In an certian exam the no who passed is three times no who were rejected if there had been 16 fewer candidates and if 6 more had been rejected the number those who passed and of no those who failed would have been 2:1 find the no of candidates

Let x is the number of candidates, and y is the number of candidates who were rejected. 

Then (x-y) is the number of candidates who passed the exam. 

 

Scenario 1: the no who passed is three times no who were rejected

Hence, x-y = 3y

x= 4y

 

Scenario 2: if there had been 16 fewer candidates and if 6 more had been rejected the number those who passed and of no those who failed would have been 2:1

 

So, number of candidates = x-16

No of candidates rejected = y+6

Number of candidates passed = x-16 - (y+6) = x-y-22

 

As per the question, x-y-22/ y+6 = 2/1

x-y-22 = 2(y+6)

x-y-22 = 2y+12

x = 3y+34

4y = 3y+34(Substituting x = 4y from above)

y = 34

 

x= 34*4 = 136.