We know we can use the deviation method to find the arithmetic mean of numbers. It is very useful in cases where the numbers are huge, as it considerably brings down the calculation time.
The same method can be applied to weighted averages, as well. Let’s look at an example very similar to the one we examined when we were working on deviations in the case of arithmetic means:
What is the average of 452, 452, 453, 460, 467, 480, 499, 499, 504?
What would you say the average is here? Perhaps, around 470?
We have two 452s – 452 is 18 less than 470.
453 is 17 less than 470.
460 is 10 less than 470.
467 is 3 less than 470.
Overall, the numbers less than 470 are (2*18) + 17 + 10 + 3 = 66 less than 470.
480 is 10 more than 470.
We have two 499s – 499 is 29 more than 470.
504 is 34 more than 470.
Overall, the numbers more than 470 are 10 + (2*29) + 34 = 102 more than 470.
The shortfall is not balanced by the excess; there is an excess of 102-66 = 36.
So what is the average? If we assume that the average of these 9 numbers is 470, there will be an excess of 36. We need to distribute this excess evenly among all of the numbers, and hence, the average will increase by 36/9 = 4.
Therefore, the required mean is 470 + 4 = 474. (If we had assumed the mean to be 474, the shortfall would have balanced the excess.)
This method is used in exactly the same way when we have a simple average as when we have a weighted average. The reason we are reviewing it is that it can be very handy in weighted average questions involving more than two quantities.
We often deal with questions on weighted averages involving two quantities using the scale method. Let’s see how to use the deviation method for more than 2 quantities on an official GMAT question:
Question: Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?
(A) y + 3z
(B) (y +z) / 4
(C) 2y + 3z
(D) 3y + z
(E) 3y + 4.5z
Grade 1 milk contains 1% fat. Grade 2 milk contains 2% fat. Grade 3 milk contains 3% fat. The mixture of all three contains 1.5% fat. So, grade 1 milk provides the shortfall and grades 2 and 3 milk provide the excess.
Shortfall = x*(1.5 – 1)
Excess = y*(2 – 1.5) + z*(3 – 1.5)
Since 1.5 is the actual average, the shortfall = the excess.
x*(1.5 – 1) = y*(2 – 1.5) + z*(3 – 1.5)
x/2 = y/2 + 3z/2
x = y + 3z
And there you have it – the answer is A.
We easily used deviations here to arrive at the relation. It’s good to have this method – useful for both simple averages and weighted averages – in your test toolkit.
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