Fractions — Exercise 4(D)

Find the sum :

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

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

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

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

Answer

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

$\Rightarrow \dfrac{2 + 3}{7}\\[1em] \Rightarrow \dfrac{5}{7}\\[1em]$

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

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

$\Rightarrow \dfrac{5 + 1}{8}\\[1em] \Rightarrow \dfrac{6}{8}\\[1em] \Rightarrow \dfrac{3}{4}\\[1em]$

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

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

$\Rightarrow \dfrac{7 + 4}{9}\\[1em] \Rightarrow \dfrac{11}{9}\\[1em] \Rightarrow 1\dfrac{2}{9}\\[1em]$

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

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

$\Rightarrow \dfrac{11}{6} + \dfrac{1}{6}\\[1em] \Rightarrow \dfrac{11 + 1}{6}\\[1em] \Rightarrow \dfrac{12}{6}\\[1em] \Rightarrow 2$

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

Find the sum :

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

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

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

Answer

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

$\Rightarrow \dfrac{5 + 3 + 1}{14}\\[1em] \Rightarrow \dfrac{9}{14}\\[1em]$

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

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

$\Rightarrow \dfrac{7 + 5 + 11}{12}\\[1em] \Rightarrow \dfrac{23}{12}\\[1em] \Rightarrow 1\dfrac{11}{12}\\[1em]$

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

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

$\Rightarrow \dfrac{15}{8} + \dfrac{3}{8} + \dfrac{13}{8}\\[1em] \Rightarrow \dfrac{15 + 3 + 13}{8}\\[1em] \Rightarrow \dfrac{31}{8}\\[1em] \Rightarrow 3\dfrac{7}{8}\\[1em]$

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

Find the sum :

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

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

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

Answer

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

LCM of 9 and 6 = 18

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

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

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

LCM of 12 and 16 = 48

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

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

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

LCM of 12 and 18 = 36

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

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

Find the sum :

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

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

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

Answer

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

LCM of 10, 15 and 20 = 60

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

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

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

LCM of 8, 16 and 24 = 48

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

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

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

LCM of 6, 9, 18 and 27 = 54

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

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

Find the sum :

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

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

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

Answer

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

LCM of 6, 8 and 12 = 24

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

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

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

LCM of 2, 3, 4 and 5 = 60

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

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

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

LCM of 3, 9, 2 and 18 = 18

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

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

Find the difference :

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

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

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

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

Answer

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

$\Rightarrow \dfrac{9 - 5}{11}\\[1em] \Rightarrow \dfrac{4}{11}$

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

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

$\Rightarrow \dfrac{61}{7} - \dfrac{6}{7}\\[1em] \Rightarrow \dfrac{61 - 6}{7}\\[1em] \Rightarrow \dfrac{55}{7} = 7\dfrac{6}{7}$

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

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

$\Rightarrow \dfrac{27}{8} - \dfrac{15}{8}\\[1em] \Rightarrow \dfrac{27 - 15}{8}\\[1em] \Rightarrow \dfrac{12}{8}\\[1em] \Rightarrow \dfrac{3}{2} = 1\dfrac{1}{2}$

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

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

$\Rightarrow \dfrac{7 - 3}{4}\\[1em] \Rightarrow \dfrac{4}{4}\\[1em] \Rightarrow 1$

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

Find the difference :

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

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

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

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

Answer

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

LCM of 15 and 20 = 60

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

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

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

LCM of 7 and 14 = 14

$\Rightarrow \dfrac{29}{7} - \dfrac{29}{14}\\[1em] \Rightarrow \dfrac{29 \times 2}{7 \times 2} - \dfrac{29 \times 1}{14 \times 1}\\[1em] \Rightarrow \dfrac{58}{14} - \dfrac{29}{14}\\[1em] \Rightarrow \dfrac{58 - 29}{14}\\[1em] \Rightarrow \dfrac{29}{14}\\[1em] \Rightarrow 2\dfrac{1}{14}\\[1em]$

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

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

LCM of 8 and 12 = 24

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

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

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

$\Rightarrow \dfrac{5}{1} - \dfrac{11}{4}\\[1em] \Rightarrow \dfrac{5 \times 4}{1 \times 4} - \dfrac{11}{4}\\[1em] \Rightarrow \dfrac{20}{4} - \dfrac{11}{4}\\[1em] \Rightarrow \dfrac{20 - 11}{4}\\[1em] \Rightarrow \dfrac{9}{4}\\[1em] \Rightarrow 2\dfrac{1}{4}\\[1em]$

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

Find the difference :

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

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

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

Answer

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

LCM of 8 and 12 = 24

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

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

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

LCM of 6 and 10 = 30

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

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

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

$\Rightarrow \dfrac{12}{1} - \dfrac{53}{8}\\[1em] \Rightarrow \dfrac{12 \times 8}{1 \times 8} - \dfrac{53 \times 1}{8 \times 1}\\[1em] \Rightarrow \dfrac{96}{8} - \dfrac{53}{8}\\[1em] \Rightarrow \dfrac{96 - 53}{8}\\[1em] \Rightarrow \dfrac{43}{8}\\[1em] \Rightarrow 5\dfrac{3}{8}$

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

Simplify :

$\dfrac{4}{9} - \dfrac{5}{12} + \dfrac{1}{4}$

Answer

LCM of 9, 12 and 4 = 36

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

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

Simplify :

$6\dfrac{2}{3} + 4\dfrac{1}{6} - 2\dfrac{2}{9}$

Answer

Solving,

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

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

Simplify :

$9 - 4\dfrac{1}{2} - 2\dfrac{1}{5}$

Answer

LCM of 2 and 5 = 10

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

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

Simplify :

$10\dfrac{3}{4} - 5\dfrac{1}{8} - 4\dfrac{5}{12}$

Answer

LCM of 4, 8 and 12 = 24

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

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

Simplify :

$6\dfrac{4}{5} - 3\dfrac{4}{15} + 4\dfrac{3}{10}$

Answer

LCM of 5, 15 and 10 = 30

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

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

Simplify :

$4\dfrac{5}{12} + 3\dfrac{11}{18} - 2\dfrac{7}{24}$

Answer

LCM of 12, 18 and 24 = 72

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

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

Subtract the sum of $9\dfrac{3}{4}$ and $3\dfrac{5}{6}$ from $15\dfrac{7}{12}$.

Answer

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

LCM of 4 and 6 = 12

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

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

$\Rightarrow \dfrac{187}{12} - \dfrac{163}{12}\\[1em] \Rightarrow \dfrac{187 - 163}{12}\\[1em] \Rightarrow \dfrac{24}{12}\\[1em] \Rightarrow 2$

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

Subtract the sum of $2\dfrac{5}{12}$ and $3\dfrac{3}{4}$ from the sum of $7\dfrac{1}{3}$ and $5\dfrac{1}{6}$.

Answer

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

LCM of 12 and 4 = 12

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

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

LCM of 3 and 6 = 6

$\Rightarrow \dfrac{22}{3} + \dfrac{31}{6}\\[1em] \Rightarrow \dfrac{22 \times 2}{3 \times 2} + \dfrac{31}{6}\\[1em] \Rightarrow \dfrac{44}{6} + \dfrac{31}{6}\\[1em] \Rightarrow \dfrac{44 + 31}{6}\\[1em] \Rightarrow \dfrac{75}{6}\\[1em] \Rightarrow \dfrac{25}{2}\\[1em]$

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

LCM of 6 and 2 = 6

$\Rightarrow \dfrac{25}{2} - \dfrac{37}{6}\\[1em] \Rightarrow \dfrac{25 \times 3}{2 \times 3} - \dfrac{37}{6} \\[1em] \Rightarrow \dfrac{75}{6} - \dfrac{37}{6} \\[1em] \Rightarrow \dfrac{75 - 37}{6} \\[1em] \Rightarrow \dfrac{38}{6} \\[1em] \Rightarrow \dfrac{19}{3}\\[1em] \Rightarrow 6\dfrac{1}{3}\\[1em]$

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

What should be added to $9\dfrac{4}{7}$ to get 16?

Answer

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

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

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

What must be subtracted from $9\dfrac{1}{6}$ to get $6\dfrac{1}{9}$?

Answer

Let the number that can be subtracted be x.

$\Rightarrow 9\dfrac{1}{6} - x = 6\dfrac{1}{9}\\[1em] \Rightarrow \dfrac{55}{6} - x = \dfrac{55}{9}\\[1em] \Rightarrow x = \dfrac{55}{6} - \dfrac{55}{9}\\[1em]$

LCM of 6 and 9 = 18

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

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

Of $\dfrac{17}{20}$ and $\dfrac{21}{25}$, which is greater and by how much?

Answer

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

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

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

The sum of two fractions is $14\dfrac{5}{12}$. If one of them is $7\dfrac{2}{3}$, find the other.

Answer

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

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

Other number

$\Rightarrow 14\dfrac{5}{12} - 7\dfrac{2}{3}\\[1em] \Rightarrow \dfrac{173}{12} - \dfrac{23}{3}\\[1em]$

LCM of 12 and 3 = 12.

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

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

From a piece of wire $12\dfrac{3}{4}$ m long, a small piece of length $3\dfrac{5}{6}$ m has been cut off. What is the length of the remaining piece?

Answer

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

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

Remaining piece

$\Rightarrow 12\dfrac{3}{4} - 3\dfrac{5}{6}\\[1em] \Rightarrow \dfrac{51}{4} - \dfrac{23}{6}\\[1em]$

LCM of 4 and 6 = 12

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

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

Three boxes weigh $9\dfrac{1}{2}$ kg, $14\dfrac{1}{5}$ kg and $18\dfrac{3}{4}$ kg respectively. A porter carries all the three boxes. What is the total weight carried by the porter?

Answer

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

LCM of 2, 5 and 4 = 20

Total weight

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

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

On one day, a labourer earned ₹ 125. Out of this money, he spent ₹ $68\dfrac{1}{2}$ on food, ₹ $20\dfrac{3}{4}$ on tea and ₹ $16\dfrac{2}{5}$ on other eatables. How much does he save on that day?

Answer

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

Total

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

LCM of 2, 4 and 5 = 20

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

Savings = Income − Total expenses

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

Hence, savings = $₹19\dfrac{7}{20}$.