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A watch is stopped for 15 minutes every hour on the hour. How many actual hours elapse during the interval the watch shows 12 noon to 12 midnight.

In addition to the 12 hours that take us from 12 noon to 12 midnight, we need to add the 15 minutes picked up at each hour. There are 13 such (including the minutes picked up at 12 noon and 12 midnight).

But the "back of the book" says that you count the 12 hours, and you count 11 (instead of 13) 15 minutes picked up. The fact they got 11 seems to mean that they are disregarding the 15 minutes picked up at noon and at midnight. This doesn't seem right since "the interval from when the watch shows 12 noon to 12 midnight" starts before we stop the clock for 15 minutes at 12 noon, and ends after the clock is stopped for 15 minutes at midnight.

Am I interpreting the problem correctly?

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    $\begingroup$ It's hard to know if the $15$ minutes at $12$ noon should be counted or not, due to the wording of the problem, but you would not count the $15$ minutes at midnight since you would stop counting as soon as the watch showed midnight (the fact that it freezes there for another $15$ minutes is irrelevant). $\endgroup$ – Clayton Apr 18 '18 at 19:54
  • $\begingroup$ Ah. I see what you're saying. $\endgroup$ – trynalearn Apr 18 '18 at 19:57
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    $\begingroup$ Is the watch stopped for $15$ minutes every actual hour, or only when its own minute hand is pointing at the $12$? (ETA: I guess you can tell what they meant based on the answer they get, but it's sort of unclear.) $\endgroup$ – Brian Tung Apr 18 '18 at 20:00
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    $\begingroup$ You are misinterpretting the question. The watch doesn't lose fifteen minutes every time IT reads an hour (run for an hour, stop for fifteen, run for an hour, stop for fifteen), It stops for 15 minutes at the top of a REAL hour (stop for 15 minutes, run for the remaining 45 minutes, stop for 15 minutes, run the remaining 45 minutes). $\endgroup$ – fleablood Apr 18 '18 at 22:15
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    $\begingroup$ Yeah, you seem to be correct. But the problem is badly stated. And it is wrong as if that were the case then "during the interval the watch shows 12 noon to 12 midnight" would have to include the half hour it stops for noon and midnight because those ARE in the interval that it is showing 12 noon and 12 midnight. You simply can not claim they are not. $\endgroup$ – fleablood Apr 19 '18 at 0:59
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Edit: After reading fleablood's answer, I'm rather convinced that my interpretation of the question is wrong. That first sentence, "A watch is stopped for 15 minutes every hour on the hour," makes it sound like the watch stops on hours of real time, not hours in terms of the watch's time. I assumed this second interpretation in my answer below. But it seems that the "back of the book" missed this subtlety too? Anyways, future readers should see fleablood's answer in addition to this one.


As mentioned in the comments, yeah, the puzzle statement isn't exactly clear. It makes sense to stop counting time the moment the watch strikes midnight, but we don't know whether we should start counting from the first moment the watch reads 12 noon (in which case we count that 15 minute pause), or if we should start counting from the last moment the watch reads 12 noon (once that 15 minute pause is over). So depending on your interpretation, the answer is either

$$12\,\mathrm{hrs} + (11)\frac{1}{4}\,\mathrm{hrs} = \left(14+\frac{3}{4}\right) \,\mathrm{hrs} \quad\text{or}\quad 12\,\mathrm{hrs} + (12)\frac{1}{4}\,\mathrm{hrs} = 15\,\mathrm{hrs}\,.$$

The first one is what the "back of the book" says, but I personally like the second interpretation better since you get a whole number of hours. Maybe there's a better way to phrase the puzzle to imply this second interpretation. Here's my (wordy) version:

You have an old antique stopwatch, and the hands of the stopwatch are both pointing directly to 12. You start the stopwatch, but nothing immediately happens, both hands frozen on 12 (the stopwatch is old, after all). After exactly 15 minutes pass though, the stopwatch begins working normally and the hands begin to move (thank goodness it's not completely broken). But one hour later, when the minute hand is pointing to 12 again, the stopwatch freezes for another 15 minutes. You conclude that there is some quirk about the watch that causes it to freeze up every time the minute hands points to 12. Now watching a stopwatch count time isn't exactly exciting, so naturally you fall asleep. You awaken later just in time to see the hour hand and minute hand meet up at 12 again. How long have you been asleep?

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  • $\begingroup$ I agree its confusing but the "back of the book" suggests that your interpretation is correct. I want to ask a follow up. How many hours elapse (according to the watch) from 12 noon to 12 midnight real time? $\endgroup$ – trynalearn Apr 19 '18 at 0:57
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In a real hour, the watch will be stopped for $15$ minutes and running for $45$ minutes. So in a real period of $k$ hours the watch will have been running for $\frac 34k$ hours. So in a $\frac 43*12 = 16$ hour period the watch will have run for $12$ hours.

Now there are three options. 1) If when the clock read noon, it was the top of the hour: then clock would stop for fifteen minutes. Then the watch would start at 15 past the hour. Then $16$ hours later the clock will have advanced $12$ hours. As this is fifteen minutes past the hour, the watch won't stop then. So the time elapsed is $16$ hours and $15$ minutes.

2) If the clock reached midnight but then stopped for fifteen minutes because it was the top of the hour: That would mean that it was the top of the hour $16$ hours ago when the watch first started. But that is contradictory as that means the watch started running at the top of an hour.

3) If neither when the watch noon or midnight was the top of the hour: Then the watch ran for $16$ hours.

Presumably the watch was correct when it first hit the top of an hour and lost the fifteen minutes. So from the watches point of view, it runs for 45 minutes, then stops for $15$ minutes. The mark on the watch when the watch stops can be any mulitple of $45$ minutes which, modulo $60$, can be any $15$ minute mark.

So 1) and 3) are certainly both possible.

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  • $\begingroup$ Yeah, as the problem is written you're right. I totally missed that subtlety, and it looks like the "the back of the book" missed it too. $\endgroup$ – Mike Pierce Apr 18 '18 at 23:45

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