Question 92665
When the rear wheel makes 1 rotation, the bicycle moves forward a distance equal to the 
circumference of the rear wheel. The circumference of the rear wheel is given by the equation:
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{{{C[r] = pi * D}}}
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where {{{C[r]}}} is the circumference of the rear wheel and D is the diameter of the rear wheel.
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When the bicycle moves forward an amount equal to the circumference of the rear wheel, the smaller 
front wheel must rotate enough times so that its circumference travels the same amount of 
distance as the circumference of the rear wheel.
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The circumference of the front wheel is given by the equation:
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{{{C[f] = pi * d}}}
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where d is the diameter of the front wheel and {{{C[f]}}} is the circumference of the front 
wheel. But the problem tells you that to customize the bicycle the new front wheel 
has a diameter equal to half the diameter of the rear wheel or {{{d = D/2}}}. So in the 
equation for the circumference of the front wheel we can replace d by {{{D/2}}}. When that is 
done the equation for the circumference of the front wheel becomes:
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{{{C[f] = pi * D/2}}}
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Now the problem is translated to "When the back wheel rotates once the bicycle travels forward
a distance of {{{C[r] = pi * D}}}. How many times does the front wheel have to rotate so
that the distance its circumference travels is also {{{C = pi*D}}}?"
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If the front wheel makes two rotations its circumference distance is doubled and the
distance it covers in those two rotations is:
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{{{2*C[f] = 2 * pi*D/2}}}
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Note that in the right side the multiplier of 2 cancels the denominator of 2 so the equation
becomes:
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{{{2*C[f] = pi*D}}}
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and the right side of this equation is equal to the distance that the bicycle has traveled
with one rotation of the rear wheel.  Therefore, for every rotation of the rear wheel the
front wheel must rotate twice.  
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Hope this explanation helps to clarify the problem a little bit for you.