} \\[1em] = 7\frac{13}{24}.\end{aligned}$$

Hence, the required sum = $7\frac{13}{24}$.

$2\frac{8}{9}, \frac{11}{18} \text{ and } 3\frac{5}{6}$

LCM of 9, 18 and 6 = 18.

$$\begin{aligned}\Rightarrow 2\frac{8}{9} + \frac{11}{18} + 3\frac{5}{6} = \frac{26}{9} + \frac{11}{18} + \frac{23}{6} \\[1em] = \frac{52}{18} + \frac{11}{18} + \frac{69}{18} \\[1em] = \frac{132}{18} \\[1em] = \frac{22}{3} \\[1em] = 7\frac{1}{3}.\end{aligned}$$

Hence, the required sum = $7\frac{1}{3}$.

$1\frac{11}{12} - \frac{13}{16}$

LCM of 12 and 16 = 48.

$$\begin{aligned}\Rightarrow 1\frac{11}{12} - \frac{13}{16} = \frac{23}{12} - \frac{13}{16} \\[1em] = \frac{92}{48} - \frac{39}{48} \\[1em] = \frac{53}{48} \\[1em] = 1\frac{5}{48}.\end{aligned}$$

Hence, $1\frac{11}{12} - \frac{13}{16} = 1\frac{5}{48}$.

$2\frac{3}{4} - 1\frac{5}{6}$

LCM of 4 and 6 = 12.

$$\begin{aligned}\Rightarrow 2\frac{3}{4} - 1\frac{5}{6} = \frac{11}{4} - \frac{11}{6} \\[1em] = \frac{33}{12} - \frac{22}{12} \\[1em] = \frac{11}{12}.\end{aligned}$$

Hence, $2\frac{3}{4} - 1\frac{5}{6} = \frac{11}{12}$.

$2\frac{5}{7} + \frac{3}{14} - \frac{13}{21}$

LCM of 7, 14 and 21 = 42.

$$\begin{aligned}\Rightarrow 2\frac{5}{7} + \frac{3}{14} - \frac{13}{21} = \frac{19}{7} + \frac{3}{14} - \frac{13}{21} \\[1em] = \frac{114}{42} + \frac{9}{42} - \frac{26}{42} \\[1em] = \frac{97}{42} \\[1em] = 2\frac{13}{42}.\end{aligned}$$

Hence, $2\frac{5}{7} + \frac{3}{14} - \frac{13}{21} = 2\frac{13}{42}$.

$3\frac{5}{6} - \frac{1}{6} - 1\frac{1}{12}$

LCM of 6 and 12 = 12.

$$\begin{aligned}\Rightarrow 3\frac{5}{6} - \frac{1}{6} - 1\frac{1}{12} = \frac{23}{6} - \frac{1}{6} - \frac{13}{12} \\[1em] = \frac{46}{12} - \frac{2}{12} - \frac{13}{12} \\[1em] = \frac{31}{12} \\[1em] = 2\frac{7}{12}.\end{aligned}$$

Hence, $3\frac{5}{6} - \frac{1}{6} - 1\frac{1}{12} = 2\frac{7}{12}$.

$6 + \frac{3}{10} - 1\frac{8}{15}$

LCM of 1, 10 and 15 = 30.

$$\begin{aligned}\Rightarrow 6 + \frac{3}{10} - 1\frac{8}{15} = 6 + \frac{3}{10} - \frac{23}{15} \\[1em] = \frac{180}{30} + \frac{9}{30} - \frac{46}{30} \\[1em] = \frac{143}{30} \\[1em] = 4\frac{23}{30}.\end{aligned}$$

Hence, $6 + \frac{3}{10} - 1\frac{8}{15} = 4\frac{23}{30}$.

$1\frac{3}{4} + 2\frac{5}{7} - 1\frac{3}{14}$

LCM of 4, 7 and 14 = 28.

$$\begin{aligned}\Rightarrow 1\frac{3}{4} + 2\frac{5}{7} - 1\frac{3}{14} = \frac{7}{4} + \frac{19}{7} - \frac{17}{14} \\[1em] = \frac{49}{28} + \frac{76}{28} - \frac{34}{28} \\[1em] = \frac{91}{28} \\[1em] = \frac{13}{4} \\[1em] = 3\frac{1}{4}.\end{aligned}$$

Hence, $1\frac{3}{4} + 2\frac{5}{7} - 1\frac{3}{14} = 3\frac{1}{4}$.

$4 + 3\frac{1}{8} - 3\frac{1}{6}$

LCM of 1, 8 and 6 = 24.

$$\begin{aligned}\Rightarrow 4 + 3\frac{1}{8} - 3\frac{1}{6} = 4 + \frac{25}{8} - \frac{19}{6} \\[1em] = \frac{96}{24} + \frac{75}{24} - \frac{76}{24} \\[1em] = \frac{95}{24} \\[1em] = 3\frac{23}{24}.\end{aligned}$$

Hence, $4 + 3\frac{1}{8} - 3\frac{1}{6} = 3\frac{23}{24}$.

$6 - 3\frac{1}{2} - 2\frac{1}{5}$

LCM of 1, 2 and 5 = 10.

$$\begin{aligned}\Rightarrow 6 - 3\frac{1}{2} - 2\frac{1}{5} = 6 - \frac{7}{2} - \frac{11}{5} \\[1em] = \frac{60}{10} - \frac{35}{10} - \frac{22}{10} \\[1em] = \frac{3}{10}.\end{aligned}$$

Hence, $6 - 3\frac{1}{2} - 2\frac{1}{5} = \frac{3}{10}$.

$1\frac{5}{8} - 2\frac{1}{6} + 3\frac{3}{4}$

LCM of 8, 6 and 4 = 24.

$$\begin{aligned}\Rightarrow 1\frac{5}{8} - 2\frac{1}{6} + 3\frac{3}{4} = \frac{13}{8} - \frac{13}{6} + \frac{15}{4} \\[1em] = \frac{39}{24} - \frac{52}{24} + \frac{90}{24} \\[1em] = \frac{77}{24} \\[1em] = 3\frac{5}{24}.\end{aligned}$$

Hence, $1\frac{5}{8} - 2\frac{1}{6} + 3\frac{3}{4} = 3\frac{5}{24}$.

$3\frac{1}{2} + 1\frac{2}{3} - 2\frac{1}{4}$

LCM of 2, 3 and 4 = 12.

$$\begin{aligned}\Rightarrow 3\frac{1}{2} + 1\frac{2}{3} - 2\frac{1}{4} = \frac{7}{2} + \frac{5}{3} - \frac{9}{4} \\[1em] = \frac{42}{12} + \frac{20}{12} - \frac{27}{12} \\[1em] = \frac{35}{12} \\[1em] = 2\frac{11}{12}.\end{aligned}$$

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

$4\frac{3}{5} - 2\frac{7}{9} - 1\frac{2}{15} - \frac{2}{45}$

LCM of 5, 9, 15 and 45 = 45.

$$\begin{aligned}\Rightarrow 4\frac{3}{5} - 2\frac{7}{9} - 1\frac{2}{15} - \frac{2}{45} = \frac{23}{5} - \frac{25}{9} - \frac{17}{15} - \frac{2}{45} \\[1em] = \frac{207}{45} - \frac{125}{45} - \frac{51}{45} - \frac{2}{45} \\[1em] = \frac{29}{45}.\end{aligned}$$

Hence, $4\frac{3}{5} - 2\frac{7}{9} - 1\frac{2}{15} - \frac{2}{45} = \frac{29}{45}$.

Solving,

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

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

$\frac{4}{9} \times \frac{3}{5}$

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

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

$\frac{5}{12} \times 8$

$$\begin{aligned}\Rightarrow \frac{5}{12} \times 8 = \frac{5 \times 8}{12} \\[1em] = \frac{40}{12} \\[1em] = \frac{10}{3} \\[1em] = 3\frac{1}{3}.\end{aligned}$$

Hence, $\frac{5}{12} \times 8 = 3\frac{1}{3}$.

$\frac{7}{6} \text{ of } \frac{3}{14}$

$$\begin{aligned}\Rightarrow \frac{7}{6} \text{ of } \frac{3}{14} = \frac{7}{6} \times \frac{3}{14} \\[1em] = \frac{7 \times 3}{6 \times 14} \\[1em] = \frac{21}{84} \\[1em] = \frac{1}{4}.\end{aligned}$$

Hence, $\frac{7}{6} \text{ of } \frac{3}{14} = \frac{1}{4}$.

$3\frac{3}{8} \times 3\frac{6}{7}$

$$\begin{aligned}\Rightarrow 3\frac{3}{8} \times 3\frac{6}{7} = \frac{27}{8} \times \frac{27}{7} \\[1em] = \frac{27 \times 27}{8 \times 7} \\[1em] = \frac{729}{56} \\[1em] = 13\frac{1}{56}.\end{aligned}$$

Hence, $3\frac{3}{8} \times 3\frac{6}{7} = 13\frac{1}{56}$.

$\frac{1}{2} \text{ of } \frac{1}{3} \times \frac{3}{4}$

$$\begin{aligned}\Rightarrow \frac{1}{2} \text{ of } \frac{1}{3} \times \frac{3}{4} = \frac{1}{6} \times \frac{3}{4} \\[1em] = \frac{3}{24} \\[1em] = \frac{1}{8}.\end{aligned}$$

Hence, $\frac{1}{2} \text{ of } \frac{1}{3} \times \frac{3}{4} = \frac{1}{8}$.

$\frac{3}{7} \times \frac{5}{9} \times 4\frac{1}{5}$

$$\begin{aligned}\Rightarrow \frac{3}{7} \times \frac{5}{9} \times 4\frac{1}{5} = \frac{3}{7} \times \frac{5}{9} \times \frac{21}{5} \\[1em] = \frac{3 \times 5 \times 21}{7 \times 9 \times 5} \\[1em] = \frac{315}{315} \\[1em] = 1.\end{aligned}$$

Hence, $\frac{3}{7} \times \frac{5}{9} \times 4\frac{1}{5} = 1$.

$1\frac{1}{3} \times 1\frac{2}{7} \text{ of } 1\frac{1}{4}$

$$\begin{aligned}\Rightarrow 1\frac{1}{3} \times 1\frac{2}{7} \text{ of } 1\frac{1}{4} = \frac{4}{3} \times \frac{9}{7} \times \frac{5}{4} \\[1em] = \frac{4 \times 9 \times 5}{3 \times 7 \times 4} \\[1em] = \frac{180}{84} \\[1em] = \frac{15}{7} \\[1em] = 2\frac{1}{7}.\end{aligned}$$

Hence, $1\frac{1}{3} \times 1\frac{2}{7} \text{ of } 1\frac{1}{4} = 2\frac{1}{7}$.

$\frac{2}{3} \div 1\frac{1}{5}$

$$\begin{aligned}\Rightarrow \frac{2}{3} \div 1\frac{1}{5} = \frac{2}{3} \div \frac{6}{5} \\[1em] = \frac{2}{3} \times \frac{5}{6} \\[1em] = \frac{10}{18} \\[1em] = \frac{5}{9}.\end{aligned}$$

Hence, $\frac{2}{3} \div 1\frac{1}{5} = \frac{5}{9}$.

$4\frac{1}{2} \div \frac{4}{9}$

$$\begin{aligned}\Rightarrow 4\frac{1}{2} \div \frac{4}{9} = \frac{9}{2} \div \frac{4}{9} \\[1em] = \frac{9}{2} \times \frac{9}{4} \\[1em] = \frac{81}{8} \\[1em] = 10\frac{1}{8}.\end{aligned}$$

Hence, $4\frac{1}{2} \div \frac{4}{9} = 10\frac{1}{8}$.

1 ÷ $\frac{2}{5}$

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

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

$\frac{4}{9} \div \frac{4}{9}$

$$\begin{aligned}\Rightarrow \frac{4}{9} \div \frac{4}{9} = \frac{4}{9} \times \frac{9}{4} \\[1em] = 1.\end{aligned}$$

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

$2\frac{1}{3} \div 1\frac{3}{4}$

$$\begin{aligned}\Rightarrow 2\frac{1}{3} \div 1\frac{3}{4} = \frac{7}{3} \div \frac{7}{4} \\[1em] = \frac{7}{3} \times \frac{4}{7} \\[1em] = \frac{4}{3} \\[1em] = 1\frac{1}{3}.\end{aligned}$$

Hence, $2\frac{1}{3} \div 1\frac{3}{4} = 1\frac{1}{3}$.

$\frac{1}{4} \text{ of } 2\frac{2}{7} \div \frac{3}{5}$

$$\begin{aligned}\Rightarrow \frac{1}{4} \text{ of } 2\frac{2}{7} \div \frac{3}{5} = \frac{1}{4} \times \frac{16}{7} \div \frac{3}{5} \\[1em] = \frac{4}{7} \div \frac{3}{5} \\[1em] = \frac{4}{7} \times \frac{5}{3} \\[1em] = \frac{20}{21}.\end{aligned}$$

Hence, $\frac{1}{4} \text{ of } 2\frac{2}{7} \div \frac{3}{5} = \frac{20}{21}$.

$1\frac{1}{4} \times \frac{1}{2} \div 1\frac{1}{3}$

$$\begin{aligned}\Rightarrow 1\frac{1}{4} \times \frac{1}{2} \div 1\frac{1}{3} = \frac{5}{4} \times \frac{1}{2} \div \frac{4}{3} \\[1em] = \frac{5}{8} \div \frac{4}{3} \\[1em] = \frac{5}{8} \times \frac{3}{4} \\[1em] = \frac{15}{32}.\end{aligned}$$

Hence, $1\frac{1}{4} \times \frac{1}{2} \div 1\frac{1}{3} = \frac{15}{32}$.

$6\frac{1}{7} \times 0 \times 5\frac{3}{8}$

Any number multiplied by 0 is 0.

$\Rightarrow 6\frac{1}{7} \times 0 \times 5\frac{3}{8} = 0.$

Hence, $6\frac{1}{7} \times 0 \times 5\frac{3}{8} = 0$.

$\frac{3}{4} \times 1\frac{1}{3} \div \frac{3}{7} \text{ of } 2\frac{5}{8}$

Solving 'of' first,

$$\begin{aligned}\Rightarrow \frac{3}{7} \text{ of } 2\frac{5}{8} = \frac{3}{7} \times \frac{21}{8} = \frac{9}{8}\\[1em] \Rightarrow \frac{3}{4} \times 1\frac{1}{3} \div \frac{9}{8} = \frac{3}{4} \times \frac{4}{3} \div \frac{9}{8} \\[1em] = 1 \div \frac{9}{8} \\[1em] = 1 \times \frac{8}{9} \\[1em] = \frac{8}{9}.\end{aligned}$$

Hence, $\frac{3}{4} \times 1\frac{1}{3} \div \frac{3}{7} \text{ of } 2\frac{5}{8} = \frac{8}{9}$.

$2\frac{1}{4} \div \frac{2}{7} \text{ of } 1\frac{1}{3} \times \frac{2}{3}$

Solving 'of' first,

$$\begin{aligned}\Rightarrow \frac{2}{7} \text{ of } 1\frac{1}{3} = \frac{2}{7} \times \frac{4}{3} = \frac{8}{21}\\[1em] \Rightarrow 2\frac{1}{4} \div \frac{8}{21} \times \frac{2}{3} = \frac{9}{4} \div \frac{8}{21} \times \frac{2}{3} \\[1em] = \frac{9}{4} \times \frac{21}{8} \times \frac{2}{3} \\[1em] = \frac{9 \times 21 \times 2}{4 \times 8 \times 3} \\[1em] = \frac{378}{96} \\[1em] = \frac{63}{16} \\[1em] = 3\frac{15}{16}.\end{aligned}$$

Hence, $2\frac{1}{4} \div \frac{2}{7} \text{ of } 1\frac{1}{3} \times \frac{2}{3} = 3\frac{15}{16}$.

Using BODMAS we simplify bracket first.

$5 - \left(\frac{8}{11} - 3\frac{3}{11}\right)$

$$\begin{aligned}\Rightarrow 5 - \left(\frac{8}{11} - \frac{36}{11}\right) = 5 - \left(\frac{8 - 36}{11}\right) \\[1em] = 5 - \left(\frac{-28}{11}\right) \\[1em] = 5 + \frac{28}{11} \\[1em] = \frac{55}{11} + \frac{28}{11} \\[1em] = \frac{83}{11} \\[1em] = 7\frac{6}{11}.\end{aligned}$$

Hence, $5 - \left(\frac{8}{11} - 3\frac{3}{11}\right) = 7\frac{6}{11}$.

Using BODMAS we simplify bracket first.

$\frac{1}{2} \div \left(\frac{7}{8} - \frac{3}{5}\right)$

$$\begin{aligned}\Rightarrow \frac{1}{2} \div \left(\frac{7}{8} - \frac{3}{5}\right) = \frac{1}{2} \div \left(\frac{35}{40} - \frac{24}{40}\right) \\[1em] = \frac{1}{2} \div \frac{11}{40} \\[1em] = \frac{1}{2} \times \frac{40}{11} \\[1em] = \frac{20}{11} \\[1em] = 1\frac{9}{11}.\end{aligned}$$

Hence, $\frac{1}{2} \div \left(\frac{7}{8} - \frac{3}{5}\right) = 1\frac{9}{11}$.

Using BODMAS we simplify bracket first.

$2\frac{1}{3} \div \left(5\frac{1}{2} + 3\frac{3}{4}\right)$

$$\begin{aligned}\Rightarrow 2\frac{1}{3} \div \left(\frac{11}{2} + \frac{15}{4}\right) = \frac{7}{3} \div \left(\frac{22}{4} + \frac{15}{4}\right) \\[1em] = \frac{7}{3} \div \frac{37}{4} \\[1em] = \frac{7}{3} \times \frac{4}{37} \\[1em] = \frac{28}{111}.\end{aligned}$$

Hence, $2\frac{1}{3} \div \left(5\frac{1}{2} + 3\frac{3}{4}\right) = \frac{28}{111}$.

Using BODMAS we simplify bracket first.

$\left(3\frac{7}{8} - 3\frac{3}{5}\right) \div \frac{1}{2}$

$$\begin{aligned}\Rightarrow \left(\frac{31}{8} - \frac{18}{5}\right) \div \frac{1}{2} = \left(\frac{155}{40} - \frac{144}{40}\right) \div \frac{1}{2} \\[1em] = \frac{11}{40} \div \frac{1}{2} \\[1em] = \frac{11}{40} \times 2 \\[1em] = \frac{22}{40} \\[1em] = \frac{11}{20}.\end{aligned}$$

Hence, $\left(3\frac{7}{8} - 3\frac{3}{5}\right) \div \frac{1}{2} = \frac{11}{20}$.

Using BODMAS we simplify bracket first.

$\frac{4}{7} \div \left(\frac{1}{3} \times 2\frac{4}{5}\right)$

$$\begin{aligned}\Rightarrow \frac{4}{7} \div \left(\frac{1}{3} \times \frac{14}{5}\right) = \frac{4}{7} \div \frac{14}{15} \\[1em] = \frac{4}{7} \times \frac{15}{14} \\[1em] = \frac{60}{98} \\[1em] = \frac{30}{49}.\end{aligned}$$

Hence, $\frac{4}{7} \div \left(\frac{1}{3} \times 2\frac{4}{5}\right) = \frac{30}{49}$.

$\left(\frac{1}{2} + \frac{1}{3}\right) \div \left(\frac{1}{4} - \frac{1}{6}\right)$

Using BODMAS we simplify bracket first.

$$\begin{aligned}\Rightarrow \left(\frac{3}{6} + \frac{2}{6}\right) \div \left(\frac{3}{12} - \frac{2}{12}\right) = \frac{5}{6} \div \frac{1}{12} \\[1em] = \frac{5}{6} \times 12 \\[1em] = 10.\end{aligned}$$

Hence, $\left(\frac{1}{2} + \frac{1}{3}\right) \div \left(\frac{1}{4} - \frac{1}{6}\right) = 10$.

$\left(\frac{24}{35} \div \frac{6}{7} + \frac{5}{9}\right) \times \frac{3}{4}$

Using BODMAS, we solve the division inside the bracket first,

$$\begin{aligned}\Rightarrow \frac{24}{35} \div \frac{6}{7} = \frac{24}{35} \times \frac{7}{6} = \frac{4}{5}.\\[1em] \Rightarrow \left(\frac{4}{5} + \frac{5}{9}\right) \times \frac{3}{4} = \left(\frac{36}{45} + \frac{25}{45}\right) \times \frac{3}{4} \\[1em] = \frac{61}{45} \times \frac{3}{4} \\[1em] = \frac{183}{180} \\[1em] = \frac{61}{60} \\[1em] = 1\frac{1}{60}.\end{aligned}$$

Hence, $\left(\frac{24}{35} \div \frac{6}{7} + \frac{5}{9}\right) \times \frac{3}{4} = 1\frac{1}{60}$.

$\frac{3}{4} \text{ of } 6\frac{1}{8} - \frac{2}{3} \text{ of } 2\frac{1}{4}$

Using BODMAS, we solve 'of' first,

$$\begin{aligned}\Rightarrow \frac{3}{4} \text{ of } 6\frac{1}{8} = \frac{3}{4} \times \frac{49}{8} = \frac{147}{32}.\\[1em] \Rightarrow \frac{2}{3} \text{ of } 2\frac{1}{4} = \frac{2}{3} \times \frac{9}{4} = \frac{3}{2}.\\[1em] \Rightarrow \frac{147}{32} - \frac{3}{2} = \frac{147}{32} - \frac{48}{32} \\[1em] = \frac{99}{32} \\[1em] = 3\frac{3}{32}.\end{aligned}$$

Hence, $\frac{3}{4} \text{ of } 6\frac{1}{8} - \frac{2}{3} \text{ of } 2\frac{1}{4} = 3\frac{3}{32}$.

$\frac{7}{30} \text{ of } \left(\frac{1}{3} + \frac{7}{15}\right) \div \left(\frac{5}{6} - \frac{3}{5}\right)$

Using BODMAS, we simplify the brackets first,

$$\begin{aligned}\Rightarrow \frac{1}{3} + \frac{7}{15} = \frac{5}{15} + \frac{7}{15} = \frac{12}{15} = \frac{4}{5}.\\[1em] \Rightarrow \frac{5}{6} - \frac{3}{5} = \frac{25}{30} - \frac{18}{30} = \frac{7}{30}.\\[1em] \Rightarrow \frac{7}{30} \text{ of } \frac{4}{5} \div \frac{7}{30} = \frac{7}{30} \times \frac{4}{5} \div \frac{7}{30} \\[1em] = \frac{28}{150} \div \frac{7}{30} \\[1em] = \frac{28}{150} \times \frac{30}{7} \\[1em] = \frac{4}{5}.\end{aligned}$$

Hence, $\frac{7}{30} \text{ of } \left(\frac{1}{3} + \frac{7}{15}\right) \div \left(\frac{5}{6} - \frac{3}{5}\right) = \frac{4}{5}$.

$2\frac{1}{2} - 3\frac{1}{2} \times 1\frac{3}{4} + 2\frac{1}{2}$

Using BODMAS, we solve the multiplication first,

$$\begin{aligned}\Rightarrow 3\frac{1}{2} \times 1\frac{3}{4} = \frac{7}{2} \times \frac{7}{4} = \frac{49}{8}.\\[1em] \Rightarrow 2\frac{1}{2} - \frac{49}{8} + 2\frac{1}{2} = \frac{5}{2} - \frac{49}{8} + \frac{5}{2} \\[1em] = \frac{20}{8} - \frac{49}{8} + \frac{20}{8} \\[1em] = \frac{-9}{8} \\[1em] = -1\frac{1}{8}.\end{aligned}$$

Hence, $2\frac{1}{2} - 3\frac{1}{2} \times 1\frac{3}{4} + 2\frac{1}{2} = -1\frac{1}{8}$.

$4\frac{5}{7}\left(3\frac{1}{8} \div \frac{11}{12}\right)$

Using BODMAS, we simplify the bracket first,

$$\begin{aligned}\Rightarrow 3\frac{1}{8} \div \frac{11}{12} = \frac{25}{8} \times \frac{12}{11} = \frac{300}{88} = \frac{75}{22}.\\[1em] \Rightarrow 4\frac{5}{7} \times \frac{75}{22} = \frac{33}{7} \times \frac{75}{22} \\[1em] = \frac{33 \times 75}{7 \times 22} \\[1em] = \frac{2475}{154} \\[1em] = \frac{225}{14} \\[1em] = 16\frac{1}{14}.\end{aligned}$$

Hence, $4\frac{5}{7}\left(3\frac{1}{8} \div \frac{11}{12}\right) = 16\frac{1}{14}$.

$\frac{2}{5} \text{ of } \left(\frac{1}{7} - \frac{1}{12}\right) \text{ of } 1\frac{2}{5}$

Using BODMAS, we simplify the bracket first,

$$\begin{aligned}\Rightarrow \frac{1}{7} - \frac{1}{12} = \frac{12}{84} - \frac{7}{84} = \frac{5}{84}.\\[1em] \Rightarrow \frac{2}{5} \text{ of } \frac{5}{84} \text{ of } 1\frac{2}{5} = \frac{2}{5} \times \frac{5}{84} \times \frac{7}{5} \\[1em] = \frac{2 \times 5 \times 7}{5 \times 84 \times 5} \\[1em] = \frac{70}{2100} \\[1em] = \frac{1}{30}.\end{aligned}$$

Hence, $\frac{2}{5} \text{ of } \left(\frac{1}{7} - \frac{1}{12}\right) \text{ of } 1\frac{2}{5} = \frac{1}{30}$.

$\left(\frac{1}{2} - \frac{1}{3}\right)\left(\frac{3}{4} - \frac{4}{5}\right) \div \left(\frac{1}{2} - \frac{2}{5} + \frac{1}{7}\right)$

Using BODMAS, we simplify the brackets first,

$$\begin{aligned}\Rightarrow \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}.\\[1em] \Rightarrow \frac{3}{4} - \frac{4}{5} = \frac{15}{20} - \frac{16}{20} = \frac{-1}{20}.\\[1em] \Rightarrow \frac{1}{2} - \frac{2}{5} + \frac{1}{7} = \frac{35}{70} - \frac{28}{70} + \frac{10}{70} = \frac{17}{70}.\\[1em] \Rightarrow \frac{1}{6} \times \frac{-1}{20} \div \frac{17}{70} = \frac{-1}{120} \times \frac{70}{17} \\[1em] = \frac{-70}{2040} \\[1em] = \frac{-7}{204}.\end{aligned}$$

Hence, $\left(\frac{1}{2} - \frac{1}{3}\right)\left(\frac{3}{4} - \frac{4}{5}\right) \div \left(\frac{1}{2} - \frac{2}{5} + \frac{1}{7}\right) = \frac{-7}{204}$.

$\frac{5}{6} - \frac{3}{5}\left(\frac{1}{3} + \frac{2}{11}\right)$

Using BODMAS, we simplify the bracket first,

$$\begin{aligned}\Rightarrow \frac{1}{3} + \frac{2}{11} = \frac{11}{33} + \frac{6}{33} = \frac{17}{33}.\\[1em] \Rightarrow \frac{5}{6} - \frac{3}{5} \times \frac{17}{33} = \frac{5}{6} - \frac{51}{165} \\[1em] = \frac{5}{6} - \frac{17}{55} \\[1em] = \frac{275}{330} - \frac{102}{330} \\[1em] = \frac{173}{330}.\end{aligned}$$

Hence, $\frac{5}{6} - \frac{3}{5}\left(\frac{1}{3} + \frac{2}{11}\right) = \frac{173}{330}$.

Total length of the rope = $10\frac{1}{2}$ m.

Length cut off = $4\frac{5}{8}$ m.

Length of remaining rope = Total length − Length cut off

$$\begin{aligned}= 10\frac{1}{2} - 4\frac{5}{8} \\[1em] = \frac{21}{2} - \frac{37}{8} \\[1em] \text{L.C.M. of 2 and 8 = 8}\\[1em] = \frac{84}{8} - \frac{37}{8} \\[1em] = \frac{47}{8} \\[1em] = 5\frac{7}{8} \text{ m}.\end{aligned}$$

Hence, the length of the remaining rope = $5\frac{7}{8}$ m.

Length of the cloth = 5 m.

Shrinkage = $\frac{1}{25}$ of its length

$$\begin{aligned}= \frac{1}{25} \times 5 \\[1em] = \frac{5}{25} \\[1em] = \frac{1}{5} \text{ m}.\end{aligned}$$

Length of the cloth after washing = 5 − $\frac{1}{5}$

$$\begin{aligned}= \frac{25}{5} - \frac{1}{5} \\[1em] = \frac{24}{5} \\[1em] = 4\frac{4}{5} \text{ m}.\end{aligned}$$

Hence, the length of the cloth after washing = $4\frac{4}{5}$ m.

Price of wheat = ₹ $12\frac{1}{2}$

Price of rice = ₹ $25\frac{3}{4}$

Price of vegetables = ₹ $10\frac{1}{4}$

Money given to Shopkeeper = ₹ 100

Total money spent = $12\frac{1}{2} + 25\frac{3}{4} + 10\frac{1}{4}$

$$\begin{aligned}= \frac{25}{2} + \frac{103}{4} + \frac{41}{4} \\[1em] = \frac{50}{4} + \frac{103}{4} + \frac{41}{4} \\[1em] = \frac{194}{4} \\[1em] = \frac{97}{2} \\[1em] = 48\frac{1}{2}.\end{aligned}$$

So, total money spent = ₹ $48\frac{1}{2}$.

Money returned by the shopkeeper = 100 − $48\frac{1}{2}$

$$\begin{aligned}= \frac{200}{2} - \frac{97}{2} \\[1em] = \frac{103}{2} \\[1em] = 51\frac{1}{2}.\end{aligned}$$

Hence, the shopkeeper returned ₹ $51\frac{1}{2}$.

Total oranges = 500.

Number of rotten oranges = $\frac{3}{25} \text{ of } 500 = \frac{3}{25} \times 500$ = 60.

Number of oranges kept for guests = $\frac{1}{5} \text{ of } 500 = \frac{1}{5} \times 500$ = 100.

Oranges used = 60 + 100 = 160.

Oranges left = Total oranges − Oranges used

Oranges left = 500 − 160 = 340.

Hence, the number of oranges left in the box = 340.

Total weight of the ornament = 96 g.

Weight of copper = $\frac{1}{12} \text{ of } 96 = \frac{1}{12} \times 96$ = 8 g.

Weight of gold = Total weight − Weight of copper

= 96 − 8

= 88 g.

Hence, the weight of gold = 88 g.

Let the total work = 1.

Work done on Monday = $\frac{1}{2}$.

Work done on Tuesday = $\frac{1}{3}$.

Total work done in two days = $\frac{1}{2} + \frac{1}{3}$

$$\begin{aligned}= \frac{3}{6} + \frac{2}{6} \\[1em] = \frac{5}{6}.\end{aligned}$$

Work left for Wednesday = Total work - Work done in two days

Work left for Wednesday = 1 − $\frac{5}{6}$

$$\begin{aligned}= \frac{6}{6} - \frac{5}{6} \\[1em] = \frac{1}{6}.\end{aligned}$$

Hence, she will have to do $\frac{1}{6}$ of the work on Wednesday.

Part of money spent = $\frac{3}{8}$.

Part of money left = 1 − $\frac{3}{8} = \frac{8}{8} - \frac{3}{8} = \frac{5}{8}$.

Given, $\frac{5}{8}$ of the money = ₹ 720.

$$\begin{aligned}\Rightarrow \text{Total money} = 720 \div \frac{5}{8} \\[1em] = 720 \times \frac{8}{5} \\[1em] = 144 \times 8 \\[1em] = 1152.\end{aligned}$$

Hence, the man had ₹ 1,152 at first.

Part of students who are boys = $\frac{4}{5}$.

Part of students who are girls = 1 − $\frac{4}{5} = \frac{1}{5}$.

Given, $\frac{1}{5}$ of the students = 100 girls.

$$\begin{aligned}\Rightarrow \text{Total students} = 100 \div \frac{1}{5} \\[1em] = 100 \times 5 \\[1em] = 500.\end{aligned}$$

Number of boys = $\frac{4}{5} \text{ of } 500 = \frac{4}{5} \times 500$ = 400.

Hence, the number of boys = 400.

Given, $\frac{3}{4}$ of the journey = 12 km.

$$\begin{aligned}\Rightarrow \text{Total journey} = 12 \div \frac{3}{4} \\[1em] = 12 \times \frac{4}{3} \\[1em] = 16 \text{ km}.\end{aligned}$$

Distance left = Total journey − Distance covered

= 16 − 12

= 4 km.

Hence, the distance still left to be covered = 4 km.

Part of journey left = $\frac{1}{4}$.

Part of journey travelled = 1 − $\frac{1}{4} = \frac{3}{4}$.

Given, $\frac{3}{4}$ of the journey = 15 km.

$$\begin{aligned}\Rightarrow \text{Total journey} = 15 \div \frac{3}{4} \\[1em] = 15 \times \frac{4}{3} \\[1em] = 20 \text{ km}.\end{aligned}$$

Hence, the full length of the journey = 20 km.

Total income = ₹ 7,200.

Fraction of income spent = $\frac{3}{10} + \frac{1}{4} + \frac{1}{10} + \frac{2}{25}$

LCM of 10, 4, 10 and 25 = 100.

$$\begin{aligned}= \frac{30}{100} + \frac{25}{100} + \frac{10}{100} + \frac{8}{100} \\[1em] = \frac{73}{100}.\end{aligned}$$

Fraction of income saved = 1 − $\frac{73}{100} = \frac{27}{100}$.

Money saved = $\frac{27}{100}$ of 7,200

$$\begin{aligned}= \frac{27}{100} \times 7,200 \\[1em] = 27 \times 72 \\[1em] = 1,944.\end{aligned}$$

Hence, the man saved ₹ 1,944 in that month.

On the number line, point A lies between 0 and 1. Therefore, its value is greater than 0 and less than 1.

The interval between 0 and 1 is divided into 3 equal parts, and point A lies on the second division mark.

$\therefore A = \frac{2}{3}$

Point B lies between 1 and 2. Therefore, its value is greater than 1 and less than 2.

The interval between 1 and 2 is divided into 4 equal parts, and point B lies on the third division mark after 1.

$\therefore B = 1 + \frac{3}{4} = 1\frac{3}{4}$

Hence, option 2 is the correct option.

Like fractions have the same denominator. The fractions with denominator 5 are $\frac{3}{5}, \frac{2}{5} \text{ and } \frac{4}{5}$.

Hence, option 4 is the correct option.

These fractions have the same numerator 15. Among fractions with the same numerator, the one with the smallest denominator is the largest. The smallest denominator is 7.

Hence, option 1 is the correct option.

These are like fractions (same denominator 15). Among like fractions, the one with the smallest numerator is the least. The smallest numerator is 7.

Hence, option 1 is the correct option.

$$\begin{aligned}\Rightarrow \frac{7}{16} + \frac{13}{16} - \frac{5}{16} = \frac{7 + 13 - 5}{16} \\[1em] = \frac{15}{16}.\end{aligned}$$

Hence, option 1 is the correct option.

$$\begin{aligned}\Rightarrow 31\frac{2}{3} - 16\frac{2}{3} - 15\frac{2}{3} \\[1em] \text{Changing mixed fraction into improper fraction}\\[1em] = \frac{95}{3} - \frac{50}{3} - \frac{47}{3} \\[1em] = \frac{95 - 50 - 47}{3} \\[1em] = \frac{-2}{3}.\end{aligned}$$

Hence, option 4 is the correct option.

Division by zero is not defined.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{4} \times \frac{1}{4} \div \frac{1}{4} \text{ of } \frac{1}{4} \\[1em] \Rightarrow \frac{1}{4} \times \frac{1}{4} \div \frac{1}{16} \\[1em] \Rightarrow \frac{1}{4} \times \frac{1}{4} \times 16 \\[1em] \Rightarrow \frac{16}{16} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, option 1 is the correct option.

Solving 'of' first,

$$\begin{aligned}\Rightarrow \frac{1}{4} \text{ of } \frac{1}{4} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}.\\[1em] \Rightarrow \frac{1}{16} \div \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \times \frac{4}{1} \times \frac{1}{4} \\[1em] = \frac{1}{16}.\end{aligned}$$

Hence, option 3 is the correct option.

$$\begin{aligned}\Rightarrow \frac{x}{3} = 15 \\[1em] \Rightarrow x = 15 \times 3 = 45.\\[1em] \Rightarrow \frac{x}{5} = \frac{45}{5} = 9.\end{aligned}$$

Hence, option 2 is the correct option.

According to BODMAS,

Checking A: 7 ÷ 7 x 7 - 7 = 1 x 7 - 7 = 7 - 7 = 0. So A is true.

Checking R: $7 \times \frac{1}{7} \times 7 - 7 = 7 - 7 = 0$. So R is true and correctly explains A.

Hence, option 3 is the correct option.

If $\frac{1}{2}$ of a fraction is 15, then the fraction = 15 × 2 = 30, so R is true.

Now, $\frac{1}{5} \text{ of } 30 = \frac{1}{5} \times 30$ = 6, so A is true.

Both A and R are true, and R correctly explains A.

Hence, option 3 is the correct option.

$$\begin{aligned}\Rightarrow x = 1\frac{1}{2} \div 3\frac{1}{2} \times 2\frac{1}{3} = \frac{3}{2} \div \frac{7}{2} \times \frac{7}{3} \\[1em] = \frac{3}{2} \times \frac{2}{7} \times \frac{7}{3} \\[1em] = 1.\end{aligned}$$

So, the value of $x$ is 1, and both statements are true with statement 2 showing the correct working.

Hence, option 1 is the correct option.