Fractions - Exercise 2(C)

Simplify :

$1\dfrac{2}{3} + \dfrac{5}{6}$ of $\dfrac{24}{25}$

Answer

We have:

$1\dfrac{2}{3} + \dfrac{5}{6}$ of $\dfrac{24}{25}$

= $\dfrac{5}{3}$ + $\dfrac{5}{6}$ of $\dfrac{24}{25}$ [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

$\begin{array}{ll} = \dfrac{5}{3} + \dfrac{5}{6} \times \dfrac{24}{25} \\\\ = \dfrac{5}{3} + \dfrac{1}{6} \times \dfrac{24}{5} \\\\ = \dfrac{5}{3} + \dfrac{24}{30} \\\\ = \dfrac{5}{3} + \dfrac{4}{5} & \text{[Of simplified]} \\\\ = \dfrac{25 + 12}{15} = \dfrac{37}{15} & \text{[Addition simplified]} \\\\ = 2\dfrac{7}{15} & \text{[Converting improper to mixed fraction]} \end{array}$

∴ The answer is $2\dfrac{7}{15}$

Simplify :

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

Answer

We have:

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

= $\dfrac{1}{3}$ of $\dfrac{14}{3}$ ÷ $\dfrac{7}{3}\times \dfrac{3}{2}$ [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

$\begin{array}{ll} = \dfrac{1}{3} \times \dfrac{14}{3} ÷ \dfrac{7}{3}\times \dfrac{3}{2} \\\\ = \dfrac{1 \times 14}{3 \times 3} ÷ \dfrac{7}{3}\times \dfrac{3}{2} \\\\ = \dfrac{14}{9} ÷ \dfrac{7}{3}\times \dfrac{3}{2} & \text{[Of simplified]} \\\\ = \dfrac{14}{9} \times \dfrac{3}{7}\times \dfrac{3}{2} & [\text{Reciprocal of } \dfrac{7}{3} \text{ is } \dfrac{3}{7}] \\\\ = \dfrac{14}{3} \times \dfrac{1}{7} \times \dfrac{3}{2} \\\\ = \dfrac{2}{3} \times \dfrac{1}{1} \times \dfrac{3}{2} \\\\ = \dfrac{2 \times1}{3 \times 1} \times \dfrac{3}{2} = \dfrac{2}{3} \times \dfrac{3}{2} & \text{[Division simplified]} \\\\ = \dfrac{2 \times 3}{3 \times 2} = \dfrac{6}{6}\\\\ = \dfrac{1}{1} = 1 & \text{[Multiplication simplified]} \end{array}$

∴ The answer is 1

Simplify :

$2\dfrac{1}{4} + 1\dfrac{1}{6} - 1\dfrac{2}{3} ÷ 2\dfrac{2}{3}$ of $3\dfrac{3}{4}$

Answer

We have:

$2\dfrac{1}{4} + 1\dfrac{1}{6} - 1\dfrac{2}{3} ÷ 2\dfrac{2}{3}$ of $3\dfrac{3}{4}$

= $\dfrac{9}{4}$ + $\dfrac{7}{6}$ - $\dfrac{5}{3}$ ÷ $\dfrac{8}{3}$ of $\dfrac{15}{4}$ [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

$\begin{array}{ll} = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{3} ÷ \dfrac{8}{3}\times \dfrac{15}{4} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{3} ÷ \dfrac{8 \times 15}{3 \times 4} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{3} ÷ \dfrac{120}{12} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{3} ÷ \dfrac{10}{1} & \text{[Of simplified]} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{3} \times \dfrac{1}{10} & [\text{Reciprocal of } \dfrac{10}{1} \text{ is } \dfrac{1}{10}] \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5 \times 1}{3 \times 10} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{30} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{1}{6} \\\\ = \dfrac{27 + 14}{12} - \dfrac{1}{6} \\\\ = \dfrac{41}{12} - \dfrac{1}{6} & \text{[Addition simplified]} \\\\ = \dfrac{41 - 2}{12} \\\\ = \dfrac{39}{12} \\\\ = \dfrac{13}{4} & \text{[Subtraction simplified]} \\\\ = 3\dfrac{1}{4} & \text{[Converting improper to mixed fraction]} \\\\ \end{array}$

∴ The answer is $3\dfrac{1}{4}$

Simplify :

$1\dfrac{1}{2}\times2\dfrac{3}{4} ÷ 1\dfrac{4}{7}$ of $2\dfrac{5}{8}$

Answer

We have:

$1\dfrac{1}{2}\times2\dfrac{3}{4} ÷ 1\dfrac{4}{7}$ of $2\dfrac{5}{8}$

= $\dfrac{3}{2}\times \dfrac{11}{4}$ ÷ $\dfrac{11}{7}$ of $\dfrac{21}{8}$ [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

$\begin{array}{ll} = \dfrac{3}{2}\times \dfrac{11}{4} ÷ \dfrac{11}{7} \times \dfrac{21}{8} \\\\ = \dfrac{3}{2}\times \dfrac{11}{4} ÷ \dfrac{11}{1} \times \dfrac{3}{8} \\\\ = \dfrac{3}{2}\times \dfrac{11}{4} ÷ \dfrac{11 \times 3}{1 \times 8} \\\\ = \dfrac{3}{2}\times \dfrac{11}{4} ÷ \dfrac{33}{8} & \text{[Of simplified]} \\\\ = \dfrac{3}{2}\times \dfrac{11}{4} \times \dfrac{8}{33} & [\text{Reciprocal of } \dfrac{33}{8} \text{ is } \dfrac{8}{33}] \\\\ = \dfrac{3}{2}\times \dfrac{1}{4} \times \dfrac{8}{3} \\\\ = \dfrac{3}{2}\times \dfrac{1}{1} \times \dfrac{2}{3}\\\\ = \dfrac{3}{2}\times \dfrac{1 \times 2}{1 \times 3} \\\\ = \dfrac{3}{2}\times \dfrac{2}{3} & \text{[Division simplified]} \\\\ = \dfrac{6}{6} = 1 & \text{[Multiplication simplified]} \\\\ \end{array}$

∴ The answer is 1

Simplify :

$\Big(2\dfrac{3}{4} + 1\dfrac{5}{6}\Big)$ ÷ $2\dfrac{1}{5}$ of $3\dfrac{1}{3}$

Answer

We have:

$\Big(2\dfrac{3}{4} + 1\dfrac{5}{6}\Big)$ ÷ $2\dfrac{1}{5}$ of $3\dfrac{1}{3}$

= $\Big(\dfrac{11}{4} + \dfrac{11}{6}\Big)$ ÷ $\dfrac{11}{5}$ of $\dfrac{10}{3}$ [Converting mixed to improper fraction]

According to BODMAS rule, we simplify brackets first

$\begin{array}{ll} = \Big(\dfrac{33 + 22}{12}\Big) ÷ \dfrac{11}{5} \text{ of } \dfrac{10}{3} \\\\ = \dfrac{55}{12} ÷ \dfrac{11}{5} \text{ of } \dfrac{10}{3} & \text{[Brackets simplified]} \\\\ = \dfrac{55}{12} ÷ \dfrac{11}{5} \times \dfrac{10}{3} \\\\ = \dfrac{55}{12} ÷ \dfrac{110}{15} \\\\ = \dfrac{55}{12} ÷ \dfrac{22}{3} & \text{[Of simplified]} \\\\ = \dfrac{55}{12} \times \dfrac{3}{22} & [\text{Reciprocal of } \dfrac{22}{3} \text{ is } \dfrac{3}{22}] \\\\ = \dfrac{5}{12} \times \dfrac{3}{2} & \text{[Dividing 55 and 22 by 11]} \\\\ = \dfrac{15}{24} = \dfrac{5}{8} & \text{[Division simplified]} \end{array}$

∴ The answer is $\dfrac{5}{8}$

Simplify :

$\dfrac{7}{15}$ of $\Big(\dfrac{2}{3} + \dfrac{7}{12}\Big)$ ÷ $\Big(\dfrac{5}{6}-\dfrac{3}{5}\Big)$

Answer

We have:

$\dfrac{7}{15}$ of $\Big(\dfrac{2}{3} + \dfrac{7}{12}\Big)$ ÷ $\Big(\dfrac{5}{6}-\dfrac{3}{5}\Big)$

According to BODMAS rule, we simplify brackets first

$\begin{array}{ll} = \dfrac{7}{15}\text{ of }\Big(\dfrac{8 + 7}{12}\Big) ÷ \Big(\dfrac{5}{6}-\dfrac{3}{5}\Big) \\\\ = \dfrac{7}{15}\text{ of } \Big(\dfrac{15}{12}\Big) ÷ \Big(\dfrac{5}{6}-\dfrac{3}{5}\Big) \\\\ = \dfrac{7}{15}\text{ of } \dfrac{15}{12} ÷ \Big(\dfrac{5}{6}-\dfrac{3}{5}\Big) & \text{[First bracket simplified]} \\\\ = \dfrac{7}{15}\text{ of } \dfrac{15}{12} ÷ \Big(\dfrac{25 - 18}{30}\Big) \\\\ = \dfrac{7}{15}\text{ of } \dfrac{15}{12} ÷ \dfrac{7}{30} & \text{[Second bracket simplified]} \\\\ = \dfrac{7}{15} \times \dfrac{15}{12} ÷ \dfrac{7}{30} \\\\ = \dfrac{7}{12} ÷ \dfrac{7}{30} & \text{[Of simplified]} \\\\ = \dfrac{7}{12} \times \dfrac{30}{7} & [\text{Reciprocal of } \dfrac{7}{30} \text{ is } \dfrac{30}{7}] \\\\ = \dfrac{1}{12} \times \dfrac{30}{1} & \text{[Dividing 7 and 7 by 7]} \\\\ = \dfrac{1}{2} \times \dfrac{5}{1} & \text{[Dividing 30 and 12 by 6]} \\\\ = \dfrac{5}{2} & \text{[Division simplified]} \\\\ = 2\dfrac{1}{2} \end{array}$

∴ The answer is $2\dfrac{1}{2}$

Simplify :

$\Big(22÷5\dfrac{1}{2}\Big)$ ÷ $2\dfrac{1}{5}$ of $3\dfrac{1}{3} + 1\dfrac{5}{11}$

Answer

We have:

$\Big(22÷5\dfrac{1}{2}\Big)$ ÷ $2\dfrac{1}{5}$ of $3\dfrac{1}{3} + 1\dfrac{5}{11}$

= $\Big(22÷\dfrac{11}{2}\Big)$ ÷ $\dfrac{11}{5}$ of $\dfrac{10}{3}$ + $\dfrac{16}{11}$ [Converting mixed to improper fraction]

According to BODMAS rule, we simplify brackets first

$\begin{array}{ll} = \Big(22 \times \dfrac{2}{11}\Big) ÷ \dfrac{11}{5}\text{ of }\dfrac{10}{3} + \dfrac{16}{11} & [\text{Reciprocal of } \dfrac{11}{2} \text{ is } \dfrac{2}{11}] \\\\ = \Big(\dfrac{44}{11}\Big) ÷ \dfrac{11}{5}\text{ of }\dfrac{10}{3} + \dfrac{16}{11} & \text{[Divide 44 and 11 by 11]} \\\\ = \dfrac{4}{1} ÷ \dfrac{11}{5}\text{ of }\dfrac{10}{3} + \dfrac{16}{11} & \text{[Bracket simplified]} \\\\ = \dfrac{4}{1} ÷ \dfrac{11}{5} \times \dfrac{10}{3} + \dfrac{16}{11} \\\\ = \dfrac{4}{1} ÷ \dfrac{110}{15} + \dfrac{16}{11} & \text{[Divide 110 and 15 by 5]} \\\\ = \dfrac{4}{1} ÷ \dfrac{22}{3} + \dfrac{16}{11} & \text{[Of simplified]} \\\\ = \dfrac{4}{1} \times \dfrac{3}{22} + \dfrac{16}{11} & [\text{Reciprocal of } \dfrac{22}{3} \text{ is } \dfrac{3}{22}] \\\\ = \dfrac{12}{22} + \dfrac{16}{11} & \text{[Divide 12 and 22 by 2]} \\\\ = \dfrac{6}{11} + \dfrac{16}{11} & \text{[Division simplified]} \\\\ = \dfrac{6 + 16}{11} = \dfrac{22}{11} & \text{[Divide 22 and 11 by 11]} \\\\ = 2 & \text{[Division simplified]} \end{array}$

∴ The answer is 2

Simplify :

$6\dfrac{1}{3}$ ÷ $\Big(2\dfrac{1}{5} + 3\dfrac{1}{2}\Big)$ of $3\dfrac{1}{3}$

Answer

We have:

$6\dfrac{1}{3}$ ÷ $\Big(2\dfrac{1}{5} + 3\dfrac{1}{2}\Big)$ of $3\dfrac{1}{3}$

= $\dfrac{19}{3}$ ÷ $\Big(\dfrac{11}{5} + \dfrac{7}{2}\Big)$ of $\dfrac{10}{3}$ [Converting mixed to improper fraction]

According to BODMAS rule, we simplify brackets first

$\begin{array}{ll} = \dfrac{19}{3} ÷ \Big(\dfrac{22 + 35}{10}\Big)\text{ of }\dfrac{10}{3} \\\\ = \dfrac{19}{3} ÷ \dfrac{57}{10}\text{ of }\dfrac{10}{3} & \text{[Bracket simplified]} \\\\ = \dfrac{19}{3} ÷ \dfrac{57}{10} \times \dfrac{10}{3} \\\\ = \dfrac{19}{3} ÷ \dfrac{570}{30} \\\\ = \dfrac{19}{3} ÷ \dfrac{19}{1} & \text{[Of simplified]} \\\\ = \dfrac{19}{3} \times \dfrac{1}{19} & [\text{Reciprocal of } \dfrac{19}{1} \text{ is } \dfrac{1}{19}] \\\\ = \dfrac{1}{3} & \text{[Division simplified]} \end{array}$

∴ The answer is $\dfrac{1}{3}$