MGMAT

raju232007 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

This question was asked in one of the MGMAT cats..The solution for this problem has already been posted in this forum.But I am not convinced with the explanation.So can anyone provide an easy alternate solution?

The ans is C..

Let's start by figuring out the greatest possible number of passengers.

Breaking down our equation, P = -2(S – 4)^2 + 32, into two terms, we get:

-2(s-4)^2

and

+ 32

From these two terms, we can see that the first is negative and the second is positive, so if we want to maximize the number of passengers, we need to minimize the first term.

So, when will -2(s-4)^2 be minimized? (s-4)^2 is never going to be negative, so when s=4 it will have its minimum value of 0.

Therefore, our greatest number of passengers, 32, occurs at stop 4.

Now the question simply is:

How many passengers does the bus have at stop 6?

Plugging in s=6, we get:

P = -2*(6-4)^2 + 32
P = -2*(2)^2 + 32
P = -2*(4) + 32
P = -8 + 32 = 24

choose (C).