How many times between $2$ pm and $4$ pm does the minute hand coincides with second hand.?


$a.)\quad 118 \\ b.)\quad 119\\ c.)\quad 120\\ d.)\quad 121\\$

Number of rounds of full circle by second's hand in 2 hours =$120$.

Number of rounds of full circle by minutes's hand in 2 hours =$2$.

so is the right answer =$120-2=118$. ?

The right answer given in the book is $121$.

Update :Book's Solution

"In $2$ hours the minute completes $2$ rounds around the circumference of the clock's dial.In the same time, the second hand covers $120$ rounds/ If we count $2$ pm coincedence as the first one , the $4$ pm coincidence would be the last one.Their would be the total of $121$ coincidences in $2$ hours.

I still doubt on book.

  • $\begingroup$ Do you mean how many times does the second hand coincide with the minute hand? $\endgroup$ – Arthur Mar 18 '15 at 22:09
  • $\begingroup$ @arthur :In the book the question is exactly same as written in the title.edited it now. $\endgroup$ – R K Mar 18 '15 at 22:14
  • $\begingroup$ The way I look at it, there is a coincidence every 61 seconds. $\endgroup$ – Ron Gordon Mar 18 '15 at 22:21
  • $\begingroup$ @ron If it is $61$ then $2\times 60\times \frac{60}{61}\approx 118$. $\endgroup$ – R K Mar 18 '15 at 22:27
  • $\begingroup$ However that works out. $\endgroup$ – Ron Gordon Mar 18 '15 at 22:28

It is best to think about this problem discretely. The clock is divided into 60 positions where the two hands could possibly coincide. Lets call the start time the 0th second, where all the hands are coinciding.

Unfortunately there are two different types of clock. In one type, the minute hand moves smoothly with the second hand. Lets consider this case first.

After 60 seconds the minute hand has moved to position 1. This means that after 60 seconds there has still only been 1 counted coincidence (the starting position). After 120 seconds there has been a second counted coincidence, and the minute hand has moved to position 2. This pattern continues for 3540 seconds without deviation and we have 59 counted coincidences, with the minute hand on the 59th position and the second hand on the 0th position. We note that in 60 more seconds we will have the starting condition again; so after a second iteration of the above, we will have 118 counted coincidences, with the minutes hand on the 59th position and the second hand at the 0th position. After 60 more seconds, we have our final coincidence at exactly 4pm, and the final count is 119.

However, there is a second type of clock where the minute hand stays perfectly still until the second hand strikes the 0th position, after which point, the minute hand jumps to the next position. On this type of clock, the two hands coincide at the 0th second, then the minute hand jumps to position 1, and the two hands coincide again at the 1st second. We then follow the logic of the previous case, and we find that each iteration offers exactly one more coincidence. Since we have two iterations to consider, we have a final count of 121. (And at 4pm and 1 second we would have a count of 122, but this is outside of the range of counts). If your book offers 121 as an answer, it is probably thinking about this style of clock.

PS. It also depends on what is meant by 'between': usually this either includes both endpoints or it doesnt. If the coincidences at exactly 2pm and 4pm are disallowed, then either 117 or 119 are acceptable, depending on the type of clock.


We want the angles of each hand to line up, and we want to count how many times this happens after 2pm until 4pm. This solutions uses the concept of radians, period, and angular velocity to tackle this.

The periods of the second and minutehand respectively are $T_{s}=60s$ and $ T_{m}=3600s$. Angular velocity $\omega = \frac{2\pi}{T}$ gives $\omega_{s}=\frac{2\pi}{60}$ and $ \omega_{m}=\frac{2\pi}{3600}$.

So, the angle of each hand equals angular velocity times time $\theta = \omega t$. Let's let zero degrees be the starting position, so that angle increases literally clockwise. This gives $$\theta_{s} = \frac{2\pi}{60} t$$ $$\theta_{m} = \frac{2\pi}{3600} t$$

We want the two angles to coincide. That means that the difference of the two has to be some integer multiple $k$ of $2\pi$.

$$\theta_{s} - \theta_{m} = 2\pi k$$ $$\frac{2\pi}{60} t - \frac{2\pi}{3600} t= 2\pi k$$ $$2\pi t(\frac{1}{60} - \frac{1}{3600} )= 2\pi k$$ $$t\frac{59}{3600}= k$$

At time $t=0$, the hands coincide. At time $t=7200 s$ or two hours, we see that $$k= (7200)\frac{59}{3600}= 118$$

Thus the hands have been coincident 118 times after the starting position.

Counting the starting position, the answer is 119. But let's discount this, which is what I'm guessing is intended. Thus the answer is 118.


Assuming you want to include both 2pm and 4pm as times when the two hands line up, the answer is $119$. Here's a way to see it with a minimum of arithmetic. (The most you need to calculate is that the second hand goes around the dial $120$ times in $2$ hours.)

To make sure we count both 2pm and 4pm, let's start at one second before 2 and end at one second after 4. So with that small buffer, we can think of what we're counting as the number of times the second hand passes the minute hand as both go round in "clockwise" fashion.

Or, more exactly, the number of times the second hand passes the minute hand minus the number of times the minute hand passes the second hand.

That may seem like a silly way to put it, but consider this: It doesn't matter at what rate the two hands move, it merely matters how many times each goes around the dial.

In particular, if we "freeze" the minute hand (pointing straight up) for the moment, and let the second hand do its thing, it will go fully around $120$ times, passing the minute hand each time, and then the final two seconds, passing the minute hand once more, for a total of $121$ passes, ending at one second after the hour. If we now let the minute hand move, it goes around the dial twice, passing the (now frozen) second hand twice. Subtraction gives $121-2=119$.


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