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Re D0124 [#permalink]
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16 Sep 2014, 00:12
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Official Solution:A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\). The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point \(P\), what is the distance between the two cities? A. 48 B. 72 C. 96 D. 120 E. 192 The buses travel at the same constant speed. It would take one bus to travel 4 hours to cover the distance between the cities \(M\) and \(N\) (two buses drove for 2 hours each). We need to find the speed of the bus. If the first bus was delayed by 24 minutes and the second one left 36 minutes earlier, it makes the second bus \(24+36=60\) minutes ahead of the first bus. The meeting point was 24 miles away on this second day. We know the distance difference between the two meeting points, but we also need to find difference in time those 24 miles were covered. If the second bus drove for 1 hour before the first one departed, each of them had to go for another 1.5 hour to meet (1.5 hour + 1.5 hour + 1 hour). The second bus traveled for 2.5 hours and the first one for 1.5 hour. Therefore the meeting point on the second day was 30 minutes away from that of the previous day. So the second bus covered 24 miles in 30 minutes, which gives us the speed of the bus, 48 mph. We can calculate the distance as we already know the speed: \(4*48=192\) miles. Alternative Explanation Say the distance between the cities is \(d\) miles. Since both buses travel at the same constant speed and leave the cities at the same time then they meet at the halfway, so the first meeting point \(P\), is \(\frac{d}{2}\) miles away from \(M\) (and \(N\)). Next, since the buses meet in 2 hours then the total time to cover \(d\) miles for each bus is 4 hours. Now, on the second day one bus traveled alone for 1 hour (36min +24min), hence covered \(0.25d\) miles, and \(0.75d\) miles is left to cover. The buses meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\): Hence, \(\frac{d}{2}24=\frac{0.75d}{2}\), which gives \(d=192\) miles. Answer: E
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Re: D0124 [#permalink]
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23 Dec 2014, 16:30
Bunuel wrote: Official Solution:
A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\). The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point \(P\), what is the distance between the two cities?
A. 48 B. 72 C. 96 D. 120 E. 192
The buses travel at the same constant speed. It would take one bus to travel 4 hours to cover the distance between the cities \(M\) and \(N\) (two buses drove for 2 hours each). We need to find the speed of the bus. If the first bus was delayed by 24 minutes and the second one left 36 minutes earlier, it makes the second bus \(24+36=60\) minutes ahead of the first bus. The meeting point was 24 miles away on this second day. We know the distance difference between the two meeting points, but we also need to find difference in time those 24 miles were covered. If the second bus drove for 1 hour before the first one departed, each of them had to go for another 1.5 hour to meet (1.5 hour + 1.5 hour + 1 hour). The second bus traveled for 2.5 hours and the first one for 1.5 hour. Therefore the meeting point on the second day was 30 minutes away from that of the previous day. So the second bus covered 24 miles in 30 minutes, which gives us the speed of the bus, 48 mph. We can calculate the distance as we already know the speed: \(4*48=192\) miles. Alternative Explanation Say the distance between the cities is \(d\) miles. Since both buses travel at the same constant speed and leave the cities at the same time then they meet at the halfway, so the first meeting point \(P\), is \(\frac{d}{2}\) miles away from \(M\) (and \(N\)). Next, since the buses meet in 2 hours then the total time to cover \(d\) miles for each bus is 4 hours. Now, on the second day one bus traveled alone for 1 hour (36min +24min), hence covered \(0.25d\) miles, and \(0.75d\) miles is left to cover. The buses meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\): Hence, \(\frac{d}{2}24=\frac{0.75d}{2}\), which gives \(d=192\) miles.
Answer: E Hey Bunue, I didn't get why the buses meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\). May u help to explain?



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The buses are travelling at the same speed. One bus is at the point 0.75d when the other bus starts. So, the distance between them is 0.75d and they are travelling at the same speed. Therefore, they will meet halfway through the journey. Hope that helps.



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Re: D0124 [#permalink]
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24 Dec 2014, 13:43
Thanks Kritiu. Finally I've understood.



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Re: D0124 [#permalink]
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25 Dec 2014, 02:09
kritiu wrote: The buses are travelling at the same speed. One bus is at the point 0.75d when the other bus starts. So, the distance between them is 0.75d and they are travelling at the same speed. Therefore, they will meet halfway through the journey. Hope that helps. Thanks kritiu, I've got it!



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Re: D0124 [#permalink]
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26 Dec 2014, 10:40
total distance can be covered in 4 hrs(Given). Let it be x. After one hour: bus A covers x/4 , bus B covers 0 distance, as there is a delay of one hour between the buses. After two hours: bus A covers x/4 + x/4=x/2( bus A is at point p), bus B covered x/4. Note, at this point of time both are equidistant from the point they met on the round trip i.e 24 kms. In next one hour they will cover 48kms. distance covered in one hour=48 kms, distance covered in four hour = 48*4=192kms. **With diagram it becomes easy to understand.



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Re: D0124 [#permalink]
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06 Jan 2015, 11:23
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MN BusA BusB Both at constant equal speed meet at point P which is 2 hrs away so total time is 4 hours for each bus Now on Second day BusA departs 36 mins early and BusB departs 24 minutes late that means when A traveled for 1 hour, B just left N 1hr. MN A B
Now time left is 3 hrs. Both travel at same constant speed so they will meet after 1.5 hrs In this time A has traveled for 2.5 hours (.5 hour more than yesterday which represents 24miles.) So 4 hours = .5 * 8 is equal to distance of 24*8 = 192miles



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Re D0124 [#permalink]
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19 Feb 2015, 00:09
I think this question is good and helpful.



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Re: D0124 [#permalink]
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19 Feb 2015, 06:18
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its given that 1 bus starts 36 minutes late and the other 24 minutes early which means effectively 1 bus had head start of 1 hr. Since they are traveling at same speed and they meet 24 miles from where they met yesterday then the bus which had head start of 1 hr traveled 24 miles in 1 hr. Thus its speed is 24 miles. The speed of 2nd bus is also 24 (since both bus travels at constant speed).. Thus both bus travels for 4 hrs hence distance traveled by 1 bus =24*4
Total distance= 2*24*4= 192.. Answer



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Re: D0124 [#permalink]
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24 Feb 2015, 20:12
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Bunuel wrote: A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\). The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point \(P\), what is the distance between the two cities?
A. 48 B. 72 C. 96 D. 120 E. 192 Bunuel, This question is quite confusing. Can you explain your logic as to why you thought the busses were going in the same direction on the return trip? The question says "The following day the buses do the return trip at the same constant speed." Well if a bus leaves City M to go to City N and another leaves City N to go to City M, the "return trip" must mean that each bus returns to the point of origin; City M and City N respectively, and if that happens after meeting between City M and City N, they can never meet again, which would make this problem impossible to solve. Sorry, I am just so confused..



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Re D0124 [#permalink]
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19 Mar 2015, 02:12
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I think this question is good and helpful. However," The following day the buses do the return trip at the same constant speed." is confusing as they go in opposite directions so they do not meet. I think rewording is necessary.



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Re D0124 [#permalink]
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19 Mar 2015, 02:13
I think this question is good and helpful. However," The following day the buses do the return trip at the same constant speed." is confusing as they go in opposite directions so they do not meet. I think rewording is necessary.



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Re: D0124 [#permalink]
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21 May 2015, 12:51
Agree with above. Needs minor rewording. As is, one assumes they spent the night in P and then set off the next day in opposite directions.



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Re: D0124 [#permalink]
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21 Sep 2015, 04:43
Bunuel wrote: Official Solution:
A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\). The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point \(P\), what is the distance between the two cities?
A. 48 B. 72 C. 96 D. 120 E. 192
The buses travel at the same constant speed. It would take one bus to travel 4 hours to cover the distance between the cities \(M\) and \(N\) (two buses drove for 2 hours each). We need to find the speed of the bus. If the first bus was delayed by 24 minutes and the second one left 36 minutes earlier, it makes the second bus \(24+36=60\) minutes ahead of the first bus. The meeting point was 24 miles away on this second day. We know the distance difference between the two meeting points, but we also need to find difference in time those 24 miles were covered. If the second bus drove for 1 hour before the first one departed, each of them had to go for another 1.5 hour to meet (1.5 hour + 1.5 hour + 1 hour). The second bus traveled for 2.5 hours and the first one for 1.5 hour. Therefore the meeting point on the second day was 30 minutes away from that of the previous day. So the second bus covered 24 miles in 30 minutes, which gives us the speed of the bus, 48 mph. We can calculate the distance as we already know the speed: \(4*48=192\) miles. Alternative Explanation Say the distance between the cities is \(d\) miles. Since both buses travel at the same constant speed and leave the cities at the same time then they meet at the halfway, so the first meeting point \(P\), is \(\frac{d}{2}\) miles away from \(M\) (and \(N\)). Next, since the buses meet in 2 hours then the total time to cover \(d\) miles for each bus is 4 hours. Now, on the second day one bus traveled alone for 1 hour (36min +24min), hence covered \(0.25d\) miles, and \(0.75d\) miles is left to cover. The buses meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\): Hence, \(\frac{d}{2}24=\frac{0.75d}{2}\), which gives \(d=192\) miles.
Answer: E Sorry, I am still unable to understand direction of the travel. In beginning both buses move towards each other and meet at point P In return journey they move away from each other...so I am not getting how they meet again? I think I completely missed what question is talking about.



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Re: D0124 [#permalink]
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21 Sep 2015, 08:23
anupamadw wrote: Bunuel wrote: Official Solution:
A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\). The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point \(P\), what is the distance between the two cities?
A. 48 B. 72 C. 96 D. 120 E. 192
The buses travel at the same constant speed. It would take one bus to travel 4 hours to cover the distance between the cities \(M\) and \(N\) (two buses drove for 2 hours each). We need to find the speed of the bus. If the first bus was delayed by 24 minutes and the second one left 36 minutes earlier, it makes the second bus \(24+36=60\) minutes ahead of the first bus. The meeting point was 24 miles away on this second day. We know the distance difference between the two meeting points, but we also need to find difference in time those 24 miles were covered. If the second bus drove for 1 hour before the first one departed, each of them had to go for another 1.5 hour to meet (1.5 hour + 1.5 hour + 1 hour). The second bus traveled for 2.5 hours and the first one for 1.5 hour. Therefore the meeting point on the second day was 30 minutes away from that of the previous day. So the second bus covered 24 miles in 30 minutes, which gives us the speed of the bus, 48 mph. We can calculate the distance as we already know the speed: \(4*48=192\) miles. Alternative Explanation Say the distance between the cities is \(d\) miles. Since both buses travel at the same constant speed and leave the cities at the same time then they meet at the halfway, so the first meeting point \(P\), is \(\frac{d}{2}\) miles away from \(M\) (and \(N\)). Next, since the buses meet in 2 hours then the total time to cover \(d\) miles for each bus is 4 hours. Now, on the second day one bus traveled alone for 1 hour (36min +24min), hence covered \(0.25d\) miles, and \(0.75d\) miles is left to cover. The buses meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\): Hence, \(\frac{d}{2}24=\frac{0.75d}{2}\), which gives \(d=192\) miles.
Answer: E Sorry, I am still unable to understand direction of the travel. In beginning both buses move towards each other and meet at point P In return journey they move away from each other...so I am not getting how they meet again? I think I completely missed what question is talking about. I hope alternative solutions help: abusfromcitymistravelingtocitynataconstantspeed86478.html
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Re: D0124 [#permalink]
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17 Oct 2015, 13:45
Is this a question from OG?



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Re: D0124 [#permalink]
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18 Oct 2015, 11:12



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Re: D0124 [#permalink]
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08 Dec 2015, 19:25
Alternative solution: as 2 buses meat at the middle by driving 2 hours each. It means that the whole time each buses spends is 4 hours. If one bus delayed 24 min and the other leaves 36 min earlier, it means that the 2nd bus drives 1 hour alone. In 1 hour it will cover 1/4 of the distance. Then the rest of the distance 2 buses travel together and they will meet in the middle of the rest distance, thus 11/4=3/4 middle is 3/4/2=3/8. The distance between two points will be 1/23/8=1/8 which equals to 24 miles. Thus the whole distance is 8*24=192 miles.



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Re D0124 [#permalink]
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15 Dec 2015, 19:46
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I think this is a highquality question and I don't agree with the explanation. In second scenario, when one bus is delayed for 24 minutes and another bus leaves 36 minutes , it is clear that one bus has traveled for one hour extra before they meet. Had they traveled for same time they would meet at P but they meet 24 miles from point P. This means this distance is traveled in one hour by one of the bus that has traveled one hour more than other by the time they meet. So speed of bus would be 24 miles per hour. since it takes 4 hours so distance would be 96 miles.







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