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

                                                                                           6