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P = -2 (S – 4) ^2 + 32

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P = -2 (S – 4) ^2 + 32

by sanju09 » Fri May 13, 2011 2:36 am
The number of passengers on a certain bus at any given time is given by the equation

P = -2 (S - 4) ^2 + 32,

where P is the number of passengers and S is the number of stops the bus has made since beginning its route. If the bus begins its route with no passengers, how many passengers will be on the bus two stops after the stop where it has its greatest number of passengers?
(A) 32
(B) 30
(C) 24
(D) 14
(E) 0
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by mba4viplav » Fri May 13, 2011 3:08 am
Hmm... Quite a good one.

I solved it this way:

Note: I hope ^ means "raised to the power".

The solution is based on equation on the left side of the + sign.

If you plug in values 0, 1, 2, 3, and 4 in the place of S, the corresponding values of the equation are 0, 14, 24, 30, and 32.

Now if you go on plugging in 5, 6, 7, and 8 in the place of S, the values of the equation are 30, 24, 14, and 0.

If you still go on plugging in values beyond 8, the equation throws negative values that are not valid for counting number of passengers.

So, the maximum number of people at any stop are 32, i.e at stop number 4.

So, the number of passengers on the bus two stops after the stop where it has greatest number of passengers is 24 i.e at stop number 6.

The correct answer is (C).
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by djiddish98 » Fri May 13, 2011 4:25 am
This seems like a derivatives question, which might be out of scope on the gmat.

From what little I remember of calc, the max or min of the equation will be equal to

0 = -2(2) (s-4)

which equals

-16 = -4s -> s = 4.

Plugging in 6 for s should give the answer.
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by GMATGuruNY » Fri May 13, 2011 4:52 am
sanju09 wrote:The number of passengers on a certain bus at any given time is given by the equation

P = -2 (S - 4) ^2 + 32,

where P is the number of passengers and S is the number of stops the bus has made since beginning its route. If the bus begins its route with no passengers, how many passengers will be on the bus two stops after the stop where it has its greatest number of passengers?
(A) 32
(B) 30
(C) 24
(D) 14
(E) 0
A knowledge of derivatives is not needed.

-2(S-4)² = negative * (0 or positive).
Thus, the greatest possible value of -2(S-4)² is 0.
When -2(S-4)² = 0, S=4.
Thus, the greatest number of passengers occurs when S=4.
Two stops later, S=6:
P = -2(6-4)² + 32 = 24.

The correct answer is C.
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by djiddish98 » Fri May 13, 2011 4:58 am
GMATGuruNY wrote:
sanju09 wrote:The number of passengers on a certain bus at any given time is given by the equation

P = -2 (S - 4) ^2 + 32,

where P is the number of passengers and S is the number of stops the bus has made since beginning its route. If the bus begins its route with no passengers, how many passengers will be on the bus two stops after the stop where it has its greatest number of passengers?
(A) 32
(B) 30
(C) 24
(D) 14
(E) 0
A knowledge of derivatives is not needed.

-2(S-4)² = negative * (0 or positive).
Thus, the greatest possible value of -2(S-4)² is 0.
When -2(S-4)² = 0, S=4.
Thus, the greatest number of passengers occurs when S=4.
Two stops later, S=6:
P = -2(6-4)² + 32 = 24.

The correct answer is C.
Ah, I made it out to be more difficult than it was. Thanks!
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