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

Redeem

Help on Rate Problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
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
Top
Source: — Problem Solving |
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
Top
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
Top
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.
Top
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.
Top
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
Top
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.
Top
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.
Top
Post new topic Post Reply
• Page 1 of 1