Problems range from easy to hard.

**Q1**. On a ferry, there are

(A)

(B)

(C)

(D)

(E)

**Q2**. In Plutarch Enterprises,

(A)

(B)

(C)

(D)

(E)

**Q3**. At Didymus Corporation, there are just two classes of employees: silver and gold. The average salary of gold employees is

(A)

(B)

(C)

(D)

(E)

**Q4**. In a company of only

(A)

(B)

(C)

(D)

(E)

**Q5**. In a certain apartment building, there are one-bedroom and two-bedroom apartments. The rental prices of the apartment depend on a number of factors, but on average, two-bedroom apartments have higher rental prices than do one-bedroom apartments. Let R be the average rental price for all apartments in the building. If

(A)

(B)

(C)

(D)

(E)

**Q6**. At a certain upscale restaurant, there just two kinds of food items: entrees and appetizers. Each entrée item costs

(A)

(B)

(C)

(D)

(E)

Solutions will appear at the end of the article.

For the purposes of the QUANT, the weighted average situations occurs when we combine groups of different sizes and different group averages. For example, suppose in some parameter, suppose the male employees in a company have one average score, and the females have another average score. If there were an equal number of males and females, we could just average the two separate gender averages: that would be ridiculously easy, which is precisely why the QUANT will never present you with two groups of equal size in such a question. Instead, the number of male employees and number of female employees will be profoundly different, one significantly outnumbering the other, and then we will have to combine the individual gender averages to produce a total average for all employees. That's a weighted average.

That example, included just two groups, which is common, but sometimes there are three, and conceivably, on a very hard problem, there could be four groups. Each group is a different size, each has its own average, and the job is to find the average of everyone all together. Or, perhaps the question will give us most of the info for the individual groups, and give us the total average for everyone, and then ask us to find the size or average of a particular group.

We have three basic approaches we can take to these question.

Remember that, even with ordinary average questions, thinking in terms of the sum can often be helpful. We ** can't** add or subtract averages, but we

Some weighted average problems give percents, not actual counts, of individual groups. In that case, we could simply pick a convenient number for the size of the population, and use the sums method from there. For example, if

This method always works, although is not always the most convenient in more advanced problems. This will be demonstrated in a few of the answers below.

Sometimes the information about the sizes of individual groups is given, not in absolute counts of members, but simply in percents or proportions. In the problems above, question #2 simply gives percents, and question #5 asks for a percent. Yes, we could use Approach I, but there's a faster way.

In Approach II, we simply multiply each group average by the percent of the population, expressed as a decimal, which that group occupies. When we add these products up, the sum is magically the total average for everyone. Suppose we have three groups,

The percent have to be expressed in decimal form, so that:

This approach will be demonstrated in #2 below.

This final method can be hard to understand at first, but if you appreciate it, it is an incredibly fast time-saver. This approach only works if there are exactly two groups, no more.

Suppose there are two groups in a population,

Now, think about a number line, with the two individual group averages and the total averages indicated on the number line.

On that number line, I have labeled

Let’s say

This approach is hard to explain clearly in words. You really have to see it demonstrated in the solutions below to understand it fully. Once you understand it, though, this is an extremely fast method to solve many problems.

If the above article gave you any insights, you may want to give the practice problems another look. Remember, in your practice problems on weighted averages, practice all three of these methods. The more ways you have to understand any problems, the more options you will have on test day!

**Q1**.

**Method I: using sums**

We will divide the two masses by

To find the total average, we need to divide this total sum by the total number of vehicles,

Since we divided masses by

**Answer = (B)**

**Method II: proportional placement of the total average**

Cars to trucks is

Well, the difference in the two group averages is **part**” is

**Answer = (B)**

**Q2**. We will approach this use the proportion & percents approach. Divide all dollar amounts by

where

Now, multiply the

**Answer = (D)**

**Q3**.

**Method I: using sums**

We can use this if we pick a value for the average salary for the silver employees. It actually doesn’t matter what value we pick, because averages will fall in the same relative places regardless of whether all the individual values are slid up and down the number line. The easiest value by far to pick is zero. Let’s pretend that the silver folks make

Now, we also have to simplify the numbers of employees. We could reduce the number of employees as long as we preserve the relative ratio.

So everything would be the same if we just had

I am not even going to bother to multiply that out, because we know that the next step is to divide by

The average salary is

**Answer = (C)**

**Method II: proportional placement of the total average**

The ratio of silver employees to gold employees is

If we divide the distance between the two averages by **four parts**” away from the total average, and gold will be “**three parts**” away.

Well, the difference is

**Answer = (C)**

**Q4**. We will approach this using sums. The individual employee numbers are small. We will divide all dollar amounts by

Individual sums must add up to the total sum.

The salary of each of those four highest paid employees is

**Answer = (E)**

**Q5**. This question is designed for an analysis involving proportional placement of the mean. First, observe that

The ratio of the distances to

Cancel a factor of

One-bedroom apartments are “** \bold{13} parts**” of the building, and two-bedroom apartments are “

**Answer = (B)**

**Q6**.

**Method I: using sums**

First, last year. Let

Notice the use of the Doubling and Halving trick in the second and third lines. The two individual sums should add up to the total sum.

They start out with

Let

Again, the two individual sums should add up to the total sum.

They added

Method I was do-able, but we had to solve for many values.

**Answer = (E)**

**Method II: proportional placement of the total average**

Originally, the entrée price was

The number of entrees doesn’t change. The average drops to

If you know how to employ this method, it is much more elegant.

**Answer = (E)**