Simultaneous Motion Problems

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sunsphere314: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Mon Mar 21, 2011 1:56 pm

Simultaneous Motion Problems

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

Thank you!

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Source: Beat The GMAT — Problem Solving | Wed Sep 07, 2011 1:18 pm

pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pm; Thanked: 158 times; Followed by:21 members

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!

Success doesn't come overnight!

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Wed Sep 07, 2011 4:41 pm

I posted a solution here:

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sunman: Master | Next Rank: 500 Posts; Posts: 165; Joined: Thu Feb 17, 2011 5:05 am; Location: San Diego, CA; Thanked: 14 times; Followed by:9 members; GMAT Score:750

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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sl750: Master | Next Rank: 500 Posts; Posts: 496; Joined: Tue Jun 07, 2011 5:34 am; Thanked: 38 times; Followed by:1 members

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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navami: Legendary Member; Posts: 540; Joined: Sat Dec 20, 2008 7:24 pm; Thanked: 37 times; Followed by:6 members

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

This time no looking back!!!
Navami