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The minute hand of a clock is $15$ cm long. The distance moved by the tip of the hand in $35$ minutes is

$a.)\ 35\pi \\ \color{green}{b.)\ \dfrac{35\pi}{2}} \\ c.)\ \dfrac{5\pi}{4} \\ d.)\ \dfrac{5\pi}{2} $

For minute hand,

$12\ hrs =360^{\circ} \\ 35\ min=\left(\dfrac{35}{2}\right)^{\circ} $

Distance$=\dfrac{2\pi 15\times 35}{360\times 2}=\dfrac{35\pi}{24}\ cm $

But that is not in options

I look for a short and simple way.

I have studied maths upto $12th$ grade.

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  • $\begingroup$ The minute hand takes only one hour, not twelve, to make a complete circuit around the face of the clock. So in 35 minutes it moves 12 times as far as your calculations showed. Except for that mistake, your calculations are correct. $\endgroup$ – David K Jan 15 '16 at 18:36
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It has moved $\frac{35}{60}$ of a full circle (a full circle consists of $60$ minutes, and it has moved $35$ of those). A full circle is $2\pi\cdot15cm=30\pi cm$. It has therefore moved a total of $$ \frac{35}{60}\cdot30\pi cm=\frac{35\pi}{2}cm $$

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  • $\begingroup$ what is mistake in my answer ? $\endgroup$ – R K Jan 15 '16 at 18:16
  • $\begingroup$ @RK Your mistake is that 35 minutes doesn't translate to $\left( \frac{35}{2} \right)^\circ$. That's only $17.5^\circ$, while 35 minutes is more than half-way around the clock: Definitely at least $180^\circ$. $\endgroup$ – pjs36 Jan 15 '16 at 18:56
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First you can set up a proportion to find the central angle it has moved,

35/60=x/360→x=210

Then you can use the arc length formula arc length = 2πr(x/360) = 2π(15)(210/360)=35π/2.

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The question is referring to the MINUTE hand, NOT hour hand. Read it properly. The MINUTE hand does not only move 360 degrees in 12 hours. That's the HOUR hand. The MINUTE hand moves 360 degrees in an hour. This is because the MINUTE cycle restarts after an hour. It moves the full clock around after ONE hour, NOT TWELVE hours. So, there lies your mistake.

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Your mistake ? You are missing one fact, the minute hand has already went through $\left(360\cdot 12\right)^{\circ}$ every 12 hours.

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  • $\begingroup$ This is a very terse answer. There is room for improvement in connecting the given calculation of the OP with what you would do instead. $\endgroup$ – hardmath Jan 15 '16 at 20:16
  • $\begingroup$ So what? We are talking here of an interval of only $35$ minutes, so I don't see how your answer is relevant to the question. $\endgroup$ – Alex M. Jan 16 '16 at 14:18
  • $\begingroup$ @AlexM. from what I see, he directly did the computation with $\frac{35}{12\cdot 60}\cdot\left(360\right)^{\circ}$ instead of $\frac{35}{12\cdot 60}\cdot\left(360\cdot 12\right)^{\circ}$, that is the reason why he got $\left(\frac{35}{2}\right)^{\circ}$ instead of 210$^{\circ}$. I'm just trying to let him know where he got it wrong rather than changing the way he solved this question. $\endgroup$ – Farisan Dary Jan 16 '16 at 14:35

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