# Watch gaining and losing time

• December 23rd 2013, 02:48 AM
Natasha1
Watch gaining and losing time
My watch (which is a 12 hour watch) GAINS 3 mins every 2 hours.

a) I set my watch to the correct time at noon on 1st January. If I don't reset it, when will it next show the correct time?

I got:

real time vs gain time
2hrs vs 3 mins gain
40hrs vs 60 mins (x20)
960 hrs vs 24 hrs (x24)
= 40 days (divided 960 hrs by 24hrs)

Is this correct please for a?

b) Mrs Warma's watch (also 1 12 hour watch LOSES 5 mins every 2 hours. She also sets her watch to the correct time at noon on 1st January. When will our two watches next show the same time?

I got:

real time vs lose time
2hrs vs 5 mins lose

I am stuck after this....

c) When will our watches next show the same, correct time?

I am stuck with this

• December 23rd 2013, 03:56 AM
romsek
Re: Watch gaining and losing time
Quote:

Originally Posted by Natasha1
My watch (which is a 12 hour watch) GAINS 3 mins every 2 hours.

a) I set my watch to the correct time at noon on 1st January. If I don't reset it, when will it next show the correct time?

I got:

real time vs gain time
2hrs vs 3 mins gain
40hrs vs 60 mins (x20)
960 hrs vs 24 hrs (x24)
= 40 days (divided 960 hrs by 24hrs)

Is this correct please for a?

b) Mrs Warma's watch (also 1 12 hour watch LOSES 5 mins every 2 hours. She also sets her watch to the correct time at noon on 1st January. When will our two watches next show the same time?

I got:

real time vs lose time
2hrs vs 5 mins lose

I am stuck after this....

c) When will our watches next show the same, correct time?

I am stuck with this

Not quite. It's a 12 hr watch so it will be correct again in 12 hrs, not 24 as you've used. So it will be 20 days not 40.

b)

$\mbox{in minutes your watch shows }m1=t(1+3/120)=t(1+1/40)=\frac{41}{40}t\mbox{ minutes.}$

$\mbox{her's shows }m2=t(1-5/120)=t(1-1/24)=\frac{23}{24}t\mbox{ minutes.}$

$\mbox{The watches are able to display } t \bmod 720 \mbox{ minutes uniquely.}$

$\mbox{So we want to solve for }\left(m1 \bmod 720\right)=\left(m2 \bmod 720\right)$

see if you can work it from there.

c) you need to calculate how often the lady's watch is correct similar to how you did in (a). Then find the least common multiple of the three times, how often your watch is right, how often hers is, how often they are the same.
• December 23rd 2013, 04:13 AM
Natasha1
Re: Watch gaining and losing time
i don't understand what mod is, never done this at school
• December 23rd 2013, 04:29 AM
romsek
Re: Watch gaining and losing time
a mod b = remainder of a/b

for example (25 mod 7) = 4 because 25 - 3x7 = 4

I'll give you a hint solving this.

Try solving $\frac{41}{40}t-720=\frac{23}{24}t$
• December 23rd 2013, 09:45 AM
Soroban
Re: Watch gaining and losing time
Hello, Natasha1!

Quote:

My watch (which is a 12-hour watch) GAINS 3 mins every 2 hours.

a) I set my watch to the correct time at noon on 1st January.
If I don't reset it, when will it next show the correct time?

Your watch gains 1.5 minutes every hour.

To show the correct time again, it must gain 12 hours = 720 minutes.

. . $\frac{720\text{ min.}}{1} \times \frac{1\text{ hr.}}{1.5\text{ min.}} \:=\:480\text{ hours} \:=\:20\text{ days}$

It will show the correct time at noon on 21 January.

• December 23rd 2013, 11:43 AM
SlipEternal
Re: Watch gaining and losing time
Quote:

Originally Posted by Natasha1
b) Mrs Warma's watch (also 1 12 hour watch LOSES 5 mins every 2 hours. She also sets her watch to the correct time at noon on 1st January. When will our two watches next show the same time?

Your two watches will show the same time at the moment when your watch gains $t$ minutes and Mrs. Warma's watch loses 12 hours minus $t$ minutes. So, let $h$ be the number of periods of two hours that must pass before your watch gains $t$ minutes (and Mrs. Warma's watch loses $720-t$ minutes). Then $3h = t$ and $5h = 720-t$. This means $t=3h$ and $t=720-5h$. So, $3h = 720-5h$ shows that $8h = 720$, and $h = 90$. That means their watches will show the same time every $2h = 180$ hours. So, their watches will show the same time midnight, January 9th (that is the midnight that falls between the 8th and the 9th).
• December 23rd 2013, 03:42 PM
Soroban
Re: Watch gaining and losing time
Hello again, Natasha1!

Quote:

b) Mrs Warma's watch (a 12-hour watch) LOSES 5 mins every 2 hours.
She sets her watch to the correct time at noon on 1st January.
When will our two watches next show the same time?

Your watch gains 1.5 minutes every hour.
To show the correct time again, your watch must gain 12 hours = 720 minutes.
. . $\frac{720\text{ min.}}{1} \times \frac{1\text{ hr.}}{1.5\text{ min.}} \:=\:480\text{ hours} \:=\:20\text{ days}$

Her watch loses 2.5 minutes every hour.
To show the correct time again, her watch must lose 12 hours = 720 minutes.
. . $\frac{720\text{ min.}}{1} \times \frac{1\text{ hr.}}{2.5\text{ min.}} \:=\:288\text{ hours} \:=\:12\text{ days}$

The two watches will agree in: $LCM(20,12) = 60$ days . . . Noon on March 30.
• December 23rd 2013, 04:35 PM
SlipEternal
Re: Watch gaining and losing time
Quote:

Originally Posted by Soroban
Hello again, Natasha1!

Your watch gains 1.5 minutes every hour.
To show the correct time again, your watch must gain 12 hours = 720 minutes.
. . $\frac{720\text{ min.}}{1} \times \frac{1\text{ hr.}}{1.5\text{ min.}} \:=\:480\text{ hours} \:=\:20\text{ days}$

Her watch loses 2.5 minutes every hour.
To show the correct time again, her watch must lose 12 hours = 720 minutes.
. . $\frac{720\text{ min.}}{1} \times \frac{1\text{ hr.}}{2.5\text{ min.}} \:=\:288\text{ hours} \:=\:12\text{ days}$

The two watches will agree in: $LCM(20,12) = 60$ days . . . Noon on March 30.

This is the answer to part (c), not part (b).
• December 23rd 2013, 05:02 PM
Natasha1
Re: Watch gaining and losing time
Many thanks to all of you :). Merry Christmas everyone!