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Shawn is planning a bus trip across town that involves three buses. Bu

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Shawn is planning a bus trip across town that involves three buses. Bu  [#permalink]

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18 Jul 2011, 13:09
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Difficulty:

95% (hard)

Question Stats:

45% (03:29) correct 55% (03:02) wrong based on 216 sessions

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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 1.2 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) 12 minutes
(B) 18 minutes
(C) 48 minutes
(D) 1 hour, 12 minutes
(E) 1 hour, 20 minutes

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Re: Shawn is planning a bus trip across town that involves three buses. Bu  [#permalink]

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18 Jul 2011, 15:10
1
Second bus service starts 10 mins earlier than first bus service and both have same frequency (every 30 mins). First bus take 1hr 12 mins.
Thus second bus one can get has waiting time 30 - (10+12)=8 mins
It is same always.
If you write time arrival time for first bus(8:32,9:02,9:32, add 30mins) and departure time for second bus(8:40,9:10,9:40 etc), you will get the same.

Similarly, third bus leaves every hour and service starts exactly on hour clock, a 10 mins early(7:10am and 9am, just mins since buses have frequency of either half or one hour)
Also second bus takes 2 hour 40 mins
Waiting time = 60-(10-40) =10 mins or 10+30=40 mins (since second bus leaves every 30 mins and third every 1hr)

First waiting time is always 8 mins, so take least time 10 mins for second waiting time.
Thus, least waiting time = 8+10 = 18 mins.

It did take more than 2 mins though.
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Re: Shawn is planning a bus trip across town that involves three buses. Bu  [#permalink]

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18 Jul 2011, 23:31
1
valencia wrote:
X is planning a bus trip across town that involves three buses. Bus 1 travels between house and downtown, and leaves every half-hour starting at 7:20am. X will need to be on Bus 1 for 1 1/5 hours. Bus 2 travels between downtown and uptown every half-hour starting at 7:10am. X will have to be on Bus 2 for 2/3 hour. Lastly, Bus 3 travels between uptown and destination every hour starting at 9 Am. what is the least total amount of time X wmust spend waiting for buses, assuming all buses stay on schedule?

A. 12 mins
B. 18 mins
C. 48 Mins
D. 1 hour 12 mins
E. 1 hour 20 mins

I'd solve it like this:

1 starts at 7:20, 2@7:10, 3@9:00

Start after 9, no risks:

1: 9:20, 9:50, 10:20, 10:50, 11:20, 11:50, 12:20, 12:50, 1:20, 1:50
2: 9:10, 9:40, 10:10, 10:40, 11:10, 11:40, 12:10, 12:40, 1:10, 1:40
3: 9:00, 10:00, 11:00, 12:00, 1:00, 2:00

Test for two start time:

10:20 AND 10:50; One got to be correct

10:20
Travel: 72min
Arrive:11:32
Next:11:40(Wait=8)
Travel: 40min
Arrive:12:20
Next:1:00(Wait=40)
Total=48

10:50
Travel: 72min
Arrive:12:02
Next:12:10(Wait=8)
Travel: 40min
Arrive:12:50
Next:1:00(Wait=10)
Total=18

Ans: "B"
*******************
It will take about 2min to 2.25min max.
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Re: Shawn is planning a bus trip across town that involves three buses. Bu  [#permalink]

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19 Jul 2011, 00:39
So, we have to try 2 cenarois then, dint look at the Q stem, it states least wait time, how could one possibly think different.
Fluke - you are an eye opener
Thank You

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Re: Shawn is planning a bus trip across town that involves three buses. Bu  [#permalink]

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29 Jul 2015, 04:38
1
1
Here's my solution. My first approach was to brute force the objective function in this optimization problem, minimize time spent waiting between buses

Givens:

1.2 hour=1 hour,12 minutes
2 2/3 hour=2 hour,40 minutes

Once you start filling out a table, you realize that there's only 2 solutions to this problem, leaving at the 20 minute mark of any hour versus leaving at the 50 minute mark of any hour.

Alternatively, you can deduce that there is always a bus 2 that departs an hour and 20 minutes after Bus 1 departs. This means that you can lump in the 2 bus ride lengths and the wait between bus 1 and 2 together into a 4 hour ride to get to the bus stop for bus 3.

Depart on bus 1: 1 hour, 12 minute ride
Wait 8 minutes
Depart on bus 2: 2 hour, 40 minute ride
Total: 4 hours

This means that if you depart on the 20 minute mark of every hour, you have to wait 40 minutes for bus 3 that arrives on the hour. If you depart on the 50 minute mark of every hour, you have you wait 10 minutes for bus 3. 8 + 40 = 48, 8 + 10 = 18. 18 < 48
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Re: Shawn is planning a bus trip across town that involves three buses. Bu  [#permalink]

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29 Jul 2015, 05:05
Bunuel 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 1.2 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) 12 minutes
(B) 18 minutes
(C) 48 minutes
(D) 1 hour, 12 minutes
(E) 1 hour, 20 minutes

Kudos for a correct solution.

Time to be spent on Bus 1 = 1.2 Hour = 1 Hour and 12 mins
Time at which Bus 1 leaves are {7:20, 7:50, 8:20, 8:50, 9:20, 9:50, 10:20, 10:50, 11:20, 11:50, 12:20, 12:50, 1:20, 1:50, 1:20, 1:50 etc )

Time to be spent on Bus 2 = 2/3 Hour = 40 mins
Time at which Bus 2 leaves are {7:10, 7:40, 8:10, 8:40, 9:10, 9:40, 10:10, 10:40, 11:10, 11:40, 12:10, 12:40, 1:10, 1:40, 1:10, 1:40 etc )

Time at which Bus 3 leaves are {9:00, 10:00, 11:00, 12:00, 1:00, 2:00, 3:00, 4:00, 5:00 etc )

The least amount of waiing time will be achieved when the he reaches for Bus 2 and spends least time and subsequently he reaches for Bus 3 and waits for least time

Looking at table he takes bus 1 at 7:50 reached destination at 9:02
Waits for 8 mins
Then takes bus 2 at 9:10 reached destination at 9:50
Waits for 10 mins
Then takes bus 3 at 10:00

Total Wait time = 8+10 = 18 Mins

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Re: Shawn is planning a bus trip across town that involves three buses. Bu  [#permalink]

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29 Jul 2015, 05:20
1
1
Bunuel 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 1.2 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) 12 minutes
(B) 18 minutes
(C) 48 minutes
(D) 1 hour, 12 minutes
(E) 1 hour, 20 minutes

Kudos for a correct solution.

IMO: B

Bus 1.................Bus 2..............Bus 3
7.20AM............7:10AM............9:00AM
7:50AM............7:40AM............10:00AM
8:20AM............8:10AM............11:00AM
8:50AM............8:40AM............12:00AM

Bus 1 --> Bus 2 travel time = 1hr 12 min(given)
Bus 2 --> Bus 3 travel time = 40min (given)

Waiting time for Bus 2 is always 8 min
--> consider Boards Bus 1 at 7:20AM --> 1hr 12 min(travel time) --> 8:32AM(time at departure) --> Bus 2 at 8:40AM => Thus waiting time is 8min
--> consider Boards Bus 1 at 7:50AM --> 1hr 12 min(travel time) --> 9:02AM(time at departure) --> Bus 2 at 9:10AM => Thus waiting time is 8min

Waiting time for Bus 3 will be 10 min or 40 min
--> consider Boards Bus 2 at 8:40AM --> 40 min(travel time) --> 9:20AM(time at departure) --> Bus 3 at 10:00AM => Thus waiting time is 40 min
--> consider Boards Bus 2 at 9:10AM --> 40 min(travel time) --> 9:50AM(time at departure) --> Bus 3 at 10:00AM => Thus waiting time is 10 min

Now since minimum waiting time is asked
8 min + 10 min = 18 min
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Shawn is planning a bus trip across town that involves three buses. Bu  [#permalink]

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20 Jul 2016, 03:33
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Top Contributor
valencia 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 1.2 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) 12 minutes
(B) 18 minutes
(C) 48 minutes
(D) 1 hour, 12 minutes
(E) 1 hour, 20 minutes

The last bus runs every hour.

For example if the last bus starts at 10 am, the second bus could have arrived at 9:20 am or 9:50 am.
Now consider the second bus could have started at 8:40 am or 9: 10 am. The first bus could have arrived at 8:32 am or 9:02 am

So if Shawn arrived at 9:02 from the first bus and waited for the second bus for 8 min and started again at 9:10 and arrived at 9:50 and waited for the third bus for 10 min.

Total wait time would be 8+10=18min
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Re: Shawn is planning a bus trip across town that involves three buses. Bu  [#permalink]

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08 Nov 2017, 16:28
valencia 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 1.2 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) 12 minutes
(B) 18 minutes
(C) 48 minutes
(D) 1 hour, 12 minutes
(E) 1 hour, 20 minutes

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.

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Re: Shawn is planning a bus trip across town that involves three buses. Bu  [#permalink]

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28 Nov 2017, 02:45
colorbrandon wrote:
Here's my solution. My first approach was to brute force the objective function in this optimization problem, minimize time spent waiting between buses

Givens:

1.2 hour=1 hour,12 minutes
2 2/3 hour=2 hour,40 minutes

Once you start filling out a table, you realize that there's only 2 solutions to this problem, leaving at the 20 minute mark of any hour versus leaving at the 50 minute mark of any hour.

Alternatively, you can deduce that there is always a bus 2 that departs an hour and 20 minutes after Bus 1 departs. This means that you can lump in the 2 bus ride lengths and the wait between bus 1 and 2 together into a 4 hour ride to get to the bus stop for bus 3.

Depart on bus 1: 1 hour, 12 minute ride
Wait 8 minutes
Depart on bus 2: 2 hour, 40 minute ride
Total: 4 hours

This means that if you depart on the 20 minute mark of every hour, you have to wait 40 minutes for bus 3 that arrives on the hour. If you depart on the 50 minute mark of every hour, you have you wait 10 minutes for bus 3. 8 + 40 = 48, 8 + 10 = 18. 18 < 48

Hi colorbrandon

I solved the question, but I had problem with time
the question says " for 2/3 hour." how could you see 2 2/3 hour??
Re: Shawn is planning a bus trip across town that involves three buses. Bu &nbs [#permalink] 28 Nov 2017, 02:45
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