Question

The A.M of two numbers is $34$ and GM is $16$, the numbers are

• A. $2$ and $64$
• B. $64$ and $3$
• C. $64$ and $4$
• D. None of these

Answer 1

Answer: (C)

$64$ and $4$

Let $a,b$ be the numbers.

Then according to the problem,

$\frac{a+b}{2}=34$ ....... [Arithmetic mean]

or, $a+b=68$ ........ $\left(1\right)$

And $\sqrt{ab}=16$ ........ [Geometric mean]

or, $ab=256$

or, $a(68-a)=256$ ........ Using $\left(1\right)$

or, $a^2-68a+256=0$

or, $(a-64)(a-4)=0$

or, $a=64,4$ then $b=4,64$.

So the required numbers are $64$ and $4$.