A museum offers four video programs that run continuously throughout the
day, each program starting anew as soon as it is finished. The first program
runs every 15 minutes, the second every 30 minutes, the third every 45
minutes, and the fourth every 40 minutes; the first show of each program
starts at 10:00 A.M. and the last showing of each program ends at 4:00 P.M.
If a tour group can watch the programs in any order, but needs at least ten
minutes between programs to regroup, what is the least amount of time
the group can take to watch all four programs?
Manhattan Minority Problems
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To minimize the total time, as many breaks as possible must last only 10 minutes each.finance wrote:A museum offers four video programs that run continuously throughout the
day, each program starting anew as soon as it is finished. The first program
runs every 15 minutes, the second every 30 minutes, the third every 45
minutes, and the fourth every 40 minutes; the first show of each program
starts at 10:00 A.M. and the last showing of each program ends at 4:00 P.M.
If a tour group can watch the programs in any order, but needs at least ten
minutes between programs to regroup, what is the least amount of time
the group can take to watch all four programs?
10:00-10:30 = 30-minute show.
10-minute break.
10:40-11:20 = 40-minute show.
10-minute break.
11:30-11:45 = 15-minute show.
15 minute break.
12:00-12:45 = 45 minute show.
Total time = 2 hours, 45 minutes.
No way to minimize the time further. At least one of the breaks must last 15 minutes.
Last edited by GMATGuruNY on Thu Jul 28, 2011 3:35 am, edited 1 time in total.
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Watch the first show (15 mins) at 10.15 AM = 15 mins (10.30 AM)
Followed by regrouping = 10 mins (10.40 AM)
Watch the fourth show (40 mins) at 10.40 AM = 40 mins (11.20 AM)
Followed by regrouping = 10 mins (11.30 AM)
Watch the second show (30 mins) at 11.30 AM = 30 mins (12 PM)
Followed by regrouping = 10 mins (12.10 PM)
Wait for the third show (45 mins) = 5 mins
Watch 3rd show (45 mins) = 45 mins (1 PM)
So total time = 2hours and 45 minutes.
Is this the right answer and more importantly the right/best approach?
Followed by regrouping = 10 mins (10.40 AM)
Watch the fourth show (40 mins) at 10.40 AM = 40 mins (11.20 AM)
Followed by regrouping = 10 mins (11.30 AM)
Watch the second show (30 mins) at 11.30 AM = 30 mins (12 PM)
Followed by regrouping = 10 mins (12.10 PM)
Wait for the third show (45 mins) = 5 mins
Watch 3rd show (45 mins) = 45 mins (1 PM)
So total time = 2hours and 45 minutes.
Is this the right answer and more importantly the right/best approach?
Yes, the right answer is 2hrs 45 minutes..Here is the solution:
The first program plays every hour, hour: 15, hour:30, and
hour:45; the second plays every hour and half-hour; the third plays at 10:00, 10:45, 11:30, 12:15, and so
on; the fourth plays at 10:00, 10:40, 11:20, and so on. A look at this list shows that only the fourth show
is separated from any of the others by exactly 10 minutes. Thus, the fourth show should be viewed second
or third (i.e., somewhere in the middle) and at one of the times that does not start or end on the hour, so
that it is flanked by 2 waiting times of exactly 10 minutes each. The third waiting time must be at least 15
minutes, so the total time is at least the cumulative length of all four shows plus 35 minutes in waiting
time, or 165 minutes. (One sample schedule that achieves this minimum: 30-minute show 10:00--10:30,
40-minute show 10:40-11:20, 45-minute show 11:30-12:15, 15-minute show 12:30-12:45.)
The first program plays every hour, hour: 15, hour:30, and
hour:45; the second plays every hour and half-hour; the third plays at 10:00, 10:45, 11:30, 12:15, and so
on; the fourth plays at 10:00, 10:40, 11:20, and so on. A look at this list shows that only the fourth show
is separated from any of the others by exactly 10 minutes. Thus, the fourth show should be viewed second
or third (i.e., somewhere in the middle) and at one of the times that does not start or end on the hour, so
that it is flanked by 2 waiting times of exactly 10 minutes each. The third waiting time must be at least 15
minutes, so the total time is at least the cumulative length of all four shows plus 35 minutes in waiting
time, or 165 minutes. (One sample schedule that achieves this minimum: 30-minute show 10:00--10:30,
40-minute show 10:40-11:20, 45-minute show 11:30-12:15, 15-minute show 12:30-12:45.)
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I did the same way you did with the exception of switching 15 min + 45 minute shows, but it made no difference 2:45 minutes was the total time I got.GMATGuruNY wrote:To minimize the total time, as many breaks as possible must last only 10 minutes each.finance wrote:A museum offers four video programs that run continuously throughout the
day, each program starting anew as soon as it is finished. The first program
runs every 15 minutes, the second every 30 minutes, the third every 45
minutes, and the fourth every 40 minutes; the first show of each program
starts at 10:00 A.M. and the last showing of each program ends at 4:00 P.M.
If a tour group can watch the programs in any order, but needs at least ten
minutes between programs to regroup, what is the least amount of time
the group can take to watch all four programs?
10:00-10:30 = 30-minute show.
10-minute break.
10:40-11:20 = 40-minute show.
10-minute break.
11:30-11:45 = 15-minute show.
15 minute break.
12:00-12:45 = 45 minute show.
Total time = 2 hours, 45 minutes.
No way to minimize the time further. At least one of the breaks must last 15 minutes.
Seems too easy of a question, what level question is this?
-ECM