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.