Wrong clocks - Revision history

Oscarlevin at 20:41, 6 July 2013

2013-07-06T20:41:03Z

← Older revision		Revision as of 14:41, 6 July 2013
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	{{Hint\| When will the fast clock be one hour ahead? What will the slow clock display then?}}		{{Hint\| When will the fast clock be one hour ahead? What will the slow clock display then?}}
	{{Answer\| It will take 360 hours (15 days) for the clocks to display the same time as each other again.}}		{{Answer\| It will take 360 hours (15 days) for the clocks to display the same time as each other again.}}
	{{Solution\| Each hour, the fast clock gains one minute. So in 60 hours, the fast clock will be one hour ahead. The slow clock will be one hour behind. If we repeat this five more times, the fast clock will be 6 hours ahead, and the slow clock will be six hours behind. So both will display that it is exactly 6:00. This will take $60\times 6 = 360$ hours.}}		{{Solution\| Each hour, the fast clock gains one minute. So in 60 hours, the fast clock will be one hour ahead. The slow clock will be one hour behind. If we repeat this five more times, the fast clock will be 6 hours ahead, and the slow clock will be six hours behind. So both will display that it is exactly 6:00. This will take <m>60\times 6 = 360</m> hours.}}

	==References==		==References==

Oscarlevin at 20:40, 6 July 2013

2013-07-06T20:40:30Z

← Older revision		Revision as of 14:40, 6 July 2013
Line 8:		Line 8:

	{{Hint\| When will the fast clock be one hour ahead? What will the slow clock display then?}}		{{Hint\| When will the fast clock be one hour ahead? What will the slow clock display then?}}

	{{Answer\| It will take 360 hours (15 days) for the clocks to display the same time as each other again.}}		{{Answer\| It will take 360 hours (15 days) for the clocks to display the same time as each other again.}}

	{{Solution\| Each hour, the fast clock gains one minute. So in 60 hours, the fast clock will be one hour ahead. The slow clock will be one hour behind. If we repeat this five more times, the fast clock will be 6 hours ahead, and the slow clock will be six hours behind. So both will display that it is exactly 6:00. This will take $60\times 6 = 360$ hours.}}		{{Solution\| Each hour, the fast clock gains one minute. So in 60 hours, the fast clock will be one hour ahead. The slow clock will be one hour behind. If we repeat this five more times, the fast clock will be 6 hours ahead, and the slow clock will be six hours behind. So both will display that it is exactly 6:00. This will take $60\times 6 = 360$ hours.}}

Oscarlevin: Created page with 'Saw this one on Professor Layton 2. ==Puzzle== To analogue clocks hang side by side on a wall. One of the clocks is slightly fast, the other slightly slow. One day at noon, y...'

2010-06-10T13:26:19Z

Created page with 'Saw this one on Professor Layton 2. ==Puzzle== To analogue clocks hang side by side on a wall. One of the clocks is slightly fast, the other slightly slow. One day at noon, y...'

New page

Saw this one on Professor Layton 2.

==Puzzle==

To analogue clocks hang side by side on a wall. One of the clocks is slightly fast, the other slightly slow. One day at noon, you set both clocks correctly. After one hour you notice that the fast clock displays 1:01 and the slow clock reports the time as 12:59. Assuming the clocks continue to be fast and slow at the same rate, how long before both clocks agree on the time again?

==Help==

{{Hint| When will the fast clock be one hour ahead? What will the slow clock display then?}}

{{Answer| It will take 360 hours (15 days) for the clocks to display the same time as each other again.}}

{{Solution| Each hour, the fast clock gains one minute. So in 60 hours, the fast clock will be one hour ahead. The slow clock will be one hour behind. If we repeat this five more times, the fast clock will be 6 hours ahead, and the slow clock will be six hours behind. So both will display that it is exactly 6:00. This will take $60\times 6 = 360$ hours.}}

==References==

{{Professor Layton 2}}

[[Category: Short puzzles]]
[[Category: Number theory]]