A new high school opens in a thriving suburban community. The number of students signed up to attend the school in its first yea
2 answers:
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It has not been indicated whether the figure in the questions is a triangle or a quadrilateral. Irrespective of the shape, this can be solved. The two possible shapes and angles have been indicated in the attached image.
Now, from the information given we can infer that there is a line BD that cuts angle ABC in two parts: angle ABD and angle DBC
⇒ Angle ABC = Angle ABD + Angle DBC
Also, we know that angle ABC is 1 degree less than 3 times the angle ABD, and that angle DBC is 47 degree
Let angle ABD be x
⇒ Angle ABC = 3x-1
Also, Angle ABC = Angle ABD + Angle DBC
Substituting the values in the above equations
⇒ 3x-1 = x+47
⇒ 2x = 48
⇒ x = 24
So angle ABD = 24 degree, and angle ABC = 3(24)-1 = 71-1 = 71 degree
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.