Question

Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 borders and Rs. 600 when there are 50 boarders. What is the average expense per border when there are 100 borders?
A. 550
B. 580
C. 540
C. 570

Answer 1

Hint: Let a, b, c, … represent ‘n’ number of observations, then the average of these observations will be given by
$\Rightarrow \text{Average value = }\dfrac{(a+b+c+............)}{n}$ , where n = total number of observations.
If it is given that a variable x varies linearly with the variable y
$\begin{align}
  & \Rightarrow y\propto x \\
 & \Rightarrow y=kx \\
\end{align}$
Here k is the proportionality constant.

Complete step by step answer:
From our question we have that, the total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders
$\Rightarrow $ If the total expense is assumed to be T, the partly fixed is assumed to be x and the variable to be y, n be the number of borders.
$\Rightarrow T=\text{ }x+ny$
Now, firstly it is given that the average expense is Rs. 700 if the numbers of borders are 25.
Using, $\text{Average value = }\dfrac{(a+b+c+............)}{n}$
$\Rightarrow 700=\dfrac{\text{Total expense}}{25}$
Hence, the total expense for 25 borders is equal to \[700\times 25\] which is 17500.
Substituting the above values in \[T=x+ny\] with n = 25 and T=17500, we get
$\Rightarrow 17500\text{ }=\text{ }x\text{ }+\text{ }25y~~~$ ------equation (1)
Secondly it is also given in the question that the average expense is Rs. 600 if the number of boarders is 50.
$\begin{align}
  & \text{Average value = }\dfrac{(a+b+c+............)}{n} \\
 & \Rightarrow 600=\dfrac{\text{Total expense}}{50} \\
\end{align}$
Hence, the total expense for 50 borders in the second case is equal to \[600\times 50\] which is 30000.
Substituting the above values in \[T=x+ny\] with n = 50 and T = 30000, we get
$\Rightarrow 30000\text{ }=\text{ }x\text{ }+\text{ }50y$ -------equation (2)
Subtracting equation (1) from equation (2), we get
\[\begin{array}{*{35}{l}}
   \Rightarrow \left( \text{ }x\text{ }+\text{ }50y \right)\text{ }\text{ }\left( \text{ }x\text{ }+\text{ }25y\text{ } \right)\text{ }=\text{ }30000\text{ }\text{ }17500 \\
   \Rightarrow 50y\text{ }\text{ }25y\text{ }=\text{ }12500 \\
   \Rightarrow 25y\text{ }=\text{ }12500 \\
   ~\Rightarrow y\text{ }=\text{ }500 \\
\end{array}\]
Substituting the value of y in equation (1), we get
\[\begin{array}{*{35}{l}}
   \Rightarrow 17500\text{ }=\text{ }x\text{ }+\text{ }25\left( 500 \right) \\
   \Rightarrow 17500\text{ }\text{ }12500\text{ }=\text{ }x \\
   \Rightarrow x\text{ }=\text{ }5000 \\
\end{array}\]
Finally, we have been asked to find the average expense per boarder when there are 100 boarders
Here we have n = 100, x = 5000 and y = 500. Substituting these values in \[T=x+ny\], we get
\[\begin{align}
  & ~\Rightarrow T\text{ }=\text{ }5000\text{ }+\text{ }100\left( 500 \right) \\
 & ~\Rightarrow T\text{ }=\text{ }5000\text{ }+\text{ }50000~ \\
 & \Rightarrow T\text{ }=\text{ }55000 \\
\end{align}\]

The total expense is Rs. 55000.
Using the formula $\text{Average=}\dfrac{\text{Total Expense}}{\text{Number of Boarders}}$, we get
 $\Rightarrow \text{Average}=\dfrac{55000}{100}=550$
$\therefore $ The average expense per boarder when there are 100 boarders is Rs. 550.
Hence, Option A is correct.

Note: In this particular question it is given that the expenses are varying linearly with the number of borders but it might not be the same case with every problem. Quite a few times it may be varying with a cube root or inversely or any other case. Students should be accurate while solving the question.