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.