Fractions — Exercise 3(B)

For each pair, given below, state whether it forms like fractions or unlike fractions:

(i) $\dfrac{5}{8} \text{ and } \dfrac{7}{8}$

(ii) $\dfrac{8}{15} \text{ and } \dfrac{8}{21}$

(iii) $\dfrac{4}{9} \text{ and } \dfrac{9}{4}$

Answer

(i) $\dfrac{5}{8} \text{ and } \dfrac{7}{8}$ have the same denominator (8), so they are Like fractions.

(ii) $\dfrac{8}{15} \text{ and } \dfrac{8}{21}$ have different denominators (15 and 21), so they are Unlike fractions.

(iii) $\dfrac{4}{9} \text{ and } \dfrac{9}{4}$ have different denominators (9 and 4), so they are Unlike fractions.

Convert given fractions into fractions with equal denominators:

(i) $\dfrac{5}{6} \text{ and } \dfrac{7}{9}$

(ii) $\dfrac{2}{3}, \dfrac{5}{6} \text{ and } \dfrac{7}{12}$

(iii) $\dfrac{4}{5}, \dfrac{17}{20}, \dfrac{23}{40} \text{ and } \dfrac{11}{16}$

Answer

(i) LCM of 6 and 9 = 18.

$\dfrac{5}{6} = \dfrac{5 \times 3}{6 \times 3} = \dfrac{15}{18} \\[1em] \dfrac{7}{9} = \dfrac{7 \times 2}{9 \times 2} = \dfrac{14}{18}$

Hence, the required fractions are $\dfrac{15}{18} \text{ and } \dfrac{14}{18}$.

(ii) LCM of 3, 6 and 12 = 12.

$\dfrac{2}{3} = \dfrac{2 \times 4}{3 \times 4} = \dfrac{8}{12} \\[1em] \dfrac{5}{6} = \dfrac{5 \times 2}{6 \times 2} = \dfrac{10}{12} \\[1em] \dfrac{7}{12} = \dfrac{7}{12}$

Hence, the required fractions are $\dfrac{8}{12}, \dfrac{10}{12} \text{ and } \dfrac{7}{12}$.

(iii) LCM of 5, 20, 40 and 16 = 80.

$\dfrac{4}{5} = \dfrac{4 \times 16}{5 \times 16} = \dfrac{64}{80} \\[1em] \dfrac{17}{20} = \dfrac{17 \times 4}{20 \times 4} = \dfrac{68}{80} \\[1em] \dfrac{23}{40} = \dfrac{23 \times 2}{40 \times 2} = \dfrac{46}{80} \\[1em] \dfrac{11}{16} = \dfrac{11 \times 5}{16 \times 5} = \dfrac{55}{80}$

Hence, the required fractions are $\dfrac{64}{80}, \dfrac{68}{80}, \dfrac{46}{80} \text{ and } \dfrac{55}{80}$.

Convert given fractions into fractions with equal numerators:

(i) $\dfrac{8}{9} \text{ and } \dfrac{12}{17}$

(ii) $\dfrac{6}{13}, \dfrac{15}{23} \text{ and } \dfrac{12}{17}$

(iii) $\dfrac{15}{19}, \dfrac{25}{28}, \dfrac{9}{11} \text{ and } \dfrac{45}{47}$

Answer

(i) LCM of the numerators 8 and 12 = 24.

$\dfrac{8}{9} = \dfrac{8 \times 3}{9 \times 3} = \dfrac{24}{27} \\[1em] \dfrac{12}{17} = \dfrac{12 \times 2}{17 \times 2} = \dfrac{24}{34}$

Hence, the required fractions are $\dfrac{24}{27} \text{ and } \dfrac{24}{34}$.

(ii) LCM of the numerators 6, 15 and 12 = 60.

$\dfrac{6}{13} = \dfrac{6 \times 10}{13 \times 10} = \dfrac{60}{130} \\[1em] \dfrac{15}{23} = \dfrac{15 \times 4}{23 \times 4} = \dfrac{60}{92} \\[1em] \dfrac{12}{17} = \dfrac{12 \times 5}{17 \times 5} = \dfrac{60}{85}$

Hence, the required fractions are $\dfrac{60}{130}, \dfrac{60}{92} \text{ and } \dfrac{60}{85}$.

(iii) LCM of the numerators 15, 25, 9 and 45 = 225.

$\dfrac{15}{19} = \dfrac{15 \times 15}{19 \times 15} = \dfrac{225}{285} \\[1em] \dfrac{25}{28} = \dfrac{25 \times 9}{28 \times 9} = \dfrac{225}{252} \\[1em] \dfrac{9}{11} = \dfrac{9 \times 25}{11 \times 25} = \dfrac{225}{275} \\[1em] \dfrac{45}{47} = \dfrac{45 \times 5}{47 \times 5} = \dfrac{225}{235}$

Hence, the required fractions are $\dfrac{225}{285}, \dfrac{225}{252}, \dfrac{225}{275} \text{ and } \dfrac{225}{235}$.

Compare the given fractions by making the denominators equal:

(i) $\dfrac{2}{5} \text{ and } \dfrac{4}{9}$

(ii) $\dfrac{5}{7} \text{ and } \dfrac{8}{11}$

(iii) $\dfrac{7}{15} \text{ and } \dfrac{9}{20}$

Answer

(i) LCM of 5 and 9 = 45.

$\dfrac{2}{5} = \dfrac{2 \times 9}{5 \times 9} = \dfrac{18}{45} \\[1em] \dfrac{4}{9} = \dfrac{4 \times 5}{9 \times 5} = \dfrac{20}{45}$

Since 18 < 20, $\dfrac{18}{45} \lt \dfrac{20}{45}$

Hence, $\dfrac{2}{5} \lt \dfrac{4}{9}$

(ii) LCM of 7 and 11 = 77.

$\dfrac{5}{7} = \dfrac{5 \times 11}{7 \times 11} = \dfrac{55}{77} \\[1em] \dfrac{8}{11} = \dfrac{8 \times 7}{11 \times 7} = \dfrac{56}{77}$

Since 55 < 56, $\dfrac{55}{77} \lt \dfrac{56}{77}$

Hence, $\dfrac{5}{7} \lt \dfrac{8}{11}$

(iii) LCM of 15 and 20 = 60.

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

Since 28 > 27, $\dfrac{28}{60} \gt \dfrac{27}{60}$

Hence, $\dfrac{7}{15} \gt \dfrac{9}{20}$

Compare the given fractions by making the numerators equal:

(i) $\dfrac{4}{9} \text{ and } \dfrac{2}{5}$

(ii) $\dfrac{5}{12} \text{ and } \dfrac{8}{19}$

(iii) $\dfrac{5}{7} \text{ and } \dfrac{9}{14}$

Answer

(i) LCM of the numerators 4 and 2 = 4.

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

Since the numerators are equal and 9 < 10, the fraction with the smaller denominator is greater.

Hence, $\dfrac{4}{9} \gt \dfrac{2}{5}$

(ii) LCM of the numerators 5 and 8 = 40.

$\dfrac{5}{12} = \dfrac{5 \times 8}{12 \times 8} = \dfrac{40}{96} \\[1em] \dfrac{8}{19} = \dfrac{8 \times 5}{19 \times 5} = \dfrac{40}{95}$

Since the numerators are equal and 96 > 95, $\dfrac{40}{96} \lt \dfrac{40}{95}$

Hence, $\dfrac{5}{12} \lt \dfrac{8}{19}$

(iii) LCM of the numerators 5 and 9 = 45.

$\dfrac{5}{7} = \dfrac{5 \times 9}{7 \times 9} = \dfrac{45}{63} \\[1em] \dfrac{9}{14} = \dfrac{9 \times 5}{14 \times 5} = \dfrac{45}{70}$

Since the numerators are equal and 63 < 70, $\dfrac{45}{63} \gt \dfrac{45}{70}$

Hence, $\dfrac{5}{7} \gt \dfrac{9}{14}$

Compare the given fractions by cross-multiplication method:

(i) $\dfrac{2}{5} \text{ and } \dfrac{4}{9}$

(ii) $\dfrac{3}{8} \text{ and } \dfrac{6}{11}$

(iii) $\dfrac{5}{18} \text{ and } \dfrac{11}{21}$

Answer

(i) For $\dfrac{2}{5} \text{ and } \dfrac{4}{9}$, cross-multiplying gives 2 × 9 = 18 and 5 × 4 = 20.

Since 18 < 20,

Hence, $\dfrac{2}{5} \lt \dfrac{4}{9}$

(ii) For $\dfrac{3}{8} \text{ and } \dfrac{6}{11}$, cross-multiplying gives 3 × 11 = 33 and 8 × 6 = 48.

Since 33 < 48,

Hence, $\dfrac{3}{8} \lt \dfrac{6}{11}$

(iii) For $\dfrac{5}{18} \text{ and } \dfrac{11}{21}$, cross-multiplying gives 5 × 21 = 105 and 18 × 11 = 198.

Since 105 < 198,

Hence, $\dfrac{5}{18} \lt \dfrac{11}{21}$

Arrange the given fractions in ascending order by making the denominators equal:

(i) $\dfrac{1}{3}, \dfrac{2}{5}, \dfrac{3}{4} \text{ and } \dfrac{1}{6}$

(ii) $\dfrac{5}{6}, \dfrac{7}{8}, \dfrac{11}{12} \text{ and } \dfrac{3}{10}$

(iii) $\dfrac{5}{7}, \dfrac{3}{8}, \dfrac{9}{14} \text{ and } \dfrac{20}{21}$

Answer

(i) LCM of 3, 5, 4 and 6 = 60.

$\dfrac{1}{3} = \dfrac{1 \times 20}{3 \times 20} = \dfrac{20}{60} \\[1em] \dfrac{2}{5} = \dfrac{2 \times 12}{5 \times 12} = \dfrac{24}{60} \\[1em] \dfrac{3}{4} = \dfrac{3 \times 15}{4 \times 15} = \dfrac{45}{60} \\[1em] \dfrac{1}{6} = \dfrac{1 \times 10}{6 \times 10} = \dfrac{10}{60}$

Arranging the numerators in ascending order: 10 < 20 < 24 < 45

Hence, the ascending order is $\dfrac{1}{6}, \dfrac{1}{3}, \dfrac{2}{5}, \dfrac{3}{4}$.

(ii) LCM of 6, 8, 12 and 10 = 120.

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

Arranging the numerators in ascending order: 36 < 100 < 105 < 110

Hence, the ascending order is $\dfrac{3}{10}, \dfrac{5}{6}, \dfrac{7}{8}, \dfrac{11}{12}$.

(iii) LCM of 7, 8, 14 and 21 = 168.

$\dfrac{5}{7} = \dfrac{5 \times 24}{7 \times 24} = \dfrac{120}{168} \\[1em] \dfrac{3}{8} = \dfrac{3 \times 21}{8 \times 21} = \dfrac{63}{168} \\[1em] \dfrac{9}{14} = \dfrac{9 \times 12}{14 \times 12} = \dfrac{108}{168} \\[1em] \dfrac{20}{21} = \dfrac{20 \times 8}{21 \times 8} = \dfrac{160}{168}$

Arranging the numerators in ascending order: 63 < 108 < 120 < 160

Hence, the ascending order is $\dfrac{3}{8}, \dfrac{9}{14}, \dfrac{5}{7}, \dfrac{20}{21}$.

Arrange the given fractions in descending order by making the numerators equal:

(i) $\dfrac{5}{6}, \dfrac{4}{15}, \dfrac{8}{9} \text{ and } \dfrac{1}{3}$

(ii) $\dfrac{3}{7}, \dfrac{4}{9}, \dfrac{5}{7} \text{ and } \dfrac{8}{11}$

(iii) $\dfrac{1}{10}, \dfrac{6}{11}, \dfrac{8}{11} \text{ and } \dfrac{3}{5}$

Answer

(i) LCM of the numerators 5, 4, 8 and 1 = 40.

$\dfrac{5}{6} = \dfrac{5 \times 8}{6 \times 8} = \dfrac{40}{48} \\[1em] \dfrac{4}{15} = \dfrac{4 \times 10}{15 \times 10} = \dfrac{40}{150} \\[1em] \dfrac{8}{9} = \dfrac{8 \times 5}{9 \times 5} = \dfrac{40}{45} \\[1em] \dfrac{1}{3} = \dfrac{1 \times 40}{3 \times 40} = \dfrac{40}{120}$

With equal numerators, the fraction with the smaller denominator is greater. Arranging the denominators in ascending order: 45 < 48 < 120 < 150

Hence, the descending order is $\dfrac{8}{9}, \dfrac{5}{6}, \dfrac{1}{3}, \dfrac{4}{15}$.

(ii) LCM of the numerators 3, 4, 5 and 8 = 120.

$\dfrac{3}{7} = \dfrac{3 \times 40}{7 \times 40} = \dfrac{120}{280} \\[1em] \dfrac{4}{9} = \dfrac{4 \times 30}{9 \times 30} = \dfrac{120}{270} \\[1em] \dfrac{5}{7} = \dfrac{5 \times 24}{7 \times 24} = \dfrac{120}{168} \\[1em] \dfrac{8}{11} = \dfrac{8 \times 15}{11 \times 15} = \dfrac{120}{165}$

With equal numerators, the fraction with the smaller denominator is greater. Arranging the denominators in ascending order: 165 < 168 < 270 < 280

Hence, the descending order is $\dfrac{8}{11}, \dfrac{5}{7}, \dfrac{4}{9}, \dfrac{3}{7}$.

(iii) LCM of the numerators 1, 6, 8 and 3 = 24.

$\dfrac{1}{10} = \dfrac{1 \times 24}{10 \times 24} = \dfrac{24}{240} \\[1em] \dfrac{6}{11} = \dfrac{6 \times 4}{11 \times 4} = \dfrac{24}{44} \\[1em] \dfrac{8}{11} = \dfrac{8 \times 3}{11 \times 3} = \dfrac{24}{33} \\[1em] \dfrac{3}{5} = \dfrac{3 \times 8}{5 \times 8} = \dfrac{24}{40}$

With equal numerators, the fraction with the smaller denominator is greater. Arranging the denominators in ascending order: 33 < 40 < 44 < 240

Hence, the descending order is $\dfrac{8}{11}, \dfrac{3}{5}, \dfrac{6}{11}, \dfrac{1}{10}$.