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

Asked by Shubham Verlekar | 1st Jul, 2013, 07:26: PM

Expert Answer:

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.  

Answered by  | 1st Jul, 2013, 10:29: PM

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