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by manik11 » Sat Feb 20, 2016 9:37 pm
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%

OA : B
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by MartyMurray » Sat Feb 20, 2016 11:26 pm
I bet you will get at least two or three different ways to answer this one. Here's one ultra simple way.

Moe needs a total of 60 overall percentage points.

The exam is worth up to 40 overall percentage points.

He has already done work representing 60 of the available overall percentage points.

45% x 60 = (4 x 6) + (1 x 3) = 27 overall percentage points earned so far

So he has 27 our of the 60 overall points needed in order to pass.

He needs 60 - 27 = 33 more overall percentage points to get to 60 overall percentage points.

33/40 = .825

The correct answer is B.
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by GMATGuruNY » Sun Feb 21, 2016 3:04 am
manik11 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%
sum = (weight)(average).

Let the total weight of the course = 100.
Since the final constitutes 40% of the course, the weight of the final = 40.
Thus, the pre-final weight = 100-40 = 60.

To yield a course average of 60, the sum for the course = (course weight)(course average) = (100)(60) = 6000.
Since the pre-final average = 45, the pre-final sum = (pre-final weight)(pre-final average) = (60)(45) = 2700.
Thus:
Sum for the final = (course sum) - (pre-final sum) = 6000-2700 = 3300.
Average for the final = (sum for the final)/(weight of the final) = 3300/40 = 82.5.

The correct answer is B.
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by GMATGuruNY » Sun Feb 21, 2016 3:06 am
The following approach is known as 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
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by [email protected] » Sun Feb 21, 2016 12:39 pm
Hi manik11,

This question can be solved in a variety of ways, depending on how you choose to set up the math (you could also TEST THE ANSWERS here, but you'd likely find the right algebraic approach to be pretty easy).

Since we know that the final exam is 40% of the final grade, we know that it is 2/5 of the final grade. This means that Moe's work so far (the 45% grade) accounts for 3/5 of the final grade. Since we're looking for the minimal score needed on the final exam to raise his grade to 60%, we can set up the following equation:

(3/5)(45%) + (2/5)(X%) = 60%

27 + 2X/5 = 60
2X/5 = 33
X = (5)(33)/2
X = 165/2
X = 82.5

Final Answer: B

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by Brent@GMATPrepNow » Sun Feb 21, 2016 1:49 pm
manik11 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%
Let's use the weighted averages formula:
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

We'll let x = the score on Moe's final exam.
So, we get: Overall grade = (60/100)(45) + (40/100)(x)
Or... 60 = (60/100)(45) + (40/100)(x)
Simplify: 60 = (3/5)(45) + (2/5)(x)
Simplify: 60 = 27 + (2/5)(x)
Subtract 27 from both sides: 33 = (2/5)(x)
Multiply both sides by 5/2 to get: 82.5 = x
Answer: B

For more information on weighted averages, you can watch this free GMAT Prep Now video: https://www.gmatprepnow.com/module/gmat- ... ics?id=805

Here are some additional practice questions related to weighted averages: Cheers,
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