To earn grade B in her class Deepthi must have an average

Not a fan of how this question was written, a bit vague imo...that being said, I think author is testing ratios/%'s:

all exams weighted equally-
4(71/100) + 1(x/100) >= 5(75/100)
x>=91

Exams 1-4 individual weights : Exam 5 Weight
1:2

4(71/100) + 2(x/100) >=6(75/100)
x>=83

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!

To earn grade B in her class Deepthi must have an average score of 75% in examination. She has taken 4 examinations so far. What is the minimum percentage she must get in the fifth examination to earn grade B?
1. She averaged 71% in her first 4 exminations
2. All tests carry equal marks.

Solution:
Total marks obtained in 4 exam = A
Max marks in 4 exam = B
Marks obtained in Fifth Exam = F
Max marks of Fifth Exam = Y

1. 71% in 4 exam :

\(\frac{A}{B} * 100 = 71\)

To get B grade overall percentage should be 75%.

\(\frac{A + F}{B+Y} * 100 = 75\)
Can not obtain the value for \(\frac{F}{Y}\)
Not suff.

2. All test carry equal marks.
Total marks obtained in 4 exam = A
Max marks in 4 exam = 4M
Marks obtained in Fifth Exam = F
Max marks of Fifth Exam = M

\(\frac{A + F}{5M} * 100 = 75\)
Can not obtain the value for \(\frac{F}{M}\)
Not suff.

COMBINE 1 AND 2:

From Stmt 2:
\(\frac{A + F}{5M} * 100 = 75\)
\(\frac{A}{M} + \frac{F}{M} = 0.75*5\) ------------(A)

From Stmt 1. \(\frac{A}{4M} * 100 = 71\) << now we know that all test carries equal marks we can say B=4M to bring variables in sync to statement 2.

We can substitute \(\frac{A}{M}\) in equation (A) and get F/M

After reading the explanations, I unfortunately don't understand how we get the answer.
Is there another way to explain (even more dumbed down)?

If the weighted ratio between four tests changes then we cant find a sole answer to answer this question. Thats y combining the statements will give a sole and precise answer than banking only on option A

Can anyone please explain this one ... not able to get it correct ..

Updated my solution above I hope now it is clear what I did.

X needs to score 75% marks overall.
1.) avg for 4 tests is 71%.
a.) Assumption that fifth test has max marks 100.
(4x71+1xfifth)/5 = 75
fifth >= 91

b.) Assumption that fifth test has max marks of 20.
total max marks of 4 tests = 400
avg marks of 4 tests = 71*4 = 284
75% of 420 marks = 75/100*420 = 315
So x needs to score 315-284 = 31 marks to make overall avg at least 75%
But it is not possible because max marks for fifth test is 20.

So, all tests must have same max marks to obtain a proper solution. Hence C is correct option.

Thank you for reporting. Marked the question as Poor Quality. It won't be including in daily practice questions pool anymore.