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?

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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.

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!

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!

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

1 576 960

2 1152 1920

3 1728

4 2304 3840

5

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

John P. |

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.

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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## Comments

1 576 960

2 1152 1920

3 1728

28804 2304 3840

5

288048006 3456 5760

7 4032 6720

8 4608 7680

9 5184 8640