Respuesta :
Answer:
a). Cost of 44 pupils = $14265
b). Least number of pupils = 31
Step-by-step explanation:
The given question is incomplete; here is the complete question.
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?
Let the equation representing the total cost of maintaining a school is,
C = ax + b
Where C = Total cost of maintaining a school
a = Fee per pupil
b = Fixed running cost
x = number of pupils
a). Cost of 50 pupils = $15705
  Equation will be,
  15705 = 50a + b -------(1)
  Cost of 40 pupils = $13305
  Equation will be,
  13305 = 40a + b --------(2)
  By subtracting equation (2) from equation (1),
  15705 - 13305 = (50a + b) - (40a + b)
  2400 = 10a
  a = 240
  From equation (1),
  b = 3705
  Equation representing the total cost will be,
  C = 240x + 3705
  If x = 44
  C = 240(44) + 3705
  C = $14265
b). If the fee per pupil 'a' = $360
  Let the number of pupils = p
  Total fee of 'p' pupils = $360p
  Total cost to run the school will be = 3705 + 240p
  For the school not to be in the loss,
  360p ≥ 3705 + 240p
  360p - 240p ≥ 3705
  120p ≥ 3705
  p ≥ [tex]\frac{3705}{120}[/tex]
  p ≥ 30.875
  Therefore, to run the school without loss, number pupils should be at least 31.
