There are three statistical values typically known as *measures of central tendency*: mean, median, and mode. These three quantities are so named because they are ways of representing a group of values by means of one particular value. Of these three, mean is the most common on the CLT. Although the mean is more formally known as the *arithmetic mean*, the CLT will typically refer to it by the informal term “average.”

Ivy fervently hopes to finish her precalculus class with at least a 90 test average. Entering her last test of the year, her test grades in precalculus are $91$, $81$, $84$, $92$, $88$, and $95$. Assuming all tests are weighted equally and there is no rounding on her final test grade, what is the minimum score she must achieve on this last test in order to achieve at least a $90$ test average for the entire year?

A. 87

B. 91

C. 95

D. 99

The CLT Quantitative Reasoning section tends to demand a level understanding deeper than simple formula memorization. This is not to say that formulas aren’t important on the CLT; they can be vital*, but they usually must be applied with a solid understanding of how the parts of the formula relate.

Take the mean, or average, formula, for example: average = sum/number. In our experience, most students don’t think of this concept as a formula; rather, they carry around a strategy something like “to find an average, add up the numbers and divide by the number of numbers.” The problem with this superficial understanding is that it doesn’t readily supply a solution strategy for a question like the Approach Question above, which is not asking for the average but rather for one of the individual values that makes up the total.

The average formula helps us here, as long as we understand that any formula can be manipulated algebraically. The average formula above can be rewritten as sum = (average)(number) (or, for that matter, number = sum/average, though this is not commonly used). If we approach an average-related word problem with readiness to solve for the *sum* rather than the average, we find ourselves much further down the road of the problem-solving process.

How shall we use the sum in this problem? If we know that the sum is the average times the number, that’s helpful because we already know the desired average ($90$) and we can arrive at the number of values simply by counting the six tests Ivy has already taken and adding the upcoming (and last) test to make seven. So multiplying $7×90$ tells us that Ivy needs to total $630$ points on all her tests combined in order to reach her goal. Do you see what comes next? Adding up all her previous test scores tells us exactly how far she has to go! Put another way, we need to add up her scores and subtract that total from 630.

Since her previous scores add up to $531$, we can subtract that from $630$ and learn that Ivy must get a rather high A-plus to hit her score! The answer is **D**.

*This is especially the case with formulas that are not part of the list provided during the test, and the average formula is one of those formulas!

CLT problems about the average may ask you to calculate the average or to calculate the sum or a portion of the sum. More rarely, they may ask you to calculate the number of values based on the sum and the average.

- It’s worth putting the average formula on a flashcard and then what you could call the “sum formula” on a different flashcard. As a reminder:

$average=sum/number$ so

$sum=average×number$.

The average of a set of consecutive integers is $9.5$. If the smallest number in the set is $5$, what is the largest number in the set?

A. 8

B. 11

C. 14

D. 17

(spoiler)

The answer is **C**. Calling numbers “consecutive” means they must be next to each other in order. So we can imagine this sequence starting $5$, $6$, $7$ … and averaging $9.5$. Common sense tells us we’ll need to have some numbers above $9.5$, but how many? We could plug in the answers and work backward to check, but a smart approach here emphasizes the fact that, in an arithmetic (evenly spaced) sequence, the median always equals the mean. So the middle number will be the same as the average, and therefore the average will have an equal number of integers below it as above it. If we start counting at $5$, we’ll find that there are five integers below $9.5$, so there must also be five integers above it: $10$, $11$, $12$, $13$, and $14$.

You can also solve this problem by recognizing that in an arithmetic sequence, the average is equal to the smallest number plus the largest number divided by two. So in this case we could set up an equation, using $f$ for “first number” and $l$ for “last number”, as follows:

$2f+l 25+l 5+ll =average=9.5=19=14 $

In his lab, Josh determined that a molecule of water weighs 18.0 grams per mole, a molecule of hydrochloric acid weighs 36.5 grams per mole, and a molecule of ammonia weighs 17.0 grams per mole (all rounded to the nearest tenth). If Josh’s calculations are correct what is the average weight, in grams per mole, of these three molecules, rounded to the nearest tenth?

A. 17.5

B. 23.8

C. 27.0

D. 28.3

(spoiler)

The answer is **B**. The CLT doesn’t allow a calculator because the makers of the test want you to exercise your mathematical reasoning and logic. We should feel free to estimate on this question, especially because most of the choices are relatively far apart from each other. Taking a moment to observe the three values we’re averaging, we notice that the lower two are quite close to each other, while the highest is much higher. We can rule out choice A is much too low, given that the much higher value of $36.5$ is involved.

Let’s round the highest answer up to $37$ so we can add whole numbers. Adding the three numbers we have now results in a sum of $72$. Happily, that number is divisible by $3$, and we need to divide by $3$ since there are three total values. This gives us $24$, and answer B is far closer to that value than are either choices C or D.

One other note: you might realize that the two smaller values are going to pull down the answers quite a bit in this scenario. After all, there are too smaller numbers and only one larger one. Think of this like a weighted average, a “tug of war” between the smaller numbers and the larger one. The result should certainly be closer to $17$ and $18$ than it is to $36.5$. In this light, the two largest answer choices appear considerably too high.

The average age of a group of 8 students is 16 years old. If a 15-year-old and an 18-year-old join the group, what is the new average age?

A. 15.9

B. 16.0

C. 16.1

D. 16.2

(spoiler)

The answer is **C**. Like our approach question to this lesson, this question shows the importance of calculating the *sum* in an average question. Using the sum formula given in this module, we multiply $8×16$ to get the current sum of $128$. This puts in a position to accept new entries into the system and simply add them to our existing total. The 15-year-old and the 18-year-old bring a total of $33$ years, so our new sum is $161$. Fortunately, our new number of students is $10$, and dividing by $10$ only involves moving the decimal place left by one. $161÷10=16.1$.

In a set of nine different positive integers, the fourth-smallest integer is doubled. Which of the following

mustchange as a result of this doubling?I. Mean

II. Median

III. ModeA. I only

B. I and III only

C. II and III only

D. None of the above must change.

(spoiler)

The answer is **A**. To answer this challenging question we need to keep our sights on the word “must”; any of the statistical values that do not *have* to change can be eliminated. Starting with the mean, or average, we find perhaps the most straightforward conclusion: doubling one of the numbers *must* change the mean because it changes the sum. As long as the number of elements in the set doesn’t change, the mean will be larger. Before we continue, note that we can already eliminate choices C and D because statement I is true. Interestingly, we don’t even have to test the median, because the only thing we need to distinguish between choices A and B is a conclusion about the mode!

To address the mode, we recall that the mode is the value that appears the most in a list. In this question, each number is different, so each number is evenly the mode. If doubling the fourth-lowest number makes the result the same as another number in the list, then the mode *would* change: the number that appears twice would be the new mode. The problem is that we don’t know whether doubling the fourth-smallest number makes it equal to another number or not. Since we can’t know for sure, we eliminate statement III and therefore answer B.
If you’re curious about why statement II doesn’t have to be true (the inevitably conclusion if A is the answer!), consider this: we know that the *fifth* number in this list would be the median since it’s the middle of nine numbers. The number we’re doubling is just one before the median in order. So wouldn’t the median change because the fourth-smallest number would hop over the fifth number in value when doubled? We don’t know that for sure because there could be a large gap between the fourth and fifth numbers. For example, if the fourth-smallest number was $10$ and the median was $30$, doubling the fourth-lowest number would still leave it smaller than the median, so the median wouldn’t change.

In a class of 20 students, the average score on a test was 75. If the highest score of 94 is removed, what is the new average score on this test?

A. 73

B. 74

C. 76

D. 77

(spoiler)

The answer is **B**. By now you may be getting used to the first step on challenging average problems: *find the sum*. Since $20×75=1500$, we know that the total score of all 20 tests was $1,500$. We can now subtract $94$ and arrive at our new total of $1,406$. We know this total needs to be divided by the number of tests, but remember that this number is now $19$, not $20$.

Dividing such large numbers might look challenging, but keep in mind that the question doesn’t say anything about rounding, so the answer must be exact. Furthermore, we can use our math reasoning to conclude that, if we remove a number higher than the average, the average must go down. Focusing on choices A and B, we see that choice A is an odd number. Is that number multiplied by $19$ going to give us an even number result? No, it never could. So it must be the case that $1,406÷19=74$. Feel free to check our logic by working out the math!

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