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?
(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?
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Correct Answer: Option n
Explanation:
Let P represent the cost ; n, the number of pupils and k and c, constant terms.
\(\therefore P = k + cn\)
\(15705 = k + 50c ... (1)\)
\(13305 = k + 40c .... (2)\)
(1) - (2) :
\(2400 = 10c \implies c = \frac{2400}{10} = 240\)
Put c = 240 in (1), so that
\(15705 = k + (50)(240)\)
\(15705 = k + 12000 \implies k = 3705\)
\(P = 3705 + 240n\) ------ formula connecting price and number of pupils.
(a) When number of pupils, n = 44
\(P = 3705 + 240(44) = 3705 + 10560 = $14265\)
(b) For school not to be at loss:
\(360 \geq k + cn\)
\(360n \geq 3705 + 240n\)
\(360n - 240n \geq 3705 \implies 120n \geq 3705\)
\(n \geq \frac{3705}{120}\)
\(n \geq 30.875 \)
\(n \geq 31\)
The pupils should be 31 at least for the school not to run at loss.
Let P represent the cost ; n, the number of pupils and k and c, constant terms.
\(\therefore P = k + cn\)
\(15705 = k + 50c ... (1)\)
\(13305 = k + 40c .... (2)\)
(1) - (2) :
\(2400 = 10c \implies c = \frac{2400}{10} = 240\)
Put c = 240 in (1), so that
\(15705 = k + (50)(240)\)
\(15705 = k + 12000 \implies k = 3705\)
\(P = 3705 + 240n\) ------ formula connecting price and number of pupils.
(a) When number of pupils, n = 44
\(P = 3705 + 240(44) = 3705 + 10560 = $14265\)
(b) For school not to be at loss:
\(360 \geq k + cn\)
\(360n \geq 3705 + 240n\)
\(360n - 240n \geq 3705 \implies 120n \geq 3705\)
\(n \geq \frac{3705}{120}\)
\(n \geq 30.875 \)
\(n \geq 31\)
The pupils should be 31 at least for the school not to run at loss.