Al, Pablo and Marsha shared the driving on a 1,500-mile : GMAT Data Sufficiency (DS)
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# Al, Pablo and Marsha shared the driving on a 1,500-mile

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Manager
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Al, Pablo and Marsha shared the driving on a 1,500-mile [#permalink]

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11 Jul 2006, 04:51
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Difficulty:

55% (hard)

Question Stats:

67% (02:08) correct 33% (01:11) wrong based on 404 sessions

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Al, Pablo and Marsha shared the driving on a 1,500-mile trip. Which of the three drove the greatest distance of the trip?

(1) Al drove 1 hour longer than Pablo but at an average rate of 5 miles per hour slower than Pablo.
(2) Marsha drove 9 hours and averaged 50 miles per hour.

OPEN DISCUSSION OF THIS QUESTION IS HERE: al-pablo-and-marsha-shared-the-driving-on-a-1-500-mile-65090.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Oct 2013, 13:56, edited 1 time in total.
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11 Jul 2006, 05:32
E.

It is "easy" to see, that both (1) and (2) alone are not sufficient.

Are (1) and (2) together sufficient?

I construct a counterexample.

We know from (2) that Al and Paul drove 1050 miles.

Notation: A = distance driven by Al, P = distance driven by Paul. h(A): hours driven by Al, h(P) = hours driven by Paul. S(A) = speed (Al is the driver), S(P) = speed (when P is the driver).

A+P=1050 [1]
h(A) = h(P) + 1
s(A) = s(P) - 5
A = h(A) * s(A), P = h(P) + s(P)

Putting this together (in [1]):
h(A)*s(A) + [h(A)-1] * [S(A)+5] = 1050

Denote: h(A)=h and s(A)=s
By rearranging the terms, we get:

hs+(h-1)*(s+5) = 1050

Ass.: h=10
then: 10s + 9(s+5) = 1050 <=> 19s = 1005 <=> s=1005/19
hs = 1005*10/19 > 525 (1005/19 = 950/19 + 55/19) (~53)

Therefore, Al drove more than P.

Now ass.: h=20
then 20s + 19(s+5)=1050 <=> 39s = 955 <=> s=955/39
=> hs=955*20/39 <525

=> Paul drove more than Al.

Hence, (1) and (2) together are not sufficient.

[since you have only 5 equations but 6 variables, this result isn't really surprising]
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11 Jul 2006, 07:02
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Combining (1) and (2)

Let t and r be the time and rate corresponding to Paco
Then t+1 and r-5 are the time and rate corresponding to Al

Paco's distance t*r
Al's distance t*r + r- 5t -5

So Al drove the furthest if r-5t is greater than 5
But Paco drove the furthest if r-5t is less than 5

As we are given no info about r and t, both scenarios are possible!

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11 Jul 2006, 10:00
I immediately got (E) too, but then I started doubting myself. I was trying to find a trap ... and I kept looking and kept looking and kept looking...

I guess I've seen too many kevincan's tricky questions lately, so my instincts are starting to fail me.
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11 Jul 2006, 14:05
Getting E...

It seems like (1) & (2) are sufficient until you try to solve, which gives you an equation in Sp& Tp (Speed of Paul and Time for Paul) which is not solvable.
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11 Jul 2006, 15:07
E

there is not enough information to fully develop a relationship for Al, Pablo, and Marsha
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Re: Al, Pablo and Marsha shared the driving on a 1,500-mile [#permalink]

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12 Oct 2013, 12:22
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You don't necessarily need too much math for this one.

From (2) we know that Marsha drove 450 miles, meaning that Al and Pablo drove 1050 miles. This tells us that Marsha wasn't the one who drove the greatest distance. But still not sufficient to know who between Al and Pablo drove the most miles.

From (1), consider two extreme scenarios:
[Scenario One]: Al drove 1 mph, whereas Pablo drove 6 mph. In this case, even if Al drove an hour longer than Pablo, it was Pablo who drove the greatest number of miles.
[Scenario Two]: Al drove 200 mph, whereas Pablo drove 205 mph. In this case, an hour extra driving from Al makes him the one who drove the most distance.

I hope that is clear for the ones who avoid math for DS.
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Re: Al, Pablo and Marsha shared the driving on a 1,500-mile [#permalink]

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12 Oct 2013, 13:56
OPEN DISCUSSION OF THIS QUESTION IS HERE: al-pablo-and-marsha-shared-the-driving-on-a-1-500-mile-65090.html
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Re: Al, Pablo and Marsha shared the driving on a 1,500-mile [#permalink]

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11 Nov 2014, 15:38
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Re: Al, Pablo and Marsha shared the driving on a 1,500-mile   [#permalink] 11 Nov 2014, 15:38
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