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Frank and Edna running on a track

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Frank and Edna running on a track

by Brent@GMATPrepNow » Sat Jan 31, 2009 10:13 am
Frank and Edna are running on the same circular track. Edna is running clockwise, and Frank is running counterclockwise. The two runners pass each other every 20 seconds. If it takes Frank 45 seconds to run around the track once, how many seconds does it take Edna to run around the track once?
A) 30
B) 32
C) 35
D) 36
E) 40

Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
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by ontopofit » Sat Jan 31, 2009 10:35 am
D?

let full rev. be R
Then F(speed of frank) = R/45
E(Edna's speed) = R/x;x is time taken for full rev. by edna

now with concept of relative speed

R/(F+E) = 20
i.e
R/(R/45 + R/x) = 20
solving for x we get x = 36.

is that correct?
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by Brent@GMATPrepNow » Sat Jan 31, 2009 10:39 am
ontopofit wrote:D?

let full rev. be R
Then F(speed of frank) = R/45
E(Edna's speed) = R/x;x is time taken for full rev. by edna

now with concept of relative speed

R/(F+E) = 20
i.e
R/(R/45 + R/x) = 20
solving for x we get x = 36.

is that correct?
That's right. The answer is D. Nice approach.

Here's mine:

Start from a point when the runners cross paths. After 20 seconds, when the two runners meet, the respective distances that each person has run will add to one lap.

If it takes Frank 45 seconds to run one lap then, after 20 seconds, Frank will have completed 20/45 of one lap.
If, after 20 seconds, Frank has completed 20/45 (4/9) of one lap, then Edna has completed 25/45 (5/9) of one lap.

If it takes Edna 20 seconds to run 5/9 of one lap, we need to determine how long it takes to run one complete lap.

We can determine this using equivalent ratios: 20 seconds : 5/9 laps = x second : 1 lap

In other words: 20/(5/9) = x/1
Cross multiply to get: 5x/9 = 20
5x = 180
X=36 (answer D)
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by ontopofit » Sat Jan 31, 2009 10:59 am
oh wow
that was a new dimension to think.

Brent, I have just started to visit the forum today.I joined 2 mons back.Realised how must time i have wasted.
I have learnt a lot in a single day.
Afraid that I will forget what I have learnt.
Any suggestions?
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by gaggleofgirls » Sat Jan 31, 2009 8:57 pm
In 20 sec, Frank goes 20/45 of a revolution, so Edna must go 25/45 of the revolution in same 20 seconds in order to pass Frank then.

So, we know Edna's distance (25/45) and Time (20 sec) therefore we can figure our her rate = distance/time = (25/45)/20 (25/45) * (1/20)

25/45 reduces to 5/9 so 5/9 * 1/20. This can reduce further to 1/9/ * 1/4 = 1/36. So Edna's rate is 1 revolution in 36 seconds.

Answer = D.

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by hypik21 » Mon Feb 02, 2009 11:37 pm
wow seemed like an impossible problem to solve but so easy with the explanations
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by dendude » Tue Feb 03, 2009 12:03 pm
Here's how I approached it,

C/Sf = 45 [Sf - Frank's Speed; C - Circumference of Path]
And we need to determine, C/Se

We know,
x/Se = (C - x)/Sf = 20 [because they cross each other every 20 secs and they are running on a circular path]

Use this and the value of C/Sf to arrive at,
C/Se = 36.
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by Uri » Tue Feb 03, 2009 12:25 pm
My approach is a bit different. I have solved it in the way that we use to calculate the time taken by two individual to complete a single task.

they meet every 20 seconds. so together they cover 1/20 lap every second. let the unknown speed be x. then,
1/45 + 1/x = 1/20
solvong, we get x=36
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