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The short and the long hands of a wall clock are $8$ cm and $12$ cm respectively. Find the sum of the distance traveled by their tips in $3$ days. Give your answer in terms of $\pi$.

My solution:

Short hand:

Distance traveled in $12$ hours $= 2πr = 16π$ cm

$\Rightarrow$ Distance traveled in $3$ days$ = 3 \times 2 \times 16π = 96π$ cm

Long hand:

Distance traveled in $12$ hours $= 2πr = 24π$ cm

$\Rightarrow$ Distance traveled in $3$ days $= 3 \times 2 \times 24π = 144π$ cm

Sum of distances = $240π$ cm

But the correct answer is $1824π$ cm. How?

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The long (minute) hand travels much faster, makes one revolution per hour. – André Nicolas Aug 2 '12 at 17:16
@ André Nicolas oh, that's right. i completely overlooked that – Nirvan Aug 2 '12 at 17:18
up vote 3 down vote accepted

Short Hand

One full rotation in 12 hours $\implies 2 \pi r = 16 \pi$ cm traversed every 12 hours.

For one day, we have $32 \pi$ cm, twice that of a 12 hour period.

For 3 days, we then have $3\cdot32 \pi = 96 \pi$ cm traversed.

Long Hand

One full rotation in 1 hour $\implies 2 \pi r = 24 \pi$ cm traversed every hour.

For one day, we have $24\cdot24 \pi = 576$ cm.

For 3 days, we then have $3\cdot576 \pi = 1728\pi$ cm traversed.

Total Distance

For the total, we have $96 \pi + 1728 \pi = 1824 \pi$ cm.

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ok thanks, i got it – Nirvan Aug 2 '12 at 17:48

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