Multiply :

(i) $\dfrac{2}{3}$ by $\dfrac{4}{5}$

(ii) $\dfrac{7}{6}$ by $\dfrac{9}{2}$

(iii) $\dfrac{5}{6}$ by 30

(iv) $\dfrac{-3}{4}$ by $\dfrac{8}{7}$

(v) $\dfrac{-16}{9}$ by $\dfrac{12}{-5}$

(vi) $\dfrac{35}{-8}$ by $\dfrac{12}{-5}$

(vii) $\dfrac{-3}{10}$ by $\dfrac{-40}{9}$

(viii) $\dfrac{-32}{5}$ by $\dfrac{15}{-16}$

(ix) $\dfrac{-8}{15}$ by $\dfrac{-25}{32}$

(i) $\dfrac{2}{3}$ by $\dfrac{4}{5}$

We have:

$\phantom{=} \dfrac{2}{3} \times \dfrac{4}{5} \\[1em] = \dfrac{2 \times 4}{3 \times 5} \\[1em] = \dfrac{8}{15}$

Hence the answer is $\dfrac{8}{15}$

(ii) $\dfrac{7}{6}$ by $\dfrac{9}{2}$

We have:

$\phantom{=} \dfrac{7}{6} \times \dfrac{9}{2} \\[1em] = \dfrac{7 \times 9}{6 \times 2} \\[1em] = \dfrac{63}{12} \\[1em] = \dfrac{21}{4} \quad \text{[Dividing both by 3]}$

Hence the answer is $\dfrac{21}{4}$

(iii) $\dfrac{5}{6}$ by 30

We have:

$\phantom{=} \dfrac{5}{6} \times 30 \\[1em] = \dfrac{5}{6} \times \dfrac{30}{1} \\[1em] = \dfrac{5 \times 30}{6 \times 1} \\[1em] = \dfrac{150}{6} \\[1em] = 25$

Hence the answer is 25

(iv) $\dfrac{-3}{4}$ by $\dfrac{8}{7}$

We have:

$\phantom{=} \dfrac{-3}{4} \times \dfrac{8}{7} \\[1em] = \dfrac{-3 \times 8}{4 \times 7} \\[1em] = \dfrac{-24}{28} \\[1em] = \dfrac{-6}{7} \quad \text{[Dividing both by 4]}$

Hence the answer is $\dfrac{-6}{7}$

(v) $\dfrac{-16}{9}$ by $\dfrac{12}{-5}$

We have:

$\phantom{=} \dfrac{-16}{9} \times \dfrac{12 \times (-1)}{-5 \times (-1)} \quad \text{[Making denominator positive]} \\[1em] = \dfrac{-16}{9} \times \dfrac{-12}{5} \\[1em] = \dfrac{(-16) \times (-12)}{9 \times 5} \\[1em] = \dfrac{192}{45} \\[1em] = \dfrac{64}{15} \quad \text{[Dividing both by 3]}$

Hence the answer is $\dfrac{64}{15}$

(vi) $\dfrac{35}{-8}$ by $\dfrac{12}{-5}$

We have:

$\phantom{=} \dfrac{35 \times (-1)}{-8 \times (-1)} \times \dfrac{12 \times (-1)}{-5 \times (-1)} \quad \text{[Making denominator positive]} \\[1em] = \dfrac{-35}{8} \times \dfrac{-12}{5} \\[1em] = \dfrac{35 \times 12}{(8) \times (5)} \\[1em] = \dfrac{420}{40} \\[1em] = \dfrac{42}{4} \\[1em] = \dfrac{21}{2} \quad \text{[Dividing both by 2]}$

Hence the answer is $\dfrac{21}{2}$

(vii) $\dfrac{-3}{10}$ by $\dfrac{-40}{9}$

We have:

$\phantom{=} \dfrac{-3}{10} \times \dfrac{-40}{9} \\[1em] = \dfrac{(-3) \times (-40)}{10 \times 9} \\[1em] = \dfrac{120}{90} \\[1em] = \dfrac{12}{9} \\[1em] = \dfrac{4}{3} \quad \text{[Dividing both by 3]}$

Hence the answer is $\dfrac{4}{3}$

(viii) $\dfrac{-32}{5}$ by $\dfrac{15}{-16}$

We have:

$\phantom{=} \dfrac{-32}{5} \times \dfrac{15 \times (-1)}{-16 \times (-1)} \quad \text{[Making denominator positive]} \\[1em] = \dfrac{(-32) \times -15}{5 \times (16)} \\[1em] = \dfrac{480}{80} \\[1em] = 6 \hspace{2cm}\text{[Dividing both by 80]}$

Hence the answer is 6

(ix) $\dfrac{-8}{15}$ by $\dfrac{-25}{32}$

We have:

$\phantom{=} \dfrac{-8}{15} \times \dfrac{-25}{32} \\[1em] = \dfrac{(-8) \times (-25)}{15 \times 32} \\[1em] = \dfrac{200}{480} \\[1em] = \dfrac{20}{48} \\[1em] = \dfrac{5}{12} \quad \text{[Dividing both by 4]}$

Hence the answer is $\dfrac{5}{12}$