Lets assume total grade = 100
Max marks for Final grade is 40
Max marks for Other Grades total 60

Given: Moe has 45% marks out of 60 that is 45% of 60 = 27
To get the passing grade of 60% overall (i.e. 60 out of 100), he needs 60 - 27 = 33 more marks in the final grade.
so 33/40 * 100 = 82.5%
And B

guerrero wrote:The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class?

A. 75%
B. 82.5%
C. 85%
D. 90%
E. 92.5%

can this be solved using allegation method ? Somehow I am stuck !!

Appreciate some help .

OA B

Here's how we could solve with alligation.

Ingredients in the mixture:
F = Final grade = F.
M = midterm grade = 45.
C = course grade = 60.

Ratio of the ingredients:
Since the final is worth 40% of the course grade and the midterm is worth 60%, we get:
F:M = 40:60 = 2x:3x.

Step 1: Plot the 3 grades on a number line, with the grades for F and M on the ends and the course grade in the middle.
F -------------C=60------------45 M

Step 2: Plot the distances between the percentages.
(distance between F and C) : (distance between C and M) is equal to the RECIPROCAL of the ratio of F to M in the mixture.
Since F:M = 2x:3x, we get:
F -----3x-----C=60-----2x-----45 M

Step 3: Determine the required final grade.
The number line indicates that the distance between C=60 and M=45 is equal 2x:
60-45 = 2x
x = 7.5.

Since the distance between F and C is equal to 3x, we get:
F-60 = 3(7.5)
F = 82.5.

The correct answer is B.

Other problems solved with alligation:

https://www.beatthegmat.com/mixture-problem-t250678.html
https://www.beatthegmat.com/ratios-fract ... 15365.html