Total number of questions 
In a certain test, there are $n$ questions. In the test $2^{n-i}$ students gave wrong answers to at least $i$ questions, where $i=1,2,3,\ldots,n$. If the total number of wrong answers given is $2047$, then $n$ is equal to? 

I am unable to comprehend the language of the question especially because of "at least ". I want to understand the problem and how do I approach it? 
 A: As lulu suggests, you know that the number of students who answered precisely $i$ questions incorrectly is
$$2^{n-i}-2^{n-i-1}.$$
Indeed, the number of students that answered precisely $i$ questions incorrectly is equal to the number of those who answered at least $i$ questions incorrectly minus the number of those who answered at least $i+1$ questions incorrectly. This is a general trick that is often used. Note that there is an exception: for $i=n$, the number of students who got all answers wrong is of course just $2^{n-n}=1$. What have you gained? Well, now you know that the total number of wrong answers is
$$ 2047 = n +\sum_{i=1}^{n-1} i \cdot(2^{n-i}-2^{n-i-1}).$$
Can you then find out how big $n$ is?
If your stuck, here it is how you can go on (probably with more steps than necessary):
$$n +\sum_{i=1}^{n-1} i \cdot(2^{n-i}-2^{n-i-1}) = n+\sum_{i=1}^{n-1} i \cdot2^{n-i} - \sum_{i=1}^{n-1} i \cdot2^{n-i-1} = $$
$$= n+\sum_{i=1}^{n-1} i \cdot2^{n-i} - \sum_{i=2}^{n} (i-1) \cdot2^{n-i} = n+\sum_{i=1}^{n-1} i \cdot2^{n-i} - \sum_{i=2}^{n-1} (i-1) \cdot2^{n-i} - (n-1) = $$
$$= 1 + \sum_{i=1}^{n-1} i \cdot2^{n-i} - \sum_{i=2}^{n-1} (i-1) \cdot2^{n-i} = 1 + \sum_{i=1}^{n-1} i \cdot2^{n-i} - \sum_{i=1}^{n-1} (i-1) \cdot2^{n-i} = $$
$$=1+\sum_{i=1}^{n-1} 2^{n-i} = 1+\sum_{i=1}^{n-1} 2^i = \sum_{i=0}^{n-1} 2^{i} = \frac{1-2^n}{1-2} = 2^n-1.$$
Hence we get
$$ 2^n = 2048 \quad \text{or equivalently} \quad n= \log_2(2048) = 11.$$
