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

Question

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.

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

fluke · Accepted Answer

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.

hellishbrain · Answer

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.

vinayreguri · Answer

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

Posted from iPhone 3GS

Posted from my mobile device

colorbrandon · Answer

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.

Answer, B

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

GMATinsight · Answer

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

Answer: option B

VenoMfTw · Answer

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

SVaidyaraman · Answer

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

JeffTargetTestPrep · Answer

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

soodia · Answer

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??

dabaobao · Answer

Veritas Prep Official Solution The first thing that comes to mind is that we can just plug in the numbers and find the time it takes to wait for the buses (or that Shawn should just get a car). We can figure out the timing from 7:20 AM and take it down the line from there. Let’s do that for completion’s sake, but it doesn’t mean that this is the best course of action by any means. If Shawn gets on the first bus at 7:20, then he’ll spend 1.2 hours (or 1 hour and 12 minutes) on the bus before getting off at 8:32. It’s important to note that fractions of hours are converted into decimal by dividing by 60, not 100. The second bus comes every half hour starting at 7:10, so Shawn will assuredly miss the first three and only get on the bus that comes at 8:40. He’s waited for 8 minutes up until this point. Bus 2 will take 40 minutes to reach its destination, dropping Shawn off at 9:20 AM. From there, bus 3 will be around every hour, so he’ll have to wait until 10 AM, an additional wait of 40 minutes. Thus, if Shawn gets on the first bus and all buses stick to their schedules, he’ll wait 48 minutes. This is the answer a calculator would get, and as long as no analysis is done, it is a reasonable answer. However, we’ve all experienced situations like this in our daily lives. If the bus is coming for a specific time, your goal is usually to minimize the wait time and arrive at the bus stop slightly before the bus is due. This will minimize your wait time. If the bus will be at the stop at 10 AM, there isn’t much point in being there at 9:01 waiting (although you may break your record at Angry Birds) when you can be there at 9:55 instead. Doing some analysis of this situation, the first bus comes every 30 minutes, meaning the bus always shows up twenty minutes past the hour or ten minutes to the hour. Within each hour, there are two choices you can make: the first bus or the second bus. After that, the choice returns with only the hour hand increasing by one. We thus need to figure out what will happen if we hop on the 7:50 bus instead of the 7:20 bus. Recalculating, we’re on the first bus for 1.2 hours, meaning we get on at 7:50 AM and get off at 9:02. The second bus still comes every half hour starting at 7:40, so we can jump on the 9:10 bus after waiting 8 minutes, just like in the first example. This bus takes us 40 minutes, and therefore drops us off at 9:50. We’re 10 minutes early for the last bus, which is still scheduled at 10 AM, bringing the total amount of time waiting to 18 minutes. Taking bus 1 at 7:50 instead of 7:20 gets us to the destination at the same time but reduces the wait time by 30 minutes, and is therefore preferable. Q.png The table above highlights the repetitive nature of problems like these. Every bus that comes at twenty past the hour will lead to a 48 minute total wait time, while every bus that comes at ten to the hour will lead to an 18 minute total wait time, regardless of the hour. (again assuming that the buses always run on time) ANSWER: B General Advice Credit: Veritas Prep On GMAT problems, it’s important to take a few seconds to understand what is being asked in the problem. Rushing headlong into a solution will work on many questions, but on tricky questions, a strong analysis of the situation is required to make the most effective decision. Despite the many tricks and gimmicks touted to solve GMAT problems more efficiently, the underlying goal of this test is to gauge your ability to analyze situations and apply logic. Being able to optimize a given scenario is important not only when in business, but also when in line for a bus.

Archit3110 · Answer

the best way to solve is to take two cases 
target is to find what is the  least amount of time Shawn must spend waiting for buses 
given bus 1 ; timing of departure every 30 mins ; i.e 7:20 ; 7:50: 8:20....
time spent travelling in bus 1 ; 1.2 hours or say ; 72 mins
so Shawn will reach to catch Bus 2 least by ; either  8:32 mins or 9:02 mins 
for bus 2 frequency is every 1 hour ; 7:10; 8:10; 9:10 ; 10:10
Considering shawn taking bus at two given intervals starting from bus 1 ; he can board the bus at 9:10 am and his waiting time falls ;  38 mins and 8 mins  respectively
given time it takes to travel from bus 2 to catch bus 3 is 2/3 * 60 ; 40 mins
so he would reach by  9:10 + 40 ; 9:50 am  and every 1 hour starting 9 am bus leaves from Bus 3 ; 
so shawn at given rate will be able to catch bus 3 at 10 am respectively
i.e he will have to wait for 38+10 mins i.e 48 mins or 8+10 mins i.e 18 mins

least time taken will be 18 mins
OPTION B

Adarsh_24 · Answer

A little bit more detailed explanation of hellishbrain.
Hope I didn’t make any mistake.

Consider first(F) and second (S)
Assume both start at same time and it takes first bus 30 min to travel. Then there wouldn’t be any time wasted between f and s.
In reality F is always 10 min late. So it will arrive 20 mins earlier than next S or 10 mins later than nearby S. we only care about earlier since we have to catch the bus. 
Also in reality F takes 1h 12 min. I.e 30+30+12
So bus arrives 12 mins later than earlier found. So it’s 20-12=8 mins

Similarly S and Third(T). If they were starting at same time and second took 1hour. No time lost.
Reality S is 10 min late. So it will arrive 50 mins earlier or 10 mins later than nearby T. We care about earlier since we have to catch the bus. 
Also in reality S takes 40 min.
So bus arrives 50-40=10 mins earlier than nearby.

So 10+8=18

Kinshook · Answer

valencia

Given: 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.

Asked: Assuming all buses stay on schedule, what is the least amount of time Shawn must spend waiting for buses?

Bus 1 (house to downtown) starting times = {7:20, 7:50, 8:20, 8:50,....}
1.2 hours = 1 hours 12 minutes
Bus 1 (house to downtown) reaching times = {8:32, 9:02, 9:32, 10:02,....}

Bus 2 (downtown to uptown) starting times = {7:10, 7:40, 8:10, 8:40,.....}
2/3 hours = 40 minutes 
Bus 2 (downtown to uptown) reaching times = {7:50, 8:20, 8:50, 9:20,.....}

Bus 3 (uptown to destination) starting times = {9:00, 10:00, 11:00, 12:00,....}

Option 1: 7:20-8:32 wait -0:08 8:40-9:20 wait - 0:40 10:00 : Total wait time = 8 + 40 = 48 minutes
Option 2: 7:50-9:02 wait 0:08 9:10 - 9:50 wait 0:10 10:00 : Total wait time = 8 + 10 = 18 minutes
Option 3: 8:20-9:32 wait -0:08 9:40-10:20 wait 0:40 11:00 : Total wait time = 8 + 40 = 48 minutes
Option 4: 8:50-10:02 wait 0:08 10:10 - 10:50 wait 0:10 11:00 : Total wait time = 8 + 10 = 18 minutes

The least amount of time Shawn must spend waiting for buses = 18 minutes

iMO B
 valencia

Given: 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.

Asked: Assuming all buses stay on schedule, what is the least amount of time Shawn must spend waiting for buses?

valencia

Given: 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.

Asked: Assuming all buses stay on schedule, what is the least amount of time Shawn must spend waiting for buses?

Bus 1 (house to downtown) starting times = {7:20, 7:50, 8:20, 8:50,....}
1.2 hours = 1 hours 12 minutes
Bus 1 (house to downtown) reaching times = {8:32, 9:02, 9:32, 10:02,....}

Bus 2 (downtown to uptown) starting times = {7:10, 7:40, 8:10, 8:40,.....}
2/3 hours = 40 minutes 
Bus 2 (downtown to uptown) reaching times = {7:50, 8:20, 8:50, 9:20,.....}

Bus 3 (uptown to destination) starting times = {9:00, 10:00, 11:00, 12:00,....}

To minimize wait time for Bus 2

valencia

Given: 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.

Asked: Assuming all buses stay on schedule, what is the least amount of time Shawn must spend waiting for buses?

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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

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