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It took the bus 4 hours to get from town A to town B. What [#permalink]
07 Sep 2008, 22:43

It took the bus 4 hours to get from town A to town B. What was the average speed of the bus for the trip? 1. In the first 2 hours the bus covered 100 miles 2. The average speed of the bus for the first half of the distance was twice its speed for the second half

Re: speed and distance [#permalink]
08 Sep 2008, 12:49

It should be C.

Statement 2: Let the speed for the 1st half distance d be S1 and that for the 2nd half distance d be S2. S1 = d/t and S2 = d/2t where t is the time taken to travel each half. Therefore total time taken = 3t. 3t=4 hrs. t = 4/3hrs. From St. 1 we know that the bus travels 100 miles in 2 hrs. For 1+1/3 hr the bus travels at a certain speed and for 2-4/3=2/3 hrs, the bus travels at 1/2 that speed (by St. 2). The bus covers distance d in 4/3 hrs. It would cover d/2 in the next 4/3 hrs.Therefore it would cover d/4 in the next 2/3 hrs. Therefore, total distance covered = d + d/4 = 5d/4 = 100 Hence d = 100*4/5 = 80 miles. Therefore, total distance = 80 + 80 = 160 miles and ave. speed = 160/4 = 40 mph

Re: speed and distance [#permalink]
08 Sep 2008, 13:06

KASSALMD wrote:

It should be C.

Statement 2: Let the speed for the 1st half distance d be S1 and that for the 2nd half distance d be S2. S1 = d/t and S2 = d/2t where t is the time taken to travel each half. Therefore total time taken = 3t. 3t=4 hrs. t = 4/3hrs. From St. 1 we know that the bus travels 100 miles in 2 hrs. For 1+1/3 hr the bus travels at a certain speed and for 2-4/3=2/3 hrs, the bus travels at 1/2 that speed (by St. 2). The bus covers distance d in 4/3 hrs. It would cover d/2 in the next 4/3 hrs.Therefore it would cover d/4 in the next 2/3 hrs. Therefore, total distance covered = d + d/4 = 5d/4 = 100 Hence d = 100*4/5 = 80 miles. Therefore, total distance = 80 + 80 = 160 miles and ave. speed = 160/4 = 40 mph

I don't agree with you . You are assuming that speed is constant which may not be the case.

Will stick with E.

What is OA. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: speed and distance [#permalink]
08 Sep 2008, 13:24

No, I am not assuming the speed to be constant. In fact the speeds are in the ratio of 1:2 AND THAT IS WHAT MAKES IT EASIER TO SOLVE THE PROBLEM. If the ratio of times taken is given and the distance traveled is the same, it is easy to calculate speeds (in this question). According to what you are saying, it will never be possible to calculate exact speed because a speed of 50 mph may or may not be evenly distributed over 60 mins! (even on cruise control, because the speed varies uphill!). You will be assuming that the car moved at 5/6 miles per min. Of course you are assuming that! In my opinion, the purpose of the questioner is to see if information from the 2 statements can be assimilated to derive a LOGICAL answer or not.

Re: speed and distance [#permalink]
08 Sep 2008, 17:40

KASSALMD wrote:

It should be C.

Statement 2: Let the speed for the 1st half distance d be S1 and that for the 2nd half distance d be S2. S1 = d/t and S2 = d/2t where t is the time taken to travel each half. Therefore total time taken = 3t. 3t=4 hrs. t = 4/3hrs. From St. 1 we know that the bus travels 100 miles in 2 hrs. For 1+1/3 hr the bus travels at a certain speed and for 2-4/3=2/3 hrs, the bus travels at 1/2 that speed (by St. 2). The bus covers distance d in 4/3 hrs. It would cover d/2 in the next 4/3 hrs.Therefore it would cover d/4 in the next 2/3 hrs. Therefore, total distance covered = d + d/4 = 5d/4 = 100 Hence d = 100*4/5 = 80 miles. Therefore, total distance = 80 + 80 = 160 miles and ave. speed = 160/4 = 40 mph

the statement in blue is ok but i dont agree with the statement in red suppose that he travels d/6 or d/8 etc etc in the next 4/3 hrs and makes up for it in the next 4/3 hrs so that his avg speed for 8/3 hrs is d/2...then? see, the point here is that his avg speed for 8/3 time of the journy is s/2. there are so many ways to achieve it. what you have considered is ok, BUT ITS NOT THE ONLY POSSIBLITY.There are other ways also of achieving it.

Re: speed and distance [#permalink]
08 Sep 2008, 23:24

Another E here.

Statement 1 is not useful by itself. Statement 2 says that the rates have a ratio of 2:1 and the distances are the same. therefore, the times have a ratio of 1:2. The first half of the distance was covered in 4/3 hours. This is fine, but we are not able to use Statement 1 in any way, because the first 2 hours of the trip correspond to the whole first half of the fistance, and then 2/3 hours of the second half. _________________

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