# The average (arithmetic mean) of 100 measurements is 23, and the

The average (arithmetic mean) of \$100\$ measurements is \$23\$, and the average of \$50\$ additional measurements is \$27\$.

 Quantity A \$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\$ Quantity B \$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\$ The average of the \$150\$ measurements \$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\$ \$25\$
1. Quantity A is greater.
2. Quantity B is greater.
3. The two quantities are equal.
4. The relationship cannot be determined from the information given

So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 6 of Section 4 of Practice Test 1. Those questions testing our Numerical Methods for Describing Data knowledge can be kind of tricky, but never fear, PrepScholar has got your back!

## Survey the Question

Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on your paper.

Three times we see the average of different sets of numbers mentioned. Whenever we see averages of numbers, we should expect the question to test our Numerical Methods for Describing Data math knowledge. Now that we know what this question likely tests, we’ll be ready for whatever it throws our way!

## What Do We Know?

Let’s carefully read through the question and make a list of the things that we know.

1. We have \$100\$ numbers with an average of \$23\$
2. We also have \$50\$ other numbers with an average of \$27\$
3. We want to compare the average of all \$150\$ numbers to the value \$25\$

## Develop a Plan

Since the question asks for averages of sets of numbers, let’s think of the equation we have memorized for averages:

\$\$\Average \of \Set \of \Values = {\Sum \of \the \Values} / {\Number \of \Values}\$\$

So for example, if the sum of \$10\$ numbers is \$200\$, then the average of those \$10\$ numbers is \$200/10\$, or \$20\$. Viewed a slightly different way, if the average of \$10\$ numbers is \$20\$, then the sum of those \$10\$ numbers is \$200\$. We should keep our minds flexible when it comes to sums and averages. Some questions it’ll be easier to work with sums, whereas for other questions it’ll be easier to work with averages.

We want to know the average of all \$100\$ numbers in this question:

\$\$\Average \of \All 100 \Numbers = {\Sum \of \All 100 \Numbers}/{100}\$\$

Logically it makes sense that the sum of all \$150\$ numbers is the sum of the \$100\$ numbers and the other \$50\$ numbers:

\$\$\Sum \of \All 150 \Numbers = {\Sum \of 100 \Numbers}+{\Sum \of 50 \Numbers}\$\$

Also, the sums of the sets of \$100\$ and \$50\$ numbers can be found from our equation with sum and average in it:

\$\$\;\;\Sum \of 100 \Numbers = 100 ·\Average \of 100 \Numbers\$\$
\$\$\Sum \of 50 \Numbers = 50 ·\Average \of 50 \Numbers\$\$

We know that we could then combine these equations to get the average of all of the \$150\$ numbers:

 \$\Average \of \All 150 \Numbers\$ \$=\$ \${\Sum \of \All 150 \Numbers}/150\$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$\Average \of \All 150 \Numbers\$ \$=\$ \${\Sum \of 100 \Numbers + \Sum \of 50 \Numbers}/150\$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$ \$\Average \of \All 150 \Numbers\$ \$=\$ \${100 *\Average \of 100 \Numbers + 50 *\Average \of 50 \Numbers}/150\$

We know that if we calculate this value, we could then compare it to Quantity B (\$25\$). Let’s get to it!

## Solve the Question

Calculating the average of the \$150\$ numbers we get:

 \$\Average \of \All 150 \Numbers\$ \$=\$ \${100 *23 + 50 *27}/150\$ \$\Average \of \All 150 \Numbers\$ \$=\$ \${2{,}300 + 1{,}350}/150\$ \$\Average \of \All 150 \Numbers\$ \$=\$ \${3{,}650}/150\$

Hmmm…\$3{,}650\$ divided by \$150\$…looks as if we’ll need a calculator. Let’s pull out our measly GRE calculator here, dividing \$3{,}650\$ by \$150\$

\$\$\Average \of \All 150 \Numbers = {3{,}650}/{150}\$\$
\$\$\;\Average \of \All 150 \Numbers = 24.333\$\$

So Quantity A is \$24.333\$. This is less than Quantity B \$(25)\$, so the correct answer is B, Quantity B is greater.

Well, that wasn’t too bad. Though we did need to use our calculator at the end, and we know that most problems shouldn’t require a calculator. It might be more challenging to solve this question in a logical way without a calculator. Challenge accepted! Plus we never know if it might be fun, educational, or both!

## Develop a Plan and Solve: Logically

We know that solving math problems logically can be very rewarding. However, they can often times have more abstract solutions that require more thought and planning when compared to solutions where we just plug and chug numbers into equations. So we should always take in all of the numbers and concepts the question presents, then see if anything jumps out at us. Here, our paper is our friend, so let’s put it to good use!

Starting with a top-down approach, we know that we want to compare the average of \$150\$ numbers to the value \$25\$. Let’s write that on our paper, with a “?” in between until we figure out which side is greater.

\$\$\Average \of \All 150 \Numbers \;\;\;? \;\;\;25\$\$

If we could estimate the average of the \$150\$ numbers just to determine if it was equal to \$25\$, above \$25\$, or below \$25\$, we could finish the question. Let’s start working on a way to estimate this average!

Let’s also pay attention to any coincidental number relationships that we see, since we know to be suspicious of any coincidental number relationships. We have one set of numbers with an average of \$23\$ and a smaller set numbers with an average of \$27\$. Quantity B, \$25\$, seems coincidentally placed right smack in the middle between \$23\$ and \$27\$. Seems too coincidental to not be important, so let’s explore that a bit more.

When would a set of numbers with an average of \$23\$ and a set of numbers with an average of \$27\$ have a combined average of \$25\$? Ah, when we have half of the numbers as \$23\$ and the other half as \$27\$. That’s it! For \$150\$ numbers, if half of them (\$75\$ numbers) were \$23\$ and the other half of them were \$27\$, then the combined average for all of them would be \$25\$.

Now, for our question we know that we have more numbers with an average of \$23\$ than we have numbers with an average of \$27\$. Changing the numbers should definitely change their average, but in which direction? If we start with half of the numbers as \$23\$ (giving us an average of \$25\$), then it makes sense that having more than half of the numbers as \$23\$ would move the average closer to \$23\$. So it makes sense that the average for the \$150\$ numbers would be closer to \$23\$ than it would be to \$27\$. And since \$25\$ is right smack in the middle between \$23\$ and \$27\$, the combined average MUST be less than \$25\$ . Bingo!

Alright, since we figured out that Quantity A, the average of all \$150\$ numbers, must be less than \$25\$, then Quantity B (\$25\$) must be greater than Quantity A. The correct answer is B, Quantity B is greater.

So it looks as if we found two ways to solve this question. We could just math it out and get the average of all \$150\$ values. This does give us good peace of mind if we’re not comfortable with averages, since we can almost physically touch the value and know exactly what it is, even if it takes a bit longer to find out. Or we could logically deduce that the average for two sets of values must be closer to the average for the set of values that has more members in it. Seems as if it’s similar to voting in politics: more votes equals more influence, so having more numbers on one side moves the average closer to that side. This second solution is more abstract but can definitely save some calculation time if we’re comfortable with averages. Plus, we can give ourselves TWO gold stars for finding two different ways to solve this question.

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