A math teacher had chartered a 20 seater bus as she plans for an excursion to the zoo. The number of seats were just right for her 19 students and herself. However at last hour before they set-off, the school principal and his wife and a senior teacher wanted to join in for a free ride, knowing that this 5 hours journey would cost them if they were to drive their own SUVs'. The math teacher was informed and asked to make arrangement for some students stand while all teachers including the principal and his wife to be remained seated at all time. The math teacher sheepishly obliged welcoming her boss and the two additional free riders. Without giving much thought and desperately trying to boast her forte, she balatantly promised the principal that there ain't gonna cause her any problem and each student would be given a fair share of seating and standing time.
Can you help this teacher? How's she gonna do her maths for this problem that she promises so inadventantly? No trickery
2006-11-17 15:11:20 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Other - Science
5 hour trip times 3 students standing at all time = 900 standing-minutes. 900 min divided by 19 students = 37 min 22 s.
At departure student 1 2 and 3 is standing. When a third of student 1's standing-time is over he/she sits down on student 4's seat, who just stood up. When two thirds of student 2's standing-time is over he/she sits down on student 5's seat, who just stood up. Student 3 stands for his fair share of time, and switch place with student 6. All students from #3 to #19 do their standingtime in a single streach. When student 17 is done he/she sits down on student 1's current seat. When student 18 is done he/she sits down on student 2's current seat, and when student 19 is done, they are at their destination!
The math teacher only has to shout "switch!" every 12min 42sec.
I know you said no trickery, but she could also force the students to all stand for 5 straight hours...
2006-11-17 15:35:49 · answer #1 · answered by venkeper 1 · 0⤊ 0⤋
ok. so there are 19 students and they are the only ones that need to worry about standing. And these 19 students have to stand for 5 hours. The first thing I did was separate the 19 students into groups of at least 3 (to accomodate the principle, wife, and teacher). Because 19 is not perfectly divisible by 3, one group will have four students (so there will be an extra seat when the group of 4 stand up). Ok so there are 6 groups (5 groups of 3 and 1 group of 4). and all of these 6 groups must stand for a span of 5 hours. So each group must stand for 5/6 hours (50 minutes each). And if each group does this, every student will stand for the same amount of time for 5 hours while the cheap principal, wife, and old teacher sit comfortably!
2006-11-17 15:40:08 · answer #2 · answered by NOOOWAAY 2 · 0⤊ 0⤋
There are 16 seats available for students to sit in. The trip lasts 5 hours. So there are 80 "person-hours" or 4,800 "person-minutes" or 288,000 "person-seconds" of sitting time. There are 19 students so each should get 15,157 seconds of sitting time and should stand for 2,843 seconds (4 hours, 7 minutes, 23 seconds).
2006-11-17 15:21:40 · answer #3 · answered by Anonymous · 0⤊ 0⤋
assume 3 students must stand at any one time.total of 5 hr per 20/3 students. 45 minutes standing per student and the teacher
2006-11-17 15:21:45 · answer #4 · answered by telzey 2 · 0⤊ 0⤋
vankeper got it right.
Use his solution.
2006-11-17 16:15:56 · answer #5 · answered by actuator 5 · 1⤊ 0⤋