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# I am having difficulty with a particular style problem: Joe

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I am having difficulty with a particular style problem: Joe [#permalink]

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04 Nov 2006, 02:23
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I am having difficulty with a particular style problem:

Joe runs at a rate of 30 miles per hour, while Matt runs at a rate of 40 miles per hour. Starting from an identical starting point, if Joe starts running at 4:30, and Matt begins running at 5:20, at what time will they meet up?

A. 9:30
B. 9:00
C. 8:40
D. 7:50
E. 7:00

I eventually get to the right answer, but it is taking me way too long. What is the easiest/quickest way to solve these? Thanks!
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04 Nov 2006, 02:52
Matt covers additional 10 miles per hour as compared to Joe

Joe starts 50 minutes earlier than Matt and covers a distance of =30*50/60=25 miles

Matt in one hour covers 10 miles more than Joe. So it will take 2.30 for Matt to cover 25 miles. Thus time will be 5:20+2:30=7:50
Director
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04 Nov 2006, 08:20
I do it this way...

Find the rate for each runners....

Joe's rate = 30miles/60minutes=1mile/2minutes
Mat's rate= 40miles/60minutes=2miles/3minutes

Joe started running 50 minutes earlier.... find out what distance he covered in that time....

r*t=d
1/2*50=d
d=25

Find out how much faster Mat is running.... 2/3-1/2=1/6

Find out how many minutes will be needed to deplete the distance between Joe and Mat....

r*t=d
1/6*t=25
t=150 minutes..... or 2h 30 minutes....
Matt started running at 5:20, so he will catch up with Joe +2h 30minutes = 7:50
Director
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04 Nov 2006, 08:45
Algebraically:
Let x = Matt's time
Let x + 5/6 = Joe's time (since Joe started out 50 minutes, or 5/6 of an hour, before Matt)

40x=30(x+5/6) (They traveled equal distances)
x=2.5, the time it took Matt to catch up with Joe.

Once you get this formula in your head, I find this by far the easiest way to solve these suckers!
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...there ain't no such thing as a free lunch...

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05 Nov 2006, 00:39
Thanks for all the help...all of these explanations were great!

Second question...

What if two people are coming at each other....

Here is one I have seen before...

Steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per minute. At the same time Joyce gets on an elevator on the 51st floor of the same building and rides down at a rate of 63 floors per minute. If they continue traveling at these rates, at which floor will their paths cross?

A. 19
B. 28
C. 30
D. 32
E. 44

Thanks for any help/explanations.
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05 Nov 2006, 00:48
Good one.....

I would find the common rate of both of the travelers....

Steve rate: 57floors/60seconds
Joyce rate: 63floos/60seconds

common rate: 57/60+63/60=2floors/1second

There are 40 floors between them.

r*t=d
2*t=40
t=20.... not exactly 19
I choose A
Director
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05 Nov 2006, 00:54
Alternatevely you can see that both are travelling at approximatelly 1floor a second.... there are 40 floors so they will need approximatelly 20 seconds to meet each other.... No need for special calculations....
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05 Nov 2006, 02:39
hey,

what's the answer for this one?
i got 30???

cheers.
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05 Nov 2006, 07:57
Oh yeah, i calculated the seconds, needed the floor Sure the answer should be 30....
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05 Nov 2006, 14:35

Could you possibly finish off the calculation to show how you get to 30?

Thanks!
05 Nov 2006, 14:35
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