A) \[12\]
B) \[11\]
C) \[10\]
D) None of these
Correct Answer: B
Solution :
The number of students giving exactly \[i\] wrong answers\[={{2}^{n-i}}-{{2}^{n-i-1}}\] Hence, total number of wrong answers \[=\sum\limits_{i=1}^{n}{i\times }({{2}^{n-i}}-{{2}^{n-i-10}})+n\times 1\] \[\therefore \] \[\sum\limits_{i=1}^{n}{i}\times ({{2}^{n-i}}-{{2}^{n-i-1}})+n=2047\] \[\Rightarrow \] \[1({{2}^{n-1}}-{{2}^{n-2}})+2({{2}^{n-2}}-{{2}^{n-3}})\] \[+...+(n-1)(2-1)+n=2047\] \[\Rightarrow \] \[{{2}^{n-1}}+{{2}^{n-2}}+...+2-(n-1)+n=2047\] \[\Rightarrow \] \[1+2+{{2}^{2}}+...+{{2}^{n-1}}=2047\] \[\Rightarrow \] \[\frac{{{2}^{n}}-1}{2-1}=2047\] \[\left[ \because \,\,sum\,\,of\,\,GP\,\,series,\,\,{{S}_{n}}=a\frac{({{r}^{n}}-1)}{r-1},\,\,when\,\,r>1 \right]\]\[\Rightarrow \] \[{{2}^{n}}=2048\] \[\Rightarrow \] \[{{2}^{n}}={{2}^{11}}\] \[\Rightarrow \] \[n=11\]
You need to login to perform this action.
You will be redirected in
3 sec
