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NotSoSmart  May 25, 2017
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Let the original tire be 100cm circumference and the new one be 120cm circumference

(that is 20% =bigger)

Say you want to go 1200cm

that would be 120  rotations for the original one and 100 rotations for the new one.

 

\(\frac{120-100}{120}*100\% =\frac{200}{12} \%=16.\dot6\%\)

 

 

The number of tire rotations needed has decreased by  16 and 2/3 %

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Explaining it a different way :)

If the circumference has increased by 20% i.e. it is now 120% of the original

then

the gound covered by the tire will be  120%

so the rtations needed will be 100%/120% = 10/12 = 83.3% of the original.

which is 100-83.3 = 16.6% less than the original.   (that is .6repeater)

Melody  May 26, 2017

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