Kindly help please.?
2018-07-17 10:02 am
Red buys a bottle of 150 pills. Using a standard 12 hour clock, he sees that the clock reads exactly 3 o'clock when he takes the first pill. If he takes one pill every four hours, what hours will the clock read when he takes the last pill in the bottle?
回答 (4)
2018-07-17 10:17 am
✔ 最佳答案
Every cycle of the clock, number of pills taken = 12/4 = 3i.e. 3 o'clock, 3 + 4 = 7 o'clock, 7 + 4 = 11 o'clock
Number of cycles of the clock to take the pill = 150/3 = 50
The clock runs 50 "complete cycles".
When he takes the lost pill of the bottle, the clock will read 11 o'clock.
2018-07-17 11:49 am
The pattern is 3 , 7 , 11 , 3 ...
Repeats every three times.
150/3 = 50 no Remainder
So 11
Repeats every three times.
150/3 = 50 no Remainder
So 11
2018-07-17 10:21 am
Time_of_last_pill = Time_of_first_pill + Time_between_pills (Number_of_pills)
..........Time_of_first_pill = 3:00
..........Time_between_pills = 4 hours
..........// The 1st pill is taken at 3:00 so there are only
..........// 150-1 = 149 pills left to take
..........Number_of_pills = 149
So
Time_of_last_pill = Time_of_first_pill + Time_between_pills (Number_of_pills)
Time_of_last_pill = 3:00 + 4 hours (149)
............................= 3:00 + 596 hours
..........// Note that after every 12 hours, the time will be 3:00 again
..........// so we need to figure out how many 12 hour periods there
..........// are in 596 hours
..........Number_of_12_hour_periods = 596/12
..........Number_of_12_hour_periods = 49
..........Remaining_hours_not_in_12_hour_period = 596-49(12) = 596-588 = 8
Therefore,
Time_of_last_pill = 3:00 + 4 hours (149)
............................= 3:00 + 596 hours
............................= 3:00 + 8 hours
............................= 11:00...........ANS
..........Time_of_first_pill = 3:00
..........Time_between_pills = 4 hours
..........// The 1st pill is taken at 3:00 so there are only
..........// 150-1 = 149 pills left to take
..........Number_of_pills = 149
So
Time_of_last_pill = Time_of_first_pill + Time_between_pills (Number_of_pills)
Time_of_last_pill = 3:00 + 4 hours (149)
............................= 3:00 + 596 hours
..........// Note that after every 12 hours, the time will be 3:00 again
..........// so we need to figure out how many 12 hour periods there
..........// are in 596 hours
..........Number_of_12_hour_periods = 596/12
..........Number_of_12_hour_periods = 49
..........Remaining_hours_not_in_12_hour_period = 596-49(12) = 596-588 = 8
Therefore,
Time_of_last_pill = 3:00 + 4 hours (149)
............................= 3:00 + 596 hours
............................= 3:00 + 8 hours
............................= 11:00...........ANS
2018-07-17 10:16 am
3, 7, 11, 14 ... etc
a(150) = 3 + 4(150 - 1) = 599
599/12 = 49 R 11
... last pill at 11 o'clock
a(150) = 3 + 4(150 - 1) = 599
599/12 = 49 R 11
... last pill at 11 o'clock
收錄日期: 2021-04-24 01:08:33
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