Question

Find two positive numbers whose difference is $12$ and whose $A.M$. exceeds their $G.M$. by $2$.

• A. $15,3$
• B. $14,2$
• C. $16,4$
• D. $17,5$

Answer 1

Answer: (C)

Let the two numbers be $a$ and $b$. Then,
$a-b=12...\left(i\right)$
It is given that $A.M.-G.M.=2$
$\Rightarrow \frac{a+b}{2}-\sqrt{ab}=2$
$\Rightarrow a+b-2\sqrt{ab}=4$
$\Rightarrow {\left(\sqrt{a}-\sqrt{b}\right)}^2=4$
$\Rightarrow \sqrt{a}-\sqrt{b}=\pm 2...\left(ii\right)$
Now, $a-b=12$
$\Rightarrow \left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=12$
$\Rightarrow \left(\sqrt{a}+\sqrt{b}\right)\times \left(\pm 2\right)=12$ [Using $(ii)$]
$\Rightarrow \sqrt{a}+\sqrt{b}=\pm 6...\left(iii\right)$
Solving $\left(ii\right)$ and $\left(iii\right)$, we get $a=16,b=4$