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?
Al, Pablo, and Marsha driving
This topic has expert replies
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.
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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Combining the 2 statements: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?
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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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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