Question

Two positive numbers $x$ and $y$ are such that $x>y$. If the difference of these numbers is $5$ and their product is $24$, find sum of these numbers.

• A. $13$
• B. $23$
• C. $11$
• D. $16$

Answer 1

The correct option is C $11$

Given,

$x>y$

Thus, $x-y=5$ and $xy=24$

$x-y=5$

Squaring both sides,

$=>(x-y{)}^2=5^2$

$=>x^2+y^2-2xy=25$

$=>x^2+y^2=25+2\left(24\right)$

$=>x^2+y^2=73$

Now,

$(x+y{)}^2=x^2+y^2+2xy$

$=73+2\left(24\right)$

$=73+48$

$=121$

$=>x+y=\sqrt{121}$

$=\pm 11$

Given, $x$ and $y$ are positive numbers,

Therefore, $x+y=11$