Rorschach1337 wrote:

Is there a way to solve this problem using algebra:

"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?"

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%.

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!