Question

Average of 8 numbers is 20, that of the first two is $15.5$ and that of next three is $21\dfrac{1}{3}$, the ${{6}^{th}}$ is less than the ${{7}^{th}}$ by 4 and 7 less than the ${{8}^{th}}$. The last number is
A. 25
B. 28
C. 35
D. 32

Answer 1

Hint: We first find the mathematical form of average for the given information. From the given forms we find the sum of the last three terms. Then we use the given relations among those numbers and find the value of the last term.

Complete step-by-step solution:
It is given that Average of 8 numbers is 20, that of the first two is $15.5$ and that of the next three is $21\dfrac{1}{3}$.
We assume the numbers as ${{a}_{i}},i=1\left( 1 \right)8$.
We know that the average of n numbers is $\dfrac{1}{n}\sum{{{a}_{i}}}$.
We try to find the mathematical form of the average for 8 numbers and the first two numbers and its next three numbers.
Therefore, $\dfrac{1}{8}\sum\limits_{i=1}^{8}{{{a}_{i}}}=20,\dfrac{1}{2}\sum\limits_{i=1}^{2}{{{a}_{i}}}=15.5,\dfrac{1}{3}\sum\limits_{i=3}^{5}{{{a}_{i}}}=21\dfrac{1}{3}$.
From the average form we find the sum by multiplying with the constant term where
$\sum\limits_{i=1}^{8}{{{a}_{i}}}=160,\sum\limits_{i=1}^{2}{{{a}_{i}}}=31,\sum\limits_{i=3}^{5}{{{a}_{i}}}=64$.
We can find the sum of last 3 numbers by subtraction
Therefore, \[\sum\limits_{i=6}^{8}{{{a}_{i}}}=\sum\limits_{i=1}^{8}{{{a}_{i}}}-\sum\limits_{i=1}^{2}{{{a}_{i}}}-\sum\limits_{i=3}^{5}{{{a}_{i}}}\].
Putting the values, we get \[\sum\limits_{i=6}^{8}{{{a}_{i}}}=160-31-64=65\].
We have that ${{6}^{th}}$ is less than the ${{7}^{th}}$ by 4 and 7 less than the ${{8}^{th}}$.
Let the ${{8}^{th}}$ be $x$. Then the ${{6}^{th}}$ is $x-7$ and ${{7}^{th}}$ is $x-7+4=x-3$.
The addition will give $x+x-7+x-3=65$.
The simplification gives $3x-10=65$.
We find the value of $x$ as
$\begin{align}
  & 3x-10=65 \\
 & \Rightarrow 3x=65+10=75 \\
 & \Rightarrow x=\dfrac{75}{3}=25 \\
\end{align}$
Therefore, the ${{8}^{th}}$ is 25. Then the ${{6}^{th}}$ is $25-7=18$ and ${{7}^{th}}$ is $x-3=25-3=22$.
The correct option is A.

Note: As the average number of numbers is different to each other we cannot complete the problem using only average form. We have to convert them to summation form to find the solution.