Problems range from easy to hard.

1) On a ferry, there are 50 cars and 10 trucks. The cars have an average mass of 1200 kg and the trucks have an average mass of 3000 kg. What is the average mass of all 60 vehicles on the ferry?

- (A) 1200 kg

(B) 1500 kg

(C) 1800 kg

(D) 2100 kg

(E) 2400 kg

2) In Plutarch Enterprises, 70% of the employees are marketers, 20% are engineers, and the rest are managers. Marketers make an average salary of $50,000 a year, and engineers make an average of $80,000. What is the average salary for managers if the average for all employees is also $80,000?

- (A) $80,000

(B) $130,000

(C) $240,000

(D) $290,000

(E) $320,000

3) At Didymus Corporation, there are just two classes of employees: silver and gold. The average salary of gold employees is $56,000 higher than that of silver employees. If there are 120 silver employees and 160 gold employees, then the average salary for the company is how much higher than the average salary for the silver employees?

- (A) $24,000

(B) $28,000

(C) $32,000

(D) $36,000

(E) $40,000

4) In a company of only 20 employees, 10 employees make $80,000/yr, 6 employees make $150,000/yr, and the 4 highest-paid employees all make the same amount. If the average annual salary for the 20 employees is $175,000/yr, then what is the annual salary of each highest-paid employee?

- (A) $250,000

(B) $300,000

(C) $350,000

(D) $400,000

(E) $450,000

5) 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 R is $5,600 higher than the average rental price for all one-bedroom apartments, and if the average rental price for all two-bedroom apartments is $10,400 higher that R, then what percentage of apartments in the building are two-bedroom apartments?

- (A) 26%

(B) 35%

(C) 39%

(D) 42%

(E) 52%

6) At a certain upscale restaurant, there just two kinds of food items: entrees and appetizers. Each entrée item costs $30, and each appetizer item costs $12. Last year, it had a total of 15 food items on the menu, and the average price of a food item on its menu was $18. This year, it added more appetizer items, and the average price of a food item on its menu dropped to $15. How many appetizer items did it add this year?

- (A) 3

(B) 6

(C) 9

(D) 12

(E) 15

Solutions will appear at the end of the article.

## Weighted average

For the purposes of the GMAT, 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 GMAT 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.

## Approach I: averages & sums

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 *can* add or subtract sums! Right there is the key to one approach to the weighted average situation. If we calculate the sums for each separate group, we can simply add these sums to get the sum of the whole group. Alternately, if we know the size of the total group and the total average for everyone, we can figure out the total sum for everyone, and simply subtract the sums of the individual groups in order to find what we need.

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 group A has 20% of the population, group B has 40% of the population, and group C also has 40% of the population, then we could just pretend that group A has one person, groups B & C each have two people, and total population has five people. From this, we could calculate all our sums.

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.

## Approach II: proportional & percentages

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, groups J and K and L which together constitute the entire population. Suppose this summarizes their separate information:

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

Then, the total average is simply given by

This approach will be demonstrated in #2 below.

## Approach III: proportional placement of total average

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, group 1 and group 2, and suppose that group 1 is bigger than group 2. Let’s also suppose that group 1 has a lower group average, and group 2 has the higher group average. Of course, the total combined average of the two groups together will be between the two individual group averages. In fact, because group 1 is bigger, the combined average will have to be closer to group 1’s average, and further away from group 2’s average.

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 d1, the distance from the average of group #1 to the combined average, and d2, the distance from the combined average to the average of group 2. The ratio of these two distances is equal to a ratio of the size of the two individual groups. Let’s think about this. The bigger group, here group 1, will have more of an effect on the combined average and therefore will be closer to the combined average—a smaller distance. Therefore, the ratio of the distances must equal the reciprocal of the same ratio of the sizes of the groups:

Let’s say group 1 is 3 times larger than group 2. This means d2, the distance from group 2’s average to the combined average, would have to be 3 times bigger than d1, the distance from group 1’s average to the combined average. If the latter is x, then the former is 3x, and the total distance is 4x. If we know the averages of the two groups individually, we would simply have to divide the difference between those group averages by 4: the combined average would be one part away from group 1’s average, or three parts away from group 2’s average.

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.

## Summary

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!

## Practice problem explanations

1) **Method I: using sums**

We will divide the two masses by 1000, 1.2 and 3 respectively, to simplify calculations. Note the use of the Doubling and Halving trick in the first multiplication.

sum for cars = 50*1.2 = 100*0.6 = 60

sum for trucks = 10*3 = 30

total sum = 60 + 30 = 90

To find the total average, we need to divide this total sum by the total number of vehicles, 60.

Since we divided masses by 1000 earlier, we need to multiply by 1000 to get the answer. Total average = 1500 kg. Answer =** (B)**

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

Cars to trucks is 5:1, so if the distance between the car’s average and truck’s average were divided into 6 parts, the car’s average is 1 part away from the total average, and the truck’s average is 5 parts away.

Well, the difference in the two group averages is 3000 – 1200 = 1800 kg. Divide that by six: each “part” is 300 kg. Well, the total average must be 300 kg bigger than 1200 kg, or 5*300 kg smaller than 3000 kg. Either way, that’s 1500 kg. Answer =** (B)**

2) We will approach this use the proportion & percents approach. Divide all dollar amounts by 1000 for smaller numbers. Multiply each group average by the percent expressed as a decimal:

marketers = 0.70*50 = 35

engineers = 0.20*80 = 16

managers = 0.10*x = 0.1x

where x is the average salary for the managers. These three should add up to the average for all employees:

35 + 16 + 0.1x = 80

0.1x = 80 – 35 – 16

0.1x = 29

x = 290

Now, multiply the 1000 again, to get back to real dollar amounts. The average salary for managers is $290,000. Answer = **(D) **

3) **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 $0, and the gold folks make $56. (I divided dollars by 1000 for simplicity).

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.

silver : gold = 120:160 = 12:16 = 3:4

So everything would be the same if we just had 3 silver employees and 4 gold employees. OK, now find the sums.

silver = 3*0 = 0

gold = 4*56

I am not even going to bother to multiply that out, because we know that the next step is to divide by 7, the total number of employees.

total average = 4*56/7 = 4*8 = 32

The average salary is $32,000, which is $32,000 higher than the average for the silver employees. Answer = **(C)**

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

The ratio of silver employees to gold employees is

silver : gold = 120:160 = 12:16 = 3:4

If we divide the distance between the two averages by 7, then silver will be “four parts” away from the total average, and gold will be “three parts” away.

Well, the difference is $56,000, so that divided by 7 is $8000. That’s one part. Four parts would be $32,000, which has to be the distance from the silver average to the total average.

Answer = **(C) **

4) We will approach this using sums. The individual employee numbers are small. We will divide all dollar amounts by 1000, for easier calculations. Call the highest-paid employee salary x. Then the sums are

lowest = 10*80 = 800

middle = 6*150 = 3*300 = 900

highest = 4x

Individual sums must add up to the total sum.

800 + 900 + 4x = 3500

4x = 1800

2x = 900

x = 450

The salary of each of those four highest paid employees is $450,000.

Answer = **(E)**

5) This question is designed for an analysis involving proportional placement of the mean. First, observe that R is much closer to the average for one-bedroom apartments, so there must be more one-bedroom apartments and fewer two-bedroom apartments.

The ratio of the distances to R is

5600:10400 = 56:104

Cancel a factor of 8 from both 56 and 104

56:104 = 7:13

One-bedroom apartments are “13 parts” of the building, and two-bedroom apartments are “7 parts.” That’s a total of 7 + 13 = 20 parts in the building. Two-bedroom apartments constitute 7/20 of the apartments in the building. Since 1/20 = 5%, 7/20 = 35%.

Answer = **(B)**

6) **Method I: using sums**

First, last year. Let x be the number of entrees. Then (15 – x) is the number of appetizers. The sums are:

entrees = 30x

appetizers = (15 – x)*12 = 12*15 – 12x = 6*30 – 12x = 180 – 12x

total = 15*18 = 30*9 = 270

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.

30x + 180 – 12x = 270

18x = 90

x = 5

They start out with 5 entrees and 10 appetizers.

Let N be the number of appetizers added, so now there are 5 entrees and (10 + N) appetizers. We need to solve for N. Again, the sums:

entrees = 5*30 = 150

appetizers = (10 + N)*12 = 120 + 12N

total = (15 + N)*15 = 225 + 15N

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

150 + 120 + 12N = 225 + 15N

270 = 225 + 3N

45 = 3N

15 = N

They added 15 more appetizers. Answer = **(E)**

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

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

Originally, the entrée price was 30 – 18 = 12 from the total average, and the appetizer price was 18 – 12 = 6. This means there must have been twice as many appetizers as entrees. Therefore , with 15 items, there must have been 10 appetizers and 5 entrees.

The number of entrees doesn’t change. The average drops to $15, so the distance from the entrée price is now 30 – 15 = 15, and the distance from the appetizer price is now 15 – 12 = 3. That’s a 5-to-1 ratio, which means there must be 5x as many appetizers as entrees. Since there still are 5 entrees, there must now be 25 appetizers, so 15 have been added.

Answer = **(E)**

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

Many thanks for sharing. It’s very useful for those who are preparing for the test. Just a question, Q5 and Q6 are 700-level questions or below? I found it difficult at the beginning but once you understand the rationale behind that, it turns out to be considerably less difficult.

It’s hard to gauge whether an individual question is truly “700-level,” since getting a 700+ score is really about how well you answer a certain mix of questions. With that said, I would say that yes– these questions are more advanced, at or near the 700-level. They have enough complexity to be difficult for many test takers. You are correct that if you carefully look at the steps, these questions do get a lot easier. And it’s important to remember that many harder GMAT Quant questions do not have exceedingly difficult math; they simply have more math steps than the average problem. 🙂

Hi can you explain me Q 3 clearly

Hi Akhil,

I’m happy to provide an explanation, but can you please let me know what you didn’t understand from the explanation we included in this blog post? It’s very helpful when students provide some details about their thought process and what they did and did not understand. That way, we can make sure that our explanations are more clear and concise!

Hi!

I am confused about the answer of Q5. I did´t get why the ratio of the distances to R are 5600:10400, since 10400 is the distance of R2 to R1. Why are we not using the distance of both R1 and R2 from Total Average with would be 5600:4800?

Thanks!

Great question, Angelica. What’s important to understand here is that there is no Average 1 and Average 2. You’re not given specific separate figures for an average R1 for one bedroom apartments, and an average R2 for two bedroom apartments. These averages are only mentioned in terms of their distance from the one R that’s actually in the problem. So 10,400 isn’t the distance between the average for two bedroom apartments and the average foe one bedroom apartments. It’s just the distance from R, the group average. This is clearly stated when the problem says “the average rental price for all two-bedroom apartments is $10,400 higher that R.”

Moreover, since the the average for one bedroom apartments is 5600

belowthe overall average, and the average for two bedroom apartments is 10,400abovethe overall average, if we were going to set up the subgroup averages as R1 and R2 and then calculate the distance between them, the distance would be 10,400 PLUS 5600 = 16,000, not the 10,400 MINUS 5600 = 4800 figure that you were using.But again, we’re not actually using the distance between the averages for the two subgroups. Instead, we’re looking at the distance of each subgroup average from the overall average, expressed as R, and then using those two separate distances to calculate what this says about each group as a percentage of the whole group. So we use the ratio of 6500:10,400, the distance of each subgroup from the overall average, to get proportional percentages based on the ratio of distance, as described in Approach III above.

Does this make sense? Let me know if you still have doubts or questions.

I didn’t understand on how the price differences of appetizer and entree from the total average price affect the quantity of these two items in the menu (Q6, method 2). Pls explain.

As Mike mentioned, the “proportional placement of the total average” method can be difficult to understand at first, but is a real time saver if you can learn and understand it. So I’m glad you asked a little more about how this works! 🙂

Basically, this approach works when there is a total group average, but the total group is divided into sub-groups, each of which have their own average. In the case of Q6, we have this. At first, we have a total group average of $18 per item, and once more appetizers are added, we have a total group average of $15 per item. In either case, we have two subgroups with the same averages, before and after the appetizers were added. Subgroup 1 is entrees. The average price there is $30, since

everyentree costs $30. Subgroup 2 is appetizers. The average price for subgroup 2 is $12, since every appetizer costs$12.So here’s how this approach works: the distances of the two subgroup averages from the whole group averages form a ratio, and this ratio represents the size of the two subgroups in proportion to each other. Here’s where it gets tricky, though– the subgroup with the shorter distance from the average actually must have the larger number of things in it. Since its average is closer to the whole-group average, the subgroup whose average is less distant from the overall average must have a bigger influence– a heavier weight– on the average. So once you have your ratio, you invert it.

If this sounds a little confusing, don’t worry— I’ll explain it in depth, with examples form Q6. Read on!

So at first, the whole group average price is $18. The entree average is $30, which is $12 more than the whole group average of $18, for a distance of 12 from the average. The appetizer average is $12, which is $6 less than 18, for a distance of 6 from the group average. This means the ratio of of the two groups must is 12/6. Now, here’s the tricky part, so pay attention: since the appetizer subgroup average is closer to the whole group average, appetizers must actually be the larger, more influential subgroup. So, while the distance of 12 was taken from entrees, it’s assigned to appetizers in the ratio, with entrees taking the smaller number–6– that was the distance of the appetizers. The distances of each subgroup are inverted to get the numbers for each subgroup. The formula for this (mentioned under the heading entitled

“Approach III: proportional placement of total average”) is (d1/d2)=(N2/N1)In other words:

(Entrees = 12 distance) and (appetizers = 6 distance)

becomes a ratio of:

12 appetizers to 6 entrees, or 12a/6e, or 12/6, or 2/1. So appetizers/entrees = 2/1

And we already know that there were 15 food items total. This means there must be 10 appetizers and 5 entrees. 10/5 is the only ratio that gives us a total of 15, and is equal to 2/1.

Next, the average price drops to $15. Per the word problem, we know there are still just five entrees. And we know that the number of appetizers has increased. But what has it increased to? To figure this out, we just check the ratio.

The average price for entrees is still $30. The average price for appetizers is still $12. But now, entrees are a distance of 15 from the average (30-15=15), and appetizers are a distance of just 3 (15-12)=3. So d1(entrees) = 15 and d2(appetizers)= 3. Now we invert the numbers. N1(entrees) is 3, and N2(appetizers) is 15. There is a ratio of 15 appetizers to 3 entrees. This is also a ratio of 5/1. We know there are still just 5 entrees. So there must now be 25 appetizers, since 25/5 = 5/1. Originally there were 10 appetizers, and now there are 25. 15 appetizers were added.

Hi!

I am confused about the answer of Q5. I did´t get why the one-bedroom get 13 pieces and the two-bedroom get 7, since the problem says is the other way around.

Thanks!

Hi Estela,

Great question! Think of it this way, we know that R (average rental price of all apt) is closer to the average price of one-bedroom apts. This means that there MUST be more one-bedrooms than two-bedrooms to pull the total average price closer towards the one-bedroom price. Once you determine the ratio of distance, you can use this ratio for weights. However, as a rule, it has to be the opposite! If the distance is “one-bed:two-bed” = “7:13”, then we know that the two-bedroom is further from the mean (and has less weight). To distribute weight, we weigh one-bedrooms as 13/20, and two-bedrooms as 7/20 (utilizing the proportions we just determined). The further away, the less weight (aka opposite weights).

Hey,

You guys should launch an app for your blog also.

Loved the concepts and the presence you have over android market.

Respond if u liked my idea :p

Hello,

Number 4 was a bit confusing as the problem did not specify that the salary/group identified was the average. Will this be the case on the GMAT as well or will they always call out an average amount as an average?

Thank you.

Hi Lena,

Maybe I’ve misunderstood your question, but this problem does tell you the salary and group identified as average. Let’s look at the prompt again, first!

4) In a company of only 20 employees, 10 employees make $80,000/yr, 6 employees make $150,000/yr, and the 4 highest-paid employees all make the same amount. If the average annual salary for the 20 employees is $175,000/yr, then what is the annual salary of each highest-paid employee?This tells us we have 20 employees.

Employees 1 through 10make $80,000/yr,employees 11 through 16make $150,000/yr, andthe last 4, although we don’t know their salary yet, all make the same amount. Finally, we know thatthe average of the entire group is a salary of $175,000/yr.I would argue that this amount is definitely signaled as the average, but if you feel differently, let us know and we can talk about it more. 🙂

Hi!

Thank you for responding! My confusion was where it says “10 employees make $80,000/yr”… I read that as the total salary for the 10 employees not that 80,000 was the average salary of the group of 10 employees and the same for the 6 employees. So when I went to try the strategy of multiplying the percent of 50% * the average which I calculated to be 8,000/yr, I got 4,000 and when I did the same for the group of 6 employees then added the products and deducted the sum from the total average salary of 175,000 the difference did not allow me to find the proper answer. Once I realized that the 80,000 and 150,000 were the averages for each group, the proportional/percentage method worked. I may be the only one who didn’t realize the individual group salaries were not the group averages.

You’re very welcome, Lena 🙂

I’m glad to hear that everything is more clear now!

Happy studying!