Question

Four of the six numbers 1867, 1993, 2019, 2025, 2109 and 2121 have a mean of 2008. What is the mean of the other two numbers:
(a) 1994
(b) 2006
(c) 2022
(d) 2051

Answer 1

Hint: We have given the mean of any four numbers from 1867, 1993, 2019, 2025, 2109 and 2121 as 2008. We know that the formula of a mean is equal to $\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$. In this problem, we have given the sum of 4 observations so substitute the value of total number of observations as 4 and equate the formula to 2008. From this, we will get the sum of 4 observations. Now, we have to know the 4 given numbers whose sum we know, so first of all we will find the sum of all 6 numbers and then subtract any two numbers till we get the summation of 4 numbers that we have just found.

Complete step by step answer:
The 6 numbers we have given above as follows:
1867, 1993, 2019, 2025, 2109 and 2121
Now, we have given the mean of any four numbers as of 2008.
We know the formula for mean as:
$Mean=\dfrac{\text{Sum of observations}}{\text{Total number of observations}}$
We have given the 4 observations mean so substituting 4 in the number of observations in the above formula and also substituting the mean as 2008 we get,
$2008=\dfrac{\text{Sum of observations}}{4}$
Cross multiplying in the above equation we get,
$\begin{align}
  & 2008\left( 4 \right)=\text{Sum of observations} \\
 & \Rightarrow 803\text{2}=\text{Sum of observations} \\
\end{align}$
Now, we got the sum of 4 observations as 8032.
We are asked to find the mean of any two numbers. Let us assume that mean of these remaining two numbers as x then:
$\begin{align}
  & \dfrac{\text{Sum of two numbers}}{2}=x \\
 & \Rightarrow \text{Sum of two numbers}=2x \\
\end{align}$
Now, we have sum of two numbers as 2x and sum of the other 4 numbers as 8032 so adding the sum of 4 numbers and sum of remaining two numbers is equal to sum of all 6 numbers.
Sum of given 6 numbers is equal to:
$\begin{align}
  & 1867+1993+2019+2025+2109+2121 \\
 & =12134 \\
\end{align}$
Adding sum of 4 numbers and the remaining two numbers is equal to:
$8032+2x$
Equating the above expression to 12134 we get,
$\begin{align}
  & 8032+2x=12134 \\
 & \Rightarrow 2x=12134-8032 \\
 & \Rightarrow 2x=4102 \\
\end{align}$
Dividing 2 on both the sides we get,
$\begin{align}
  & x=\dfrac{4102}{2} \\
 & \Rightarrow x=2051 \\
\end{align}$
From the above calculations, we got the mean of the remaining two numbers as 2051.

So, the correct answer is “Option D”.

Note: You can check the mean of the remaining two numbers by multiplying the mean of two numbers by 2 and then add this sum of two numbers to the sum of the other 4 numbers.
Mean of two numbers obtained is 2051. Multiplying this mean by 2 we get,
$\begin{align}
  & 2051\times 2 \\
 & =4102 \\
\end{align}$
In the above solution, we have got the sum of 4 numbers as 8032 so adding the above result to 8032 we get,
$\begin{align}
  & 8032+4102 \\
 & =12134 \\
\end{align}$
In the above solution, we have found the sum of all 6 numbers as 12134 which is matching with the above result. This means, we have calculated the right mean of two numbers.