11) The Cost of maintaing a school is party
Constant and partly varies as the number of
Pupils with 50 pupils the cost is $15705 with 40 pupils, it is $13305
find the cost when there are 44 pupils
B if the fee per pupil is 360 what is the
Least number of pupils for which the school can be run without loss

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 ≥

    p ≥ 30.875

    Therefore, to run the school without loss, number pupils should be at least 31.

do you have an easier question cause this shh is hard like theheck

step-by-step explanation:

it would be 3 ÷by 6 because the number on each side