Question

In a certain test, there are n questions. In the test \[{{2}^{n-i}}\] students gave wrong answers to at least i question, where i \[=\] 1,2,….., n. If the total number of wrong answers given is 2047, then n is equal to
(a) 10
(b) 11
(c) 12
(d) 13

Answer 1

Hint: For every value of ‘i’ find the number of students in power of 2 then add all the cases to make it 2047. Then apply formula for the sum of the cases of geometric progression, then find the value of n which is asked.

Complete step-by-step answer:
In the question, we are told that in a certain test, there are total n questions. Now, we are informed that for at least ‘i’ questions, where I contain all the values from ‘1’ to ‘n’, total \[{{2}^{n-i}}\] students gave wrong answers. Now, we are given a total number of wrong answers given is 2047, so we have to find a value of n.
So, according to the given question for at least 1 question ${{2}^{n-1}}$ students gave wrong answers, for at least 2 questions ${{2}^{n-2}}$ students gave wrong answer similarly, for at least 3 questions the number of students will be ${{2}^{n-3}}$ ways and so on, and finally for at least n questions the number of students will be ${{2}^{n-n}}$ or ${{2}^{0}}$ or 1 way.
Let now sum up all the ways for all the cases given so we get,
${{2}^{n-1}}+{{2}^{n-2}}+{{2}^{n-3}}+.......+{{2}^{0}}$
We are given the value of the total sum in the question which is 2047.
The given sum of series is represented in geometric progression or G.P. with the first term ${{2}^{n-1}}$ , ratio as $\dfrac{1}{2}$ and number of terms as ‘n’ .
So, now we will use the formula for sum of terms of geometric progression which is $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ where a is the first term, r is the ratio and n is number of terms.
So, we can write,
$\dfrac{{{2}^{n-1}}\left( 1-{{\left( \dfrac{1}{2} \right)}^{n}} \right)}{1-\dfrac{1}{2}}=2047$
$\Rightarrow {{2}^{n}}\left( 1-{{\left( \dfrac{1}{2} \right)}^{n}} \right)=2047$
$\Rightarrow {{2}^{n}}-1=2047$
$\Rightarrow {{2}^{n}}=2048$
Now we know that ${{2}^{11}}$ is 2048
So, we can write 2048 as ${{2}^{11}}$.
So, ${{2}^{n}}={{2}^{11}}$ $\Rightarrow n=11$
Hence, the correct option is (b).

Note: Student’s should be careful while considering all the cases which add up to 2047 as missing out any case can result in a change of value n.