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Bill and Sam both rode their bikes from their school to...

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Joined: 03 Jan 2013
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Bill and Sam both rode their bikes from their school to... [#permalink]

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03 Jan 2013, 10:09
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Bill and Sam both rode their bikes from their school to the public library. They traveled the exact same route. It took Bill 12 minutes. How long did it take Sam?

(1) Sam’s average speed was 80% of Bill’s.

(2) The distance from the school to the library is two miles.

I thought I had the correct answer here with needing both pieces of information but Kaplan explains you only need statement 1...Can anyone explain this one to me? I know they both travel the same distance but that distance could be 1 mile or 20 miles which would change Sam's travel time correct?
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Re: Bill and Sam both rode their bikes from their school to... [#permalink]

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03 Jan 2013, 11:28
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nemix wrote:
Bill and Sam both rode their bikes from their school to the public library. They traveled the exact same route. It took Bill 12 minutes. How long did it take Sam?

(1) Sam’s average speed was 80% of Bill’s.

(2) The distance from the school to the library is two miles.

I thought I had the correct answer here with needing both pieces of information but Kaplan explains you only need statement 1...Can anyone explain this one to me? I know they both travel the same distance but that distance could be 1 mile or 20 miles which would change Sam's travel time correct?

Dear nemix,
Both travel a distance ---call it D. Bill took 12 minutes, and Sam took a time --- call it T. Bill traveled at speed VB and Sam traveled at speed VS.

D = RT for each person ----
For Bill: D = (VB)*12
For Sam: D = (VS)*T
Since these both equal the same distance, we can equate them.
(VB)*12 = (VS)*T

Now, statement #1 says ---- "Sam’s average speed was 80% of Bill’s." In math, VS = 0.8*VB. Plug this into the purple equation ---

(VB)*12 = [0.8*VB]*T
divide by VB
12 = 0.8*T
divide by 0.8
T = 12/0.8 = 15

It took Sam 15 minutes. Statement #1, alone and by itself, was sufficient for answering the prompt question.

As a general rule, if two people or cars travel the same distance, then their velocities are inversely proportional to the times it takes them. If, starting from one speed, you multiply by something to get the second speed --- then whatever you multiplied the speed by, you need to divide the time by that same factor.

Does all this make sense?

Mike
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Mike McGarry
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Joined: 06 May 2012
Posts: 65
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Concentration: General Management, Finance
GMAT 1: 670 Q48 V34
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Re: Bill and Sam both rode their bikes from their school to... [#permalink]

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03 Jan 2013, 11:23
Hi nemix,

You can ofcourse solve using both statements but 2nd is redundant.

Lets approach systematically.. First 1.. then 2.. then if necessary 1+2:

Case1: you are already given that Bill took 12mins.. and the Sam's avg speed relative to bill..

that is all u need to know to get speed.. SUFFICIENT
If sam had the same avg speed as bill --> they would take same time.. we dont care how much.. just some single val x
If sam had the more avg speed than bill--> sam would take less time..
If sam had the less avg speed as bill--> sam will take more time.. sams will take 12/0.8

D=s*t
s*t=const so if speed is less time is more.. time is less speed is more.

Case2: only the distance is given.. no use INSUFFICIENT

PS: Please consider kudos if u find it useful
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Joined: 03 Jan 2013
Posts: 5
Re: Bill and Sam both rode their bikes from their school to... [#permalink]

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03 Jan 2013, 13:53
mikemcgarry wrote:
nemix wrote:
Bill and Sam both rode their bikes from their school to the public library. They traveled the exact same route. It took Bill 12 minutes. How long did it take Sam?

(1) Sam’s average speed was 80% of Bill’s.

(2) The distance from the school to the library is two miles.

I thought I had the correct answer here with needing both pieces of information but Kaplan explains you only need statement 1...Can anyone explain this one to me? I know they both travel the same distance but that distance could be 1 mile or 20 miles which would change Sam's travel time correct?

Dear nemix,
Both travel a distance ---call it D. Bill took 12 minutes, and Sam took a time --- call it T. Bill traveled at speed VB and Sam traveled at speed VS.

D = RT for each person ----
For Bill: D = (VB)*12
For Sam: D = (VS)*T
Since these both equal the same distance, we can equate them.
(VB)*12 = (VS)*T

Now, statement #1 says ---- "Sam’s average speed was 80% of Bill’s." In math, VS = 0.8*VB. Plug this into the purple equation ---

(VB)*12 = [0.8*VB]*T
divide by VB
12 = 0.8*T
divide by 0.8
T = 12/0.8 = 15

It took Sam 15 minutes. Statement #1, alone and by itself, was sufficient for answering the prompt question.

As a general rule, if two people or cars travel the same distance, then their velocities are inversely proportional to the times it takes them. If, starting from one speed, you multiply by something to get the second speed --- then whatever you multiplied the speed by, you need to divide the time by that same factor.

Does all this make sense?

Mike

The more I read it the more it makes sense. Thank you sir
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Joined: 13 May 2013
Posts: 469
Re: Bill and Sam both rode their bikes from their school to... [#permalink]

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01 Aug 2013, 11:24
Bill and Sam both rode their bikes from their school to the public library. They traveled the exact same route. It took Bill 12 minutes. How long did it take Sam?

If they traveled the same route then we can use a hypothetical distance to solve for Sam's time.

(1) Sam’s average speed was 80% of Bill’s.

Bill's time:
Time=Distance/Speed
12=x/s
12=48/4
x=48, s=4

Sam's Time:
x=48
s=4*(.8)
T=48/3.2
T = 15
SUFFICIENT

(2) The distance from the school to the library is two miles.
We know nothing about Sam's speed. Even if we know the distance we don't know his speed.
INSUFFICIENT

(A)
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Joined: 09 Sep 2013
Posts: 15978
Re: Bill and Sam both rode their bikes from their school to... [#permalink]

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22 Aug 2015, 13:33
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Re: Bill and Sam both rode their bikes from their school to...   [#permalink] 22 Aug 2015, 13:33
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