Fractions — Exercise 6(B)

Reduce the given fractions to their lowest terms:

$\dfrac{8}{10}$

Answer

$\dfrac{8}{10}$

By prime factorisation,

$\Rightarrow \dfrac{8}{10} = \dfrac{2 \times 2 \times 2}{2 \times 5} \\[1em] = \dfrac{2 \times 2}{5} \\[1em] = \dfrac{4}{5}.$

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

Reduce the given fractions to their lowest terms:

$\dfrac{50}{75}$

Answer

$\dfrac{50}{75}$

By prime factorisation,

$\Rightarrow \dfrac{50}{75} = \dfrac{2 \times 5 \times 5}{3 \times 5 \times 5} \\[1em] = \dfrac{2}{3}.$

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

Reduce the given fractions to their lowest terms:

$\dfrac{18}{81}$

Answer

$\dfrac{18}{81}$

By prime factorisation,

$\Rightarrow \dfrac{18}{81} = \dfrac{2 \times 3 \times 3}{3 \times 3 \times 3 \times 3} \\[1em] = \dfrac{2}{3 \times 3} \\[1em] = \dfrac{2}{9}.$

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

Reduce the given fractions to their lowest terms:

$\dfrac{40}{120}$

Answer

$\dfrac{40}{120}$

By prime factorisation,

$\Rightarrow \dfrac{40}{120} = \dfrac{2 \times 2 \times 2 \times 5}{2 \times 2 \times 2 \times 3 \times 5} \\[1em] = \dfrac{1}{3}.$

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

Reduce the given fractions to their lowest terms:

$\dfrac{105}{70}$

Answer

$\dfrac{105}{70}$

By prime factorisation,

$\Rightarrow \dfrac{105}{70} = \dfrac{3 \times 5 \times 7}{2 \times 5 \times 7} \\[1em] = \dfrac{3}{2} \\[1em] = 1\dfrac{1}{2}.$

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

State whether true or false:

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

Answer

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

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

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

Hence, the statement is False.

State whether true or false:

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

Answer

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

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.

State whether true or false:

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

Answer

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

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

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

Hence, the statement is False.

State whether true or false:

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

Answer

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

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

Hence, the statement is False.

State whether true or false:

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

Answer

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

Converting, improper fraction into mixed fraction:

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

Hence, the statement is False.

Which fraction is greater?

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

Answer

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

LCM of 5 and 3 = 15.

$\Rightarrow \dfrac{3}{5} = \dfrac{3 \times 3}{5 \times 3} = \dfrac{9}{15} \\[1em] \Rightarrow \dfrac{2}{3} = \dfrac{2 \times 5}{3 \times 5} = \dfrac{10}{15}$

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

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

Which fraction is greater?

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

Answer

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

LCM of 9 and 4 = 36.

$\Rightarrow \dfrac{5}{9} = \dfrac{5 \times 4}{9 \times 4} = \dfrac{20}{36} \\[1em] \Rightarrow \dfrac{3}{4} = \dfrac{3 \times 9}{4 \times 9} = \dfrac{27}{36}$

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

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

Which fraction is greater?

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

Answer

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

LCM of 14 and 35 = 70.

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

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

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

Which fraction is smaller?

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

Answer

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

LCM of 8 and 5 = 40.

$\Rightarrow \dfrac{3}{8} = \dfrac{3 \times 5}{8 \times 5} = \dfrac{15}{40} \\[1em] \Rightarrow \dfrac{4}{5} = \dfrac{4 \times 8}{5 \times 8} = \dfrac{32}{40}$

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

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

Which fraction is smaller?

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

Answer

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

LCM of 15 and 7 = 105.

$\Rightarrow \dfrac{8}{15} = \dfrac{8 \times 7}{15 \times 7} = \dfrac{56}{105} \\[1em] \Rightarrow \dfrac{4}{7} = \dfrac{4 \times 15}{7 \times 15} = \dfrac{60}{105}$

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

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

Which fraction is smaller?

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

Answer

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

LCM of 26 and 39 = 78.

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

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

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

Arrange the given fractions in descending order of magnitude:

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

Answer

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

LCM of 16, 24 and 8 = 48.

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

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

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

Arrange the given fractions in descending order of magnitude:

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

Answer

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

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

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

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

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

Arrange the given fractions in descending order of magnitude:

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

Answer

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

LCM of 7, 8 and 11 = 616.

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

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

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

Arrange the given fractions in ascending order of magnitude:

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

Answer

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

LCM of 16, 12 and 4 = 48.

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

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

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

Arrange the given fractions in ascending order of magnitude:

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

Answer

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

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

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

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

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

Arrange the given fractions in ascending order of magnitude:

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

Answer

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

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

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

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

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

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

$\dfrac{6}{11} \text{...............} \dfrac{5}{9}$

Answer

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

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

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

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

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

$\dfrac{3}{7} \text{...............} \dfrac{9}{13}$

Answer

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

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

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

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

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

$\dfrac{56}{64} \text{...............} \dfrac{7}{8}$

Answer

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

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

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

Insert the symbol '=' or '>' or '<' between each of the pairs of fractions given below:

$\dfrac{5}{12} \text{...............} \dfrac{8}{33}$

Answer

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

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

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

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