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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] New post 14 Mar 2012, 14:19
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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\frac{1}{2} times as fast, the trip would have taken 2 hours.

(2) Betty's average speed for the trip was 50 miles per hour.
[Reveal] Spoiler: OA
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Re: PT #13 DS 22 [#permalink] New post 14 Mar 2012, 15:34
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Hi, there. I'm happy to help with this.

Prompt: How long did it take Betty to drive nonstop on a trip from her home to Denver, Colorado?

To apply D = RT, we would have to know both the distance and the speed to find the time.

Statement #1: If Betty's average speed for the trip had been 1 and 1/2 times as fast, the trip would have taken 2 hours.

Let D be the distance, and R be her speed. What this is saying is:

D = (1.5R)*2 = 3*R

This implies T = D/R = 3 --- the trip took her 3 hours. Statement #1, by itself, is sufficient.

Statement #2: Betty's average speed for the trip was 50 miles per hour.

Well, we know the average speed, but we don't know where Betty lives --- we have no idea of the distance from her home to Denver. Since we know R and don't know D, we can't find T. Statement #2, by itself, is insufficient.

[Reveal] Spoiler:
Answer = A


Here's a similar DS question, for more practice:
http://gmat.magoosh.com/questions/927
When you submit your answer to that, the next page will have a complete video explanation.

Does all this make sense? Please let me know if you have any questions.

Mike :)
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Re: PT #13 DS 22 [#permalink] New post 28 Nov 2012, 18:40
mikemcgarry wrote:
Hi, there. I'm happy to help with this.

Prompt: How long did it take Betty to drive nonstop on a trip from her home to Denver, Colorado?

To apply D = RT, we would have to know both the distance and the speed to find the time.

Statement #1: If Betty's average speed for the trip had been 1 and 1/2 times as fast, the trip would have taken 2 hours.

Let D be the distance, and R be her speed. What this is saying is:

D = (1.5R)*2 = 3*R

This implies T = D/R = 3 --- the trip took her 3 hours. Statement #1, by itself, is sufficient.

Statement #2: Betty's average speed for the trip was 50 miles per hour.

Well, we know the average speed, but we don't know where Betty lives --- we have no idea of the distance from her home to Denver. Since we know R and don't know D, we can't find T. Statement #2, by itself, is insufficient.

[Reveal] Spoiler:
Answer = A


Here's a similar DS question, for more practice:
http://gmat.magoosh.com/questions/927
When you submit your answer to that, the next page will have a complete video explanation.

Does all this make sense? Please let me know if you have any questions.

Mike :)

D = (1.5R)*2 = 3*R

Hi,
Should the above not be D = (2.5R)*2 = 3*R?

doesnt '1.5 times as fast ' imply 2.5R?
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Re: PT #13 DS 22 [#permalink] New post 28 Nov 2012, 21:45
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Sachin9 wrote:
Hi,
Should the above not be D = (2.5R)*2 = 3*R? Doesn't '1.5 times as fast ' imply 2.5R?

No, it doesn't. This is a very subtle verbal thing.

Suppose I have some beginning speed V.

If I have a speed that is 1.5 time faster than this, that's 1.5*V

BUT, if the speed increases by 1.5 this value, then that's V + 1.5*V = 2.5*V

Moving at a speed 1.5 faster than the original is not the same as increasing your speed by 1.5 times beyond what the original was.

Does this distinction make sense?
Mike :-)
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Re: PT #13 DS 22 [#permalink] New post 29 Nov 2012, 18:59
mikemcgarry wrote:
Sachin9 wrote:
Hi,
Should the above not be D = (2.5R)*2 = 3*R? Doesn't '1.5 times as fast ' imply 2.5R?

No, it doesn't. This is a very subtle verbal thing.

Suppose I have some beginning speed V.

If I have a speed that is 1.5 time faster than this, that's 1.5*V

BUT, if the speed increases by 1.5 this value, then that's V + 1.5*V = 2.5*V

Moving at a speed 1.5 faster than the original is not the same as increasing your speed by 1.5 times beyond what the original was.

Does this distinction make sense?
Mike :-)

Thanks Mike, I suppose this is how it should be

If I have a speed that is 1.5 times faster than this, that's 2.5*V
If I have a speed that is 1.5 times as fast as this, that's 1.5*V

This is how we apply for percentage based concepts..

Is my understanding correct?

Regards,
Sachin
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Re: PT #13 DS 22 [#permalink] New post 30 Nov 2012, 11:55
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Sachin9 wrote:
Thanks Mike, I suppose this is how it should be

If I have a speed that is 1.5 times faster than this, that's 2.5*V
If I have a speed that is 1.5 times as fast as this, that's 1.5*V
Is my understanding correct?

Dear Sachin

We are getting into some really subtleties of grammar and phrasing here.

If I have a speed that is 1.5 times as fast as V ----> that clearly should be 1.5*V

If my speed was V, but increased by 1.5 this value ----> that clearly should be 2.5*V

If I have a speed that is 1.5 times faster than V -----> not 100% percent clear, but I would say most people would interpret this as 1.5*V. The GMAT is always crystal clear in its phrasing. There is no way they would use this phrasing and expect you to come up with 2.5*V.

Part of what is unusual about this conversation is we are talking about the increases of speed. Almost everyone compares speed by saying "n times faster." In real life, no one ever talks about percentage increase or decreases for speed.

Let's change the subject to something like profits. You are perfectly correct ---- If I say "profits increased by 150%", that's 2.5 times the original profits.

Does all this make sense?

Mike :-)
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Re: PT #13 DS 22 [#permalink] New post 30 Nov 2012, 18:24
mikemcgarry wrote:
Sachin9 wrote:
Thanks Mike, I suppose this is how it should be

If I have a speed that is 1.5 times faster than this, that's 2.5*V
If I have a speed that is 1.5 times as fast as this, that's 1.5*V
Is my understanding correct?

Dear Sachin

We are getting into some really subtleties of grammar and phrasing here.

If I have a speed that is 1.5 times as fast as V ----> that clearly should be 1.5*V

If my speed was V, but increased by 1.5 this value ----> that clearly should be 2.5*V

If I have a speed that is 1.5 times faster than V -----> not 100% percent clear, but I would say most people would interpret this as 1.5*V. The GMAT is always crystal clear in its phrasing. There is no way they would use this phrasing and expect you to come up with 2.5*V.

Part of what is unusual about this conversation is we are talking about the increases of speed. Almost everyone compares speed by saying "n times faster." In real life, no one ever talks about percentage increase or decreases for speed.

Let's change the subject to something like profits. You are perfectly correct ---- If I say "profits increased by 150%", that's 2.5 times the original profits.

Does all this make sense?

Mike :-)


You Mike, It does :) Thanks a lot :)
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How long did it take Betty to drive nonstop on a trip from [#permalink] New post 21 Jan 2013, 01:34
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alimad 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 1/2 times as fast, the trip would have taken 2 hours.

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


The distance is similar in this question

equation 1 X be speed, t is the time and d is the distance

X*t=d

equation 2

1.5X*2=d >> 3X=d

substitute in 1

x*t=3X

t=3

is this approach correct?
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Re: How long did it take Betty to drive nonstop on a trip from [#permalink] New post 21 Jan 2013, 02:53
Expert's post
fozzzy wrote:
alimad 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 1/2 times as fast, the trip would have taken 2 hours.

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


The distance is similar in this question

equation 1 X be speed, t is the time and d is the distance

X*t=d

equation 2

1.5X*2=d >> 3X=d

substitute in 1

x*t=3X

t=3

is this approach correct?


Yes, it is. Even without any formula: since moving 1.5 times as fast as the actual speed would take 2 hours to cover the distance, then moving at actual speed would take 2*1.5=3 hours to cover the same distance.
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Re: How long did it take Betty to drive nonstop on a trip from [#permalink] New post 01 Aug 2013, 10:14
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.
Time = Distance/speed

Time (faster trip):
2 = d/1.5s
3s=d

Time (slower trip):
t=D/S
t=3s/s
t=3
SUFFICIENT

(2) Betty's average speed for the trip was 50 miles per hour.
No information given about distance.
INSUFFICIENT

(A)
Re: How long did it take Betty to drive nonstop on a trip from   [#permalink] 01 Aug 2013, 10:14
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