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HARD DS speed problem [#permalink]
 01 Oct 2007, 22:35
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
Hello to All!
i encountered this very hard DS question on the official GMAC practice test:
DS
Jane walked for 4 miles. What was her average speed for the first 2 miles?
1. Janes average speed for the 4 miles was 3.2 miles per hour.
2. It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles.
The official answer is C.
However, from my readings, i have learned that stopovers can mess up DS speed problems. EX Jane could have had a 5 minute stopover or a 10 minute stopover. And I believe that the answer should be E.
Also, I have read that a test-taker must not assume anything for DS. So, I must not assume that there was no stopover. Or should I say that I cannot assume that there was a stopover? Please help!
I am new to this site, so thank you in advance for the help.
Dan
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Re: HARD DS speed problem [#permalink]
01 Oct 2007, 22:56
danielwaugh wrote: Hello to All!
i encountered this very hard DS question on the official GMAC practice test:
DS
Jane walked for 4 miles. What was her average speed for the first 2 miles?
1. Janes average speed for the 4 miles was 3.2 miles per hour.
2. It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles.
The official answer is C.
However, from my readings, i have learned that stopovers can mess up DS speed problems. EX Jane could have had a 5 minute stopover or a 10 minute stopover. And I believe that the answer should be E.
Also, I have read that a test-taker must not assume anything for DS. So, I must not assume that there was no stopover. Or should I say that I cannot assume that there was a stopover? Please help!
I am new to this site, so thank you in advance for the help.
Dan
I agree the OA should be E, unless something about stopovers is mentioned. In fact, in many distance, speed, time problems it is specificly mentioned that objects did or did not stop. So why this problem should be any exception.
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fyi: going fwd, please let us know when you give the answer (spoil) in the post. some of us like the challenge of trying to solve and then giving feedback. if you choose to take my suggestion please hit the return key a few times before posting the answer.
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Re: HARD DS speed problem [#permalink]
01 Oct 2007, 23:08
danielwaugh wrote: Hello to All!
i encountered this very hard DS question on the official GMAC practice test:
DS
Jane walked for 4 miles. What was her average speed for the first 2 miles?
1. Janes average speed for the 4 miles was 3.2 miles per hour.
2. It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles.
The official answer is C.
However, from my readings, i have learned that stopovers can mess up DS speed problems. EX Jane could have had a 5 minute stopover or a 10 minute stopover. And I believe that the answer should be E.
Also, I have read that a test-taker must not assume anything for DS. So, I must not assume that there was no stopover. Or should I say that I cannot assume that there was a stopover? Please help!
I am new to this site, so thank you in advance for the help.
Dan
Here we have "average speed", so stopover doesn't make any difference. Question is fine IMO.
(1) This gves time taken for whole trip..i.e. 75 min.
INSUFF
(2) x = time taken for first 2 miles
total time take x+x+15
INSUFF
From 1 & 2
2x+15 = 75
=> x = 30 min
Average speed for first 2 miles = 2/(30/60) = 4 miles/hr
SUFF...Ans C.
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according to reviewers, stopovers make a difference. [#permalink]
01 Oct 2007, 23:25
EX
At 7:00am, Claire leaves the spa, driving her car at 80kph she reaches her house at 7:15am, packs her artwork, and at 7:45am, leaves for the design lab, which is 10km from her house. If she drives her car at 40kph in going to the design lab, what will her average speed be from the spa to the lab ?
O 30 kph
O 40 kph
O 53.3 kph
O 60 kph
O 66.7 kph
The answer is below.
AA
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Re: according to reviewers, stopovers make a difference. [#permalink]
01 Oct 2007, 23:31
danielwaugh wrote: EX
At 7:00am, Claire leaves the spa, driving her car at 80kph she reaches her house at 7:15am, packs her artwork, and at 7:45am, leaves for the design lab, which is 10km from her house. If she drives her car at 40kph in going to the design lab, what will her average speed be from the spa to the lab ?
O 30 kph
O 40 kph
O 53.3 kph
O 60 kph
O 66.7 kph
In this case, total time taken = 1 hr
Distance covered = 80 * 1/4 + 10 = 30 km
Avg speed = 30 km/h
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so, it could be like this: [#permalink]
01 Oct 2007, 23:42
Juaz wrote:
Here we have "average speed", so stopover doesn't make any difference. Question is fine IMO.
(1) This gves time taken for whole trip..i.e. 75 min.
INSUFF
(2) x = time taken for first 2 miles
total time take x+x+15
INSUFF
From 1 & 2
2x+15 = 75
=> x = 30 min
Average speed for first 2 miles = 2/(30/60) = 4 miles/hr
SUFF...Ans C.
----------------------------------------------------------------
I agree that 1. gives total time = 75 minutes.
However, 2. can give (x) + (stopover time) + (x+15).
So, combining 1. and 2., you get:
2x + 15 + (stopover time) = 75.
2x + (stopover time) = 60.
This is why I think that the answer should be E. But of course, I may be wrong in my assumptions (should I consider stopovers?)
NOTE
Average speed in traveling from point A to point B is defined to be the total distance from point A to point B, divided by the total time it takes to go from point A to point B, and this means that the stopover time will be important.
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I think the question is fine. we should just be relying on the info given. I won't lie, some of these probs get me too.
back to the orig prob. I was toying around w/this a bit and don't understand why this approach won't work (admittedly, Juaz's is much cleaner and quicker). Can anyone tell me why?
Quote: Jane walked for 4 miles. What was her average speed for the first 2 miles?
1. Janes average speed for the 4 miles was 3.2 miles per hour.
2. It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles.
1) t = 4/3.2
2)
1st leg:
rt = 2
t = 2/r
2nd leg:
r(t+15) = 2
rt + 15r = 2
t = 2 - 15r/r
2 - 15r/r + 2/r = 4/3.2 =
4-15r/r = 4/3.2
I tried solving for r and kept coming up empty. what gives???
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ggarr wrote: I think the question is fine. we should just be relying on the info given. I won't lie, some of these probs get me too. back to the orig prob. I was toying around w/this a bit and don't understand why this approach won't work (admittedly, Juaz's is much cleaner and quicker). Can anyone tell me why? Quote: Jane walked for 4 miles. What was her average speed for the first 2 miles?
1. Janes average speed for the 4 miles was 3.2 miles per hour.
2. It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles. 1) t = 4/3.2 2) 1st leg: rt = 2 t = 2/r 2nd leg: r(t+15) = 2 rt + 15r = 2 t = 2 - 15r/r 2 - 15r/r + 2/r = 4/3.2 = 4-15r/r = 4/3.2 I tried solving for r and kept coming up empty. what gives??? I understand now why stopovers do not count, cuz the stem says "It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles." which means stopovers already excluded, if there were any.
As for the last post, since it is DS, no need for calculations. in fact quite few DS`s have unrealistic answers, if you try to solve them.
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IrinaOK wrote: ggarr wrote: I think the question is fine. we should just be relying on the info given. I won't lie, some of these probs get me too. back to the orig prob. I was toying around w/this a bit and don't understand why this approach won't work (admittedly, Juaz's is much cleaner and quicker). Can anyone tell me why? Quote: Jane walked for 4 miles. What was her average speed for the first 2 miles?
1. Janes average speed for the 4 miles was 3.2 miles per hour.
2. It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles. 1) t = 4/3.2 2) 1st leg: rt = 2 t = 2/r 2nd leg: r(t+15) = 2 rt + 15r = 2 t = 2 - 15r/r 2 - 15r/r + 2/r = 4/3.2 = 4-15r/r = 4/3.2 I tried solving for r and kept coming up empty. what gives??? I understand now why stopovers do not count, cuz the stem says "It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles." which means stopovers already excluded, if there were any. As for the last post, since it is DS, no need for calculations. in fact quite few DS`s have unrealistic answers, if you try to solve them. In principle, I did the same thing that Juaz did (at least I think I did). but his worked. why didn't the above?
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ggarr wrote: IrinaOK wrote: ggarr wrote: I think the question is fine. we should just be relying on the info given. I won't lie, some of these probs get me too. back to the orig prob. I was toying around w/this a bit and don't understand why this approach won't work (admittedly, Juaz's is much cleaner and quicker). Can anyone tell me why? Quote: Jane walked for 4 miles. What was her average speed for the first 2 miles?
1. Janes average speed for the 4 miles was 3.2 miles per hour.
2. It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles. 1) t = 4/3.2 2) 1st leg: rt = 2 t = 2/r 2nd leg: r(t+15) = 2 rt + 15r = 2--------------> t=(2-15r)/rt = 2 - 15r/r 2 - 15r/r + 2/r = 4/3.2 = 4-15r/r = 4/3.2 I tried solving for r and kept coming up empty. what gives??? I understand now why stopovers do not count, cuz the stem says "It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles." which means stopovers already excluded, if there were any. As for the last post, since it is DS, no need for calculations. in fact quite few DS`s have unrealistic answers, if you try to solve them. In principle, I did the same thing that Juaz did (at least I think I did). but his worked. why didn't the above? yep, it is just miscalculation. Look at the text in red.
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and from the OFFICICAL PRACTICE SOFTWARE again [#permalink]
03 Oct 2007, 22:16
Susan drove at an average speed of 30 miles per hour for the first 30 miles of a trip and then at an average speed of 60 miles per hour for the remaining 30 miles of the trip. If she made no stops during the trip, what was Susan's average speed, in miles per hour, for the entire trip?
There is a reason for them to STATE
"If she made no stops during the trip,", right?
I really find it confusing. Could it be that they made a mistake, a typo, if you will, in writing the answer key to the DS item above (choice C)?
Please vote.
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IrinaOK wrote: ggarr wrote: IrinaOK wrote: ggarr wrote: I think the question is fine. we should just be relying on the info given. I won't lie, some of these probs get me too. back to the orig prob. I was toying around w/this a bit and don't understand why this approach won't work (admittedly, Juaz's is much cleaner and quicker). Can anyone tell me why? Quote: Jane walked for 4 miles. What was her average speed for the first 2 miles?
1. Janes average speed for the 4 miles was 3.2 miles per hour.
2. It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles. 1) t = 4/3.2 2) 1st leg: rt = 2 t = 2/r 2nd leg: r(t+15) = 2 rt + 15r = 2--------------> t=(2-15r)/rt = 2 - 15r/r 2 - 15r/r + 2/r = 4/3.2 = 4-15r/r = 4/3.2 I tried solving for r and kept coming up empty. what gives??? I understand now why stopovers do not count, cuz the stem says "It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles." which means stopovers already excluded, if there were any. As for the last post, since it is DS, no need for calculations. in fact quite few DS`s have unrealistic answers, if you try to solve them. In principle, I did the same thing that Juaz did (at least I think I did). but his worked. why didn't the above? yep, it is just miscalculation. Look at the text in red. IrinaOK,
please look at my text below your first edit. isn't that the same as what you did? t = 2 - 15r/r
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close
