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Shawn is planning a bus trip across town

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Shawn is planning a bus trip across town

by gmatdriller » Tue Jul 07, 2015 6:31 am
Shawn is planning a bus trip across town that involves three buses. Bus 1 travels between Shawn's house and downtown, and it leaves every half-hour starting at 7:20 AM. Shawn will need to be on Bus 1 for 1 1/5 hours. Bus 2 travels between downtown and uptown every half-hour starting at 7:10 AM. Shawn will need to be on Bus 2 for 2/3 hour. Lastly, Bus 3 travels between uptown and Shawn's destination every hour starting at 9 AM. Assuming all buses stay on schedule, what is the least amount of time Shawn must spend waiting for buses?

A: 12mins B: 18mins C: 48mins D: 1Hourr 12mins E: 1Hour 20mins
OA is B...Your approach on organizing work/solution on the pad
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by seandcarey » Tue Jul 07, 2015 12:14 pm
I will give my best shot at this one.

The first thing that stands out to me is the third bus. The time in-between departures is the highest at 1 hour compared to 30 minutes for buses 1 and 2. Let's take a look to see what the timing is like if Shawn gets on the first possible bus right at 7:20am.

If Shawn gets on at 7:20 am
Bus 1 - 1 and 1/5 hour = 72 minutes. It is now 8:32 and he is waiting for the second bus.
Bus 2 - The next bus departs at 8:40, so he waits 8 minutes between bus 1 and 2. At 8:40 he gets on bus two and will travel on its for 2/3 an hour = 40 minutes.
Bus 3 - He arrives waiting for bus 3, and it is now 9:20. The next bus is 40 minutes away. So in this case we learned he will waiting 48 minutes if he takes the earliest bus, and there is NO way for him to catch the 9 o'clock bus.

If we think about all of the above, we should work backwards. Since we no 9am is unattainable, and bus 3 has the longest time in-between departures. We should go backwards in order get as close to 10AM as possible. For this I made a little chart.

Bus| departure times | Times Spent on each bus
1 | 7:20 7:50 8:20 8:50 9:20 | 1 1/5 hr = 72 min
2 | 7:10 7:40 8:10 8:40 9:10 | 2/3 hr = 40 min
3 | 9:00 10:00 11:00 | NA

Working backwards now. Its 10am, the first of bus #3 that Shawn can possibly get on. Bus 2 takes 40 mins, so the first bus he can get on has to be before 9:20. The closest departure to 9:20 is 9:10. So there is 10 minutes of wait time. From the 9:10 bus we need to take out his bus 1 travel time to see what time he should get on the first bus. 9:10 - 72 minutes = 7:58am. He needs to get on the first bus before 7:58 to makes his second bus to ultimately make his third bus. He takes the 7:50 #1 bus. There is 8 minutes of wait time.

The grand total of wait time for Shawn is 18 minutes. The answer is B.

Now you might think, well can you shave wait times by doing 11am for the third bus? The answer is no. Since the intervals of B 1 and 2 are the same, and are unchanged throughout the morning, due to symmetry of the times it will always be 18 minutes.

Scarey
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by gmatdriller » Tue Jul 07, 2015 12:16 pm
Any response pls?

Kindly assist with a good layout of how to plan answering this question.

Regards.
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by seandcarey » Tue Jul 07, 2015 12:17 pm
gmatdriller wrote:Any response pls?

Kindly assist with a good layout of how to plan answering this question.

Regards.
See above.
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by Jeff@TargetTestPrep » Wed Nov 08, 2017 4:28 pm
gmatdriller wrote:Shawn is planning a bus trip across town that involves three buses. Bus 1 travels between Shawn's house and downtown, and it leaves every half-hour starting at 7:20 AM. Shawn will need to be on Bus 1 for 1 1/5 hours. Bus 2 travels between downtown and uptown every half-hour starting at 7:10 AM. Shawn will need to be on Bus 2 for 2/3 hour. Lastly, Bus 3 travels between uptown and Shawn's destination every hour starting at 9 AM. Assuming all buses stay on schedule, what is the least amount of time Shawn must spend waiting for buses?

A: 12mins B: 18mins C: 48mins D: 1Hourr 12mins E: 1Hour 20mins
If Shawn catches bus 1 at 7:20 a.m., after a travel time of 1.2 hours = 72 minutes, he will arrive downtown at 8:32 a.m. After waiting 8 minutes for the uptown bus, he will take bus 2 at 8:40 a.m., and after 2/3 hour = 40 minutes, he will arrive uptown at 9:20 a.m. After waiting 40 minutes for the final bus, he will take bus 3 at 10 a.m. to his destination. We can see that his total waiting time is 8 + 40 = 48 minutes.

However, if Shawn catches bus 1 at 7:50 a.m., he will arrive downtown at 9:02 a.m., and after an 8-minute wait, he will catch bus 2 at 9:10 a.m. After the 40-minute bus ride uptown, he will arrive uptown at 9:50 a.m. After this point, he has to wait only 10 minutes for bus 3 to his destination, at 10:00 a.m., waiting a total of 8 + 10 = 18 minutes.

Eighteen minutes is the shortest total amount of time Shawn can spend waiting for the buses during his trip, because for every downtown bus after 7:50 a.m., the pattern of waiting times will repeat.

We can observe this with the scenario in which he catches the 8:20 a.m. downtown bus. After 72 minutes, he arrives downtown at 9:32 a.m., and the next uptown bus is at 9:40 a.m., which gives a waiting time of 8 minutes. After a 40-minute bus ride on the 9:40 uptown bus, he will arrive uptown at 10:20 a.m. The next bus is to his destination and departs at 11:00 a.m., so the total waiting time is 8 + 40 = 48 minutes. As we can see, the waiting time for the scenario in which he catches the 8:20 a.m. downtown bus is the same as it is for the scenario in which he catches the 7:20 a.m. downtown bus.

Answer: B

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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