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This problem is on the practice problems for my math club, but I'm not sure how to solve it.

What's the exact time that the hour and minute hands overlap each other between 3 and 4 o clock on an analog clock?

Thanks!

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  • $\begingroup$ Find a function $h(t)$ that gives the angle of hour handle with 12 at time $t$ minutes past 3. Do the same for the minute handle to get an $m(t)$. Then you need to find out how your problem relates $h$ and $m$ to each other ... $\endgroup$ – Maesumi Aug 4 '13 at 2:59
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    $\begingroup$ Danny Cheuk's answer seems sufficient here, but my answer to this other clock-related question may also be helpful. $\endgroup$ – MJD Aug 4 '13 at 3:11
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Let $x$ be the number of minutes after 3:00 that the hour and minute hands overlap.

Since the hour hand moves $360^\circ$ over a $24$-hour period, so it moves $\frac{360^\circ}{12\times60}=0.5^\circ$ per minute. The hour hand would be at $90^\circ$ at time 3:00, therefore, it would be at $90^\circ+0.5^\circ\times x$ at the time that the hands overlap.

For the minute hand: it moves $360^\circ$ every hour, so it moves $\frac{360^\circ}{60}=6^\circ$ every minute. And since it starts at an angle of $0^\circ$ at 3:00, the minute hand should be at the angle of $6^\circ\times x$ when they overlap.

Now, set them equal to each other, you get:

\begin{align*} 90+0.5x&=6x\\ 180+x&=12x\\ 180&=11x\\ x&=\frac{180}{11} \end{align*}

$\therefore\text{The hour and minute hands overlap each other }180/11\text{ minutes after 3:00}.$

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big hand moves 12 times as fast as little hand so minutes after 3:15 (when minute hand will be on top of hour hand) equals $X$. Therefore $$12X - X = 11X;$$ $$11X = 15;$$ so $$X = \frac{15}{11};$$ so time will be 3:16 and $\frac{4}{11}$.

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