Question

The mean of $a,b,c,d$ and $e$ is $28$. If the mean of $a,c$ and $e$ is $24$, what is the mean of $b$ and $d$?

• A. $31$
• B. $32$
• C. $33$
• D. $34$

Answer 1

Answer: (D)

$34$

Formula used:

$\text{Arithmetic mean}=\frac{\text{Sum of given numbers}}{\text{Total numbers}}$

Given $\Rightarrow$ $\frac{a+b+c+d+e}{5}=28$ and $\frac{a+c+e}{3}=24$

To find $\Rightarrow$ $\frac{b+d}{2}$ which is nothing but the mean of $b$ and $d$.

$\begin{array}{cc}\frac{a+b+c+d+e}{5}& =28\\ a+b+c+d+e& =140\dots \left(i\right)\end{array}$

Also,

$\begin{array}{cc}\frac{a+c+e}{3}& =24\\ a+c+e& =72\dots \left(ii\right)\end{array}$

By using $\left(i\right)$ and $\left(ii\right)$,

$\begin{array}{cc}a+b+c+d+e& =140\\ b+d+72& =140\\ b+d& =68\\ \frac{b+d}{2}& =34\end{array}$

Hence, option $D$ is the correct answer.