Question

The sum of one-half, one third and one-fourth of a number exceeds the number by $12$. What is the number?

• A. $72$
• B. $144$
• C. $98$
• D. $156$

Answer 1

Answer: (B)

$144$

Let the number be $x$.

Then, $(\frac{x}{2}+\frac{x}{3}+\frac{x}{4})-x=12$,

L.C.M of $2,3,4=12$,

Thus multiplying above equation with $12$ both sides,
$\Rightarrow 12\times \frac{x}{2}+12\times \frac{x}{3}+12\times \frac{x}{4}-12\times x-12\times 12$

$\Rightarrow 6x+4x+3x-12x=144$

$\Rightarrow 13x-12x=144\Rightarrow x=144$.

Hence, the required number is $144.$