Question
Arrange the given fraction in descending order by making numerator equal:
$\frac{3}{7}, \frac{4}{8}, \frac{5}{7} \text { and } \frac{8}{11}$
$\frac{3}{7}, \frac{4}{8}, \frac{5}{7} \text { and } \frac{8}{11}$
✓
Answer
$ \frac{3}{7}, \frac{4}{8}, \frac{5}{7} \text { and } \frac{8}{11} $
L.C.M. of numerators $3,4,5$ and $8=120$
$ \begin{aligned} & \therefore \frac{3}{7}=\frac{3 \times 40}{7 \times 40}=\frac{120}{280} \\ & \frac{4}{9}=\frac{4 \times 30}{9 \times 30}=\frac{120}{270} \\ & \frac{5}{7}=\frac{5 \times 24}{7 \times 24}=\frac{120}{168} \\ & \frac{8}{11}=\frac{8 \times 15}{11 \times 15}=\frac{120}{165} \end{aligned} $
It is clear from the above that
$ \begin{aligned} & \frac{120}{165}>\frac{120}{168}>\frac{120}{270}>\frac{120}{280} \\ & \text { or } \frac{8}{11}>\frac{5}{7}>\frac{4}{9}>\frac{3}{7} \end{aligned} $
Hence, $\frac{8}{11}, \frac{5}{7}, \frac{4}{9}, \frac{3}{7}$ are in descending order.
L.C.M. of numerators $3,4,5$ and $8=120$
$ \begin{aligned} & \therefore \frac{3}{7}=\frac{3 \times 40}{7 \times 40}=\frac{120}{280} \\ & \frac{4}{9}=\frac{4 \times 30}{9 \times 30}=\frac{120}{270} \\ & \frac{5}{7}=\frac{5 \times 24}{7 \times 24}=\frac{120}{168} \\ & \frac{8}{11}=\frac{8 \times 15}{11 \times 15}=\frac{120}{165} \end{aligned} $
It is clear from the above that
$ \begin{aligned} & \frac{120}{165}>\frac{120}{168}>\frac{120}{270}>\frac{120}{280} \\ & \text { or } \frac{8}{11}>\frac{5}{7}>\frac{4}{9}>\frac{3}{7} \end{aligned} $
Hence, $\frac{8}{11}, \frac{5}{7}, \frac{4}{9}, \frac{3}{7}$ are in descending order.
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