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# How long did it take betty to drive nonstop on a trip from

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Director
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How long did it take betty to drive nonstop on a trip from [#permalink]

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04 May 2008, 08:46
How long did it take betty to drive nonstop on a trip from her home to Denver, Colorado?

(1). If Betty's average speed for the trip had been 1 1/2 times as fast, the trip would have taken 2 hours.

(2). Betty's average speed for the trip was
50 miles / hour.

A. t = ?
d is distance.
r is rate.

d/r = t

d/3/2(r) = 2

d/3r = 4
d = 12r
d/r = 12 - time it took. A is sufficient

B - insufficient.

A.

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04 May 2008, 18:13
A for me as well, by just thinking about relationship btwn speed and time.
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05 May 2008, 04:52
How long did it take betty to drive nonstop on a trip from her home to Denver, Colorado?

(1). If Betty's average speed for the trip had been 1 1/2 times as fast, the trip would have taken 2 hours.

(2). Betty's average speed for the trip was
50 miles / hour.

A. t = ?
d is distance.
r is rate.

d/r = t

d/3/2(r) = 2

d/3r = 4
d = 12r
d/r = 12 - time it took. A is sufficient

B - insufficient.

A.

I don't think d/r=12 though. I believe its 3.
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08 Jul 2011, 01:49
How to do this using smart numbers?
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08 Jul 2011, 01:57
siddhans wrote:
How to do this using smart numbers?

Assume distance as 30
original speed = 10 .. time = 30/10 = 3 hrs.
1.5 times speed = 15 so time = 30/15 = 2 hrs
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08 Jul 2011, 02:10
sudhir18n wrote:
siddhans wrote:
How to do this using smart numbers?

Assume distance as 30
original speed = 10 .. time = 30/10 = 3 hrs.
1.5 times speed = 15 so time = 30/15 = 2 hrs

In what cases can we assume the distance? In case the distance remains the same ? i.e any variable which remains constant we can choose a random number? correct?
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11 Jul 2011, 03:17
bkk145 wrote:
A is correct to me.

Using formula s=v times t thus 2 = s\1,5v

We have two unknowns in this equitation, thus we need either v or s. number two gives us v, logically answer should be C.
Re: distance   [#permalink] 11 Jul 2011, 03:17
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