Question

Find the least number of candidates in an examination so that the percentage of successful candidates should be 76.8%:

Answer 1

in 100% candidate ... 76.8% successful
no of student can not be point value ..take fraction
= 76.8/100 = 768/1000 = 384/500 = 96/125
total student should be 125

Answer 2

HEY DEAR ... ✌

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suppose no. of total candidates are 'y' and successful are 'x'.
(x/y)*(100)=76.8 or x/y=0.768
or y=x/0.768
here we need minimum value of y, which is obtained by minimum value of x.
if put x=768 then y=1000
now we will try decrease value of x than we will find less value of y. so factorized 768 by 2 further.
if put x=384 then y=500
if put x=192 then y=250
if put x=96 then y=125(answer)
if put x=48 then y=62.5(not integer)
so ans is 125(minimum)


HOPE , IT HELPS ... ✌