# clock problem

• Aug 19th 2009, 06:54 PM
Vicky1997
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.
• Aug 19th 2009, 08:10 PM
Gamma
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?
• Aug 19th 2009, 09:11 PM
Vicky1997
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.
• Aug 19th 2009, 10:05 PM
eXist
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.
• Aug 19th 2009, 10:40 PM
Hello Vicky
Quote:

Originally Posted by Vicky1997
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 ($\displaystyle = 24 \times 60 = 1440$ minutes) it loses $\displaystyle 25$ minutes. So it is running at $\displaystyle \frac{1440-25}{1440}=\frac{283}{288}$ of the correct speed.

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

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

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

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

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

$\displaystyle 41472$ minutes $\displaystyle = \frac{41472}{1440}=28.8$ 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.

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