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If a bus took 4 hours to get from town a to town b, what was

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If a bus took 4 hours to get from town a to town b, what was [#permalink]

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26 Apr 2007, 22:46
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If a bus took 4 hours to get from town a to town b, what was avg speed of bus for the trip?
1. in first 2 hours bus covered 100 miles.
2. avg speed of bus for first half of the distance was twice it's speed for the second half.

Let me know what you guys think, OA to follow.
TIA.
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26 Apr 2007, 23:38
ayl989 wrote:
If a bus took 4 hours to get from town a to town b, what was avg speed of bus for the trip?
1. in first 2 hours bus covered 100 miles.
2. avg speed of bus for first half of the distance was twice it's speed for the second half.

Let me know what you guys think, OA to follow.
TIA.

I think the asnwer is (C) - let me explain

dividing the trip into 2 parts where x = total road and y = speed

first part

100 miles = distance , 2y = speed , 100/2y = time

second part

x-100 miles = distance , y = speed , x-100/y = time

Statement 1

100/2y = 2
100 = 4y
y = 25

but we don't know what x is - and we can't find the average for the whole trip.

Insufficient

Statement 2

(100/2y) + (x-100)/y = 4
100/2y + (x-100)/25 = 4
50/y + (x-100)/25 = 4

we are still left with x,y

Insufficient

Statements 1&2

Only when we use statement 1 to find y and statement 2 to find x
we can calculate the average speed for the trip

50/y + (x-100)/25 = 4
50/25 + (x-100)/25 = 4
2*25 + (x-100) = 100
x = 150

the average for the trip is

(50*2+25*2)/4 = 37.5 mph

Sufficient - the answer is (C)
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27 Apr 2007, 09:27
I got the same answer, C, but a slightly different way....

I used the simple d=rt equation.

[1] d=100 for the first 2 hrs --> therefore, 100=r2 --> r=50mph

However, we don't know whether the rate was constant for the second 2 hours and we don't know the distance for the second 2 hrs. Therefore, insufficient.

[2] this info just tells us that the rate (d/t) for the first 2 hours was twice the second 2 hours --> d/t=2d/t Insufficient.

[1]+[2] putting both statements together gives us the following equation:

50=2d/t

This mean the rate of the first 2 hours was 50 and the rate of the second 2 hours was 25.

The average of these is 75/2=37.5

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27 Apr 2007, 11:26
OA is E. I got this problem from challenge 16 question 8.
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28 Apr 2007, 13:56
ayl989 wrote:
OA is E. I got this problem from challenge 16 question 8.

Hmm.. Could you explain E? I got C as well..
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28 Apr 2007, 14:14
For the whole trip, let avgV = average velocity; T = total time; D = total distance.

1. In the 1st 2 hs, 100 mi. We donÂ´t know what happened in the following (T-2) hs. Insuff => B, C, or E.

2. First half: T1 = (D/2)/2V = D/4V. Second half: T2 = (D/2)/V = D/2V. T = 3D/4V.
avgV = D/(3D/4V) = 4V/3. Insuff => C or E.

1&2. (1) only gives info on the distance covered for the 1st 2 hs; having this at hand actually doesnÂ´t help with what weÂ´ve got in (2), as what we are looking for, avgV, is in function of the average speed of each half distance of the trip. We still cannot find V, or T, or D, so the answer should be E.

ayl989 wrote:
If a bus took 4 hours to get from town a to town b, what was avg speed of bus for the trip?
1. in first 2 hours bus covered 100 miles.
2. avg speed of bus for first half of the distance was twice it's speed for the second half.
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28 Apr 2007, 14:33

Statement ! - Insufficient... just tells us about av speed of the first part of the trip . Distance unknown

Statement 2- Insufficient
by calculating you get a equation where X= 4Y/3, Where X= total distance and Y= speed in the second part of the journey.

together statement 1&2 still insufficient as
100 miles might not be the first half of the trip.

Considering 100 miles to be the first half of the distance, makes you arrive at a wrong answer
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28 Apr 2007, 21:59
forgmat wrote:

Statement ! - Insufficient... just tells us about av speed of the first part of the trip . Distance unknown

Statement 2- Insufficient
by calculating you get a equation where X= 4Y/3, Where X= total distance and Y= speed in the second part of the journey.

together statement 1&2 still insufficient as
100 miles might not be the first half of the trip.

Considering 100 miles to be the first half of the distance, makes you arrive at a wrong answer

100 miles is the first part of the trip not the first half - see my earlier post - could anyone go over my solution and tell me where is the mistake ?
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29 Apr 2007, 07:16
boy..i went for C too but then..i realized..the first 2 hours are not the same as the first HALF of the trip..we dont know if the first 2 hours are the first half or not..

say..the first 2 hours the bus goes 50 mph to 100 miles..but then got a jet engine-turbo bost turned on and went at 200 mph for the remaining 2 hours.. which means the bus covered 400 miles in the next 2 hours..which means the first HALF (i.e distance) is 250 mile marker..
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29 Apr 2007, 08:14
KillerSquirrel wrote:

100 miles is the first part of the trip not the first half - see my earlier post - could anyone go over my solution and tell me where is the mistake ?

sorry if it was not clear. I wanted to say that as per statement 2 -avg speed of bus for first half of the distance was twice it's speed for the second half.

you have chosen for the first part and not the first half of the jouney, the av speed to be 2y , which is not correct.
( as stated in st 2). the av speed is 2y only when the distance covered is half . and here we are not sure that the distance "100" miles covered in 2hrs is half the distance.

I hope its clear now
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29 Apr 2007, 09:20
forgmat wrote:
KillerSquirrel wrote:

100 miles is the first part of the trip not the first half - see my earlier post - could anyone go over my solution and tell me where is the mistake ?

sorry if it was not clear. I wanted to say that as per statement 2 -avg speed of bus for first half of the distance was twice it's speed for the second half.

you have chosen for the first part and not the first half of the jouney, the av speed to be 2y , which is not correct.
( as stated in st 2). the av speed is 2y only when the distance covered is half . and here we are not sure that the distance "100" miles covered in 2hrs is half the distance.

I hope its clear now

clear , thanks !
29 Apr 2007, 09:20
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