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

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

by II » Sun Mar 16, 2008 11:27 am
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

I am more interested in your logic than the final answer. Thanks in advance.
Last edited by II on Mon May 05, 2008 1:53 am, edited 1 time in total.
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Re: Distance / Rates

by gabriel » Mon Mar 17, 2008 1:11 am
II 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

I am more interested in your logic than the final answer. Thanks in advance.
This is an excellent question to understand the relation between Distance, Speed and Time.

The Equation used is D=S*T .. D=Distance, S=Speed, T= Time

Now, if D is held constant then we get T = D/S or S=D/T, this means that when D is constant Speed and Time are inversely proportional to each other.

So, if T increases by a certain amount then S has to decrease by the same amount and vice versa.

According to the first statement, if the speed increases by 50% (it becomes 1.5 times the original speed) then the time taken would be 2 hrs. But remember that if speed increases by 50% and the distance traveled is constant then the time should reduce by 50%. This means that the original time taken is 4 hours. So this statement is sufficient.

Clearly B is not the answer

So answer should be A

Regards
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by tmmyc » Tue Mar 18, 2008 10:01 am
We know the distance from Betty's home to Denver is equal to her rate times her time.

d= r*t

The question wants us to find t.


Statement 1: If her rate was 1.5 times as fast, then her time would be 2 hours.

d = [1.5*r]*(2)

The distance between her home and Denver didn't change, so the two d's are actually the same. Combine the two equations:

d= r*t
d = [1.5*r]*(2)

so

r*t = (1.5)r*2
t = 3
Sufficient

Statement 2: Her rate was 50 mph.

d=50*t

From this we are still unable to find t.


Hence, the answer is A.
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Is time = 3 really sufficient?

by isaaclin » Tue Jul 29, 2008 3:28 pm
But can't t be any multiple of 3 -- like 6, 9, 12...? The question didn't ask whether Betty's time can be determined. Instead, it asked, "How long did it take..." The question seems to be asking for an exact number.

Shouldn't then (1) be insufficient?

Thanks
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Re: Rates: How long did it take Betty to drive nonstop on a

by rolodex » Tue Jul 29, 2008 10:06 pm
II 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

I am more interested in your logic than the final answer. Thanks in advance.
T = D/S
S1. D/1.5S = T-2
AD out; 2 eq three unknowns


S2. S = 50
T = D/50 B OUT

COMBINE: 50T = D; 1.5S = 75

50T = 75T - 150
25T = 150; T = 6
SO IMO C

HOWEVER: IF The statement did not tell you whether Betty stopped anywhere so you have to take the answer as E
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by fajoni » Wed Jul 30, 2008 4:33 pm
This question comes from the GMAT Quant Review Book (it's Q# 69.)

There's a good explination in the book.

The question is asking you to solve for Time. Therefore, the form of the equation you need to use is T=(D/S)

Statement one tells you that if S is increased by 1.5, then T will be 2hrs. This can be translated and put into the formula: 2=(D/1.5S)

From here, we can manipulate the equation to read 3S=D. Plugging 3S in for D in the first equation yields T=(3S/S). Thus, the S's cancel out and T=3hrs.

Statement one is sufficient and the answer (the OA from the GMAT Quant. book) is A.
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Re: Distance / Rates

by II » Fri Nov 07, 2008 5:30 pm
gabriel wrote: This is a excellent question to understand the relation between Distance, Speed and Time.

The Equation used is D=S*T .. D=Distance, S=Speed, T= Time

Now, if D is held constant then we get T = D/S or S=D/T, this means that when D is constant Speed and Time are inversely proportional to each other.

So, if T increases by a certain amount then S has to decrease by the same amount and vice versa.

According to the first statement, if the speed increases by 50% (it becomes 1.5 times the original speed) then the time taken would be 2 hrs. But remember that if speed increases by 50% and the distance traveled is constant then the time should reduce by 50%. This means that the original time taken is 4 hours. So this statement is sufficient.

Clearly B is not the answer

So answer should be A
Regards
Thanks Gabriel ... you are right ... I think this is the key learning from this question. If the Distance is kept constant, and if you increas one of the other components (eg rate), then the time has to decrease (is inversely proportional). And vica versa ... if the time had increased, then the rate would have had to decrease.

I think the other key take-away here is that you when you look at statement 1 you have can rewrite the information into an equation:
1.5*r*2 = d
we already have r*t=d from the question stem. Since d is the same, we end up with 2 equations and 2 unknowns (r and t). We have to find t, and hence can solve for t ! So sufficient ... well explained above by tmmyc !
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Re: Rates: How long did it take Betty to drive nonstop on a

by logitech » Fri Nov 07, 2008 5:57 pm
II 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

I am more interested in your logic than the final answer. Thanks in advance.
Let's keep it simple:

If you walk home everyday, you know how long it will take you to get home. T minutes

If you walk 2 times faster ? You will walk home in T/2 minutes

As long as the distance is SAME

Distance = Distance

V1.T1=V2T2

So T2= T1 ( V1/V2)
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Re: Rates: How long did it take Betty to drive nonstop on a

by II » Sat Nov 08, 2008 11:05 am
logitech wrote: Let's keep it simple:

If you walk home everyday, you know how long it will take you to get home. T minutes

If you walk 2 times faster ? You will walk home in T/2 minutes

As long as the distance is SAME

Distance = Distance

V1.T1=V2T2

So T2= T1 ( V1/V2)
Hey ... love the thinking ! keep it simple !

By the way, I was in Vegas a few weeks ago ... great city ! :-)
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by missrochelle » Sat Aug 28, 2010 11:13 am
tmmyc wrote:

Statement 1: If her rate was 1.5 times as fast, then her time would be 2 hours.

d = [1.5*r]*(2)

The distance between her home and Denver didn't change, so the two d's are actually the same. Combine the two equations:

d= r*t
d = [1.5*r]*(2)

so

r*t = (1.5)r*2
t = 3
Sufficient

Hence, the answer is A.

I don't understand how we are able to get rid of R in that equation, even if you set them equal you have 2 unknowns.

RT=1.5R(2)
How is it that you are able to isolate T?


Also, if you follow this logically -- if rate goes up by 3/2, then time goes down by 2/3.... So the 2 hour time is 2/3 less than what it wouldve been.... That gives me T=6. What am I missing in this logic?
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by diebeatsthegmat » Sat Aug 28, 2010 12:27 pm
II 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

I am more interested in your logic than the final answer. Thanks in advance.
the answer is A, isnt it?

from 2 we can say nothing cos we dont know the distance
from 1 we know D=S*T and D=1.5S*2=3S thus T=3
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by siddhans » Mon Oct 03, 2011 9:33 pm
tmmyc wrote:We know the distance from Betty's home to Denver is equal to her rate times her time.

d= r*t

The question wants us to find t.


Statement 1: If her rate was 1.5 times as fast, then her time would be 2 hours.

d = [1.5*r]*(2)

The distance between her home and Denver didn't change, so the two d's are actually the same. Combine the two equations:

d= r*t
d = [1.5*r]*(2)

so

r*t = (1.5)r*2
t = 3
Sufficient

Statement 2: Her rate was 50 mph.

d=50*t

From this we are still unable to find t.


Hence, the answer is A.
St1 says : If her rate was 1.5 times as fast, then her time would be 2 hours.

So doesnt it means r = r +1.5 and not just 1.5 r??
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by sl750 » Tue Oct 04, 2011 5:34 am
siddhans wrote:
tmmyc wrote:We know the distance from Betty's home to Denver is equal to her rate times her time.

d= r*t

The question wants us to find t.


Statement 1: If her rate was 1.5 times as fast, then her time would be 2 hours.

d = [1.5*r]*(2)

The distance between her home and Denver didn't change, so the two d's are actually the same. Combine the two equations:

d= r*t
d = [1.5*r]*(2)

so

r*t = (1.5)r*2
t = 3
Sufficient

Statement 2: Her rate was 50 mph.

d=50*t

From this we are still unable to find t.


Hence, the answer is A.
St1 says : If her rate was 1.5 times as fast, then her time would be 2 hours.

So doesnt it means r = r +1.5 and not just 1.5 r??
That is 3/2 times, which indicates multiplication. Hence it is 3/2*r
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by siddhans » Wed Oct 26, 2011 10:39 pm
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


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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by GMATGuruNY » Thu Oct 27, 2011 3:47 am
II 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.

I am more interested in your logic than the final answer. Thanks in advance.
2 times as fast = 1/2 the time.
3 times as fast = 1/3 the time.
Note the RECIPROCAL relationship:
x times as fast = 1/x the time.

To illustrate:
Let d = 60 miles and r = 2 miles per hour.
t = 60/2 = 30 hours.
2 times as fast = 4 miles per hour.
t = 60/4 = 15 hours, 1/2 the regular time.
3 times as fast = 6 miles per hour.
t = 60/6 = 10 hours, 1/3 the regular time.

Statement 1: If Betty's average speed for the trip had been 1.5 times as fast, the trip would have taken 2 hours.
3/2 times as fast = 2/3 the time.
Thus:
2/3t = 2.
t = 3.
SUFFICIENT.

Statement 2: Betty's average speed for the trip was 50 miles per hour.
Without knowing the distance, we cannot determine the time.
INSUFFICIENT.

The correct answer is A.
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