The cost of maintaining a school is partly constant and partly varies as the number of pupils with 50 pupils the cost is $15705.00 and with 40 pupils it is $13705.00
A) find the cost when there are 44 pupils
B) if the fee per pupil is $360.00 what is the least number of pupils for which the school can be run without loss
Let's call the constant cost c and the variable cost v.
In this problem, we're given that the cost with 50 pupils is $15,705 and the cost with 40 pupils is $13,705.
Using this information, we can set up two equations:
c + 50v = $15,705
c + 40v = $13,705
We now have a system of equations. We can solve this system using the method of substitution.
First, subtract the second equation from the first equation:
(c + 50v) - (c + 40v) = $15,705 - $13,705
10v = $2,000
Divide both sides by 10:
v = $200
Now that we know the variable cost per pupil, we can substitute this value back into one of the original equations to find the constant cost:
c + 40v = $13,705
c + 40($200) = $13,705
c + $8,000 = $13,705
c = $13,705 - $8,000
c = $5,705
Now, for part A, we need to find the cost when there are 44 pupils. We can use the variable cost to calculate this:
Cost = c + 44v
Cost = $5,705 + 44($200)
Cost = $5,705 + $8,800
Cost = $14,505
So, the cost with 44 pupils is $14,505.
For part B, we're given that the fee per pupil is $360. Let's call the number of pupils x. We can set up an equation to find the total revenue:
Revenue = Number of pupils x Fee per pupil
Revenue = x($360)
To break even, the revenue must be equal to the total cost:
Revenue = Total cost
x($360) = c + xv
Now we can substitute the values we found earlier for c and v:
x($360) = $5,705 + x($200)
Simplify the equation:
$360x = $5,705 + $200x
Subtract $200x from both sides:
$360x - $200x = $5,705
Simplify again:
$160x = $5,705
Divide both sides by $160:
x = $5,705 / $160
x = 35.65625
Since we can't have a fraction of a person, the least number of pupils for which the school can be run without loss is 36.