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jjykim
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Post subject: Rate problem from old GMAT exam Posted: Sun Aug 24, 2008 1:00 pm |
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This is a problem I got from one of their official tests that I purchased from mba.com
16. A Hiker walked for two days. On the second day the hiker walked 2 hours longer and at an average speed 1mile per hour faster than he walked on the first day. If during the two days he walked a total of 64 miles and spent a total of 18 hours walking, what was his average speed on the first day?
I know this is a rate problem, so the first thing I did was set up a chart, that contained the RTD for Day 1, Day 2, and the total for the 2 days. Using the relationship between the number of hours spent hiking, I solved for t and got 8. So now my question is, what step do I need to take next? Is the "average rate" for day 1 the same thing as its rate? I just don't know what to do from here....Any help would be appreciated. Thanks.
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guest612
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Post subject: solve Posted: Sun Aug 24, 2008 4:34 pm |
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I set up the grid like you said with the top being D1 & D2:
1 2 Total
R R R+1
T T T+2 18
D d (64-d) 64
as you said, t=8.
Thus, let's simply Day 2 first:
(R+1)10=64-d
10R+10=64-d
d=54-10R
Set the two distances equal to each other:
RT=54-10R
Plug in t=8
8R=54-10R
18R=54
R=3
Can you please confirm the answer? R=3
Also, can you please provide the answer choices?
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jjykim
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Post subject: Posted: Mon Aug 25, 2008 10:23 am |
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Here are the answer choices :)
and thanks for your explanation
(A) 2 mph
(B) 3 mph
(C) 4 mph
(D) 5 mph
(E) 6 mph
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jjykim
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Post subject: Posted: Mon Aug 25, 2008 10:26 am |
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oh, and B is correct. Quick question though. Why is the rate for Day 2 r? I thought it should be r+1 since the hiker walked an average of 1 mile more than the previous day, which had a rate of r. I think that's what messed me up while working on the problem.
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jjykim
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Post subject: Posted: Mon Aug 25, 2008 10:34 am |
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actually...sorry. Disregard my last message.
Thanks for your help!
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guest612
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Post subject: great! Posted: Mon Aug 25, 2008 12:32 pm |
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