Fractions — Exercise 3(C)

Reduce to a single fraction:

$\dfrac{1}{2} + \dfrac{2}{3}$

Answer

LCM of 2 and 3 = 6.

Solving,

$\Rightarrow \dfrac{1}{2} + \dfrac{2}{3} \\[1em] = \dfrac{1 \times 3}{6} + \dfrac{2 \times 2}{6} \\[1em] = \dfrac{3 + 4}{6} = \dfrac{7}{6} \\[1em] = 1\dfrac{1}{6}$

Hence, the required fraction is $1\dfrac{1}{6}$.

Reduce to a single fraction:

$\dfrac{3}{5} - \dfrac{1}{10}$

Answer

LCM of 5 and 10 = 10.

$\Rightarrow \dfrac{3}{5} - \dfrac{1}{10} = \dfrac{3 \times 2}{10} - \dfrac{1}{10}\\[1em] = \dfrac{6 - 1}{10} = \dfrac{5}{10}\\[1em] = \dfrac{1}{2}$

Hence, the required fraction is $\dfrac{1}{2}$.

Reduce to a single fraction:

$\dfrac{2}{3} - \dfrac{1}{6}$

Answer

LCM of 3 and 6 = 6.

$\Rightarrow \dfrac{2}{3} - \dfrac{1}{6} = \dfrac{2 \times 2}{6} - \dfrac{1}{6}\\[1em] = \dfrac{4 - 1}{6} = \dfrac{3}{6}\\[1em] = \dfrac{1}{2}$

Hence, the required fraction is $\dfrac{1}{2}$.

Reduce to a single fraction:

$1\dfrac{1}{3} + 2\dfrac{1}{4}$

Answer

$1\dfrac{1}{3} + 2\dfrac{1}{4} = \dfrac{4}{3} + \dfrac{9}{4}$

LCM of 3 and 4 = 12.

$= \dfrac{4 \times 4}{12} + \dfrac{9 \times 3}{12}\\[1em] = \dfrac{16 + 27}{12} \\[1em] = \dfrac{43}{12}\\[1em] = 3\dfrac{7}{12}$

Hence, the required fraction is $3\dfrac{7}{12}$.

Reduce to a single fraction:

$\dfrac{1}{4} + \dfrac{5}{6} - \dfrac{1}{12}$

Answer

LCM of 4, 6 and 12 = 12.

$\Rightarrow \dfrac{1}{4} + \dfrac{5}{6} - \dfrac{1}{12}\\[1em] = \dfrac{1 \times 3}{12} + \dfrac{5 \times 2}{12} - \dfrac{1}{12} \\[1em] = \dfrac{3 + 10 - 1}{12} \\[1em] = \dfrac{12}{12} = 1$

Hence, the required fraction is 1.

Reduce to a single fraction:

$\dfrac{2}{3} - \dfrac{3}{5} + 3 - \dfrac{1}{5}$

Answer

LCM of 1, 3 and 5 = 15.

$\Rightarrow \dfrac{2}{3} - \dfrac{3}{5} + 3 - \dfrac{1}{5} \\[1em] = \dfrac{2 \times 5}{15} - \dfrac{3 \times 3}{15} + \dfrac{3 \times 15}{15} - \dfrac{1 \times 3}{15} \\[1em] = \dfrac{10 - 9 + 45 - 3}{15} = \dfrac{43}{15} = 2\dfrac{13}{15}$

Hence, the required fraction is $2\dfrac{13}{15}$.

Reduce to a single fraction:

$\dfrac{2}{3} - \dfrac{1}{5} + \dfrac{1}{10}$

Answer

LCM of 3, 5 and 10 = 30.

$\Rightarrow \dfrac{2}{3} - \dfrac{1}{5} + \dfrac{1}{10}\\[1em] = \dfrac{2 \times 10}{30} - \dfrac{1 \times 6}{30} + \dfrac{1 \times 3}{30} \\[1em] = \dfrac{20 - 6 + 3}{30} \\[1em] = \dfrac{17}{30}$

Hence, the required fraction is $\dfrac{17}{30}$.

Reduce to a single fraction:

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

Answer

$2\dfrac{1}{2} + 2\dfrac{1}{3} - 1\dfrac{1}{4} = \dfrac{5}{2} + \dfrac{7}{3} - \dfrac{5}{4}$

LCM of 2, 3 and 4 = 12.

$= \dfrac{5 \times 6}{12} + \dfrac{7 \times 4}{12} - \dfrac{5 \times 3}{12} \\[1em] = \dfrac{30 + 28 - 15}{12} \\[1em] = \dfrac{43}{12}\\[1em] = 3\dfrac{7}{12}$

Hence, the required fraction is $3\dfrac{7}{12}$.

Reduce to a single fraction:

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

Answer

$2\dfrac{5}{8} - 2\dfrac{1}{6} + 4\dfrac{3}{4} = \dfrac{21}{8} - \dfrac{13}{6} + \dfrac{19}{4}$

LCM of 8, 6 and 4 = 24.

$= \dfrac{21 \times 3}{24} - \dfrac{13 \times 4}{24} + \dfrac{19 \times 6}{24} \\[1em] = \dfrac{63 - 52 + 114}{24} = \dfrac{125}{24} = 5\dfrac{5}{24}$

Hence, the required fraction is $5\dfrac{5}{24}$.

Simplify:

$\dfrac{3}{4} \times 6$

Answer

Solving,

$\Rightarrow \dfrac{3}{4} \times 6 \\[1em] = \dfrac{3 \times 6}{4} \\[1em] = \dfrac{18}{4} \\[1em] = \dfrac{9}{2} \\[1em] = 4\dfrac{1}{2}$

Hence, the required value is $4\dfrac{1}{2}$.

Simplify:

$\dfrac{2}{3} \times 15$

Answer

Solving,

$\Rightarrow \dfrac{2}{3} \times 15 \\[1em] = \dfrac{2 \times 15}{3} \\[1em] = \dfrac{30}{3} \\[1em] = 10$

Hence, the required value is 10.

Simplify:

$\dfrac{3}{4} \times \dfrac{1}{2}$

Answer

Solving,

$\Rightarrow \dfrac{3}{4} \times \dfrac{1}{2} \\[1em] = \dfrac{3 \times 1}{4 \times 2} \\[1em] = \dfrac{3}{8}$

Hence, the required value is $\dfrac{3}{8}$.

Simplify:

$\dfrac{9}{12} \times \dfrac{4}{7}$

Answer

Solving,

$\Rightarrow \dfrac{9}{12} \times \dfrac{4}{7} \\[1em] = \dfrac{9 \times 4}{12 \times 7} \\[1em] = \dfrac{36}{84} \\[1em] = \dfrac{3}{7}$

Hence, the required value is $\dfrac{3}{7}$.

Simplify:

$45 \times 2\dfrac{1}{3}$

Answer

Solving,

$\Rightarrow 45 \times 2\dfrac{1}{3} \\[1em] = 45 \times \dfrac{7}{3} \\[1em] = \dfrac{45 \times 7}{3} \\[1em] = \dfrac{315}{3} \\[1em] = 105$

Hence, the required value is 105.

Simplify:

$36 \times 3\dfrac{1}{4}$

Answer

Solving,

$\Rightarrow 36 \times 3\dfrac{1}{4} \\[1em] = 36 \times \dfrac{13}{4} \\[1em] = \dfrac{36 \times 13}{4} \\[1em] = \dfrac{468}{4} \\[1em] = 117$

Hence, the required value is 117.

Simplify:

$2 \div \dfrac{1}{3}$

Answer

Solving,

$\Rightarrow 2 \div \dfrac{1}{3} \\[1em] = 2 \times \dfrac{3}{1} \\[1em] = 6$

Hence, the required value is 6.

Simplify:

$3 \div \dfrac{2}{5}$

Answer

Solving,

$\Rightarrow 3 \div \dfrac{2}{5} \\[1em] = 3 \times \dfrac{5}{2} \\[1em] = \dfrac{15}{2} \\[1em] = 7\dfrac{1}{2}$

Hence, the required value is $7\dfrac{1}{2}$.

Simplify:

$1 \div \dfrac{3}{5}$

Answer

Solving,

$\Rightarrow 1 \div \dfrac{3}{5} \\[1em] = 1 \times \dfrac{5}{3} \\[1em] = \dfrac{5}{3} \\[1em] = 1\dfrac{2}{3}$

Hence, the required value is $1\dfrac{2}{3}$.

Simplify:

$\dfrac{1}{3} \div \dfrac{1}{4}$

Answer

Solving,

$\Rightarrow \dfrac{1}{3} \div \dfrac{1}{4} \\[1em] = \dfrac{1}{3} \times \dfrac{4}{1} \\[1em] = \dfrac{4}{3} \\[1em] = 1\dfrac{1}{3}$

Hence, the required value is $1\dfrac{1}{3}$.

Simplify:

$-\dfrac{5}{8} \div \dfrac{3}{4}$

Answer

Solving,

$\Rightarrow -\dfrac{5}{8} \div \dfrac{3}{4} \\[1em] = -\dfrac{5}{8} \times \dfrac{4}{3} \\[1em] = -\dfrac{5 \times 4}{8 \times 3} \\[1em] = -\dfrac{20}{24} \\[1em] = -\dfrac{5}{6}$

Hence, the required value is $-\dfrac{5}{6}$.

Simplify:

$3\dfrac{3}{7} \div 1\dfrac{1}{14}$

Answer

Solving,

$\Rightarrow 3\dfrac{3}{7} \div 1\dfrac{1}{14} \\[1em] = \dfrac{24}{7} \div \dfrac{15}{14} \\[1em] = \dfrac{24}{7} \times \dfrac{14}{15} \\[1em] = \dfrac{24 \times 14}{7 \times 15} \\[1em] = \dfrac{336}{105} \\[1em] = \dfrac{16}{5} \\[1em] = 3\dfrac{1}{5}$

Hence, the required value is $3\dfrac{1}{5}$.

Simplify:

$3\dfrac{3}{4} \times 1\dfrac{1}{5} \times \dfrac{20}{21}$

Answer

Solving,

$\Rightarrow 3\dfrac{3}{4} \times 1\dfrac{1}{5} \times \dfrac{20}{21} \\[1em] = \dfrac{15}{4} \times \dfrac{6}{5} \times \dfrac{20}{21} \\[1em] = \dfrac{15 \times 6 \times 20}{4 \times 5 \times 21} \\[1em] = \dfrac{1800}{420} \\[1em] = \dfrac{30}{7} \\[1em] = 4\dfrac{2}{7}$

Hence, the required value is $4\dfrac{2}{7}$.

Subtract:

2 from $\dfrac{2}{3}$

Answer

Solving,

$\Rightarrow \dfrac{2}{3} - 2 \\[1em] \Rightarrow \dfrac{2}{3} - \dfrac{6}{3} \\[1em] \Rightarrow \dfrac{2 - 6}{3} \\[1em] \Rightarrow -\dfrac{4}{3} \\[1em] \Rightarrow -1\dfrac{1}{3}$

Hence, the required value is $-1\dfrac{1}{3}$.

Subtract:

$\dfrac{1}{8} \text{ from } \dfrac{5}{8}$

Answer

Solving,

$\dfrac{5}{8} - \dfrac{1}{8} = \dfrac{5 - 1}{8} \\[1em] = \dfrac{4}{8} = \dfrac{1}{2}$

Hence, the required value is $\dfrac{1}{2}$.

Subtract:

$-\dfrac{2}{5} \text{ from } \dfrac{2}{5}$

Answer

Solving,

$\dfrac{2}{5} - \left(-\dfrac{2}{5}\right) = \dfrac{2}{5} + \dfrac{2}{5} \\[1em] = \dfrac{2 + 2}{5} = \dfrac{4}{5}$

Hence, the required value is $\dfrac{4}{5}$.

Subtract:

$-\dfrac{3}{7} \text{ from } \dfrac{3}{7}$

Answer

Solving,

$\dfrac{3}{7} - \left(-\dfrac{3}{7}\right) = \dfrac{3}{7} + \dfrac{3}{7} \\[1em] = \dfrac{3 + 3}{7} = \dfrac{6}{7}$

Hence, the required value is $\dfrac{6}{7}$.

Subtract:

0 from $-\dfrac{4}{5}$

Answer

Solving,

$-\dfrac{4}{5} - 0 = -\dfrac{4}{5}$

Hence, the required value is $-\dfrac{4}{5}$.

Subtract:

$\dfrac{2}{9} \text{ from } \dfrac{4}{5}$

Answer

$\dfrac{4}{5} - \dfrac{2}{9}$

LCM of 5 and 9 = 45.

$= \dfrac{4 \times 9}{45} - \dfrac{2 \times 5}{45} \\[1em] = \dfrac{36 - 10}{45} = \dfrac{26}{45}$

Hence, the required value is $\dfrac{26}{45}$.

Subtract:

$-\dfrac{4}{7} \text{ from } -\dfrac{6}{11}$

Answer

$-\dfrac{6}{11} - \left(-\dfrac{4}{7}\right) = -\dfrac{6}{11} + \dfrac{4}{7}$

LCM of 11 and 7 = 77.

$= -\dfrac{6 \times 7}{77} + \dfrac{4 \times 11}{77} \\[1em] = \dfrac{-42 + 44}{77} = \dfrac{2}{77}$

Hence, the required value is $\dfrac{2}{77}$.

Find the value of:

$\dfrac{1}{2}$ of 10 kg

Answer

Solving,

$\Rightarrow \dfrac{1}{2} \text{ of } 10 \text{ kg} \\[1em] \Rightarrow \dfrac{1}{2} \times 10 \\[1em] \Rightarrow 5 \text{ kg}$

Hence, the required value is 5 kg.

Find the value of:

$\dfrac{3}{5}$ of 1 hour

Answer

1 hour = 60 minutes.

Solving,

$\Rightarrow \dfrac{3}{5} \text{ of } 1 \text{ hour} \\[1em] \Rightarrow \dfrac{3}{5} \times 60 \\[1em] \Rightarrow 36 \text{ minutes}$

Hence, the required value is 36 minutes.

Find the value of:

$\dfrac{4}{7} \text{ of } 2\dfrac{1}{3}$ kg

Answer

Solving,

$\Rightarrow \dfrac{4}{7} \text{ of } 2\dfrac{1}{3} \text{ kg} \\[1em] \Rightarrow \dfrac{4}{7} \times \dfrac{7}{3} \\[1em] \Rightarrow \dfrac{4 \times 7}{7 \times 3} \\[1em] \Rightarrow \dfrac{4}{3} \\[1em] \Rightarrow 1\dfrac{1}{3} \text{ kg}$

Hence, the required value is $1\dfrac{1}{3}$ kg.

Find the value of:

$3\dfrac{1}{2}$ times of 2 metre

Answer

Solving,

$\Rightarrow 3\dfrac{1}{2} \text{ times of } 2 \text{ metre} \\[1em] \Rightarrow \dfrac{7}{2} \times 2 \\[1em] \Rightarrow 7 \text{ metres}$

Hence, the required value is 7 metres.

Find the value of:

$\dfrac{1}{2} \text{ of } 2\dfrac{2}{3}$

Answer

Solving,

$\Rightarrow \dfrac{1}{2} \text{ of } 2\dfrac{2}{3} \\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{8}{6} \\[1em] \Rightarrow \dfrac{4}{3} \\[1em] \Rightarrow 1\dfrac{1}{3}$

Hence, the required value is $1\dfrac{1}{3}$.

Find the value of:

$\dfrac{5}{11} \text{ of } \dfrac{4}{5}$ of 22 kg

Answer

Solving,

$\Rightarrow \dfrac{5}{11} \text{ of } \dfrac{4}{5} \text{ of } 22 \text{ kg} \\[1em] \Rightarrow \dfrac{5}{11} \times \dfrac{4}{5} \times 22 \\[1em] \Rightarrow \dfrac{5 \times 4 \times 22}{11 \times 5} \\[1em] \Rightarrow \dfrac{440}{55} \\[1em] \Rightarrow 8 \text{ kg}$

Hence, the required value is 8 kg.

Simplify and reduce to a simple fraction:

$\dfrac{3}{3\dfrac{3}{4}}$

Answer

Solving,

$\Rightarrow \dfrac{3}{3\dfrac{3}{4}} \\[1em] \Rightarrow 3 \div \dfrac{15}{4} \\[1em] \Rightarrow 3 \times \dfrac{4}{15} \\[1em] \Rightarrow \dfrac{12}{15} \\[1em] \Rightarrow \dfrac{4}{5}$

Hence, the required value is $\dfrac{4}{5}$.

Simplify and reduce to a simple fraction:

$\dfrac{\dfrac{3}{5}}{7}$

Answer

Solving,

$\Rightarrow \dfrac{\dfrac{3}{5}}{7} \\[1em] \Rightarrow \dfrac{3}{5} \div 7 \\[1em] \Rightarrow \dfrac{3}{5} \times \dfrac{1}{7} \\[1em] \Rightarrow \dfrac{3}{35}$

Hence, the required value is $\dfrac{3}{35}$.

Simplify and reduce to a simple fraction:

$\dfrac{3}{\dfrac{5}{7}}$

Answer

Solving,

$\Rightarrow \dfrac{3}{\dfrac{5}{7}} \\[1em] \Rightarrow 3 \div \dfrac{5}{7} \\[1em] \Rightarrow 3 \times \dfrac{7}{5} \\[1em] \Rightarrow \dfrac{21}{5} \\[1em] \Rightarrow 4\dfrac{1}{5}$

Hence, the required value is $4\dfrac{1}{5}$.

Simplify and reduce to a simple fraction:

$\dfrac{2\dfrac{1}{5}}{1\dfrac{1}{10}}$

Answer

Solving,

$\Rightarrow \dfrac{2\dfrac{1}{5}}{1\dfrac{1}{10}} \\[1em] \Rightarrow \dfrac{11}{5} \div \dfrac{11}{10} \\[1em] \Rightarrow \dfrac{11}{5} \times \dfrac{10}{11} \\[1em] \Rightarrow \dfrac{110}{55} \\[1em] \Rightarrow 2$

Hence, the required value is 2.

Simplify and reduce to a simple fraction:

$\dfrac{2}{5} \text{ of } \dfrac{6}{11} \times 1\dfrac{1}{4}$

Answer

Solving,

$\Rightarrow \dfrac{2}{5} \text{ of } \dfrac{6}{11} \times 1\dfrac{1}{4} \\[1em] \Rightarrow \dfrac{2}{5} \times \dfrac{6}{11} \times \dfrac{5}{4} \\[1em] \Rightarrow \dfrac{2 \times 6 \times 5}{5 \times 11 \times 4} \\[1em] \Rightarrow \dfrac{60}{220} \\[1em] \Rightarrow \dfrac{3}{11}$

Hence, the required value is $\dfrac{3}{11}$.

Simplify and reduce to a simple fraction:

$2\dfrac{1}{4} \div \dfrac{1}{7} \times \dfrac{1}{3}$

Answer

Solving,

$\Rightarrow 2\dfrac{1}{4} \div \dfrac{1}{7} \times \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{9}{4} \div \dfrac{1}{7} \times \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{9}{4} \times \dfrac{7}{1} \times \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{9 \times 7 \times 1}{4 \times 1 \times 3} \\[1em] \Rightarrow \dfrac{63}{12} \\[1em] \Rightarrow \dfrac{21}{4} \\[1em] \Rightarrow 5\dfrac{1}{4}$

Hence, the required value is $5\dfrac{1}{4}$.

Simplify and reduce to a simple fraction:

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

Answer

Solving,

$\Rightarrow \dfrac{1}{3} \times 4\dfrac{2}{3} \div 3\dfrac{1}{2} \times \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{14}{3} \div \dfrac{7}{2} \times \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{14}{9} \times \dfrac{2}{7} \times \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{14 \times 2 \times 1}{9 \times 7 \times 2} \\[1em] \Rightarrow \dfrac{28}{126} \\[1em] \Rightarrow \dfrac{2}{9}$

Hence, the required value is $\dfrac{2}{9}$.

Simplify and reduce to a simple fraction:

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

Answer

Solving,

Simplifying 'of' first: $\dfrac{3}{7} \text{ of } 2\dfrac{5}{8} = \dfrac{3}{7} \times \dfrac{21}{8} = \dfrac{63}{56} = \dfrac{9}{8}$

$\Rightarrow \dfrac{2}{3} \times 1\dfrac{1}{4} \div \dfrac{3}{7} \text{ of } 2\dfrac{5}{8} \\[1em] \Rightarrow \dfrac{2}{3} \times \dfrac{5}{4} \div \dfrac{9}{8} \\[1em] \Rightarrow \dfrac{2}{3} \times \dfrac{5}{4} \times \dfrac{8}{9} \\[1em] \Rightarrow \dfrac{2 \times 5 \times 8}{3 \times 4 \times 9} \\[1em] \Rightarrow \dfrac{80}{108} \\[1em] \Rightarrow \dfrac{20}{27}$

Hence, the required value is $\dfrac{20}{27}$.

Simplify and reduce to a simple fraction:

$0 \div \dfrac{8}{11}$

Answer

Solving,

$\Rightarrow 0 \div \dfrac{8}{11} \\[1em] \Rightarrow 0 \times \dfrac{11}{8} \\[1em] \Rightarrow 0$

Hence, the required value is 0.

Simplify and reduce to a simple fraction:

$\dfrac{4}{5} \div \dfrac{7}{15} \text{ of } \dfrac{8}{9}$

Answer

Solving,

Simplifying 'of' first: $\dfrac{7}{15} \text{ of } \dfrac{8}{9} = \dfrac{7}{15} \times \dfrac{8}{9} = \dfrac{56}{135}$

$\Rightarrow \dfrac{4}{5} \div \dfrac{7}{15} \text{ of } \dfrac{8}{9} \\[1em] \Rightarrow \dfrac{4}{5} \div \dfrac{56}{135} \\[1em] \Rightarrow \dfrac{4}{5} \times \dfrac{135}{56} \\[1em] \Rightarrow \dfrac{4 \times 135}{5 \times 56} \\[1em] \Rightarrow \dfrac{540}{280} \\[1em] \Rightarrow \dfrac{27}{14} \\[1em] \Rightarrow 1\dfrac{13}{14}$

Hence, the required value is $1\dfrac{13}{14}$.

Simplify and reduce to a simple fraction:

$\dfrac{4}{5} \div \dfrac{7}{15} \times \dfrac{8}{9}$

Answer

Solving,

$\Rightarrow \dfrac{4}{5} \div \dfrac{7}{15} \times \dfrac{8}{9} \\[1em] \Rightarrow \dfrac{4}{5} \times \dfrac{15}{7} \times \dfrac{8}{9} \\[1em] \Rightarrow \dfrac{4 \times 15 \times 8}{5 \times 7 \times 9} \\[1em] \Rightarrow \dfrac{480}{315} \\[1em] \Rightarrow \dfrac{32}{21} \\[1em] \Rightarrow 1\dfrac{11}{21}$

Hence, the required value is $1\dfrac{11}{21}$.

Simplify and reduce to a simple fraction:

$\dfrac{4}{5} \text{ of } \dfrac{7}{15} \div \dfrac{8}{9}$

Answer

Solving,

Simplifying 'of' first: $\dfrac{4}{5} \text{ of } \dfrac{7}{15} = \dfrac{4}{5} \times \dfrac{7}{15} = \dfrac{28}{75}$

$\Rightarrow \dfrac{4}{5} \text{ of } \dfrac{7}{15} \div \dfrac{8}{9} \\[1em] \Rightarrow \dfrac{28}{75} \div \dfrac{8}{9} \\[1em] \Rightarrow \dfrac{28}{75} \times \dfrac{9}{8} \\[1em] \Rightarrow \dfrac{28 \times 9}{75 \times 8} \\[1em] \Rightarrow \dfrac{252}{600} \\[1em] \Rightarrow \dfrac{21}{50}$

Hence, the required value is $\dfrac{21}{50}$.

Simplify and reduce to a simple fraction:

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

Answer

Solving,

Simplifying 'of' first: $\dfrac{1}{2} \text{ of } \dfrac{3}{4} = \dfrac{1}{2} \times \dfrac{3}{4} = \dfrac{3}{8}$

$\Rightarrow \dfrac{1}{2} \text{ of } \dfrac{3}{4} \times \dfrac{1}{2} \div \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{3}{8} \times \dfrac{1}{2} \div \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{3}{16} \div \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{3}{16} \times \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{9}{32}$

Hence, the required value is $\dfrac{9}{32}$.

A bought $3\dfrac{3}{4}$ kg of wheat and $2\dfrac{1}{2}$ kg of rice. Find the total weight of wheat and rice bought by A.

Answer

Given,

Weight of wheat bought = $3\dfrac{3}{4}$ kg

Weight of rice bought = $2\dfrac{1}{2}$ kg

Total weight = Weight of wheat + Weight of rice

Total weight $= 3\dfrac{3}{4} + 2\dfrac{1}{2} = \dfrac{15}{4} + \dfrac{5}{2}$

LCM of 4 and 2 = 4.

$= \dfrac{15 \times 1}{4 \times 1} + \dfrac{5 \times 2}{2 \times 2} \\[1em] = \dfrac{15}{4} + \dfrac{10}{4} \\[1em] = \dfrac{15 + 10}{4} \\[1em] = \dfrac{25}{4} \\[1em] = 6\dfrac{1}{4} \text{ kg}$

Hence, the total weight of wheat and rice is $6\dfrac{1}{4}$ kg.

Which is greater, $\dfrac{3}{5} \text{ or } \dfrac{7}{10}$ and by how much?

Answer

LCM of 5 and 10 = 10.

$\dfrac{3}{5} = \dfrac{3 \times 2}{10} = \dfrac{6}{10} \text{ and } \dfrac{7}{10} = \dfrac{7}{10}$

Since 7 > 6, $\dfrac{7}{10} \gt \dfrac{3}{5}$

Difference = $\dfrac{7}{10} - \dfrac{6}{10} = \dfrac{1}{10}$

Hence, $\dfrac{7}{10}$ is greater than $\dfrac{3}{5}\text { by } \dfrac{1}{10}$.

What number should be added to $8\dfrac{2}{3}$ to get $12\dfrac{5}{6}$?

Answer

Let the required number be x.

Then, $8\dfrac{2}{3} + x = 12\dfrac{5}{6}$

$\Rightarrow x = 12\dfrac{5}{6} - 8\dfrac{2}{3} \\[1em] = \dfrac{77}{6} - \dfrac{26}{3} \\[1em] = \dfrac{77}{6} - \dfrac{26 \times 2}{6} \\[1em] = \dfrac{77 - 52}{6} = \dfrac{25}{6} = 4\dfrac{1}{6}$

Hence, the required number is $4\dfrac{1}{6}$.

What should be subtracted from $8\dfrac{3}{4}$ to get $2\dfrac{2}{3}$?

Answer

Let the required number be x.

Then, $8\dfrac{3}{4} - x = 2\dfrac{2}{3}$

$\Rightarrow x = 8\dfrac{3}{4} - 2\dfrac{2}{3} \\[1em] = \dfrac{35}{4} - \dfrac{8}{3} \\[1em] = \dfrac{35 \times 3}{12} - \dfrac{8 \times 4}{12} \\[1em] = \dfrac{105 - 32}{12} = \dfrac{73}{12} = 6\dfrac{1}{12}$

Hence, the required number is $6\dfrac{1}{12}$.

A rectangular field is $16\dfrac{1}{2}$ m long and $12\dfrac{2}{5}$ m wide. Find the perimeter of the field.

Answer

Given,

Length of the field = $16\dfrac{1}{2}$ m

Breadth of the field = $12\dfrac{2}{5}$ m

Length = $16\dfrac{1}{2} \text { m }= \dfrac{33}{2}$ m and breadth $= 12\dfrac{2}{5}\text { m }= \dfrac{62}{5}$ m

Perimeter = 2 x (length + breadth)

$= 2 \times \left(\dfrac{33}{2} + \dfrac{62}{5}\right) \\[1em] = 2 \times \left(\dfrac{33 \times 5}{2 \times 5} + \dfrac{62 \times 2}{5 \times 2}\right) \\[1em] = 2 \times \left(\dfrac{165}{10} + \dfrac{124}{10}\right) \\[1em] = 2 \times \dfrac{165 + 124}{10} \\[1em] = 2 \times \dfrac{289}{10} \\[1em] = \dfrac{289}{5} = 57\dfrac{4}{5} \text{ m}$

Hence, the perimeter of the field is $57\dfrac{4}{5} \text { m }.$

Sugar costs ₹ $37\dfrac{1}{2}$ per kg. Find the cost of $8\dfrac{3}{4}$ kg sugar.

Answer

Given,

Cost of sugar per kg = ₹ $37\dfrac{1}{2}$

Quantity of sugar = $8\dfrac{3}{4}$ kg

Cost of $8\dfrac{3}{4}$ kg sugar = Cost per kg × Quantity

$= 37\dfrac{1}{2} \times 8\dfrac{3}{4} \\[1em] = \dfrac{75}{2} \times \dfrac{35}{4} \\[1em] = \dfrac{75 \times 35}{2 \times 4} = \dfrac{2625}{8} \\[1em] = 328\dfrac{1}{8}$

Hence, the cost of $8\dfrac{3}{4}$ kg sugar is ₹ $328\dfrac{1}{8}$.

A motor cycle runs $31\dfrac{1}{4}$ km consuming 1 litre of petrol. How much distance will it run consuming $1\dfrac{3}{5}$ litre of petrol?

Answer

Given,

Distance run on 1 litre of petrol = $31\dfrac{1}{4}$ km

Quantity of petrol = $1\dfrac{3}{5}$ litres

Distance run on $1\dfrac{3}{5}$ litres = Distance per litre × Quantity of petrol

$= 31\dfrac{1}{4} \times 1\dfrac{3}{5} \\[1em] = \dfrac{125}{4} \times \dfrac{8}{5} \\[1em] = \dfrac{125 \times 8}{4 \times 5} = \dfrac{1000}{20} \\[1em] = 50 \text{ km}$

Hence, the motor cycle will run 50 km.

A rectangular park has length = $23\dfrac{2}{5}$ m and breadth = $16\dfrac{2}{3}$ m. Find the area of the park.

Answer

Given,

Length of the park = $23\dfrac{2}{5}$ m

Breadth of the park = $16\dfrac{2}{3}$ m

Area = length × breadth

$\text{Area } = 23\dfrac{2}{5} \times 16\dfrac{2}{3} \\[1em] = \dfrac{117}{5} \times \dfrac{50}{3} \\[1em] = 39 \times 10 \\[1em] = 390 \text{ m}^2.$

Hence, the area of the park is 390 m².

Each of 40 identical boxes weighs $4\dfrac{4}{5}$ kg. Find the total weight of all the boxes.

Answer

Given,

Number of identical boxes = 40

Weight of each box = $4\dfrac{4}{5}$ kg

Total weight = Number of boxes × Weight of each box

Total weight $= 40 \times 4\dfrac{4}{5} = 40 \times \dfrac{24}{5}$

= $\dfrac{40 \times 24}{5} = \dfrac{960}{5} = 192 \text{ kg}$

Hence, the total weight of all the boxes is 192 kg.

Out of 24 kg of wheat, $\dfrac{5}{6}$ th of wheat is consumed. Find how much wheat is still left?

Answer

Given,

Total quantity of wheat = 24 kg

Fraction of wheat consumed = $\dfrac{5}{6}$

Fraction of wheat left $= 1 - \dfrac{5}{6} = \dfrac{6 - 5}{6} = \dfrac{1}{6}$

Wheat left = Fraction left × Total quantity

Wheat left $= \dfrac{1}{6} \times 24 = 4$ kg

Hence, 4 kg of wheat is still left.

A rod of length $2\dfrac{2}{5}$ metre is divided into five equal parts. Find the length of each part so obtained.

Answer

Given,

Total length of the rod = $2\dfrac{2}{5}$ metre

Number of equal parts = 5

Length of each part = Total length ÷ Number of parts

Length of each part $= 2\dfrac{2}{5} \div 5 = \dfrac{12}{5} \div 5$

= $\dfrac{12}{5} \times \dfrac{1}{5} = \dfrac{12}{25} \text{ m}$

Hence, the length of each part is $\dfrac{12}{25}$ m.

If A = $3\dfrac{3}{8}$ and B = $6\dfrac{5}{8}$, find:

(i) A ÷ B

(ii) B ÷ A.

Answer

A $= 3\dfrac{3}{8} = \dfrac{27}{8}$ and B $= 6\dfrac{5}{8} = \dfrac{53}{8}$

(i) A ÷ B $= \dfrac{27}{8} \div \dfrac{53}{8} = \dfrac{27}{8} \times \dfrac{8}{53} = \dfrac{27}{53}$

Hence, A ÷ B = $\dfrac{27}{53}$.

(ii) B ÷ A

$= \dfrac{53}{8} \div \dfrac{27}{8} \\[1em] = \dfrac{53}{8} \times \dfrac{8}{27}\\[1em] = \dfrac{53}{27} = 1\dfrac{26}{27}$

Hence, B ÷ A = $1\dfrac{26}{27}$.

Cost of $3\dfrac{5}{7}$ litres of oil is ₹ $83\dfrac{1}{2}$. Find the cost of one litre oil.

Answer

Given,

Quantity of oil = $3\dfrac{5}{7}$ litres

Cost of $3\dfrac{5}{7}$ litres of oil = ₹ $83\dfrac{1}{2}$

Cost of one litre = Total cost ÷ Quantity of oil

Cost of one litre $= 83\dfrac{1}{2} \div 3\dfrac{5}{7}$

$= \dfrac{167}{2} \div \dfrac{26}{7} \\[1em] = \dfrac{167}{2} \times \dfrac{7}{26} \\[1em] = \dfrac{167 \times 7}{2 \times 26} = \dfrac{1169}{52} \\[1em] = 22\dfrac{25}{52}$

Hence, the cost of one litre of oil is ₹ $22\dfrac{25}{52}$.

The product of two numbers is $20\dfrac{5}{7}$. If one of these numbers is $6\dfrac{2}{3}$, find the other.

Answer

Let the other number be x.

Then, $6\dfrac{2}{3} \times x = 20\dfrac{5}{7}$

$\Rightarrow x = 20\dfrac{5}{7} \div 6\dfrac{2}{3} \\[1em] = \dfrac{145}{7} \div \dfrac{20}{3} \\[1em] = \dfrac{145}{7} \times \dfrac{3}{20} \\[1em] = \dfrac{145 \times 3}{7 \times 20} = \dfrac{435}{140} \\[1em] = \dfrac{87}{28} = 3\dfrac{3}{28}$

Hence, the other number is $3\dfrac{3}{28}$.

By what number should $5\dfrac{5}{6}$ be multiplied to get $3\dfrac{1}{3}$?

Answer

Let the required number be x.

Then, $5\dfrac{5}{6} \times x = 3\dfrac{1}{3}$

$\Rightarrow x = 3\dfrac{1}{3} \div 5\dfrac{5}{6} \\[1em] = \dfrac{10}{3} \div \dfrac{35}{6} \\[1em] = \dfrac{10}{3} \times \dfrac{6}{35} \\[1em] = \dfrac{10 \times 6}{3 \times 35} = \dfrac{60}{105} \\[1em] = \dfrac{4}{7}$

Hence, the required number is $\dfrac{4}{7}$.