The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it...
Question 1 Report
The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.
(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 run without a loss?
Answer Details
Let the cost be \(C=a+bn\), where \(a\) is the constant part and \(b\) the cost per pupil.
\(50\) pupils: \(a+50b=15705\). \(40\) pupils: \(a+40b=13305\). Subtracting: \(10b=2400\Rightarrow b=240\); then \(a=15705-50(240)=3705\). So \[C=3705+240n.\]
(a) For \(n=44\): \[C=3705+240(44)=3705+10560=\$14{,}265.\]
(b) Revenue \(=360n\). To run without a loss, \(360n\ge3705+240n\Rightarrow120n\ge3705\Rightarrow n\ge30.875\). The least whole number of pupils is 31.
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