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Rate Data Sufficiency

by gmatassistance » Mon Mar 08, 2010 6:11 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.

The Ans is A - can someone please explain why?
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by bpgen » Mon Mar 08, 2010 7:42 pm
Always use 'Spoiler' to hide your answer, showing it really makes biasing towards that answer..
see previous thread at https://www.beatthegmat.com/bill-and-sally-t53271.html
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by hooliganpete » Tue Mar 09, 2010 9:14 am
The Ans is A - can someone please explain why?
The key to solving a problem like this is figuring out the variables, or pieces of information you'd need to know to solve the problem. In this case we have the distance between Bill and Sally so we only really need to know their rates of speed.

(1) To answer this question it matters less how fast each are going. The important piece of information is that no matter how fast each is running, Bill covers 1.5 times the distance of Sally. By telling us that Bill's rate is 50% greater than Sally's, or B=1.5S no matter how far the distance we can figure out how far Bill runs when they meet: 500=S+1.5S (distance = Sally's rate + Bill's rate). 500=2.5S. 500/2.5=S, S=200ft, Bill=300ft. Bill has run 300 ft when they meet at a rate that is 50% faster than Sally. SUFFICIENT

(2) Though we know that Bill's speed is 4 ft/s faster than Sally's average speed this doesnt tell us anything about how fast either is running. For example, Sally's speed could be 2 ft/s making Bill's speed 6 ft/s. Or, Sally's speed could be 6 ft/s and Bill could be 10 ft/s. We just can't tell from the information provided. NOT SUFFICIENT
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by crackthetest » Tue Mar 09, 2010 11:21 am
You could use the RTD chart for this:

1)
rate time d
B 1.5r t = 1.5rt
S r t = rt
total = 500

1.5rt+rt = 500
2.5rt = 500
rt = 200

so they meet when Sally is at 200 ft.

2) rate time d
B r+4 t = rt+4t
S r t = rt
total = 500

2rt+ 4t = 500 -- Not sufficient.

Hence A.
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