|
|
#1 (permalink) |
|
Eager!
Join Date: Aug 2008
Posts: 94
![]() |
distance and rate
Al, Pablo, and Marsha shared the driving on a 1,500-mile trip. Which of the three drove the greatest distance on 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. |
|
|
|
|
|
#2 (permalink) |
|
I JUST got here.
Join Date: Sep 2008
Posts: 10
![]() |
Al(A), Pablo(P), Marsha(M)
Distance(D) = Speed(S)* Time(T) STEM 1: TA = TP + 1, A could have driven lesser distance if P was much faster than A, and we know nothing about M SA = SP - 5, A could have driven significantly more time than P, and we know nothing about M Hence STEM1 by itself is inconclusive STEM 2: TM = 9, SM = 50 --> DM = 450 --> So DA + DP = 1050 Clearly A or P should drive more distance than M, but dodnot know the relation between distances driven by A and P. Hence STEM2 by itself is inadequate. Combining STEM1 and STEM2 DA = (TP + 1) * (SP - 5) = DP + SP - 5*TP - 5 We donot know if SP - 5*TP - 5 is positive or negative to know, which of DA or DB is greater? Another possible equation is DA + DP = 1050 --> 2*DP + SP - 5*TP - 5 = 1050. This still doesnot anser the above question. Hence answer is E. |
|
|
|
|
|
#4 (permalink) |
|
Gone in 120 sec
![]() ![]() ![]() Join Date: Mar 2008
Posts: 608
![]() |
Yes, I agree that we cannot solve for the value. But if we just try random numbers, we are getting Al has traveled the longest distance.
Al: 5m/h lower than pl but 1hr longer Al: 45m/h ;3hr => 135m pl: 50m/h;2hr => 100m Al: 120 m/h; 5hr =>600 Pl: 125 m/h; 4hr =>500 Any thoughts |
|
|
|
Contact TestMagic TestMagic Forums Archive Privacy Statement
TestMagic Locations
Legal
Privacy
SEO by vBSEO 3.2.0
Copyright © 2009 TestMagic
Ad Management by RedTyger