Chapter 9.3, Problem 71STP

To determine

To Find: The number removed.

Answer to Problem 71STP

  $(C)45$

Explanation of Solution

Given: The mean of six numbers is $20$ . If one of the number is removed, the average remaining numbers is $15$

  $\begin{array}{l}(A)42\\ (B)43\\ (C)45\\ (D)48\end{array}$

Given: Mean of six numbers is $20$

Thus

  $\begin{array}{l}Mean=\frac{\text{sum of six numbers}}{6}\\ 20=\frac{\text{sum of six numbers}}{6}\\ sum=120\end{array}$

Now

Average of five numbers is $15$

Thus

  $\begin{array}{l}Mean=\frac{\text{sum of six numbers}}{5}\\ 15=\frac{\text{sum of six numbers}}{5}\\ sum=75\end{array}$

Now

Let the number removed be $x$

Then

Sum of six numbers = sum of five numbers + number removed

We get

  $\begin{array}{l}120=75+x\\ x=45\end{array}$

Hence, $(C)45$