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

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

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New post 28 Feb 2011, 12:25
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 1,5 times as fast, the trip would have taken 2 hours
2) Betty`s average speed for the trip was 50 miles per hour

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Re: How long did it take [#permalink]

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New post 28 Feb 2011, 14:18
Madelaine88 wrote:
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 1,5 times as fast, the trip would have taken 2 hours
2) Betty`s average speed for the trip was 50 miles per hour


IMO A
From (1),
\(d/s\) -\(d/1.5s\) = \(2\)
\(d\) = \(6s\)
We know d = t*s
Hence \(t\) =\(6\) So, Sufficient
from (2) .
s= 50 So, Insufficient
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Re: How long did it take [#permalink]

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New post 28 Feb 2011, 16:28
Agree with the answer. It does not matter for DS, but its d/1.5s = 2. t = 3.

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Re: How long did it take [#permalink]

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New post 28 Feb 2011, 19:15
The answer is A :

Let x be the average speed and t be the time taken during trip, so from (1) we have :

1.5(x) * 2 = x*t
=> t = 3, sufficient.

From (2), all we have is x = 50, so not sufficient.

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Re: How long did it take [#permalink]

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New post 28 Feb 2011, 21:17
Statement 1.since the distance is equal:
V1T1=V2T2
V1T1=1.5V1*(T1-2)
T1=1.5T1-3
T1=6

Statement 1 is sufficient

Statement 2:Ony average velocity is given so nothing can be concluded.Hence Ststement 2 is insufficient.

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Re: How long did it take [#permalink]

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New post 03 Oct 2011, 22:45
subhashghosh wrote:
The answer is A :

Let x be the average speed and t be the time taken during trip, so from (1) we have :

1.5(x) * 2 = x*t
=> t = 3, sufficient.

From (2), all we have is x = 50, so not sufficient.

Regards,
Subhash

Statement 1 says that if Betty' speed has been 1 1/2 times as fast , the trip would have taken 2 hours...

Doesnt it mean Is 'S' was original speed new speed = s + 1 1/2 S = 5/2 S and not 1.5 S????

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Re: How long did it take [#permalink]

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New post 04 Oct 2011, 05:03
Statement 2 is not enough. Gives us only one variable.
Statement one gives us enough to solve with an if then scenario.
Answer A

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Re: How long did it take [#permalink]

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New post 04 Oct 2011, 05:07
Statement 2 is not enough. Gives us only one variable.
Statement one gives us enough to solve with an if then scenario.
Answer A

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Re: How long did it take [#permalink]

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New post 26 Oct 2011, 23:34
Can someone please comment if this is right approach?


Let 'S' be Speed and 'D' be distance.


st1)Original Equation: S * (t+2) = D
New Equation(If betty's speed was 1 1/2 times as fast the trip would have taken 2 hours) => (S + 3/2 S) * 2 = D

Doesnt it mean Is 'S' was original speed new speed = s + 1 1/2 S = 5/2 S and not 1.5 S????

S * (t+2) = 5/2 S * 2

Thus 3S = St

Thus t = 3


Hence A


Is this correct? I see people write 1.5 S and not 5/2 S in the new equation .... So is my approach correct or wrong ??? PLease comment :?: :?: :?:

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Re: How long did it take [#permalink]

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New post 18 Nov 2011, 07:15
siddhans wrote:
Can someone please comment if this is right approach?


Let 'S' be Speed and 'D' be distance.


st1)Original Equation: S * (t+2) = D
New Equation(If betty's speed was 1 1/2 times as fast the trip would have taken 2 hours) => (S + 3/2 S) * 2 = D

Doesnt it mean Is 'S' was original speed new speed = s + 1 1/2 S = 5/2 S and not 1.5 S????

S * (t+2) = 5/2 S * 2

Thus 3S = St

Thus t = 3


Hence A


Is this correct? I see people write 1.5 S and not 5/2 S in the new equation .... So is my approach correct or wrong ??? PLease comment :?: :?: :?:


Can someone please reply? How do we know "as fast" means s + 3/2 s or 1.5 s???

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Re: How long did it take   [#permalink] 18 Nov 2011, 07:15
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