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Suppose you have the following problem: if the hour and minute hands are opposite to each other at 6 o'clock, when will they coincide?

My attempt at a solution:

Let $x$ = the number of minute spaces moved by the hour hand.

Let $y$ = the number of minute spaces moved by the minute hand.

Since the minute hand moves twelve times as fast as the hour hand, $y=12x$

Now, this is where it can get a little tricky. According to the problem, the hands are currently opposite each other; or, in other words, $y=60+x$. This is because, to reach the hour hand, the minute hand has to move 60 minute spaces (since that is how long an hour is), but then, to be opposite the other hand, move the same amount of minute spaces as the hour hand has (i.e., in this problem, 30 minute spaces, which is half an hour).

So, to coincide, the hour and minute hands must have the same number of minute spaces. I think that this should be represented by $y=60-x$, since, in the current scenario, the minute hand is $x$ minute spaces ahead of the hour hand, and thus it must move back $x$ spaces to coincide with the other hand.

So, the two equations are $y=12x$ and $y=60-x$.

After substituting $12x$ for $y$ in the second equation, we get:

$12x=60-x$

$13x$=60

$x=\frac{60}{13}$

Substituting this value back in the 1st equation, we obtain $\frac{720}{13}=y$. After converting that to a mixed number, we get $55 \frac{5}{13}$ minutes.

My Questions

To me, it doesn't make sense how I should convert this value into an answer; I have a number of minutes, but not a specific time on the clock.

I also don't know if my second equation for when the minute and hour hands coincide is correct, and therefore I have an overall doubt with regards to the accuracy of my solution. If my solution is wrong, could someone please explain my error so that I can get the right answer?

Thanks

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    $\begingroup$ After $55$ minutes, or so, we're nearly at $7$:$00$, so the hour and should be on the $7$ and the minute hand will be nearly back to the $12$, so that doesn't work. I think you'll do better to focus on the two angles. $\endgroup$
    – lulu
    Feb 24 at 14:53

3 Answers 3

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The clock is now at 6 o'clock, so the minute hand points up and the hour hand points down. I'll use your definitions but make them a bit more precise:

Let x be the number of minute spaces moved by the hour hand between now and the moment the hands coincide.
Let y be the number of minute spaces moved by the minute hand between now and the moment the hands coincide.

As you say, the minute hand moves twelve times as fast as the hour hand, so y=12x.

The minute hand needs to travel 30 minute spaces around the clock to get where the hour hand starts off at, and then needs to travel as much as the hour hand to catch up. Basically the hour hand has a head start of 30 minute spaces that the minute hand needs to catch up on. This gives the second equation y=30+x.

Solving these two equations gives y=12*30/11 = 32.7272. This is the number of minute spaces that the minute hand has to travel, i.e. the number of minutes that have elapsed. This is 32 minutes and 43.6363 seconds after 6 o'clock.

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Write down the clockwise angle from the top that the hours hand and the minutes hand make. Let $H$ be the angle (clockwise from the top) that the hours hand makes, and let $M$ be the angle (clockwise from the top) that the minutes hand makes. Further let the time be h:m where $h$ is the hour (an integer) and $m$ the minutes (a real number), then

$H = 30 (h + m / 60 ) = 30 h + m / 2 $

and

$M = 6 m $

When the two hands coincide, we have $H = M$, therefore,

$ 30 h + m / 2 = 6 m $

Now we're interested in the first time after $\text{6:00}$, so $h = 6$, and from this we get

$ 180 + m / 2 = 6 m $

Hence,

$ m = \dfrac{180}{5.5} = 32.72 = 32 \text{ minutes}$ and $ 43 \text{ seconds} $

Hence the time at which the two hands will coincide is $\text{6:32:43}$

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Let's start with an estimate. After 30 minutes, the minute hand will be pointing straight down (toward 6) and the hour hand will have moved a little bit (half way between 6 and 7). So the hour hand is still a bit ahead of the minute hand. But this suggests that the catch should come a little after 6:30.

Now let's be more precise. At $6:00$, the minute hand is at position $0$ on the clock face (measuring position in minutes after the hour), and the hour hand is at position $30$. At $t$ minutes after $6:00$ (for $0\le t\le 60$), the minute hand is at position $t$ and the hour hand is at position $30+\frac{t}{12}$.

Setting these expressions equal to each other gives $t=30+\frac{t}{12}$ (this will give the time after $6:00$ when the hands coincide). Solving for $t$ gives $t=\frac{360}{11}$ minutes. That is, $32$ and $\frac{8}{11}$ minutes.

So the catch comes at time $6:32 \frac{8}{11}$, or if you prefer $6:32:43 \frac{7}{11}$. Note that this is in line with our estimate.

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