BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

hiker

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 345
Joined: Wed Mar 18, 2009 6:53 pm
Location: Sao Paulo-Brazil
Thanked: 12 times
GMAT Score:660

hiker

by shibal » Sun Apr 19, 2009 1:59 pm
A hiker walked for 2 days. On the 2nd day he walked 2 hrs longer and at an avg 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 hrs walking, what was his avg speed on the 1st day?
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 392
Joined: Thu Jan 15, 2009 12:52 pm
Location: New Jersey
Thanked: 76 times

by truplayer256 » Sun Apr 19, 2009 2:21 pm
Let's say the hiker walked x miles on the first day and it took him t hours to walk those x miles. On the second day, the hiker walked z miles in t+2 hours. From the given information in the problem, we can say that:

x+z=64
t+t+2=18
2t+2=18
t=8

Average speed for first day= total dist./total time= x/t=y--> x=8y (plugged in the value of t)
Average speed for second day= total dist./total time= z/t+2=y+1 z=10y+10 (plugged in the value of t)

Now, we have x=8y and z=10y+10 and we know that x+z=64, so 8y+10y+10=64 and y=3, which represents the hiker's average speed on the first day.

There's probably a much easier solution to this problem, but this is what I could think of.
Top
Post new topic Post Reply
• Page 1 of 1