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 Post subject: Maths Help
PostPosted: Mon Nov 05, 2007 8:39 pm 
Two electric clocks with hours 1 to 12 marked on their faces.
The first clock gains 20mins per day and the second gains 15mins per day. They are both set to the correct time of 8o'clock.

1. After how many days has the first clock gained 12hrs?

2. After how many days has the second clock gained 12hrs?

3. When the second clock gained 12hrs it will be showing 8o'clock, what time will then be shown on the first clock?

4. After how many days (from start) will both clocks again be showing the same time?


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 Post subject:
PostPosted: Wed Nov 07, 2007 12:51 pm 
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Joined: Wed Oct 31, 2007 3:15 pm
Posts: 78
Hi there guest27,
I think the answers are as follows:

1) The first clock gains 20mins per day, so it gains 1 hour every 3days.
Therefore it takes 3 x 12 days to gain 12 hours, so the answer is 36 days.

2) The second clock gains 15 mins per day, so it gains 1 hour every 4 days. Therefore it takes 4 x 12 days to gain 12 hours, so the answer is 48 days.

3) It takes 48 days for the second clock to gain 12 hours. So after 48 days the first clock will have gained 48 x 20 mins = 960 mins = 16 hours.
If it had gained 12 hours it would be 8 0'clock again so 4 more hours after this is 12 o'clock.

4) The first clock takes 36 days to gain 12 hours.
The second clock takes 48 days to gain 12 hours.
So we need to find the Lowest Common Multiple of 36 and 48.
You can do this either by working out the prime factors or by simply writing out the 36 and the 48 times tables. This gives:
36 - 36,72,108,144
48 - 48, 96,144

So the Lowest Common Multiple is 144, so the clocks would be showing the same time after 144 days.

I hope that these answers help you.
abcdef


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 Post subject:
PostPosted: Wed Nov 07, 2007 4:01 pm 
Thank you so much..


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 Post subject:
PostPosted: Wed Nov 07, 2007 10:07 pm 
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Joined: Fri Jul 06, 2007 8:31 pm
Posts: 1192
guest27 wrote:
Thank you so much..


That is a hard question. (Or the last part at least).

Where did it come from?

Regards
SVE

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