Answer:
Sabemos que:
y tenemos que:
Con esto podemos encontrar el valor de a:
8*a/(a+ 4) = 6
8*a = 6*(a + 4) = 6*a + 24
8a - 6a = 24
2a = 24
a = 24/2 = 12.
Tambien sabemos que:
Y de ahí podemos despejar b:
(16*b)/(b + 8) = 12
16*b = 12*(b + 8) = 12b + 96
16b - 12b = 96
4b = 96
b = 96/4 = 24
Entonces tenemos a = 12 y b = 24, y el MH de a y b es:
MH(12,24) = 2*12*24/(12 + 24) = 24*24/36 = 16
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 ≥
p ≥ 30.875
Therefore, to run the school without loss, number pupils should be at least 31.
