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Before I answer this question, I have a few questions: Does the watch instantly gain time at the end of the day (00:00, 12:00)? VS Does the watch's rate gain 10 seconds?
POTW 7/8 First, I'll list the facts (it helps). - We're going from 12th birthday to 90th - Watch gains 10 seconds at the end of every day. - How many times is the watch correct? Okay, so a watch is normally correct twice a day. However, if we add 10 seconds each time, it will ruin that pattern. Based on this, it'll have to be until the next 12 hours worth of 10 seconds to be correct. So 12 x 60 x 60 = 43200/10 = 4320. Okay so since we added 10 seconds 4320 times, it'll have to be 4320 days. For me, it's easier to work with years, so I'll convert to years (it also makes counting easier). 4320 days = 4320/365 = 11.8356164 years. So every 11.8356164 years, the time is correct. 90 - 12 = 78. The number of times 11.8356164 goes into 78 is the answer. In other words, 78/11.8356164 = 6.5902778. We round down those extra years caused by not being enough 10 seconds. The number of times the watch is correct is 6.
Grade 7/8 POTW: Info: - Gains 10 seconds everyday. - Never adjusts to right time - How many times will the watch show the righ time form 90th bday to 12th bday?
This sounds really complicated to solve, but I have an idea to solve it. Find the number of times in one year it shows the right time and multiply it by 78. Then add leap years. So for the first day, his watch is ok. On the second day, his watch is 10 seconds off. On the third day, his watch is 20 seconds off. On the fourth day, his watch will be 30 seconds off. On the fifth day, his watch will be 40 seconds off. On the sixth day, his watch will be 50 seconds off. On the seventh day, it will be on time. This means that every week, the watch will be on time, on the same day. Every 7 days, the watch will be on time. There are 365 days in a year (except for the occasional leap year), so I can multiply that by 78 (the number of years) and add the leap years. To find the leap years. 78/4=19.5 (not a whole number). 19 *4 =76. 19 leap years. (365*78)+19=28489
Leap years only add a day every 4 years. If there are 19 leap years like you say, that is only 19 days more. I'm pretty sure leap years don't make a difference for this given data. Also, at the end of the 7th day, it is 1 minute later than the right time. I think that minor error influenced your answer.
Anyways, let's get into the problem. If I were to keep going as a pattern, then first day, watch is correct(I don't think we are counting this?). Second day, ten seconds ahead. Third day, 20 seconds ahead. 4th, 30 secs. 5th, 40 secs. 6th, 50 secs. 7th, ONE MINUTE AHEAD. It will not go back to the right time, so every time the correct time shows a time, then the watch will show one minute ahead, like 4:05 and 4:06.
As far as I know, this would mean that every 12 hours would be the correct time.
Since it gains 10 seconds and I need 12 hours, then I do 12*60*6 = 4320. Every 4320 days would be one correct time. Simplified into years this would be 4320/365 = 11.835(11.84).
Every 11 years would be one correct time, so I do (90-12+1)/11.84 = 6.67 Approx.
This means that it would appear 6 times. A lot less than what I would expect, and even more less than what other people wrote as their answers, so not sure if I'm right or not. Oh well, guess we'll have to see! -Alan
Day 3: Correct time: 3:10:00 Watch time: 3:10:20 Gained another 10 seconds.
Day 4: Correct time: 3:10:00 Watch time: 3:10:30.
Day 5 watch time: 3:10:40 Day 6 watch time: 3:10:50. Correct time is the same thing: 3:10:00. Day 7 watch time: Adding another 10 seconds would be 3:11:00. Correct time would be 3:10:00. The correct time is not going to go any faster, so the watch time would be one minute ahead of the correct time.
When figuring out this question, we must consider many things: the minute hand and hour hand will also be affected when the second hand is moved, leap years, daylight savings time, and that a.m. and p.m. doesn't matter if this were an analog watch. After calculating considering everything, I found that there will be 6 times where the analog watch shows the correct time from when he is 12 to 90.
However, since it didn't specify the type of watch we are dealing with (not taking account of the image), the watch will never show the correct time. Let's say it is a digital smart watch. If we set that watch to a specific date, it will never show the correct time, since 1. it shows a.m. and p.m., 2. that usually the date will also be shown on a smart watch, 3. leap years, daylight savings time, etc. will also be pushed to different days, be completely different from the actual time.
Disclaimer: I didn't really put the realistic standards in time for this question, like leap years, daylight savings, or anything that changes the watch's time "course" if there are 365 days in a year always. Also, a "sidereal" day is about 23 hours, 56 minutes, and 4 seconds, so the watch isn't really correct in the first place if it's an analog watch with its rotations of hands.
Answer: 10 seconds gained every day 12-under 90th birthday 365*10 = 3650 seconds gained each year. 3650 to hours would then be roughly 1 hour. 90-12 = 78, which would mean 78*1 for hours turned ahead, which is 78 hours were gained over the course of his 12th birthday to just below his 90th. There’s a possible 6 times that the watch would go back to the same time then, not counting a.m or p.m. If counting, then it would be 3 times. HOWEVER, if it’s an analog watch, realistically, I doubt an analog watch would last for 78 years, unless Jeff restores the battery of his watch. If it were digital, then the watch would never truly show the same time. This is because digital ones or smart watches show the date AND time, meaning that there won’t ever be the same date and time as anytime else in the future.
Jeff's watch will show the correct time 3 times after his 12th birthday and before his 90th birthday. However, if the watch doesn't have am or pm indicators, it will show the correct time 6 times. I did my work in my math book.
I found out that the clock will tell the correct time every 12 hours its ahead. 12 hours is 43200 secs which means that the clock will be another 12 hours ahead every 4320 days or 11.8 years. There is a 78 year gap from his 12 and 90 birthdays and in those 78 years, the clock will tell the correct time 6.6 times (78/11.8).
The clock will tell the correct time every 6 years.
Before I answer this question, I have a few questions:
ReplyDeleteDoes the watch instantly gain time at the end of the day (00:00, 12:00)?
VS
Does the watch's rate gain 10 seconds?
Not sure, sorry. Just go after it! Don't worry about mistakes because that is where the best learning comes from!
DeletePOTW 7/8
DeleteFirst, I'll list the facts (it helps).
- We're going from 12th birthday to 90th
- Watch gains 10 seconds at the end of every day.
- How many times is the watch correct?
Okay, so a watch is normally correct twice a day. However, if we add 10 seconds each time, it will ruin that pattern. Based on this, it'll have to be until the next 12 hours worth of 10 seconds to be correct. So 12 x 60 x 60 = 43200/10 = 4320.
Okay so since we added 10 seconds 4320 times, it'll have to be 4320 days. For me, it's easier to work with years, so I'll convert to years (it also makes counting easier). 4320 days = 4320/365 = 11.8356164 years. So every 11.8356164 years, the time is correct. 90 - 12 = 78. The number of times 11.8356164 goes into 78 is the answer. In other words, 78/11.8356164 = 6.5902778. We round down those extra years caused by not being enough 10 seconds.
The number of times the watch is correct is 6.
One question: Do leap years count?
ReplyDeleteGREAT question. I think the solution uses 365 days a year, so no. But you raise a very valid point Fiona
DeleteGrade 7/8 POTW:
ReplyDeleteInfo:
- Gains 10 seconds everyday.
- Never adjusts to right time
- How many times will the watch show the righ time form 90th bday to 12th bday?
This sounds really complicated to solve, but I have an idea to solve it. Find the number of times in one year it shows the right time and multiply it by 78. Then add leap years.
So for the first day, his watch is ok. On the second day, his watch is 10 seconds off. On the third day, his watch is 20 seconds off. On the fourth day, his watch will be 30 seconds off. On the fifth day, his watch will be 40 seconds off. On the sixth day, his watch will be 50 seconds off. On the seventh day, it will be on time. This means that every week, the watch will be on time, on the same day. Every 7 days, the watch will be on time. There are 365 days in a year (except for the occasional leap year), so I can multiply that by 78 (the number of years) and add the leap years. To find the leap years. 78/4=19.5 (not a whole number). 19 *4 =76. 19 leap years. (365*78)+19=28489
Please correct me if I'm wrong. I added leap years, so my answer may be different than yours
DeleteLeap years only add a day every 4 years. If there are 19 leap years like you say, that is only 19 days more. I'm pretty sure leap years don't make a difference for this given data.
DeleteAlso, at the end of the 7th day, it is 1 minute later than the right time. I think that minor error influenced your answer.
POTW: I'm really not confident with my answer XD.
ReplyDeleteAnyways, let's get into the problem.
If I were to keep going as a pattern, then first day, watch is correct(I don't think we are counting this?). Second day, ten seconds ahead. Third day, 20 seconds ahead. 4th, 30 secs. 5th, 40 secs. 6th, 50 secs. 7th, ONE MINUTE AHEAD. It will not go back to the right time, so every time the correct time shows a time, then the watch will show one minute ahead, like 4:05 and 4:06.
As far as I know, this would mean that every 12 hours would be the correct time.
Since it gains 10 seconds and I need 12 hours, then I do 12*60*6 = 4320. Every 4320 days would be one correct time.
Simplified into years this would be 4320/365 = 11.835(11.84).
Every 11 years would be one correct time, so I do (90-12+1)/11.84 = 6.67 Approx.
This means that it would appear 6 times.
A lot less than what I would expect, and even more less than what other people wrote as their answers, so not sure if I'm right or not. Oh well, guess we'll have to see!
-Alan
I am still confused why the watch will reach a minute ahead
DeleteWhat do you mean?
DeleteIt won't jump back to the same time on the 7th day. For example:
DeleteCorrect time: 3:10 :00
Time on day 1: 3:10:00
Day 2:
Correct time: 3:10:00
Watch time: 3:10:10
Gained 10 seconds.
Day 3:
Correct time: 3:10:00
Watch time: 3:10:20
Gained another 10 seconds.
Day 4:
Correct time: 3:10:00
Watch time: 3:10:30.
Day 5 watch time: 3:10:40
Day 6 watch time: 3:10:50. Correct time is the same thing: 3:10:00.
Day 7 watch time: Adding another 10 seconds would be 3:11:00. Correct time would be 3:10:00. The correct time is not going to go any faster, so the watch time would be one minute ahead of the correct time.
Hope this helps for clarification!
Grade 8 POTW:
ReplyDeleteWhen figuring out this question, we must consider many things: the minute hand and hour hand will also be affected when the second hand is moved, leap years, daylight savings time, and that a.m. and p.m. doesn't matter if this were an analog watch.
After calculating considering everything, I found that there will be 6 times where the analog watch shows the correct time from when he is 12 to 90.
However, since it didn't specify the type of watch we are dealing with (not taking account of the image), the watch will never show the correct time. Let's say it is a digital smart watch. If we set that watch to a specific date, it will never show the correct time, since 1. it shows a.m. and p.m., 2. that usually the date will also be shown on a smart watch, 3. leap years, daylight savings time, etc. will also be pushed to different days, be completely different from the actual time.
It does specify which watch it is. It says a pocket watch in the question, which is the one in the picture.
DeleteDisclaimer:
ReplyDeleteI didn't really put the realistic standards in time for this question, like leap years, daylight savings, or anything that changes the watch's time "course" if there are 365 days in a year always. Also, a "sidereal" day is about 23 hours, 56 minutes, and 4 seconds, so the watch isn't really correct in the first place if it's an analog watch with its rotations of hands.
Answer:
10 seconds gained every day
12-under 90th birthday
365*10 = 3650 seconds gained each year. 3650 to hours would then be roughly 1 hour. 90-12 = 78, which would mean 78*1 for hours turned ahead, which is 78 hours were gained over the course of his 12th birthday to just below his 90th. There’s a possible 6 times that the watch would go back to the same time then, not counting a.m or p.m. If counting, then it would be 3 times.
HOWEVER, if it’s an analog watch, realistically, I doubt an analog watch would last for 78 years, unless Jeff restores the battery of his watch. If it were digital, then the watch would never truly show the same time. This is because digital ones or smart watches show the date AND time, meaning that there won’t ever be the same date and time as anytime else in the future.
Grade 8 POTW
ReplyDeleteJeff's watch will show the correct time 3 times after his 12th birthday and before his 90th birthday. However, if the watch doesn't have am or pm indicators, it will show the correct time 6 times. I did my work in my math book.
Every 4320 hours, the clock shows the right time. Therefore, It will show the correct time 158 times between his 12th and 90th birthdays.
ReplyDeleteI'm not sure about my answer, so can you specify on how you got the answer of 158 Luke? Maybe I did something incorrectly.
DeleteI found out that the clock will tell the correct time every 12 hours its ahead. 12 hours is 43200 secs which means that the clock will be another 12 hours ahead every 4320 days or 11.8 years. There is a 78 year gap from his 12 and 90 birthdays and in those 78 years, the clock will tell the correct time 6.6 times (78/11.8).
ReplyDeleteThe clock will tell the correct time every 6 years.