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17 Oct 2017, 12:51
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Difficulty:
15%
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Question Stats:
81% (01:20) correct
19%
(01:38)
wrong
based on 267
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You score 80 on your mid-term exam and 100 on your final exam. Only these two exams make up your final grade of 92. How heavily did your teacher weigh the final exam?
A. 20%
B. 30%
C. 40%
D. 50%
E. 60%
A. 20%
B. 30%
C. 40%
D. 50%
E. 60%
The formula for weighted average is (Component 1)(Weighing 1) + (Component 2)(Weighing 2). I am not sure if we can use this to solve the question. The only method discussed in the prep book is the "teeter-totter" method. I believe it would be hard to use this method on the GMAT, given the time constraints. Thank you for your help. The answer is 60%.
17 Oct 2017, 14:32
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Rorschach1337
Rorschach1337 - I cannot figure out which method you think would be hard. Teeter-totter? Regular weighted?
Well, if the teeter-totter method is anything like alligation or the scale method, I cannot help you. They both make me crazy.
But I think weighted average is perfect for this question.
If each counted 50 percent, the average would be \(\frac{80(.5) + 100(.5)}{2} = (40 + 50) = 90\)
Because the average is not dead-center (halfway between one score and another), the final exam must count more in the score's total.
You have two differently weighted quantities that contribute to an overall average, for a weighted average where the final exam "weighs" more than the midterm.
I think I use your formula (I'm not sure what "component 1" means). It will work. I would guess the highlighted part is a potential stumbling block.
On one hand, first you sum the two weighted quantities to the overall score (92) times the overall weight (x + y)
On the other hand, once you have a variable with a value in hand, they sum to 100.
Here is one iteration of the formula I use (sometimes I use "concentration" or "percent" where you see "Weight).
(Weight of A)(Quantity of A) + (Weight of B)(Quantity of B) = (Weight of A + B) (Quantity of A + B)
Let x = the weight of the mid term exam
Let y = the weight of the final exam
\(80x + 100y = 92(x + y)\)
\(80x + 100y = 92x + 92y\)
\(8y = 12x\)
\(\frac{x}{y} = \frac{2}{3}\)
We want the weight of \(y\), the final exam. Solve for \(y\)
\(x + y = 100\)
\(x = \frac{2}{3} y\)
\(\frac{2}{3 y} + y = 100\)
\(2y + 3y = 300\)
\(5y = 300\)
\(y = 60\)
You could also notice that
\(\frac{x}{y} = \frac{2}{3}\) means \(x:y = 2:3\). There are five parts total in that ratio. \(y\) is three of the five.
\(y = \frac{3}{5} = \frac{60}{100} = 60\) Percent
Hope it helps!
18 Oct 2017, 00:37
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genxer123 thank you very much for the quick reply. This makes everything so much easier. Earlier, I was referring to the Teeter-totter method being a little too difficult to use on the GMAT, given that you not only have to create the visual but also compute the values. Therefore, I was looking for a way to solve this using algebra but I could not figure it out. I appreciate that you went into great detail as well as helping me visualize it further with the ratio method. Thanks again, cheers.
