This post will introduce weighted average questions you’ll see on the GMAT.  There is one main formula you need to solve simple GMAT Average questions:

• Average = SUM / # of observations

Note that this can be rearranged to read:

• SUM = Average x (# of obs)
• # obs = SUM / Average

So, if you are given ANY 2 of the 3 values, you should be able to find the 3rd. For example:

John drinks an average of 1.5 cups of water/day. After how many days has he drank 3 gallons of water? (1 gallon = 16 cups.)

In this case, we are looking for the number of days (or observations) such that we total 48 cups (3 gallons) of water.

# = SUM / Average

• # days = 48 cups / 1.5 cups/day
• # days = 32 days

NEVER AVERAGE AVERAGES!

Class A  has 15 students and an average height of 60”. Class B has 20 students. What is class B’s average height if the average height of both classes is 65”.

One might say:  (A + B) / 2 = 65”; A = 60”; so B must be 70”. However, keep in mind:

TOTAL AVERAGE = TOTAL SUM / TOTAL OBS

CLASS A + CLASS B = BOTH

• 15 students + 20 students = 35 students
• 60” average + 68.75” average = 65” average
• 900” total in A + 1375” total in B = 2275” total in Both

The given information is in black. The necessary intermediate steps are in blue, and the red is your answer. Note that the average of the averages ≠ total average. We must calculate each average separately, and to do this we need the SUM and # of observations for each category. This brings us to the idea of WEIGHTED AVERAGES.

A WEIGHTED AVERAGE is needed when you are taking average of a large group in which there are subgroups with a different number of observations in each. Take a look at this generalized formula, assuming there are 3 groups A, B and C.

(Average of A x Obs in A) + (Average of B x Obs in B) + (Average of C x Obs in C)

(Obs in A) + (Obs in B) + (Obs in C)

Think of weighted averages like a tug of war between numbers. The “stronger” one side (dog) is, the more that weighted average (tennis ball) will be “pulled” in that direction.

In the previous question, we had:

CLASS A + CLASS B = BOTH

• 15 students + 20 students = 35 students
• 60” average + 68.75” average = 65” weighted average

Note that the weighted average is CLOSER to B’s average than it is to A’s. This is because there are 20 students in Class B compared to only 15 students in Class A.

Two More Examples

At a certain restaurant, the average (arithmetic mean) number of customers served for the past x days was 75. If the restaurant serves 120 customers today, raising the average to 90 customers per day, what is the value of x?

A. 2

B. 5

C. 9

D. 15

E. 30

WITHOUT using the formula, we can see that today the restaurant served 30 customers above the average. The total amount ABOVE the average must equal total amount BELOW the average. This additional 30 customers must offset the “deficit” below the average of 90 created on the x days the restaurant served only 75 customers per day.

30/15 = 2 days. Choice (A).

WITH the formula, we can set up the following:

• 90 = (75x + 120)/(x + 1)
• 90x + 90 = 75x + 120
• 15x = 30

x = 2  Answer Choice (A)

Use whichever makes more sense to you!

Anita spent a total of \$780 on 52 bottles of wine for her wedding. She then decided to buy 8 bottles of sparkling wine for the toasts, as well. Was the average (arithmetic mean) price per bottle of wine less than \$20?

(1) Each bottle of sparkling wine cost more than \$15.

(2) Each bottle of sparkling wine cost less than \$40.

Take another look at what exactly the question is calling for: the TOTAL average price of all the wine at the wedding. We should look at the suggested average (\$20) and use that as our threshold amount.

• 60 bottles * \$20/bottle = \$1200 total
• \$1200 total – \$780 (given) = \$420 (left for sparkling wine)
• \$420 / 8 bottles = \$52.50/bottle of sparkling wine (for the total average to equal \$20)

Which of the answer choices are conclusively above or below \$52.50/bottle of sparkling wine? Only (2). With (1), we can be below OR above the threshold, so (1) is not sufficient.