Question

Find two numbers such that the sum of twice the first and thrice the second is $89$ and four times the first exceeds five times the second by $13$

• A. The required numbers are $10$ and $19$
• B. The required numbers are $26$ and $11$
• C. The required numbers are $22$ and $15$
• D. The required numbers are $4$ and $21$

Answer 1

The correct option is C The required numbers are $22$ and $15$

Let the two numbers be $x$ and $y$.

As per the given condition, we have

$2x+3y=89$ ....(1)

and $4x=13+5y$

$\Rightarrow 4x-5y=13$ ....(2)

Multiply equation (1) by $2$, we get

$4x+6y=178$ ....(3)

Subtract equations (2) and (3), we get

$11y=165$

$\Rightarrow y=15$

Put this value of $y$ in equation (1),

$2x+3\left(15\right)=89$

$\Rightarrow 2x+45=89$

$\Rightarrow 2x=44$

$\Rightarrow x=22$

Therefore, the required numbers are $22$ and $15$.