Question

The food bill of a hostel is partly constant and partly varies as the number of days a student has dined. Yash dined for 25 days and his food bill is Rs2200. Riyan dined for 20 days and her food bill is Rs1800. Find the constant price of the food bill and the charge of dining per day.

Answer 1

Hint: The question asks for two unknown quantities whose sum is given and we have to find the two unknown quantities. So we will start by assigning a variable to each of them, that is, the constant price of the food be x rupees and the charge of dining per day be x rupees. Based on the given conditions, we will get two equations with two variables each. Solving which, we will get the values of x and y.

Complete step by step solution:
According to the question given, we have been provided with two scenarios of food bills based on which we have to find the constant price of the food bill and the charge of dining per day.
Let the constant price of the food bill be Rs.x
And let the charge of dining per day be Rs.y
So the first scenario states, Yash dined for 25 days and his food bill is Rs2200, this gives us the equation
\[x+25y=2200\]--------------(1)
And the second scenario states, Riyan dined for 20 days and her food bill is Rs1800, this gives us the equation
\[x+20y=1800\]--------------(2)
So we have two equations with two variables, now we need to solve these equations to get the values of x and y.
Equation (1) – equation (2)
We get,
\[(x+25y)-(x+20y)=2200-1800\]
Opening the brackets, we get,
\[\Rightarrow x+25y-x-20y=400\]
\[\Rightarrow 25y-20y=400\]
\[\Rightarrow 5y=400\]
\[\Rightarrow y=80\]
We now have the value of y, we will put the value of y in either of the equations and the get the value of x as well.
So, putting the value of y in equation (1) we get,
\[x+25(80)=2200\]
Solving for x we get,
\[\Rightarrow x+2000=2200\]
\[\Rightarrow x=2200-2000\]
\[\Rightarrow x=200\]
We have the value of \[x=200\] and \[y=80\]
Therefore,
the constant price of the food bill is \[Rs.200\]
And the charge of dining per day is \[Rs.80\]

Note: The constant price of the food is not multiplied with the number of days each person dined in the equation formed as the constant price of the food is independent of the number of days.
And the above equations can also be solved by substituting the value of one of the variables from an equation to another equation which is as follows:
\[x+25y=2200\]--------------(1)
\[x+20y=1800\]--------------(2)
From equation (1), the value of x is:
\[x=2200-25y\]-------------(3)
Substituting the value of x in equation (2), we get,
\[(2200-25y)+20y=1800\]
\[\Rightarrow 2200-25y+20y=1800\]
Separating the variable and the constants, we get,
\[\Rightarrow 2200-1800=25y-20y\]
\[\Rightarrow 400=5y\]
\[\Rightarrow y=80\]
Putting the value of y in equation (3), we get,
\[x=2200-25y\]
\[\Rightarrow x=2200-25(80)\]
\[\Rightarrow x=2200-2000\]
\[\Rightarrow x=200\]
We have the value of \[x=200\] and \[y=80\]
Therefore,
the constant price of the food bill is \[Rs.200\]
And the charge of dining per day is \[Rs.80\]