Question

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

• A. $46,138$
• B. $32,152$
• C. $64,120$
• D. $72,112$

Answer 1

Answer: (D)

$72,112$

Let the two parts be $x$ and $184-x.$

According to the given condition, we have
$\frac{x}{3}=\frac{1}{7}(184-x)+8\Rightarrow \frac{x}{3}=\frac{184-x+56}{7}$ .....$\left[\text{Taking LCM on RHS}\right]$

$\Rightarrow \frac{x}{3}=\frac{240-x}{7}$

$\Rightarrow 7x=3(240-x)$ .....$\left[\text{On cross-multiplication}\right]$

$\Rightarrow 7x=720-3x$

$\Rightarrow 7x+3x=720$ .....$[\text{Transposing}-3x\text{to LHS}]$

$\Rightarrow 10x=720\Rightarrow x=72$

So, one part is $72$ and other part is $184-72=112$.