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Bunuel
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The expenses of a hostel are partly fixed and partly variable. It costs $120 per head per day if there are 20 students and $110 per head per day if there are 25 students. What will be the cost per head per day if the number of students increases to 40?

A. 105
B. 100
C. 95
D. 80
E. 75

Say the fixed cost is \($x\) and the variable cost is \($y\) per head. Then:

\(cost \ per \ head=\frac{total \ cost}{# \ of \ students}=\frac{x+20y}{20}=120\) --> \(x+20y=20*120\);

\(cost \ per \ head=\frac{total \ cost}{# \ of \ students}=\frac{x+25y}{25}=110\) --> \(x+25y=25*110\);

Subtract the first equation from the second: \(5y=25*110-20*120\) --> \(y=70\) --> \(x=1,000\).

Therefore the cost per head per day for 40 students will be \(\frac{x+40y}{y}=\frac{1,000+40*70}{40}=95\).

Answer: C.

Great explanation

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+1 C

Fixed costs: $ 1,000
Variable costs per head: $70
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Bunuel
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The expenses of a hostel are partly fixed and partly variable. It costs $120 per head per day if there are 20 students and $110 per head per day if there are 25 students. What will be the cost per head per day if the number of students increases to 40?

A. 105
B. 100
C. 95
D. 80
E. 75

Say the fixed cost is \($x\) and the variable cost is \($y\) per head. Then:

\(cost \ per \ head=\frac{total \ cost}{# \ of \ students}=\frac{x+20y}{20}=120\) --> \(x+20y=20*120\);

\(cost \ per \ head=\frac{total \ cost}{# \ of \ students}=\frac{x+25y}{25}=110\) --> \(x+25y=25*110\);

Subtract the first equation from the second: \(5y=25*110-20*120\) --> \(y=70\) --> \(x=1,000\).

Therefore the cost per head per day for 40 students will be \(\frac{x+40y}{y}=\frac{1,000+40*70}{40}=95\).

Answer: C.


Can you kindly explain how 20y & 25y are taken?
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PareshGmat
Bunuel
NYC5648
The expenses of a hostel are partly fixed and partly variable. It costs $120 per head per day if there are 20 students and $110 per head per day if there are 25 students. What will be the cost per head per day if the number of students increases to 40?

A. 105
B. 100
C. 95
D. 80
E. 75

Say the fixed cost is \($x\) and the variable cost is \($y\) per head. Then:

\(cost \ per \ head=\frac{total \ cost}{# \ of \ students}=\frac{x+20y}{20}=120\) --> \(x+20y=20*120\);

\(cost \ per \ head=\frac{total \ cost}{# \ of \ students}=\frac{x+25y}{25}=110\) --> \(x+25y=25*110\);

Subtract the first equation from the second: \(5y=25*110-20*120\) --> \(y=70\) --> \(x=1,000\).

Therefore the cost per head per day for 40 students will be \(\frac{x+40y}{y}=\frac{1,000+40*70}{40}=95\).

Answer: C.


Can you kindly explain how 20y & 25y are taken?

The variable cost is \($y\) per head, thus for 20 students it's 20y and for 25 students it's 25y.

Hope it's clear.
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I solved the problem using an approach similar to Bunuel's. However, it took me 2:31 min to read the question and solve the problem. I wonder if there is a faster way to solve this problem because these additional seconds (~31 seconds) all add up.

My steps:
F + 20 * V = 120 * 20
F + 25 * V = 110 * 25

5 * V = 50 (55 - 12 * 4)
5 * V = 50 * 7
V = 70
F = 2400 - 1400 = 1000
Total = 1000 + 70 * 40 = 3800
At this time I looked at the answer choice and did not see anything matching :) and quickly realized I need per head cost
So I divided the above result by 40 to get 95.

I could potentially save few seconds if I am more alert but I still feel the above steps would take me 2 min or more.

I would like to know how much time other people are taking to solve the same problem
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Fun question if you don't overthink it! I spent way too long wondering if the variable cost had a linear relationship with the number of students.
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Total Cost for 20 students = 20*120 = 2400
Total Cost for 25 students = 25*110 = 2750

We can see that adding 5 students increases the total cost by 350 (a trap is trying to decrease the cost by 10 per, assuming the total cost stays at a fixed 2400).

So, the increase in total cost going from 20 to 40 students will be 4*350 = 1,400.

3800/40 = 380/4 = 190/2 = 95
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Given: The expenses of a hostel are partly fixed and partly variable. It costs $120 per head per day if there are 20 students and $110 per head per day if there are 25 students.

Asked: What will be the cost per head per day if the number of students increases to 40?

Let the fixed cost be $x and variable costs be $y/student per day.

120*20 = x + 20y
110*25 = x + 25y

20(120-y) = x = 25(110-y)
480 - 4y = 550 - 5y
y = 70
x = 1000

The cost per head per day if the number of students increases to 40 = (x + 40y )/40 = x/40 + y = 25 + 70 = $95

IMO C
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