Question

The expenses of a hotel consists of two parts. One part varies with the number of inmates while the other is always constant. When the number of inmates is 200 and 250 the expenses are respectively Rs. 1300 and Rs. 1600. Then the expenses for 300 inmates are
a. Rs. 1800
b. Rs. 1900
c. Rs. 2000
d. Rs. 2100

Answer 1

Hint: As the expenses of a hotel has been divided into two parts. Thus, assume that the variable part is \[x\] and the constant part is \[K\]. Form the equations using the variable, constant and the number of inmates and put it equal to the expenses to determine the values of both the parts, one is variable part and the other is constant part.

Complete step-by-step solution:
Consider the given data,
We have to let that the variable part is \[x\] and the constant part is equal to \[K\].
The variable part varies with the number of inmates and the second part remains always constant as per the question.
Form two equations between the variable, constant, number of inmates and expenses.
The data for first equation is as follows:
The number of inmates is 200 and the expense is Rs. 1300.
We will multiply the number of inmates with the variable value as the variable varies with the number of inmates.
Thus, we get the equation as follows:
\[200x + K = 1300\] ---(1)
The data for second equation is as follows:
The number of inmates is 250 and the expense is Rs. 1600.
We will multiply the number of inmates with the variable value as the variable varies with the number of inmates.
Thus, we get the equation as follows:
\[250x + K = 1600\] ---(2)
Now, solve both the equations (1) and (2) to find the values of \[x\] and \[K\].
Thus, find the value of \[K\] in terms of \[x\] from the equation (1),
We get,
\[K = 1300 - 200x\]
Substituting the obtained value of \[K\] in equation (2) and determining the value of \[x\].
\[250x + \left( {1300 - 200x} \right) = 1600\]
Now, simplify the equation,
We get,
\[
   \Rightarrow 250x - 200x = 1600 - 1300 \\
   \Rightarrow 50x = 300 \\
   \Rightarrow x = \dfrac{{300}}{{50}} \\
   \Rightarrow x = 6 \\
\]
Hence, we get the value of the variable \[x = 6\].
Substitute the value of \[x = 6\] in the expression \[K = 1300 - 200x\] to determine the value of \[K\].
\[
  K = 1300 - 200\left( 6 \right) \\
   = 1300 - 1200 \\
   = 100 \\
 \]
Thus, we get the value of the constant \[K = 100\].
Now, form another equation to find the expenses when the number of inmates is 300.
Thus, the required expenses equation is as follows,
\[300x + K\]
Now, by substituting the obtained values of \[x = 6\] and \[K = 100\] to determine the required expenses.
Thus, we get that,
\[300\left( 6 \right) + 100 = 1900\]
Hence, we can conclude that the expenses for 300 inmates are Rs. 1900.
The option (b) is the correct option as the expenses are Rs. 1900 for 300 inmates.

Note: Form the equations by multiplying the number of inmates with the variable part and then add the constant part and do not multiply the number of inmates with the constant part and after this put it equal to expenses. We can solve the equation by substitution method or elimination method. We can verify the variable and constant value by keeping it in the equation, as if we get the same expenses as given implies that the values are correct otherwise not.