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A train always travels at one of two speeds: 160 km/hr in

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A train always travels at one of two speeds: 160 km/hr in [#permalink] New post 08 Oct 2007, 10:55
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D
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A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Was its average speed from A to B greater than 100 km/hr?

(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance from A to B is more than 1000 km.

Kevin Armstrong
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Re: DS: Train [#permalink] New post 08 Oct 2007, 11:17
kevincan wrote:
A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Was its average speed from A to B greater than 100 km/hr?

(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance from A to B is more than 1000 km.

Kevin Armstrong
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Ans is A.

We can plug in #'s or solve algebraically.

I chose to plug in w/ this choice.

Avg R: tot Distance/tot Time

Say Distance is 120km.
Rural || Urban
R 160 || 40
T 1/2hr || 1hr
D 80 || 40

Rural distance is 80 b/c 2/3*120=80. (same reasoning for Urban)

So we have 120/(1/2hr+1hr) --> 120/3/2Hrs ---> 120*2/3 ->80kph

So NO the train did not exceed an average speed of 100kmph

Just looking at B it says the distance exceeds 1000km. If you don't feel comfortable w/ the above. Just try 1200km. You will see the answer is still 80kmph
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Re: DS: Train [#permalink] New post 08 Oct 2007, 11:38
kevincan wrote:
A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Was its average speed from A to B greater than 100 km/hr?

(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance from A to B is more than 1000 km.

Kevin Armstrong
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Stmtn 1 -
if total is assumed as 1 , we can find the average speed .
Comes to 80 miles/hr.

Stmtn2 - redundant information . need to know the break up of time or distance.

A
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Re: DS: Train [#permalink] New post 08 Oct 2007, 11:42
Hi Kevin,
Have you recently join Manhattan Review?
Congrats/goodluck..


kevincan wrote:
A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Was its average speed from A to B greater than 100 km/hr?

(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance from A to B is more than 1000 km.

Kevin Armstrong
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A. (160 x 2/3) + (40 x 1/3) = 360/3 = 120

so the average speed is al least 120 km/h
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 [#permalink] New post 08 Oct 2007, 11:47
Ans is A.

Logic of the question suggests that we need only know what fraction of the entire trip was rural vs not, in order to find the actual time.

1st premise provides us exactly this information.

2nd premise is a dud.

By the way, this is a great candidate for time saving by avoiding unnecessary math. If you can visualize this problem, you need not write any math down to solve it -- give yourself more time for the other calc intensive problems down the road.

Another GMAC favorite in this genre is "did the train ever exceed speed X during such and such trip?". Well, maybe for a split second the train gassed it up and went super duper fast, while noone was looking... the answer to this question is usually E, because you usually cannot know if there was a split second surge in speed while the rest of the trip was normal.
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Re: DS: Train [#permalink] New post 08 Oct 2007, 12:57
kevincan wrote:
A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Was its average speed from A to B greater than 100 km/hr?

(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance from A to B is more than 1000 km.

Kevin Armstrong
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A. If travel was 2/3 in rural, then average speed would be 80 km/h. BUT, travel is more than 2/3 in rural, so average speed is > 80 km/h
INSUFF

B. Total distance doesn't matter...INSUFF

Combining A & B doesn't help as we don't know how long train travelled in rural/urban area.

Ans E.
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 [#permalink] New post 08 Oct 2007, 13:48
from basics, what i know is that in order to find the average speed we need to find the distance and time only.
we do not have the time it took the train to travel with different rates of speed so. we can find the time using both of the statements.
i think it is C. speed is >40 mp/hour
will post the explanation if it is the correct answer.
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Re: DS: Train [#permalink] New post 08 Oct 2007, 13:53
kevincan wrote:
A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Was its average speed from A to B greater than 100 km/hr?

(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance from A to B is more than 1000 km.

Kevin Armstrong
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hmmm.........
the question doesnot seem that much difficult but i can guess a qusestion from him is not too simple as well.
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Re: DS: Train [#permalink] New post 08 Oct 2007, 13:54
Fistail wrote:
kevincan wrote:
A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Was its average speed from A to B greater than 100 km/hr?

(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance from A to B is more than 1000 km.

Kevin Armstrong
Manhattan Review


hmmm.........
the question doesnot seem that much difficult but i can guess a qusestion from him is not too simple as well.


kevin doesnt post simple questions?! :)
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Re: DS: Train [#permalink] New post 09 Oct 2007, 04:27
[quote="kevincan"]OA is not A[/quote]

The answer has to be A - the OA is wrong.

The minimum avg speed in first case is 120km/h = (160x(2/3) + 40x(1/3)

Since minimum is 120, it answers the question.

Answer is A.

I've no clue how most of you got 80 km/h
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Re: DS: Train [#permalink] New post 09 Oct 2007, 04:55
stopper5 wrote:
kevincan wrote:
OA is not A


The answer has to be A - the OA is wrong.

The minimum avg speed in first case is 120km/h = (160x(2/3) + 40x(1/3)

Since minimum is 120, it answers the question.

Answer is A.

I've no clue how most of you got 80 km/h


Ans. is E

From stat 1 we can get the min speed av but setting the distance to 2/3 when the train travels at 160 Km/h.

this gives us an av. speed of 80 Km/h; indeed av speed = distance/time.
= d / [(2d/3x160) + (d/3x40)]

Max speed is when the train travels all the time in rural area => speed = 160.

Thus stat. 1 is not sufficient

Stat. is not sufficient as well.
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Re: DS: Train [#permalink] New post 09 Oct 2007, 06:18
stopper5 wrote:
kevincan wrote:
OA is not A


The answer has to be A - the OA is wrong.

The minimum avg speed in first case is 120km/h = (160x(2/3) + 40x(1/3)

Since minimum is 120, it answers the question.

Answer is A.

I've no clue how most of you got 80 km/h


I love your confidence, but the weighted average should be in terms of time, not distance! Please don't forget this!
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Re: DS: Train [#permalink] New post 09 Oct 2007, 06:33
kevincan wrote:
stopper5 wrote:
kevincan wrote:
OA is not A


The answer has to be A - the OA is wrong.

The minimum avg speed in first case is 120km/h = (160x(2/3) + 40x(1/3)

Since minimum is 120, it answers the question.

Answer is A.

I've no clue how most of you got 80 km/h


I love your confidence, but the weighted average should be in terms of time, not distance! Please don't forget this!



Somebody is too confident here :)

distance = time x speed (dont forget that)

so av. speed = total distance / total time (what I have calculated)

however 160 x (2/3) doesn't mean anything since it's speed x distance ...
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 [#permalink] New post 09 Oct 2007, 06:45
kevincan wrote:
OA is not A


Kevin can you give OA as mine is E but really not sure
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Re: DS: Train [#permalink] New post 09 Oct 2007, 06:56
kevincan wrote:
A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Was its average speed from A to B greater than 100 km/hr?

(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance from A to B is more than 1000 km.

Kevin Armstrong
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Let's do it step-by-step.

What is average speed? It is the total distance travelled divided by the total time. If we imagine that a car traveled half the distance with the speed 50 mph, and the other half with the speed 100 mph, how to get the average speed?

Let's put 40 miles as the total distance. Hence, 20 miles is 1/2 of it.
The first half was covered in 0.4 of an hour (i.e. in 24 min.) The second one - 0.2 of an hour (12 min.). Therefore, the total time equals 0.6 of an hour, and the total distance is 40 miles.

As a result, the average speed is 40 miles/0.6 of an hour, the average speed is about 66.6(6) mph.

Now let's look at the problem. If we know that more than 2/3 of the distance is made with the speed 160, let's just put some numbers.

Distance = 300. 200 with 160mph, 100 - 40 mph. The average speed is MORE than 80 mph (it would be 80 if EXACTLY 2/3 of the route is covered with superspeed :) ) We don't know, how much is the exact number. It can be greater then 100, if, for example, the train travel 295 "rural" miles and 5 urban (average speed equals 155 mph). 295 out of 300 miles is still "more than 2/3 of the distance".

As a result, A is useless. IF exactly 2/3 of the distance mean more than 100 mph, anyway, then we would use it. However, it does not.

Finally, B is useless, since the average speed is not bound to the total distance directly, but through the time spent.

My answer is E.

Please, fell free to correct any of my grammar mistakes. I am Russian, and for the moment I am fiercely trying to remember anything about English and its grammar.
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 [#permalink] New post 09 Oct 2007, 08:35
Perhaps the nicest way is to see what the average speed would be if fraction f of the trip were through rural areas.
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Re: DS: train [#permalink] New post 09 Oct 2007, 08:43
[quote="kevincan"]Perhaps the nicest way is to see what the average speed would be if fraction f of the trip were through rural areas.[/quote]


ooops! sorry guys :)... My brain short circuited!
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Re: DS: Train [#permalink] New post 09 Oct 2007, 09:44
kevincan wrote:
A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Was its average speed from A to B greater than 100 km/hr?

(1) More than 2/3 of the distance from A to B is through rural areas
(2) The distance from A to B is more than 1000 km


I guess i got it. Suppose the distance is 1200km.
rural area = 80% of 1200km = 960km
no of hours = 6 hours
urban area = 20% of 1200km = 240km
no of hours = 6 hours
so the average speed = (960+240)/12 = 100 km/h

Suppose: the rural area = 90% of 1200km = 1080 km
no of hours = 1080/160 hours
urban area = 10% of 1200 km = 120 km
no of hours = 3 hours

so the average speed = (1080 + 120)/(1080/160 + 12) => 100 km/h

so E.
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Re: DS: Train [#permalink] New post 31 Oct 2007, 03:47
get clear E for this question,

Average speed= total distance/ total time=D/T
D=dr+du
T=total time=time in rural area+time in urban area
T=dr/40+du/160

A=D/(dr/40+du/160)

A=D*160*40/ (dr*40+du*160)=D*160*40/(dr+4*du)*40

thus

A=D*160/(dr+4*du)

(1) INSUFF

(2/3)D<dr, then (1/3)D>du, for convenience lets assume dr=(2/3)D and du=(1/3)D

A=D*160/((2/3)D+(1/3)*4*D)
A=160/(2/3+4*1/3) --------------------> whatever the total distance is THIS will be the equation for average speed

160/(2/3+4*1/3) =~27, to cancel previous assumption, lets calculate
160/(0,70+0,3*4)=~84

thus Average speed>27------------> average speed could be more than 100, could be less.

(2) INSUFF

obviously insuff

both) INSUFF

A=160/(2/3+4*1/3) --------------------> whatever the total distance is THIS will be the equation for average speed, thus yeilds the same result as A

calculations is a must for tough DS...
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Re: DS: Train [#permalink] New post 31 Oct 2007, 09:14
kevincan wrote:
A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Was its average speed from A to B greater than 100 km/hr?

(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance from A to B is more than 1000 km.

Kevin Armstrong
Manhattan Review


Suppose f is the fraction of the total distance d that was through rural areas. Total time is fd/160 + (1-f)d/40 = (fd+4(1-f)d)/160

Average speed = 160 /(f+4(1-f))= 160/(4 -3f)
For this to be greater than 100, (4-3f)/8 > 1/5

4-3f > 8/5
3f > 12/5
f > 4/5
Re: DS: Train   [#permalink] 31 Oct 2007, 09:14
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