Question

Divide 184 into two parts such that one- third of one part may exceed one-seventh of the other part by 4.

• A. 61.6 and 120.4
• B. 62.6 and 120.4
• C. 63.6 and 120.4
• D. 64.6 and 120.4

Answer 1

Answer: (C)

63.6 and 120.4

Let the two parts be $x$ and $y$
=> $x+y=184$ --- (1)
Given, one- third of one part may exceed one-seventh of the other part by $4$
$=>\frac{x}{3}-\frac{y}{7}=4$
$=>7x-3y=84$ --- (2)

Multiplying equation $\left(1\right)$ with $3$ we get, $3x+3y=552$ ----- equation $\left(3\right)$

Adding equations $2$ and $3$, we get $10x=636=>x=63.6$

Substituting
$x=63.6$ in the equation $\left(2\right)$, we get $63.6+y=184=>y=120.4$

Thus , the parts are $63.6;120.4$