shelrod007 wrote:
I am not getting the point as to why is C the answer ?
We need to find the min % of the 5th test , Deepthi has to average 75 % for final examination.
Statement A ) 1st 4 tests % is 71% which would obviously mean to get 75% she should have a minimum % of 79 in the last test . Seems sufficient to deduce .
Statement B) insufficient cannot know how much she scored in the first 4 tests
I doubt the OA of this quest . Can Bunuel comment on this ?
let me give a try:
St1: She averaged 71% in her first 4 examinations
Let us say summation of maximum marks (that can be achieved in all exams taken together) = x
total marks obtained = 0.71 x
Now, new percentage = 75%. Let us say summation of maximum marks in 5 subjects = y.
total marks to be obtained = 0.75 y
So marks obtained in last test = 0.75y - 0.71x
max marks of last test = y - x
So percentage to be achieved in last test = (0.75y - 0.71x) / (y-x) * 100
Still two variables left. We need a relation of either y in terms of x or vice versa.
Insufficient!
2. All tests carry equal marks
means : Last test max marks = max marks of each of first four
so, (y-x) = x/4
4y = 5x
y = (5/4)x
Again two variables leading to nothing.
Insufficient!
Combining:
percentage of in last test = (0.75y - 0.71x) / (y-x) * 100
&
y = (5/4)x
Solving above, we get 91 % as answer.
Answer: C
Hope this helps!