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A wheel requires certain recolutions to cover a distance.If the circumference of wheel is incresed by 2o percent what is the reduction in no
of revolutions in percent
REDUCTION IN REVOLUTION WILL BE :- 16.66% APPROX
LET THE CIRCUMMFRENCE OF THE WHEEL BE - X m
SO CIRCUMFRENCE OF THE INCREASED WHEEL - (100+20)% OF X = 6X/5 m.
No . OF REVOLUTIONS REQUIRED TO COVER 6X/5 m WILL BE = 1
No. OF REVOLUTIONS FOR X m WILL BE = X = 5/6
6X/5
SO REDUCTION IN REVOLUTION WILL BE = 1- 5/6 = 1 = 16.66% APPROX
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