math - rev word problem

[Electrons]

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( I can't use exact example, so I reworded it)
A squashed cricket sit at the edge of a 12 in diameter ancient record that is playing at 33(1/3) rpm. Approximately how fast, in feet per min is the squashed cricket moving?

Answer is 100. How so?

 

[xdianaax]

Thank you for following the copyright rules.🙂

 

[eleanor_rigby]

r = 6 in = 1/2 ft
C = 2*pi*r
= 2(3.14)(1/2 ft)
= 3.14 ft (distance covered by cricket in one revolution/min)

3.14 * 33 1/3 = 105 ft/min

I'm not getting 100 as an exact answer. 🙁

 

[PorkChop2008]

r = 6 in = 1/2 ft
C = 2*pi*r
= 2(3.14)(1/2 ft)
= 3.14 ft (distance covered by cricket in one revolution/min)

3.14 * 33 1/3 = 105 ft/min

I'm not getting 100 as an exact answer. 🙁

It's probably because they rounded pi to 3. (3 x 33 1/3 = 100)