Question

Sum of two numbers is 4 more than the twice of difference of the two numbers. If one of the two numbers is three more than the other number, then find the numbers.

• A. $(\frac{13}{2},\frac{7}{2})$
• B. (1, 3)
• C. $(\frac{4}{5},3)$
• D. (1, 2)

Answer 1

The correct option is A

$(\frac{13}{2},\frac{7}{2})$

Let one number be x and other be y.

1st case:
$x+y=4+2(x-y)$
$\Rightarrow x+y=4+2x-2y$
$\Rightarrow x-3y+4=0$ ...(i)

2nd case:
$x=3+y$ ...(ii)

On substituting (ii) in (i), we get
$(3+y)-3y+4=0$
$\Rightarrow y=\frac{7}{2}$

On substituting the value of y in (ii), we get
$x=3+\frac{7}{2}=\frac{13}{2}$

Therefore, the numbers are $\frac{13}{2}$ and $\frac{7}{2}$.