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Question: In a certain test, there are n questions. In the test \[{{2}^{n-i}}\] students gave wrong answers to...

In a certain test, there are n questions. In the test 2n−i{{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

Explanation

Solution

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 2n−i{{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 2n−1{{2}^{n-1}} students gave wrong answers, for at least 2 questions 2n−2{{2}^{n-2}} students gave wrong answer similarly, for at least 3 questions the number of students will be 2n−3{{2}^{n-3}} ways and so on, and finally for at least n questions the number of students will be 2n−n{{2}^{n-n}} or 20{{2}^{0}} or 1 way.
Let now sum up all the ways for all the cases given so we get,
2n−1+2n−2+2n−3+.......+20{{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 2n−1{{2}^{n-1}} , ratio as 12\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 a(1−rn)1−r\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,
2n−1(1−(12)n)1−12=2047\dfrac{{{2}^{n-1}}\left( 1-{{\left( \dfrac{1}{2} \right)}^{n}} \right)}{1-\dfrac{1}{2}}=2047
⇒2n(1−(12)n)=2047\Rightarrow {{2}^{n}}\left( 1-{{\left( \dfrac{1}{2} \right)}^{n}} \right)=2047
⇒2n−1=2047\Rightarrow {{2}^{n}}-1=2047
⇒2n=2048\Rightarrow {{2}^{n}}=2048
Now we know that 211{{2}^{11}} is 2048
So, we can write 2048 as 211{{2}^{11}}.
So, 2n=211{{2}^{n}}={{2}^{11}} ⇒n=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.