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Re: How long did it take Betty to drive nonstop on a trip from h [#permalink]
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Answer A is sufficient.

it is the same distance - so S1*T1=S2*T2

let us assume that the original speed is x and original time t

x*t=3/2x*2
t=3x/x
t=3

the trip took 3 hours.

The answer B is not sufficient because we only know the speed but not distance.
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Re: How long did it take Betty to drive nonstop on a trip from h [#permalink]
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Even though you can plug in numbers to solve this question, it is important to notice one important algebraic caveat in this type of problems.

Statement 1 is sufficient. However it provides only 1 equation with 2 variables.
R (1.5) (2) = D

At first glance, it looks like you cannot solve this using the Equation Rule of Sufficiency (the one that states that "you need n number of distinct, linear equations to solve for n variables..."). The catch is that all Distance problems are already giving us 1 equation and 3 variables, namely R * T = D.

So when you look at statement 1 you actually have 2 equations and 3 variables

(1.5) (2) = D/R
T = D/R

If you substitute, you kill one variable and thus you can solve.

(1.5) (2) = T
3 = T


Statement 2 is insufficient.
The equations you have are:
T = D/R
R = 50
3 Variables, 2 Equations --> Insufficient.
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Re: How long did it take Betty to drive nonstop on a trip from h [#permalink]
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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 3/2 times as fast, the trip would have taken 2 hours.
(2) Betty's average speed for the trip was 50 miles per hour.


Target question: What was Betty's travel time?

This is a great candidate for rephrasing the target question

Let D = The distance driven
Let R = Betty's average speed

Since Time = Distance/Rate, we can rephrase the target question. . .

REPHRASED target question: What is the value of D/R?

Statement 1: If Betty's avg speed for the trip had been 1.5 times as fast, the trip would have taken 2 hours.
The distance is still the same (D), but the rate is 1.5R
So, statement 1 tells us that D/1.5R = 2
Is this enough information to find the value of D/R?
Sure, just take the equation D/1.5R = 2 and multiply both sides by 1.5 to get D/R = 3
Since we're able to answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: Betty's avg speed for the trip was 50 miles per hour.
This tells us that R=50
However, we cannot find the value of D/R, since we don't know the value of D (distance)
So, statement 2 is NOT SUFFICIENT

Answer: A

RELATED VIDEO FROM OUR COURSE
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Re: How long did it take Betty to drive nonstop on a trip from h [#permalink]
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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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How long did it take Betty to drive nonstop on a trip from h [#permalink]
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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 3/2 times as fast, the trip would have taken 2 hours.
(2) Betty's average speed for the trip was 50 miles per hour.

You can also use ratios.

1.5 times as fast means that if Betty's speed was 2s actually, 1.5 times would mean 3s. The ratio of actual speed : hypothetical (increased) speed = 2:3
Therefore, ratio of actual time : hypothetical time = 3:2
(because time is inversely proportional to speed)

The hypothetical time has been given to be 2 hrs. So the actual time must be 3 hrs.
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Re: How long did it take Betty to drive nonstop on a trip from h [#permalink]
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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 3/2 times as fast, the trip would have taken 2 hours.
(2) Betty's average speed for the trip was 50 miles per hour.

In the original condition, from vt=d, there are 3 variables(v,t,d) and 1 equation(vt=d), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) &2), you can easily find out that C is the answer. However, in 1), you can get 2 equations(time and velocity). That is, use vt=d from (3/2)v*2=d and you can get t=3, which is sufficient. Therefore, the answer is A.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: How long did it take Betty to drive nonstop on a trip from h [#permalink]
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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: How long did it take Betty to drive nonstop on a trip from h [#permalink]
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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 h [#permalink]
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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 3/2 times as fast, the trip would have taken 2 hours.
(2) Betty's average speed for the trip was 50 miles per hour.



(1)
We need to know how long it took Betty to drive, so time=distance/speed.

Let's denote:
s1 is average speed
t1 is the time needed to complete the journey when travelling at speed s1
s2 is the average speed when travelling at faster speed
t2 is the time needed to complete the journey when travelling at speed s2.

Then,
t1=d/s1

But also as per (1):
s1=3/2*s2
or s2=2/3*s1 (equation #1)

And:
t2=d/s2=2hours (from statement 1)

But also as per (eq. #1):
t2= d/s2 = d/((2/3)*s1) = 2 hours --> t2 = (2/3)*(d/s1) = 2 --> (2/3)*t1 = 2 --> t1= 3 hours

Hence,
(1) is SUFFICIENT

(2) Nothing is said about distance between "her home" and Colorado, hence we have no means to determine t or d/s.
(2) IS INSUFFICIENT

ANSWER: A
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Re: How long did it take Betty to drive nonstop on a trip from h [#permalink]
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.

Answer = A


Here's a similar DS question, for more practice:
https://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: How long did it take Betty to drive nonstop on a trip from h [#permalink]
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: How long did it take Betty to drive nonstop on a trip from h [#permalink]
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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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