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# The number of passengers on a certain bus at any given time is given

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The number of passengers on a certain bus at any given time is given  [#permalink]

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Updated on: 14 Feb 2018, 07:13
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15% (low)

Question Stats:

82% (02:06) correct 18% (02:36) wrong based on 94 sessions

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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, what is the value of S when the bus has its greatest number of passengers?

a) 9
b) 6
c) 4
d) 2
e) 1

Originally posted by nekiwa07 on 14 Feb 2018, 07:07.
Last edited by Bunuel on 14 Feb 2018, 07:13, edited 1 time in total.
Renamed the topic and edited the question.
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14 Feb 2018, 07:18
nekiwa07 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, what is the value of S when the bus has its greatest number of passengers?

a) 9
b) 6
c) 4
d) 2
e) 1

$$P = –2(S – 4)^2 + 32$$:

$$–2(S – 4)^2 = (-2)*(square \ of \ a \ number)= (negative)*(non-negative) = (non-positive)$$. So, this term is negative or 0.

Therefore, P, the number of passengers, will be maximized when $$–2(S – 4)^2$$ is 0. In this case $$P_{max} = –2(S – 4)^2 + 32= 0+32=32$$. For, $$–2(S – 4)^2$$ to be 0, S should be 4.

Hope it's clear.
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Re: The number of passengers on a certain bus at any given time is given  [#permalink]

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14 Feb 2018, 08:08
Bunuel wrote:
nekiwa07 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, what is the value of S when the bus has its greatest number of passengers?

a) 9
b) 6
c) 4
d) 2
e) 1

$$P = –2(S – 4)^2 + 32$$:

$$–2(S – 4)^2 = (-2)*(square \ of \ a \ number)= (negative)*(non-negative) = (non-positive)$$. So, this term is negative or 0.

Therefore, P, the number of passengers, will be maximized when $$–2(S – 4)^2$$ is 0. In this case $$P_{max} = –2(S – 4)^2 + 32= 0+32=32$$. For, $$–2(S – 4)^2$$ to be 0, S should be 4.

Hope it's clear.

Similar question: https://gmatclub.com/forum/the-number-o ... 87963.html
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24 Feb 2018, 22:39
substitute the value of S so we get no of passengers from the equation

A) p= -2(9-4)*2 +32 = -18
B) p= -2(6-4)*2 +32 = 24
c) p= -2(4-4)*2 + 32 = 32
D)p= -2(1-4)*2 +32 = 14

ans is option C because it has highst number of pasngrs

in gmat we start substituting from option C which makes this problem much easier
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Re: The number of passengers on a certain bus at any given time is given  [#permalink]

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31 Mar 2018, 21:50
The negetive sign in the first part of the equation indicates that P will be the largest when (S-4)^2 =0

-> S-4 = 0
S = 4

Therefore Option C
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Re: The number of passengers on a certain bus at any given time is given  [#permalink]

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31 Mar 2018, 23:22
Prasanna675 wrote:
substitute the value of S so we get no of passengers from the equation

A) p= -2(9-4)*2 +32 = -18
B) p= -2(6-4)*2 +32 = 24
c) p= -2(4-4)*2 + 32 = 32
D)p= -2(1-4)*2 +32 = 14

ans is option C because it has highst number of pasngrs

in gmat we start substituting from option C which makes this problem much easier

How does it make it any easier if we start substituting from Option C ? Won't you still need to substitute in all the OCs ?
Let me know if there is a way. it would be really helpful to minimize time.
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Re: The number of passengers on a certain bus at any given time is given  [#permalink]

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01 Apr 2018, 00:07
DhruvPatelD10 wrote:
Prasanna675 wrote:
substitute the value of S so we get no of passengers from the equation

A) p= -2(9-4)*2 +32 = -18
B) p= -2(6-4)*2 +32 = 24
c) p= -2(4-4)*2 + 32 = 32
D)p= -2(1-4)*2 +32 = 14

ans is option C because it has highst number of pasngrs

in gmat we start substituting from option C which makes this problem much easier

How does it make it any easier if we start substituting from Option C ? Won't you still need to substitute in all the OCs ?
Let me know if there is a way. it would be really helpful to minimize time.

I've been told the same many times by people. I think the logic behind this rule of thumb for substitution is " most tricky substitution questions have the answer hidden in the bottom 2-3"

It is natural tendency to begin with first option and hence people would burn up their time wasting effort. But again, it's just a theory! I never use substitution ( which can be another drawback - precise solutions can be time consuming themselves !)

Best,
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Re: The number of passengers on a certain bus at any given time is given   [#permalink] 01 Apr 2018, 00:07
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