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Bill and Sally

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by gmatnmein2010 » Fri Feb 19, 2010 7:56 pm
Bill and Sally see each other across a field. They are 500 feet apart. If at the same instant of time they start running toward each other in a direct line how far will Bill have traveled when they meet?

(1) Bill ran at an average speed that was 50% greater than Sally ' s average speed.

(2) Bill ran at an average speed 4 feet per second faster than Sally ' s average speed.

OA after some discussion..

please explain ur answer
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Source: — Data Sufficiency |
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by thephoenix » Fri Feb 19, 2010 9:30 pm
gmatnmein2010 wrote:Bill and Sally see each other across a field. They are 500 feet apart. If at the same instant of time they start running toward each other in a direct line how far will Bill have traveled when they meet?

(1) Bill ran at an average speed that was 50% greater than Sally ' s average speed.

(2) Bill ran at an average speed 4 feet per second faster than Sally ' s average speed.

OA after some discussion..

please explain ur answer
IMO A
1. Lets assume sally ran at 20 and 40, so bill ran at 30 and 60 respectively

500/(20+30) = 10*30 = 300
500/(40+60)=5*60 = 300
sufficient

2. By the same reasoning as above assume sally runs at 20 and 30 and bill at 24 and 34 respectively
500/(20+24) = (500/44)*24
500/(30+34)= (500/64)*34
both give different answers
insuff
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by harsh.champ » Sat Feb 20, 2010 7:24 am
gmatnmein2010 wrote:Bill and Sally see each other across a field. They are 500 feet apart. If at the same instant of time they start running toward each other in a direct line how far will Bill have traveled when they meet?

(1) Bill ran at an average speed that was 50% greater than Sally ' s average speed.

(2) Bill ran at an average speed 4 feet per second faster than Sally ' s average speed.

OA after some discussion..

please explain ur answer
Here the concept of relative speed can be used.

Statement 1:- Let Sally's speed be x.
Then Bill's speed will be 1.5x
Hence,Relative speed will be 2.5x.
Time taken = (Relative Distance)/(Relative Speed) = 500/2.5x

Hence,Bill has traveled [500/2.5x] x 1.5x = 300 feet.Sufficient

Statement 2:-Let Sally's speed be y.
Then Bill's speed will be y+4.
Their relative speed will be 2y+4.
Time taken = (Relative Distance)/(Relative Speed) = [500/2y+4]

Hence,Bill has traveled [500/2y + 4] x (y+4)
Over here we can't find that y doesn't cancel out in the Nr. and Dr. like i the previous case.
Hence insufficient.[spoiler]
The answer would be A.[/spoiler]
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by kstv » Sat Feb 20, 2010 9:23 am
Bill and Sally see each other across a field. They are 500 feet apart. If at the same instant of time they start running toward each other in a direct line how far will Bill have traveled when they meet?

(1) Bill ran at an average speed that was 50% greater than Sally ' s average speed.

(2) Bill ran at an average speed 4 feet per second faster than Sally ' s average speed.

(1) Whatever time they take to meet, Bill will have covered 1.5 times more distance than Sally. The ratio of distance covered by Bill and Sally is 3:2. Total Distance is 500 feet. Suff.

(2) Bill ran 4 feet per second faster than Sally , but Sally's speed is not known. It can be 10, 20 fts/sec etc. Total time to meet will vary and also the distance where they meet. Insuff.
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by shashank.ism » Sat Feb 20, 2010 12:03 pm
gmatnmein2010 wrote:Bill and Sally see each other across a field. They are 500 feet apart. If at the same instant of time they start running toward each other in a direct line how far will Bill have traveled when they meet?

(1) Bill ran at an average speed that was 50% greater than Sally ' s average speed.

(2) Bill ran at an average speed 4 feet per second faster than Sally ' s average speed.

OA after some discussion..

please explain ur answer
St. 1 : let v be sally's speed so bill's speed = 1.5v
let bill traveled p distance so sally traveled 500-p distance...
p/1.5v = (500-p)/v --> p = 750 -1.5p --> 2.5p = 750 --> p = 750/2.5 = 300 so billl traveled 300m when they meet
suff.

St2. let speed of sallly be v, speed of bill be 4+v.
so p/(4+v) = (500-p)/v --> pv = (4+v) ( 500-p) --> two variables can't be calculated --- insufficient

Ans A
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