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Two watches are together at 12 O'clock. If one gains 75 seconds each hour, and the other loses 45 seconds each hour, when will they be together again at 12?

 Two watches are together at 12 o'clock.  If one gains 75 seconds each hour , and the other loses 45 seconds each hour, when will they be together again at 12?

Comments

Andre's answer is for when both clocks show 12 oclock AND when the actual time is 12 o clock. The question does not require the time to be accurate. Note the times when clocks show 12 clock and actual time is 12 oi clock follow (in hours)
75 fast 45 slow
1 576 960
2 1152 1920
3 1728 2880
4 2304 3840
5 2880 4800
6 3456 5760
7 4032 6720
8 4608 7680
9 5184 8640  
 
So 2880 hours, by least common factor, is when both clocks show 12 and the actual time is 12. 
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5 Answers

let's look at the problem from a different point of view
they both are at 12 o'clock
how many hours must go by so that both clocks are at 12 o'clock again ?
the fast clock makes a 12 hour cycle but gains 1.25 minutes times 12=15 minutes, so it is not at 12 o'clock
the slow clock makes a 12 hour cycle but loses 0.75 minutes times 12=9 minutes so it is not at 12 0'clock
in 12 hours there are 720 minutes, so the fast clock must go through 48 12 hour cycles to be at 12 0'clock
(720/15=48)
the slow clock must go through 80 12 hour cycles to be at 12 o'clock (720/9=80) 
when the fast clock goes through 48 12hr cycles it gains 720 minutes (one 12 hour cycle)
when the slow clock goes through 80 12hr cycles it loses 720 minutes (one 12 hour cycle)
now find the LCM of 48 and 80
48=2x2x2x2x3  and 80=2x2x2x2x5
the LCM=2x2x2x2x3x5=240
the two clocks must make 240 12hr cycles to both be at 12 o'clock at the same time
240x12=2880 hours
Let n be the number of actual hours that pass after 12 o’clock.

At t=n, watch 1 shows

X = n*(3600+75)/3600 mod 12 =3675/3600*n mod 12 hours

(mod 12 means subtract the appropriate multiple of 12 so that the expression is between 0 and 12; this is what a watch shows us.)

Watch 2 shows

Y=n*(3600-45)/3600 mod 12= 3555/3600*n mod 12 hours

 
The difference shown after n hours is

X-Y= 120/3600*n mod 12 = n/30 mod 12 hours = n mod 360


This difference is to be 0:

n mod 360=0,

so n is an integer multiple of 360: n=360*m.
 
For both watches to show 12 o’clock, we also require X=Y=0:

3675/3600*n mod 12 = 0, 3675/3600*360*m mod 12=0

3675/10*m mod 12=0

3675*m mod 120=0

The smallest integer solution is m=8, so after n=360*8=2880 hours both watches will show 12 o’clock again!

Comments

Andre is using least common factor method
75 fast 45 slow
1 576 960
2 1152 1920
3 1728 2880
4 2304 3840
5 2880 4800
6 3456 5760
7 4032 6720
8 4608 7680
9 5184 8640
Above hours pasts are for when the clocks read 12 oclock AND the actual time is 12 oclock. the question does not ask for the time to be accurate, just when the times of the two clocks match (regardless if the actual time is 12 oclock).
but this answer appears to be when both clocks read 12 oclock AND the actual time is 12 oclock!!!
Andre,
I believe you are correct.  I am just having a slight problem understanding everying you set forth, including the mod expression.  Is it possible you can explain it more clearly?  Thank you so much.  Ron
not sure why my comments are not showing up here
other commenters show how to approximate and solve by induction. that is, every hour 2 minutes (75 seconds and 45 seconds = 120 seconds = 2 minutes)  differ between the two clocks. So for the two clocks to have the same time both showing 12 o'clock,  the time difference needs to be 12 hours, which is 720 minutes or 43,200 seconds. Or does it? Twelve hours later, the clocks are showing the same time, but they are obviously not showing 12 o'clock. (12 hours later - both are off so neither shows twleve o'clock) 
 
you need make a system of equations, but that is your homework assignment you should do it yourself! here is some help 
Facts:
60 minutes = 3600 seconds
x = clock 1
y = clock 2
 
Equation
x = 3600a +75
y = 3600a - 45
 
 
question when will x = y = 12 o'clock?
Hi Ron;
Robert J.'s answer is extraordinary, as always.
I would like to add to it.
Your instructor would probably like to see a more extensive breakdown.
 
|75 seconds|+|-45 seconds|=120 seconds
(120 seconds)(1 minute/60 seconds)=?
Let's cancel units where appropriate...
(120 seconds)(1 minute/60 seconds)=2 minutes
 
12 hours(60 minutes/1 hour)=720 minutes
x=quantity for equalization...
2 minutes(x hours)=720 minutes
Let's cancel units where appropriate...
2 minutes(x hours)=720 minutes
2(x hours)=720
x hours=360
In each hour, the difference is 75+45 = 120 sec = 2 min
So, in 30 hours, the difference is 60 min = 1 hours, and thus in 12*30 = 360 hours, the difference is 12 hours.

Comments

After 360 hours, both watches will show 7:30. The question was: when will they both be at 12 again?
Hi Andre,
How did you get 7:30?360 is a muliple of 12. If it starts at 12, then it ends at 12.
Hi Robert,
In 360 hours, watch 1 will have gained 360*75=27,000 seconds = 7.5 hours, so it will show 12+7.5 = 7:30.
Watch 2 will have lost 360*45=16,200 seconds = 4.5 hours, so it will show 12-4.5 = 7:30.
 
I calculated the first real 12 o'clock time when the two hands meet.

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