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

Redeem

Work rate problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 1460
Joined: Tue Dec 29, 2009 1:28 am
Thanked: 135 times
Followed by:7 members

Work rate problem

by selango » Thu May 27, 2010 3:53 am
A hiker walking at a constant rate of 4 miles per hour is passed by a cyclist travelling in the same direction along the same path at a constant rate of 20 miles per hour. 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 hiker catch up?

1)6 2/3

2)15

3)20

4)25

5)26 2/3

OA C
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1560
Joined: Tue Nov 17, 2009 2:38 am
Thanked: 137 times
Followed by:5 members

by thephoenix » Thu May 27, 2010 4:00 am
rlative speed = 20-4=16miles/hr=4/15 miles/minute
in 5 minutes the cyclist cover a dis of 4/15*5 miles=4/3 miles
at a speed of 4miles / hour the hiker will need (4/3)/4 hour=1/3 hour=20 minute
Many of the great achievements of the world were accomplished by tired and discouraged men who kept on working
Top
Senior | Next Rank: 100 Posts
Posts: 85
Joined: Sat Aug 15, 2009 4:36 am
Thanked: 19 times

by gmatjedi » Thu May 27, 2010 4:33 am
my approach:

in 5 minutes, hiker's distance = 5min*1hr/60min*4mi/hr= 1/3 mile
in 5 minutes, cyclist's distance = 5 min*1hr/60min*20mi/hr=5/3 miles

difference in distance = 5/3-1/3=4/3 mile
so hiker must make up 4/3 miles

4/3 mi=4 mi/hr*t
t=1/3 hr=20 min

the above illustrates the concept of relative speed, as mentioned by phoenix, by showing that in 5 minutes both the cyclist and the hiker are advancing
Top
User avatar
GMAT Instructor
Posts: 1052
Joined: Fri May 21, 2010 1:30 am
Thanked: 335 times
Followed by:98 members

by Patrick_GMATFix » Thu May 27, 2010 6:15 am
In short, the cyclist (20mph) is traveling at 5 times the speed of the hiker (4mph). this means that the hiker will need to spend 5 times as much time traveling to cover the same distance.

The cyclist went on for 5 minutes. To cover that distance the hiker will have to walk 5*5=25 minutes.

The first 5 minutes, both people are traveling. The last 20 minutes, only the hiker is traveling while the cyclist waits for him. So the cyclist waits 20 minutes. The answer is C.

This is GMATPrep question 1248

Good luck,
-Patrick
  • Ask me about tutoring.
Top
GMAT Instructor
Posts: 1302
Joined: Mon Oct 19, 2009 2:13 pm
Location: Toronto
Thanked: 539 times
Followed by:164 members
GMAT Score:800

by Testluv » Fri May 28, 2010 12:25 am
All of the above approaches are sound (and Patrick's was creative) but I just want to elaborate on thephoenix's:
rlative speed = 20-4=16miles/hr=4/15 miles/minute
in 5 minutes the cyclist cover a dis of 4/15*5 miles=4/3 miles
at a speed of 4miles / hour the hiker will need (4/3)/4 hour=1/3 hour=20 minute
Because they are travelling in the same direction, we can simply subtract the rates to find either the rate at which the faster object "catches up" to the slower one or, once caught up, the rate at which the faster object "exceeds" the slower one:

20 - 4 = 16

So, after passing the hiker, for 5 minutes the cyclist exceeds the hiker at a rate of 16mph.

Because 5 is just 1/12 of an hour, the cyclist has gained (1/12)*16 = 4/3 of a mile on the hiker (this is just distance = rate*time). Because we also know the hiker's rate, we can now use distance = rate*time again to solve for time, as thephoenix did.

TAKEAWAY:

When two objects travel in the same direction, we can subtract the rates.
When two objects travel in opposite directions (meeting up or spreading apart), we can add the rates.
Kaplan Teacher in Toronto
Top
Post new topic Post Reply
• Page 1 of 1