*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 **cent**ury or **cent**ennial), 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.

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%

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: $104 $, $52 $, $10040 $, $0.4$. 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: $60÷4=15$. 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 “$3x$”, and our equation follows:

$x+3x=604x=60x=15 $

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

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.

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

$percentpercent 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)

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 $41 $, we can multiply the original price of $80 by $41 $ 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 $0.75×80=60$.

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 score 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: $(new−old)/old$. 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: $406 =100x $. Cross-multiply and you get $40x=600$, 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%.

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=60x $

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 $54$.

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 $41 $. 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**.

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.

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