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Simultaneous Motion Problems

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by sunsphere314 » Wed Sep 07, 2011 1:18 pm
Would someone please explain?

A hiker walked for two days. On the second day, the hiker walked two hours longer and at an average speed 1 mile 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 the average speed on the first day?

a) 2mph b) 3mph c) 4 mph d) 5mph e) 6 mph

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by pemdas » Wed Sep 07, 2011 2:05 pm
call 1st day's speed S1 and time T1, then the 2nd days speed and time will be S1+1 and T1+2
the total distance walked is S1*T1+(S1+1)*(T1+2)=64, time spent for total distance is T1+(T1+2)=18

by solving two equations with two variables above we get the answer ==>
2T1=16, T1=8
8*S1+(S1+1)(8+2)=64, 8*S1+8*S1+2*S1+8+2=64, 18*S1=54, S1=3

b
sunsphere314 wrote:Would someone please explain?

A hiker walked for two days. On the second day, the hiker walked two hours longer and at an average speed 1 mile 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 the average speed on the first day?

a) 2mph b) 3mph c) 4 mph d) 5mph e) 6 mph

Thank you!
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by GMATGuruNY » Wed Sep 07, 2011 4:41 pm
I posted a solution here:

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by sunman » Wed Sep 07, 2011 5:05 pm
Hi, this is my first time posting here. I came across this post randomly...this problem took me only about 75 - 90 seconds to solve.

Would this be an actual 700-800 level type question on the GMAT? I'd like to know if I'm in pretty good shape on the quantitative section.
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by sl750 » Thu Sep 08, 2011 4:31 am
On day 1, he walked for t hrs at a speed of s mph; On day 2 he walked t+2 hrs longer at a speed of s+1 mph faster.

Total distance covered = Distance on day1 + Distance on day2 =64
Total time covered = t+t+2=18, or t=8 hrs

s*8+10(s+1)=64, or s=3mph
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by navami » Thu Sep 08, 2011 5:46 am
This means on the second day he walked for 10 hours
ANd on teh first day hiker walked for 8 hours.

Now lets consider teh speed on first day was X Miles/hour
Speed on seconday = X + 1 Miles/Hour


now 8 * ( X) + 10 * ( X + 1 ) = 64
= 3

Hence B
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by navami » Thu Sep 08, 2011 5:47 am
@sunman : put your best effort. Leave rest on destiny. Everything will be all right.
Cheers
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