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Math Help - clock problem

  1. #1
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    clock problem

    A slow clock loses 25 minutes a day. At noon on the first of October, it is set to show the correct time. When will this clock next show the correct time?

    Vicky.

    Could I get some help pls.
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  2. #2
    Super Member Gamma's Avatar
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    does am/pm count?

    Like will it "show the correct time" after it loses 12 hours or does it need to lose a full 24 hours?
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  3. #3
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    I am not sure. But i don't think the problem mentioned anything about it. And could you pls explain why it makes a difference?

    Vicky.
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  4. #4
    Member eXist's Avatar
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    It makes a difference because on a clock with only 12 hours (no pm or am) it will be shorter since there are less hours to rotate through. With a 24 hour clock (am and pm included) there will be more hours to rotate through.
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  5. #5
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    Hello Vicky
    Quote Originally Posted by Vicky1997 View Post
    A slow clock loses 25 minutes a day. At noon on the first of October, it is set to show the correct time. When will this clock next show the correct time?

    Vicky.

    Could I get some help pls.
    I'll assume that it's a 12-hour clock - so it shows the 'correct' time if, for example, it's actually 8 pm but the clock 'thinks' it's 8 am.

    In one day ( = 24 \times 60 = 1440 minutes) it loses 25 minutes. So it is running at \frac{1440-25}{1440}=\frac{283}{288} of the correct speed.

    So x minutes after noon on the first of October, the clock 'thinks' that \frac{283x}{288} minutes have elapsed. The clock will first show the correct time, then, when the difference between these two numbers of minutes = 12 \times 60 = 720, because the clock will then be exactly 12 hours slow. (If it's a 24 hour clock, you'll have to make this 24 hours, or 1440 minutes.) So:

    x-\frac{283x}{288}=720

    \Rightarrow 288x-283x = 720\times 288

    \Rightarrow x = \frac{720\times 288}{5}= 41472

    We now need to find the date and time, 41472 minutes after noon on 1st October.

    41472 minutes = \frac{41472}{1440}=28.8 days

    and 0.8 days = 19.2 hours = 19 hours 12 minutes

    So it's 19 hours 12 minutes after noon on 29th October, or 7:12 am on 30th October.

    Grandad

    PS Here's a much more obvious way of doing the calculation.

    It takes one day for the clock to lose 25 minutes. So in \frac{720}{25}=28.8 days it loses 720 minutes. Sorry I made it seem a lot harder than it really was!

    Last edited by Grandad; August 19th 2009 at 11:36 PM. Reason: Add PS
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