Aptitude - Clock - Discussion

Why does everybody say that the minute hand does a complete circle in exactly 1 hour, when it is evidently not so (since the clock is losing time)?

The solution should be like this:

Let Wm be the angular speed of the minute hand, in radians per astronomical minute, and Wh be the angular speed of the hour hand. The constraint of the clock mechanism requires that Wh=Wm/12 (per each full revolution of the hour hand, the minute hand makes 12 full revolutions).

The requirement that the hands meet every 64 minutes is:

Wm*t = Wh*t + 2*pi, where t=64 astronomical minutes (astronomical minutes = actual minutes!).

Hence we have Wm*(11/12) = 2*pi/64,

or Wm = 2*pi*(12/11/64) rad/ast. minute, giving a period of revolution of (11/12)*64 = 58.6(6) astronomical minutes per revolution.

In 1 astronomical hour the minute hand covers Wm*60 = 2*pi*(60/64)*(12/11) radians, or simply 90/88=1+1/44 of a full revolution. Thus after each astronomical hour the minute hand shows 1/44*60 "clock minutes" more than the number of astronomical minutes passed.

In other words, each astronomical hour the clock GAINS 15/11 of a minute. Thus it GAINS (15/11)*24 = 32+8/11 clock minutes per astronomical day (i.e., each astronomical day it shows time which is 32+8/11 minutes more than the actual astronomical time).

55 min. Spaces are gained by minute hand in 60 min period.

To find how many spaces it has actually gained, we need to fix a standard point first. !

With respect to it, we need to see the difference by how much is it actually varying. !

So let us assume the standard point to be the place where the minutes hand and hours hand has been coincided. !

I may be 12:00, 1.06, 2.11, 3.17. etc. from there. 60 minutes implies the minute hand must come back to the same point where it has started. !

Is it not. ?

Now, 60 minute passed and so minute hand covers 60 minute spaces.

And the hour hand advances by 5 minute spaces. !

So from the standard point fixed initially (we assumed the standard point is where the minute point and hours hand were coinciding. Also. 60 minutes will be passed when the minute hand comes back to the same position from where it started).

Now, there is.

An absolute 60 min spaces covered by minute hand in 60 min and then there is 5 min spaces advanced by hour hand in 60 min period. !

So on total.

Total advancement is 60-5 = 55 minute spaces. !

Say its 12 o' clock:
The minute hand and the hour hand points towards 12

After an hour i.e at 1 o' clock:
The minute hand points at 12 and the hour hand points 1

The difference between the two hands after 60 min (1 hr) is 55 spaces but the hands did not coincide!

They will coincide when 60 spaces are covered

And to cover 60 spaces it takes = (60/55*60) = 65.4545 min (in a normal clock)

Now given in question is that the hands coincide at 64 min (defective clock)

So loss in time when the hands coincide is 65.4545 - 64 = 1.4545 min (this loss happens for every 64 min in our defective clock)

So for 1 min our loss is = 1.4545/64 min

For a day 1.4545/64*(24*60) = 32.7 min

@Naresh : just remember this

If coinciding time > 65(5/11) then our clock is going slow than normal (our watch is loosing time) and if coinciding time < 65(5/11) then its going fast than normal (or gaining time).

By 65(5/11) i mean 65.4545 and not (65*5)11

And we say loosing time in questions just to state the irregularity in time.

HOPE THIS CLEARS IT ALL

My 2 cents:

So far we know that it takes 60 min for the minute hand to gain a 55 min space lead over the hour hand. We assume that a 60 min space lead actually means that the two hands coincide.

The time required for a 60 min space lead would be 60*60/55 min for a normal clock. Instead of this, the error clock has a time requirement of 64 min for the min hand to gain a 60 min space lead over the hour hand.

With this new value we now try to calculate how many minutes the error clock would take for it's min hand to gain a 55 min space lead. Basically- "how many minutes of the error clock would signify i hour for the normal clock?"

This works out to be 64*55/60 min..which is 58.6666 min.

So for a normal clock which counts 60 minutes (1 hour), the error clock has clocked up 60/58.6666 min(1.022 hours).

Over a period of 24 hours, the error clock has would gain 24*1.022 hours=24.4545 hours. The difference being 24.4545-24 hr
Which is equal to .4545 hr = .4545*60 min = 32.72 min ~32 8/11 .

@Jess.

See for an hour the min hand has to travel 60 spaces(ticks) right?

And for the same hour, the hour hand travels only 5 spaces(ticks).
So the actual displacement by both in the same direction is = the difference between the spaces travelled.

then, the time spaces travelled is 60-5=55 min(representing each tick, not the actual minute) spaces.

And the total minute spaces travelled for a minute is 60/55 right?

then, for an hour it is (60/55)*60 which is equal to 720/11 = 65.4545 giving the time for each coincidence of the min and hour hand.

If the given value(coincidence time) is >65.4545 means your watch is losing time i.e. clock is slower.

If the given value is <65.4545 means your watch is gaining time i.e. clock is faster
Here the coincidence time is 64 hence, 65.4545 - 64 = 1.4545 this is for 64 mins.
Then for a minute, the time variation is 1.4545/64.
And to get the same for a day we multiply by 60 and 24.

Hence,
Time lost is (1.4545/64)*60*24 = 32.7272=32(8/11) min.

Say its 12 o' clock:
The minute hand and the hour hand points towards 12.

After an hour i.e at 1 o' clock:
The minute hand points at 12 and the hour hand points 1.

The difference between the two hands after 60 min (1 hr) is 55 spaces but the hands did not coincide.

They will coincide when 60 spaces are covered.
And to cover 60 spaces it takes = (60/55 * 60) = 65.4545 min (in a normal clock).
Now given in question is that the hands coincide at 64 min (defective clock).
So, loss in a time when the hands coincide is 65.4545 - 64 = 1.4545 min (this loss happens for every 64 min in our defective clock).

So for 1 min our loss is = 1.4545/64 min.
For a day 1.4545/64 * (24*60) = 32.7.

If coinciding time > 65(5/11) then our clock is going fast than normal (our watch is losing time) and if coinciding time < 65(5/11) then it's going slow than normal (or gaining time).

By 65(5/11) I mean 65.4545 and not (65*5)11.

And we say losing time in questions just to state the irregularity in time.

@Raman,

We know that the two hands (hour hand and min hand) of a clock coincide after 60 mins.

The question here is if the hands of the clock coincide after every 64 mins instead of 60 mins, how much time the clock/watch will lose in a day.

From the above comments, I can see many people are confused that if the hands coincide after every 64 mins instead of 60mins, then how come the clock/watch is losing time!

If the hands are coinciding after 64 mins instead of 60 mins, it means they are taking more time to coincide. The longer the hands take to coincide, the more time they lose.

Say for example, the time is 12 pm. In normal clock it will take 60 mins for the hands to coincide, that is to strike 1 pm. Now, as given in the example, if the hands coincide after 64 mins, it means the clock/watch will strike 1 pm after 64 mins instead of 60 mins. So it means where it took 60 mins to complete one hour, it is taking 64 mins. So here it is losing time.

Hope, this helps! :).

To coincide with each other space(ticks) between them should be ZERO(no space).

Consider starting from 12 noon.

in 1 hr--> min hand covers spaces(ticks) =60
in 1 hr--> hour hand covers spaces(ticks) = 5
so in 1 min --> hour hand covers 5/60=1/12 space(tick)

so after 1 hr=60 mins (1.00) spaces between them will be equal to 5
so have to cover 5 places for minute hand to meet hour hand
total time = 60+5 ... but in 5 mins hour hand will further go ahead by 5*1/12 ticks
so have to cover 5/12 ticks for minute hand to meet
total time = 60+5+5/12 ... but agin in 5/12 mins hour hand will go ahead buy 5/12*1/12= 5/12^2... and so on

So total time will be = 60+5+(5/12+5/12^2+5/12^3+.........) this GP
total time to meet = 720/11 i.e., 65.(5/11).

This time required to coincide hands in correct clock... solve accordingly.

I think the question is incorrect.

It is like this: A correct clock will have a time gap of (720/11 = 65 and 5/11) minutes between two successive overlaps of the hours hand and the minute hand. IF this time is being covered by the incorrect clock in 64 minutes (64 < 65 5/11), then the incorrect clock is gaining time, not losing time. What should have been covered in 65 5/11 mins is now being covered in 64 mins.

Thus the time gained in 65 5/11 minutes is 1 5/11 minutes (=16/11). So the time gained in 24 hrs (1440 minutes) is (1440 * (16/11) * (11/720) = 32 mins.

I believe when the incorrect clock covers a specified (correct) time in lesser (incorrect) time, then the incorrect clock is gaining time.

The answer must be : clock gains time of 32 mins per day.

@Dawa Tshering.

Total minutes per day: 24 x 60 = 1440.

Minute and hour hands coincide 22 times per day so, 1440/22.
= 720/11,
= 65 5/11.

Here in question, it coincides every 64 mins.
So, loss in every coincide is;
65 5/11 - 64 = 16/11 or 1 5/11.

Till this you are correct!

However the no of coincidence per day is 22 for a normal clock.
In this peculiar clock, the coincidence happens every 64 mins.
Therefore, in 24 hours the no of coincidence is ( 24*60)/64 which is 22.5.
That means in 24 hours, this peculiar clock coincides 22.5 times.

Now, going by your logic,
Loss per every coincide x number of coincide per day = total loss.
So, 16/11 x 22.5 = 32 8/11 mins per day.