Question

The sum of three numbers is $92$. The second number is three times the first and the third exceeds the second by $8$. The three numbers are:

• A. $14,36,42$
• B. $18,30,44$
• C. $8,38,46$
• D. $12,36,44$

Answer 1

The correct option is D $12,36,44$

Let the three numbers are $x$, $y$ and $z$.
Since, The sum of three numbers is 92.
$\therefore$ $x+y+z=92$ ...(1)
Also, The second number is three times the first.
$\therefore$ $y=3x$
$\therefore$ $x=\frac{y}{3}$ ...(2)
The third number exceeds the second by 8.
$\therefore$ $z-y=8$ ...(3)
Substituting (2) in (1) we get,
$4y+3z=276$ ...(4)
Solving (3) and (4) we get,
$y=36$ and $z=44$
Re substituting above two values in (1) we get,
$x=12$