Al, Pablo, and Marsha driving

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Al, Pablo, and Marsha driving

by fangtray » Wed Apr 04, 2012 6:01 am
Al, Pablo, and Marsha shared the driving on a 1,500 mile trip. Which of the 3 drove the greatest distance on the trip?

1. Al drove 1 hr longer than Pablo at an avg rate of 5 mph slower than pablo
2. Marsha drove 9 hrs and averaged 50 mph.

1. insufficient dont know anything about Marsha
2. dont know anything about Al and Pablo, insufficient


If you put both together, Al and Pablo drove a total of 1,050 miles with Al driving 1 hr longer than Pablo, and 5 mph less than pablo. Aren't we able to solve for Al and Pablo now?

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by Pharo » Wed Apr 04, 2012 6:22 am
I have D = 1500;
Va,Ta = Speed Al,Time Al
Vm,Tm = Speed Marsha, Time Marsha
Vp,Tp = Speed Pablo, Time Pablo

Then : D = (Va*Ta) + (Vp*Tp) + (Vm*Tm)

Statement 1 : Gives me Vm = 50, Tm = 9. ; sub them in the master equation
D = (Va*Ta) + (Vp*Tp) + 450
;Not sufficient

Statement 2 : Ta = Tp +1 and Va = Vp-5 ; sub them in the master equation
D = ((Vp-5)*(Tp+1)) + (Vp*Tp) + (Vm*Tm)
;Not sufficient

Put the two statements together :

D = ((Vp-5)*(Tp+1)) + (Vp*Tp) + 450 ; i have two variables and one equation. solution cannot be found.

; the answer is E.

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by GMATGuruNY » Wed Apr 04, 2012 10:59 am
fangtray wrote:Al, Pablo, and Marsha shared the driving on a 1,500 mile trip. Which of the 3 drove the greatest distance on the trip?

1. Al drove 1 hr longer than Pablo at an avg rate of 5 mph slower than pablo
2. Marsha drove 9 hrs and averaged 50 mph.

1. insufficient dont know anything about Marsha
2. dont know anything about Al and Pablo, insufficient


If you put both together, Al and Pablo drove a total of 1,050 miles with Al driving 1 hr longer than Pablo, and 5 mph less than pablo. Aren't we able to solve for Al and Pablo now?
Combining the 2 statements:
Distance driven by Marsha = 50*9 = 450.
Distance traveled by Al and Pablo = 1500 - 450 = 1050.
Let the time for Pablo = t and the time for Al = t+1.
The rate for Pablo is 5mph greater than the rate for Al.
TRY EXTREME CASES.

Case 1: Rate for Pablo = 10mph, rate for Al = 5mph
Since the total distance driven by Pablo and Al is 1050, we get:
10t + 5(t+1) = 1050
15t = 1045
t = 1045/15 ≈ 70, implying that the time for Al ≈ 70+1 = 71.
Distance for Pablo ≈ 10*70 = 700, distance for Al ≈ 5*71 = 355.
Pablo drives the greatest distance.

Case 2: Rate for Pablo = 1045mph, rate for Al = 1040mph
Since the total distance driven by Pablo and Al is 1050, we get:
1045t + 1040(t+1) = 1050.
2085t = 10
t = 10/2085 ≈ 1/200, implying that the time for Al ≈ 1/200 + 1 ≈ 1.
Distance for Pablo ≈ 1045(1/200) ≈ 5, distance for Al ≈ 1040*1 ≈ 1040.
Al drives the greatest distance.

Since in the first case Pablo drives the greatest distance, but in the second case Al drives the greatest distance, INSUFFICIENT.

The correct answer is E.
Last edited by GMATGuruNY on Thu Jan 30, 2020 3:51 am, edited 1 time in total.
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by Pharo » Wed Apr 04, 2012 1:06 pm
GMATGuruNY wrote:TRY EXTREME CASES.
Good idea, thank you :)