Question

Find x in each of the following :
(i) $\frac{1}{4!}+\frac{1}{5!}=\frac{x}{6!}$
(ii) $\frac{x}{10!}=\frac{1}{8!}+\frac{1}{9!}$
(iii) $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$

Answer 1

(i) $\frac{1}{4!}+\frac{1}{5!}=\frac{x}{6!}$
$\Rightarrow \frac{1}{4!}+\frac{1}{5\times 4!}=\frac{x}{6\times 5\times 4}$
$\Rightarrow 4!\times [\frac{1}{4!}+\frac{1}{5\times 4!}]=\frac{x}{30}$
$\Rightarrow 1+\frac{1}{5}=\frac{x}{30}$
$\Rightarrow \frac{6}{5}=\frac{x}{30}$
$\Rightarrow \frac{x}{30}=\frac{6}{5}\Rightarrow x=\frac{6\times 30}{5}$ $\Rightarrow x=6\times 6\Rightarrow x=36$
Hence, x= 36.
(ii) $\frac{x}{10!}=\frac{1}{8!}+\frac{1}{9!}$
$\Rightarrow x=\frac{10!}{8!}+\frac{10!}{9!}$ $\Rightarrow x=\frac{10\times 9\times 8!}{8!}+\frac{10\times 9!}{9!}$ $\Rightarrow x=10\times 9+10\Rightarrow x=100$
(iii) $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$
$\Rightarrow x=\frac{8!}{6!}+\frac{8!}{7!}$ $\Rightarrow x=\frac{8\times 7\times 6!}{6!}+\frac{8\times 7!}{7!}$ $\Rightarrow x=8\times 7+8\Rightarrow x=64$