Question

The average of $8$ Numbers is$20$, that of the first two is $15.5$ and that of the next three is$\frac{211}{3}$, The $6^{th}$ is less than the $7^{th}$ by $4$ and $7$ less than the $8^{th}.$ What is the last number?

Answer 1

Step 1: Make an equation by the given data.

Sum of $8$ numbers$=20\times 8=160$

Sum of first $5$ numbers $=15.5\times 2+\frac{211}{3}\times 3$

$$\begin{gathered} =31+64 \\ =95 \end{gathered}$$

Sum of first $5$ numbers $=95$

Sum of remaining $3$ numbers $=(160–95)=65$

Step 2: Assuming the remaining numbers for making an equation and then solving.

Let the $6^{th},7^{th}\&8^{th}$ numbers be $x,y\&z$ respectively.

$$\begin{gathered} y-x=4........\left[1\right] \\ z-x=7........\left[2\right] \end{gathered}$$

$$\begin{gathered} \therefore y=x+4 \\ \therefore z=x+7 \end{gathered}$$

But from the above condition, we get,

$x+y+z=65$

$$\begin{gathered} \therefore x+x+4+x+7=65 \\ 3x+11=65 \\ 3x=65-11 \\ 3x=54 \\ \therefore x=\frac{54}{3} \\ x=18 \end{gathered}$$

But from equation second we get,

$$\begin{gathered} z–x=7 \\ z-18=7 \\ \therefore z=7+18 \\ z=25 \end{gathered}$$

Therefore, the last number is $25.$