Q270 / 270 · NDA 2011 (Phase 2) · Mathematics

If the arithmetic and geometric means of two numbers are 10,8 respectively, then one number exceeds the other number by

  • A
    $8$
  • B
    $10$
  • Correct answer
    C
    $12$
  • D
    $16$

Solution

Let numbers are $a$ and $b$ $\mathrm{A} \cdot \mathrm{M}=\frac{a+b}{2} \Rightarrow 10=\frac{a+b}{2}$ $\Rightarrow \quad a+b=20$ ...(i) $\mathrm{G} . \mathrm{M}=\sqrt{a b} \Rightarrow 8=\sqrt{a b}$ $\Rightarrow a b=64$ $\because \quad(a-b)^2=(a+b)^2-4 a b$ $\begin{aligned} & =(20)^2-4(64) \\ & =400-256=144\end{aligned}$ $a-b=12$ ...(ii) from (i) and (ii) $\left.\begin{array}{rl}a+b & =20 \\ a-b & =12\end{array}\right] \Rightarrow a=16, b=4$ $\therefore \quad 16-4=12$ $\therefore$ one number exceeds the other number by 12 .

Hence, the correct answer is option: 'C'.

Relevant Questions