Question

Divide $184$ into two parts such that one-third of one part may exceed one-seventh of another part by $8.$

Answer 1

Let the first part be $x$. Then the other part will be $184–x$.

It is given that ${\left(\frac{1}{3}\right)}^{rd}$ of one part exceeds ${\left(\frac{1}{7}\right)}^{th}$ of another part by $8$:

$\Rightarrow$$\frac{x}{3}-\frac{(184–x)}{7}=8$
$\Rightarrow$$\frac{(7x–552+3x)}{21}=8$ [ as LCM of $3$and $7$is$21$ ]

$\Rightarrow$ $(7x–552+3x)=8\times 21$

$\Rightarrow$ $10x–552=168$

$\Rightarrow$ $10x=168+552$

$\Rightarrow$ $10x=720$

$\Rightarrow$ $x=\frac{720}{10}$

$\Rightarrow$ $x=72$

If $x=72$, then $184–x=184-72$

$=112$

Hence, $184$ can be divided into $72$ and $112$ to satisfy the given conditions.