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.
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