Percents | Math Reasoning | Achievable CLT

Introduction

Percent questions deal in one way or another with the relationship between a part and a whole. As the origin of the word “percent” implies (think century or centennial), the number 100 is at the heart of the percent calculation. A percent is a way of expressing the relationship between a part and a whole if the whole were $100$ . And because that relationship involves dividing the part by the whole, percents are closely related to fractions and ratios. In sum, a percent calculation relates a number to 100 but then lists the number without the $100$ in the final result.

Approach Question

In the town of Corinth, 40% of middle school students are homeschooled. The remainder all attend either the Lyceum or the School of Socrates, and three times as many middle school students attend the Lyceum as attend the School of Socrates. What percentage of Corinth middle schoolers attend the Lyceum?

A. 20%
B. 36%
C. 40%
D. 45%

Explanation

The UnCLES method becomes especially helpful in the case of a word problem. We need to carefully note not only the $40%$ but also the “three times as many”. The moment you encounter $40%$ , other ways of expressing this quantity might pop into your mind: $\frac{4}{10}$ , $\frac{2}{5}$ , $\frac{40}{100}$ , $0.4$ . Any of these quantities may be helpful, but since percents are often more “user-friendly” than fractions or decimals, let’s stick with $40$ .

Now we proceed to the word “remainder” and the phrase “three times as many,” both of which should draw our attention because they refer to numerical quantities. The beauty of using percents is evident in how quickly we may combine the ideas of “ $40%$ ” and “remainder” to come up with $60%$ left over after the initial $40$ . All further calculations will come out of this $60%$ .

There are two populations that make up this remaining $60%$ : the students at the Lyceum and the students at the School of Socrates. Here’s where “three times as many” comes in, and here we must be extra careful. As soon as students encounter the word “three times,” they often think that they should be dividing $60$ by three. That would mean that $20%$ of students are in the smaller population (in this case, the School of Socrates). But we can see the error in that thinking by the fact that this calculation leaves $40%$ for the Lyceum and $40%$ is not three times as much as $20%$ .

So what went wrong there? Think of it like a ratio: three parts of one thing compared to one part of another (as you might find in a recipe). If we simply divide by three, we are ignoring that “one part” altogether. In reality, there are four parts, or “shares” of the amount, in this scenario: three shares go to the Lyceum and the other share goes to the School of Socrates. We need to divide, not by three, but by four: $60 \div 4 = 15$ . If you condition yourself to think something like “four total parts” the moment you see that one quantity is three times as much as other, you will help yourself a great deal with ratios.

So the answer is $15$ , right? Not so fast. Effective word problem solvers always check back to see what the question is asking. Is it the School of Socrates or the Lyceum in the question itself? It’s the Lyceum, and that school has three parts, or shares, of that remaining $60$ percent. So we need to multiply that “share” of $15$ by $3$ , to get $45$ . The answer is D.

One further note: if you’ve received good teaching about percents in the context of algebra, you might be aware that there’s another way to do the problem. Once we have our “60 percent” in focus, we can set up an equation after labeling the two school populations in terms of $x$ . It makes sense to call the School of Socrates “ $x$ ” because the Lyceum is described as having “three times as many” students. That makes the Lyceum correspond to “ $3 x$ ”, and our equation follows:

$x + 3 x = 60 4 x = 60 x = 15$

But again, “ $x$ ” doesn’t represent our final answer; we need to plug $15$ in for $x$ in the “ $3 x$ ” expression representing the Lyceum. So once again we arrive at 45.

Definitions

Percent: Break the word into two parts: per + cent. This literally means “per hundred”. You will help yourself, as soon as you encounter a percent, to immediately put it over $100$ in your mind (or on paper)! Then this fraction becomes easier to see as a decimal as well. For example if $20% = \frac{20}{100}$ , it comes easier to reduce $\frac{20}{100}$ to $\frac{2}{10}$ and see that its decimal form is $0.2$ .
Part: We note two formulas for you to memorize in the flashcard portion of this lesson. One of them references the words “part” and “whole.” The part in a percent expression is the number that is meant to divided by the some total or “whole.” Typically, the “part” is less than the whole, which makes sense when we think of the normal meaning of these words. However, this is not always the case, as we can see a population that grows from $1, 000$ to $1, 500$ . The new population is actually $150$ % of the old population because the “part” has grown larger than the original “whole”.
Whole: The “whole” is the entire quantity to which a part is compared, naturally following the words “out of” (as in “What percent is $45$ out of $60$ ?”). As shown above, a whole is usually, but not always, greater than the part.

Topics for Cross-Reference

Ratios

Variations

On rare occasions, the CLT will ask about percent change rather than simply percentage. You can tell this is happening when the question pairs the word “percent” or “percentage” with the words “more/greater” or “less” rather than “of” or “out of.”

In addition, although probability is technically a calculation meant to be expressed as a fraction or decimal, the CLT will ask about probability (or “chance”) and offer answer choices in terms of percent.

Strategy Insights

What number do you think might be a helpful starting point for a percent question? Probably $100$ , right? If you are not given a quantity and need a place to begin, $100$ is a great place to start. This is especially helpful when the question involves increasing and then decreasing by a certain percentage (or decreasing and then increasing).
On a related note, it might make more sense in the context of some questions to simply use $10$ instead of $100$ , to keep the values smaller. Our entire mathematical system is based on place values with 10, so take advantage of that fact!
When doing mental math, break down larger numbers into $10$ % (by removing one zero) or 1% (by removing two zeroes). For example, you could find $27$ % of $300$ by thinking as follows: each $10$ % of $300$ is $30$ and I have two $10$ % “shares,” so that’s $60$ . Each $1$ % of $300$ is $3$ and I have seven of those, so that’s $21$ . $60 + 21 = 81$ .

Flashcard Fodder

There are two formulas you must know to be fully equipped for percent questions:

$percent percent change = part / whole = (new - old) / old$

(the “old” in the percent change formula is the starting point in a situation before it moves to the “new” value)

Sample Questions

Difficulty 1

A dress is originally priced at $80 but is on sale for 25% off. What is the sale price of the dress?

A. $20
B. $55
C. $60
D. $70

(spoiler)

The answer is C. Without a calculator at hand, the best approach here is probably to convert $25$ % to a fraction since it’s a common percentage. Since $25$ % is the same as $\frac{1}{4}$ , we can multiply the original price of $80 by $\frac{1}{4}$ to get $20. But be careful! The answer is not choice A because we have to subtract $20 from the original total.

One way to avoid falling for the trap answer is to change your percentage at the beginning of the process. If you realize that subtracting 25% is the same as multiplying the original by 75%, you can answer the problem in one step and get right to the answer of $60. In decimal form, this looks like $0.75 \times 80 = 60$ .

Difficulty 2

Jay got 40 questions correct on one test and then 46 correct on the next one. Jay’s number of questions correct on the second test was what percent greater than his number of questions correct on the first test?

A. 6
B. 12
C. 15
D. 20

(spoiler)

The answer is C. There is a clear trap answer here: answer A. It’s the number of questions that rises by $6$ , not the percentage. For percent change, we need to apply the formula: $(n e w - o l d) / o l d$ . The new value here is $46$ and the old (previous) is $40$ . Subtracting and dividing, we get $6/40$ .

But which of these answers is equivalent to $6/40$ ? You could use long division to create a decimal, then convert to a percentage. The algebraic approach is to set up a proportion: $\frac{6}{40} = \frac{x}{100}$ . Cross-multiply and you get $40 x = 600$ , so $x = 15$ .

But it may be quickest of all to reduce the fraction to $3/20$ by dividing by two, then compare that fraction to the answers. If $2/20$ is 10% and $4/20$ is $20%$ , it makes sense that our fraction in this case converts to $15%$ .

Difficulty 3

Leah’s score increased by $15$ percentage points from one exam to the next. If she got $45$ out of $60$ questions correct on the first exam and both exams had $60$ questions, how many questions did she answer correctly on the second exam?

A. 50
B. 54
C. 55
D. 60

(spoiler)

The answer is B. Since we have to understand percentage here in light of the proportion of questions correct, we should start by figuring out Leah’s original percentage correct. Using the percent formula, we need to divide 45 by 60. Without the benefit of a calculator, we should again turn toward a reduced fraction for help: $45/60 = 3/4$ , which is the same as $75%$ .

Once we know the original test score was $75%$ , we can now add the $15$ percent mentioned at the beginning of the question to get $90%$ . A $90%$ score needs to be fed back into the percent formula, as follows:

$0.9 = \frac{x}{60}$

Note that we’ve converted the 90% to a decimal before plugging it in. The “part” is unknown, so we’ve labeled it $x$ ; the “whole” is 60. Solving this equation for $x$ yields $54$ .

Difficulty 4

A statistician records the number of people attending a four-game series between two baseball teams. After counting the attendance on the first day, the statistician notes that attendance falls by $20%$ from the first day to the second, grows by $10%$ on the second day to the third day, and grows by $25%$ from the third day to the last day. Compared to the first day, what was the attendance on the last day, expressed as a percent?

A. 95%
B. 105%
C. 110%
D. 115%

(spoiler)

The answer is C. Although we could begin by assigning $x$ (or a different variable) to the starting amount, it’s much easier to plug in $100$ , since we’re dealing with a percent question. If we call the first day’s attendance $100$ , we can readily see that the attendance on the second day would be $80$ ( $20%$ less). From here, we have to remember that we are no longer starting with $100$ ; the next percent change, a growth of $10%$ , starts with $80$ . $10%$ of $80$ is $8$ , so adding that amount gives us $88$ . That takes us to the last day, which is a $25%$ increase from $88$ . A difficult calculation, but remember that $25%$ is the same as $\frac{1}{4}$ . One-fourth of $88$ is $22$ , and adding $22$ to $88$ yields $110$ . Since we started at $100$ , our final answer represents the actual percent we’re looking for, answer C.

Difficulty 5

A company sets a revenue target of $500,000. At the beginning of Year 1, the company’s revenue is $280,000. If the company grows steadily at $20%$ per year, during which year will the company’s revenue reach its target?

A. Year 3
B. Year 4
C. Year 5
D. Year 6

(spoiler)

The answer is B. This question would be easier if a calculator were available; lacking this tool gives us a chance to stretch and practice our estimation skills.

To begin, we’ll calculate $20%$ of $280, 000$ . To help, remember that we can always find $10%$ of a number by moving the decimal place one to the left (which in this case drops as zero). If $10%$ of $280, 000$ is therefore $28, 000$ , then $20%$ must be double that, or $56, 000$ . Adding that total to the original amount gives us $336, 000$ . This marks the end of Year 1.

Continuing to calculate $20%$ of each subsequent value is going to get complicated. Let’s round each result to the nearest $10, 000$ . Changing $336, 000$ to $340, 000$ allows us to use the same process as before and find $68, 000$ as the $20%$ result. (We’ll need to keep in mind that we’ve round up somewhat and determine whether that affects our final answer.) Adding $68, 000$ to $336, 000$ gives us us $404, 000$ (now we’re at the end of Year 2).

In this case, we’ll now round down to $400, 000$ . That makes for a cleaner calculation that $20%$ of this number is $80, 000$ . Add that to $404, 000$ and we have $484, 000$ at the end of Year 3. It should be clear now, even without calculating, that we are soon going to cross the $500.000$ threshold, and that will happen during Year 4. We first rounded up, then down, so those changes from estimation should somewhat balance each other out. Even though our result is still not exact, it’s certainly close enough to say “Year 4” with confidence.

For Reflection

How will you approach percent questions on test day? Write down at least three takeaways from this module.
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
Can you explain to yourself the difference between percent questions and percent change questions? Try to give yourself a couple of examples of each kind.