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# This is MGMAT's challenge problem of the week. Have fun !

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This is MGMAT's challenge problem of the week. Have fun ! [#permalink]

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09 Jan 2006, 00:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

This is MGMAT's challenge problem of the week. Have fun !

Alex and Brenda both stand at point X. Alex begins to walk away from Brenda in a straight line at a rate of 4 miles per hour. One hour later, Brenda begins to ride a bicycle in a straight line in the opposite direction at a rate of R miles per hour. If R > 8, which of the following represents the amount of time, in terms of R, that Alex will have been walking when Brenda has covered twice as much distance as Alex?
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09 Jan 2006, 01:23
Let time "Alex will have been walking" = t

Distance covered by Alex = 4t
Distance covered by Brenda = R (t-1)

4t * 2 = R(t-1)
t = R/(R-8)
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09 Jan 2006, 01:55
ps_dahiya wrote:
Let time "Alex will have been walking" = t

Distance covered by Alex = 4t
Distance covered by Brenda = R (t-1)

4t * 2 = R(t-1)
t = R/(R-8)

too fast dahiya!

x = 4t
2x = R(t-1) solving for t we get

t = R/(R-8)
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09 Jan 2006, 02:59
Can someone please solve the following explicitly?

4t * 2 = R(t-1)
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09 Jan 2006, 10:28
Can someone please solve the following explicitly?

4t * 2 = R(t-1)

Isn't it:

8*t = R*t - R
Since R>8,

R*t-8*t = R
t (R-8) = R
ot t = R/(R-8)
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09 Jan 2006, 10:28
Let the time reqd be t ,

2(4+ 4t) = Rt
t = 8/(R-8)
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09 Jan 2006, 10:37
Thanks giddi

Don't know why I couldn't make it too.
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09 Jan 2006, 11:50
hkm_gmat wrote:
Let the time reqd be t ,

2(4+ 4t) = Rt
t = 8/(R-8)

hkm_gmat, didn't you calculate the time that Brenda took. They have asked for time that Alex took. So 1 needs to be added to your answer.

T(Alex) = 8/(R-8) + 1
= R/(R-8)

Is my reasoning OK?
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09 Jan 2006, 18:16
I've got the same . The OE for this question is not out yet, but I'm pretty sure all of us breezed through this one.
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09 Jan 2006, 21:32
I too got R/ R-8 but took almost 4 min
09 Jan 2006, 21:32
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