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.
Hello VickyI'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.
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 $\displaystyle \frac{720}{25}=28.8$ days it loses 720 minutes. Sorry I made it seem a lot harder than it really was!