Help on Rate Problem

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 68
Joined: Mon May 04, 2009 10:01 pm

Help on Rate Problem

by DCJ » Thu Jul 23, 2009 10:26 pm
A hiker walking at a constant rate of 4mph is passed by a cyclist traveling in the same direction along the same path at a constant rate of 20mph. The cyclist stops to wait for the hiker 5 minutes after passing her, while the hiker continues to walk at her constant rate. How many minutes must the cyclist wait until the bike catches up?

A. 6 2/3
B. 15
C. 20
D. 25
E. 26 2/3

Junior | Next Rank: 30 Posts
Posts: 22
Joined: Thu Jul 02, 2009 9:55 pm
Thanked: 2 times

by Rajani » Thu Jul 23, 2009 10:51 pm
Hiker - 4m/hr
cyclist - 20 m/hr

Cyclist
-------
60 mins = 20 m
5 mins = x
x = 5/3

So the hiker has to cover 5/3 miles in y mins

60 mins = 4 miles
y mins = 5/3miles

y = 25 mins

Junior | Next Rank: 30 Posts
Posts: 22
Joined: Thu Jul 02, 2009 9:55 pm
Thanked: 2 times

by Rajani » Thu Jul 23, 2009 10:58 pm
Hiker - 4m/hr
cyclist - 20 m/hr

Cyclist
-------
60 mins = 20 m
5 mins = x
x = 5/3

So the hiker has to cover 5/3 miles in y mins

60 mins = 4 miles
y mins = 5/3miles

y = 25 mins

Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Tue Jun 02, 2009 7:13 am
Location: NYC
Thanked: 1 times

by cammijc » Fri Jul 24, 2009 6:25 am
Cyclist:

60 min = 20 m
5 min = x
x = 5/3

Hiker:

60 min = 4m
5 min = y
y = 1/3

So, during the five min after the cyclist passes the hiker, the cyclist traveled 5/3 mile while the hiker traveled 1/3 mile (because the hiker keeps going after the cyclist passes her.

This means that when the cyclist stops, the hiker has to go 4/3 mile to catch up (5/3 - 1/3).

If she goes 1/3 m in 5 min, then it will take her 20 min to go 4/3 mile (5 min*4).

IMO C.

Master | Next Rank: 500 Posts
Posts: 129
Joined: Tue May 19, 2009 9:12 am
Thanked: 8 times

by cata1yst » Fri Jul 24, 2009 6:44 am
cammijc wrote:Cyclist:

60 min = 20 m
5 min = x
x = 5/3

Hiker:

60 min = 4m
5 min = y
y = 1/3

So, during the five min after the cyclist passes the hiker, the cyclist traveled 5/3 mile while the hiker traveled 1/3 mile (because the hiker keeps going after the cyclist passes her.

This means that when the cyclist stops, the hiker has to go 4/3 mile to catch up (5/3 - 1/3).

If she goes 1/3 m in 5 min, then it will take her 20 min to go 4/3 mile (5 min*4).

IMO C.
I got the same answer the same way. You need to take into account the distance traveled by the hiker during the 5 minutes the biker uses after he passes the hiker.

C.

Master | Next Rank: 500 Posts
Posts: 338
Joined: Fri Apr 17, 2009 1:49 am
Thanked: 9 times
Followed by:3 members

by kaulnikhil » Fri Jul 24, 2009 6:51 am
it should be C

Senior | Next Rank: 100 Posts
Posts: 68
Joined: Mon May 04, 2009 10:01 pm

Help on Rate Problem

by DCJ » Fri Jul 24, 2009 9:42 am
Thanks everyone. The correct answer is C. 20. I also kept getting 25 until I realized I needed to take into account that the hiker was still walking while the cyclist waited.

Senior | Next Rank: 100 Posts
Posts: 43
Joined: Thu Mar 12, 2009 8:56 pm
Thanked: 10 times
Followed by:1 members
GMAT Score:730

by jjk » Fri Jul 24, 2009 11:39 am
Rajani wrote:Hiker - 4m/hr
cyclist - 20 m/hr

Cyclist
-------
60 mins = 20 m
5 mins = x
x = 5/3

So the hiker has to cover 5/3 miles in y mins

60 mins = 4 miles
y mins = 5/3miles

y = 25 mins
What you should have done is see that the relative speed of the cyclist is 20 - 14 = 16 mph. 5 minutes elapse after the cyclist passes the hiker, and 5 minutes equals 1/12 of an hour. So 16 X 1/12 = 16/12 = 4/3 miles.

4/3 miles is the distance between the cyclist and the hiker when the cyclist decides to stop and wait for the hiker to catch up.

From there, it's a simple d = rt problem. 4/3 = 4t, so t = 1/3 of an hour, 20 minutes.