Question

The difference between two numbers is $48$ and the difference between their arithmetic mean and their geometric mean is $18$. Then, the greater of two numbers is

• A. $96$
• B. $60$
• C. $54$
• D. $49$

Answer 1

The correct option is D

$49$

Explanation for the correct option :

Step-1 : Assumption

Let the two numbers be $x$ and $y$ where $x>y$.

Then their arithmetic mean will be $\mathit{A}\mathbf{.}\mathit{M}\mathbf{.}\mathbf{=}\frac{\mathbf{x}\mathbf{+}\mathbf{y}}{\mathbf{2}}$ and the geometric mean will be $\mathit{G}\mathbf{.}\mathit{M}\mathbf{.}\mathbf{=}\sqrt{\mathbf{x}\mathbf{y}}$.

Step-2 : Constructing two linear equations using two conditions given in the question.

Given that the difference of the two numbers is $48$. So, we have

$x-y=48$ $\left(1\right)$

Also, given that the difference between their arithmetic mean and the geometric mean is $18$. So, we obtain

$\frac{x+y}{2}-\sqrt{xy}=18$ $\left(2\right)$

Step-3 : Solving the two equations using the substitution method

From equation $\left(1\right)$, we get $x=y+48$.

Substituting $x=y+48$ in $\left(2\right)$, we get

$$\begin{gathered} \frac{y+48+y}{2}-\sqrt{(y+48)y}=18 \\ \Rightarrow y+24-18=\sqrt{(y+48)y} \\ \Rightarrow y+6=\sqrt{y^2+48y} \\ \Rightarrow {\left(y+6\right)}^2=y^2+48y \\ \Rightarrow y^2+12y+36=y^2+48y \\ \Rightarrow 36y=36 \\ \Rightarrow y=1 \end{gathered}$$

Then,

$$\begin{gathered} x=y+48 \\ =1+48 \\ =49 \end{gathered}$$

$\therefore x=49,y=1$ is the solution of $\left(1\right)$ and $\left(2\right)$.

So, the greater of the two numbers is $49$

Hence, option (D) is the correct answer.