Question

The absolute difference of the two positive numbers whose arithmetic mean is $34$ and the geometric mean is $16,$ is

• A. $60$
• B. $64$
• C. $56$
• D. $68$

Answer 1

The correct option is A $60$

Let $a$ and $b$ be two positive numbers.
Given, $\frac{a+b}{2}=34,\sqrt{ab}=16$

Now, $a+b=68,ab=256$
$(a-b{)}^2=(a+b{)}^2-4ab\Rightarrow (a-b{)}^2={68}^2-4\times 256\Rightarrow (a-b{)}^2=16({17}^2-8^2)\Rightarrow (a-b{)}^2=16\times 25\times 9\Rightarrow |a-b|=4\times 5\times 3\therefore |a-b|=60$