In a certain examination paper, there are n questions for j=1.2...n. There are 2^n-1 students who answered j or more questions wrongly. If the total numebr if wrong answers is 4095, then the value od n is?

A. 12

B. 11

C. 10

D. 9

• A. i think the answer should be 12
  Explanation:: there are 2^n-1 students who answered the questions wrongly and no. of wrong answers=4095,given that they answered j or more wrongly.. so 4095 >=2^n-1
  solving it.. max value of n can be 12.
  plz check if i am wrong.
• 11 years agoHelpfull: Yes(18) No(0)
• according to question ...2^n-1 students have done 1 or more questions wrongly,
  2^n-2 students have done 2 or more questions wrongly....so on....
  so,(2^n-1)-(2^n-2) students have done 1 question wrong exactly....(2^n-2)-(2^n-3) students have done 2 questions wrongly.....(2^n-3)-(2^n-4) students have done 3 questions wrongly......therefore total number of wrong answers are 1*(2^n-1 - 2^n-2)+2*(2^n-2 - 2^n-3)+3*(2^n-3 - 2^n-4)+.....+n-1(2^1-2^0)==>2^n-1+2^n-2+2^n-3+......+1=4095......this is GP with a=2^n-1 & r=1/2.....
  solving this we get n=12.
  
  
• 11 years agoHelpfull: Yes(8) No(1)
• total number of ways in which 0 or more question can be answered wrongly is total number of ways selection of question i.e Co+C1+C2.....Cn=2^n
  C0+C1+C2...Cn=2^n-1 toatal ways of answering 1 or more qusn wrongly
  =>2^n-1=4095
  2^n=4096
  n=12
• 11 years agoHelpfull: Yes(7) No(0)
• koi solution post kar do plz
• 11 years agoHelpfull: Yes(1) No(0)
• Since,1== 4095 (max n wrong questions-all wrong(j=n) )
  conversely,(2^n-1)*1
• 11 years agoHelpfull: Yes(0) No(0)