no of questions already attempted = x
max no of questions that one can wrong in the remaining 24 = n

25% of x + n = 25% of (x+24)
x/4 + n = x/4 + 6
n = x/4 + 6 - x/4
n = 6

Option B[/spoiler]

You could also pick numbers.

Assume she has attempted 100 questions until now and got 75 correct.

She has 24 remaining, so total = 124. 75% of 124 = 93. Therefore she has to get 18 more answers right.

So, she can get a max of 6 incorrect. B IMO

P.S Actually you dont have to calculate anything here. You have 2 things (one of which is 75% and the other some % and together they form 75% again)

So the other should also be 75% right (or) 25% wrong, (or) 1/4th of 24 (or) 6

Let x be the total number of questions answered till now.
Number of questions answered correctly = 0.75x.

Let y be the number of questions that are yet to be answered correctly so that the total number of questions answered correct will be 75%

Hence,

0.75x + y = (24+x)(0.75)
y = 18

The maximum number of questions that can go wrong are 24-18 = 6
IMO B

I prefer the weighted averages method for this. U can solve it in under 30 secs

Let the Total number of questions be x.
The questions remaining are 24 so the questions already answered is x-24

The weighted average has to be 25 meaning she can have 25 incorrect answers in the entire test

75% of x-24 have been answered correctly so it follows that 25% of x-24 have been answered incorrectly. So we can have at most z% of the remaining 24 questions answered incorrectly. We can form the equation as follows

[(x-24)/x}*25 + (24/x)*z=25 (Don't be alarmed by the two variables as x gets cancelled out)
Solving for z we get
z=25% i.e. She can get atmost 25% of the remaining questions wrong.
(25/100)24=24/4= 6 Hence B