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enter image description here

You need to find out what time it is.

The logic question allegedly asked of middle school children in Russia.

I think the question has more than one answer. So to clear my confusion, I will ask you to write down every possible answer.

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2 Answers 2

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First of all, some deduction:

As two of the hands are (basically) on the number (meaning the large 1-12 that are written on most clocks), then at least one of them must be the $12$. This is because for the minute hand to be on the number, the second hand must be on the $12$, and the same is true for the hour in terms of the minute. The second hand could be anywhere, but since two hands match the number, one must be the $12$.

A word on notation:

I will be using the same distinctions as @Prim3numbah did in his answer. That is, the top-left hand is $A$, then $B$ and $C$ continue clockwise from there.

We now have two scenarios:

Either $A$ is on the $12$, or $C$ is.

With the first scenario:

Assume $A$ is on the $12$. It cannot be the hour hand, as this would mean it was noon, which common sense should tell you it is not. If $A$ is the minute hand, $C$ would have to be the hour hand, as the time would be perfectly on the hour. The time would then be 10:00:24, and because $A$ is so close to the number, I would argue that this is not the case. If $A$ is the second hand, however, $C$ would be the minute hand, and the time would be 4:50:00, and this lines up, as $B$ is approximately $\dfrac56$ of the way between the numbers.

With the second scenario:

Assume $C$ is on the $12$. It can't be the hour hand for reasons described above. If $C$ is the minute hand, $A$ is the hour hand, and the time would be 2:00:34, which doesn't line up well. If $C$ is the second hand, $A$ would be the minute, and the time would be 6:10:00, which also does not match the picture.

This means:

As I see it, there is one time that the clock can be, which is precisely 4:50:00.

An Interesting Point:

This puzzle is interesting in that it will work whether the clock runs forwards or backwards. While clockwise is likely to be the assumption for a middle school problem, if it is not required, then as the hands are fairly symmetrical, a time of 4:50:00 will also work.

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  • $\begingroup$ Isn't it the same reason why it can't be 10:00:24 and 2:00:34? In the first because A is very close to the number, in the second because C is very close to the number. am I right? $\endgroup$ Commented Mar 14, 2022 at 9:19
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    $\begingroup$ Precisely. If the time was truly either of those, one could expect the minute hand (whichever it is for the scenario) to be about halfway between the tick marks on the clock. As this is not the case, the suggested times cannot be accurate. $\endgroup$
    – PiGuy314
    Commented Mar 14, 2022 at 17:10
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I think the answer is

There does not exist a setting of hour hand, minute hand and second hand that looks like the one in the image.

If we call each hand

enter image description here

We see that

there are 3!=6 possibilities in total where each hand is either HH, MH or SH.. Furthermore, A and C are pointing slightly to the right of a whole number 1-12 (the shift to the right looks/is identical) which means we can deduce that those hands are not hour hand AND minute hand. Meaning A and B can NOT be HH:MH or MH:HH. This leaves us with 4 possibilities left.

A=HH, B=MH and C=SH. Since A is, as mentioned pointing slightly to the right of a whole number(1-12) this means the only configuration that works is if B is pointing slightly to the right of "12", which we can see is when B≈04 and C≈30. Which gives the time 08:04:30.

A=MH, B=HH, C=SH. Now in this case, B is pointing slightly to the left of a whole number which means A should be pointing close to "12" (at least between 11 and 12) and we can see that it does if A≈55 and C≈45. Which gives the time 03:55:45

A=SH, B=HH, C=MH. Same reasoning here (as 2) but this time the only possibility is if the time is ≈ 05:55:05.

A=SH, B=MH, C=HH. Since C is pointing slightly to the left of a whole number (1-12) this means the only possibility that works is if C≈06, B≈04 and A≈40, which gives the time 06:04:40.

Now,

For 1.

Since C (the second hand) is almost exactly 30, the minute hand should be halfway between 4 and 5 but its clearly closer to 4 so this possibility can't be correct.

For 2.

Since A (the minute hand) is almost exactly 55 (somewhere between 55 and 56) this means the hour hand should be closer to 20(4) than 19, but we can see it's closer to 19 rather than 20(4) so this possibility can't be correct.

For 3.

Exact same reasoning here (as 2), so this possibility is not the correct one either.

For 4.

Since A (the second hand) is almost exactly at 40, B (the minute hand) should be closer to 5 than 4 but it's clearly not.

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  • $\begingroup$ Considering that this was allegedly presented to children (~14 years of age), I don't believe we should assume that the image is as precise as your answer implies. $\endgroup$ Commented Mar 13, 2022 at 12:26

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