In a certain examination paper there are n questions For j 1 2 hellip n there are 2n ndash 1students...........

Quantitative Aptitude ( CCS) Number System

In a certain examination paper, there are n questions. For j = 1, 2 …n, there are 2n–1students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is (129/12/3)

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Explanation:

There will many probabilitiesfrom getting n ans wrong to getting one ans …..
1+2+4+8+……+2^n=4095
But in this equation some of the student are repeated 1 student got n ans wrong and 2 student got n-1 or more ans wrong in this n-1 and n both are included.
So how will we calculate no of student who exactly got n ans wrong….?
2^(n-j) student got j or more wrong
2^(n-j-1)got j+1or more wrong
So , 2^(n-j) - 2^(n-j-1) students got exactly j wrong.
Sum….
{2^(n+1)-1}- {2^(n)-1}=2^n=4095
Approximately
n=12