# SAT Math: Problem Solving and Data Analysis

Covering 29% of the concepts on the SAT, the Problem Solving and Data Analysis section is the second most common on SAT Math, after Heart of Algebra. Keeping in line with real-world scenarios, these SAT problems will ask you to infer information based on a study with any number of participants or interpret data from a graph. This is actually beneficial for SAT students, given that you’ll be learning lots about cause and effect and inferential statistics in college.

Now, let’s talk about how to approach the various types of Problem Solving and Data Analysis questions on the SAT, plus some practice questions to get you started! Feel free to use the Table of Contents to navigate directly to the topics you want to learn.

## Quick Facts About SAT Problem Solving and Data Analysis

You can expect to see about 17 Problem Solving and Data Analysis questions on the SAT Math section, although they will not be outright labeled by question type. Some answers will be single-step questions, while others will be multistep problems. If you fear mental math, there’s no need to sweat it in this section—you’ll be allowed to use a calculator. Like the other SAT Math sections, you’ll receive a Problem Solving and Data Analysis subscore on a scale of 1 to 15.

## Ratio, Proportion, Units, and Percentage

This Problem Solving and Data Analysis question type shouldn’t come as a surprise since it has probably been part of your math courses for the last five years (yep, you most likely went over this stuff, in some form, all the way back in early middle school). I’ll start with ratios.

### Understanding Ratios

A good way to think of ratios is apples and oranges. Say I have two oranges and three apples, the ratio of oranges to apples is 2:3. Seems straightforward. What if I have four apples and 6 oranges? If you answered 4:6, that is not quite correct. You have to think of the ratio the way you would a fraction, in lowest terms. Both 4 and 6 can be divided by ‘2’, giving you 2:3. Notice how that is the same ratio as 2:3.

This highlights an important conceptual idea: ratio is not about total number. It is about the number of one thing, to the number of another thing, reduced, so that the ratio is expressed as two prime numbers.

One last thing about ratios. Let’s say you have a ratio of 1:2. This is not the same thing as ½. The bottom number in a fraction is always the total. The total of a ratio is always the parts of a ratio added together. In this case, 1:2 is 1 + 2 = 3. So if I have 1 apple to two oranges, 1/3 of the fruit are apples and 2/3 are oranges.

If you have more than two ratios, make sure to add up all of the ratios. For instance, if the ratio of blue marbles to red marbles to green marbles is 2 : 5: 7, red marbles account for 5/14 of the total (2 + 5 + 7 = 14).

Do you think you got that? Well, here are some practice questions to test your knowledge of Problem Solving and Data Analysis.

### Understanding Percentages

Percentages can be surprisingly complicated on the SAT. Part of that is because we can’t always translate them into fractions, which are easier to work with algebraically. While it’s easy enough to think of $$50%$$ as $$frac{1}{2}$$, it’s rarely so easy to make the conversion on the SAT, especially when the percentages given are, say, 35% or 15%.

Finding a percent is pretty easy, as long as you have a calculator. Just divide the part by the whole and multiply the decimal that comes out by 100. So if you ate 10 out of a serving of 12 buffalo wings, then you ate (10/12)100=83.33%. Remembering that formula can save you some grief when you have to use it algebraically.

However, the SAT won’t just test you on the simple process of finding the percentage of a number (like calculating a tip). Instead, it’ll ask you to calculate in reverse (finding the whole from the part), find a combination of percentages, find a percent change, or give some other scenario-specific piece of information.

Being prepared for percent change questions, in particular, will take you far on Problem Solving and Data Analysis.

The equation for percent increase $$= frac{text{New Number – Original Number}}{text{Original Number}}*100$$.

The equation for percent decrease is $$= frac{text{Original Number – New Number}}{text{Original Number}}*100$$.

### Practice Questions

#### Easy Questions

1. The ratio of shirts to shorts to pairs of shoes in Kevin’s closet is 5 : 2 : 3. If Kevin owns 10 shirts, how many pairs of shoes does he have to give away so that he ends up having the same number of shorts as he does pairs of shoes?

A) 1
B) 2
C) 4
D) 5

Okay, this question is slightly evil, since shorts sounds like shirts and it is easy to get the two mixed up when you are reading fast. So always pay attention, even on easier questions!

Since we know that Kevin has 10 shirts and that 10, therefore, corresponds to the number ‘5’ in the ratio, that the actual number of shorts, shirts, etc., he owns is double the number in the ratio. Thus, he owns four shirts and six pairs of shoes. So he’ll have to give away two pairs of shoes so that he’ll have the same number of shoes as he does shorts. Answer: (B).

On some questions, you’ll have to figure out the proportion between two different units.

#### Medium Difficulty Question

1. There are 200,000 eligible voters in district X, 60% of whom voted in the 2008 state election. In 2010 state election, the number of eligible voters in district X increased by 20% but if only 55% voted in this election, how many total votes were cast in the 2010 state election, assuming that no voter can cast more than one vote?

A) 12,000
B) 120,000
C) 132,000
D) 176,000

# of voters who voted in 2008 election is equal to 200,000 x 60 = 120,000

In 2010, the number of overall eligible voters increased by 20%, so 20% of 200,000 is 40,000 giving us 240,000 total voters.

55% of 240,000 gives us 132,000. Answer C).

#### Difficult Questions

1. A cartographer owns a square map in which one inch corresponds to 7/3 of a mile. What is the area of the map in square inches if the map covers a territory of 49 square miles?

SPR: ______________

We know that 7/3 of mile = one inch.

We also know that the area is 49 square miles, meaning that each side = 7: √49 = 7). To find how many inches correspond to 7 miles, we set up the following equation:

7 = 7/3x, x = 3

Here is the little twist that you want to watch out for. The question is asking for square miles in inches, so we have to take 3^2 which equals 9.

1. For the following percents, convert each to a fraction and a decimal:

5% =
26% =
37.5% =
125% =

Another possible question type, and one that most are familiar with and probably dread, is the percent question.

To reduce something by a certain percentage, either turn that percent into a ratio over 100 or convert the percent into a decimal by moving the point back two spaces. For example, 40% equals both 40/100 and 0.40. So the answers are:

5% = .05, 5/100 or 1/20 (you don’t always have to reduce for quick calculations)
26% = .26, 26/100 or 13/50
37.5% = .375, 375/1000 or 3/8
125% = 1.25, 125/100, 5/4

1. In a popular department store, a designer coat is discounted 20% off of the original price. After not selling for three months, the coat is further marked down another 20%. If the same coat sells online for 40% lower than the original department store price, what percent less would somebody pay if they were to buy the coat directly online than if they were to buy the coat after it has been discounted twice at the department store?

A) 4%
B) 6.25%
C) 16%
D) 36%

When you are not given a specific value for a percent problem, use 100 since it is easiest to increase or decrease in terms of %.

1st discount: 20% off of 100 = 80.

2nd discount: 20% off of 80 = 64.

Online, the coat sells for 40% off of the original department store price, which we assumed is 100.

Online discount: 40% of 100 = 60.

This is the tricky part. We are not comparing the price difference (which would be 4 dollars) but how much percent less 60 (online price) is than 64 (department store sale price).

Percent difference: (64 – 60)/64 = 1/16 = 6.25%. Answer B).

## Subjects and Treatments

This is not an official title but the name I’m giving to questions that deal with studies trying to determine cause and effect.

In order to understand how to approach Subjects and Treatments questions, let’s talk about randomization. The idea of randomization is the essence, the beating heart, of determining cause and effect. It helps us more reliably answer the question of whether a certain form of treatment causes a predictable outcome in subjects.

Randomization can happen at two levels. First off, when researchers select from the population in general, they have to make sure that they are not unknowingly selecting a certain type of person. Say, for instance, that I want to know what percent of Americans use Instagram. If I walk on a college campus and ask students there, I’m not taking a randomized sample of Americans (think how different my response rate would be if I decide to poll the audience at a Rolling Stones concert).

On the other hand, if I went to a city phone directory and dropped a quarter on the page, choosing the name that the center of the quarter was closest to, I would be doing a much better job of randomizing (though, one would rightly argue, I’d still be skewing to an older age-group, assuming that most young people have only cell phones, which aren’t listed in city directories). For the sake of argument, let’s say our phone directory method is able to randomly choose for all ages.

After throwing the quarter a total of a hundred times on randomly selected pages (we wouldn’t want only people whose names begin with ‘C’, because they might share some common trait), our sample size consists of 100 subjects. If we were to ask them about their Instagram use, our findings would far more likely skew with the general population. Therefore, this method would allow us to make generalizations about the population at large.

### Subjects and Treatments Practice Question

A high school track coach has a new training regimen in which runners are supposed to exercise twice a week riding a stationary bike for one hour instead of doing a one-hour run twice a week. Her theory is that, by biking, students will not overly exert their running muscles but will still exercise their cardiovascular system. To test this theory out she had her varsity athletes (the faster runners) incorporate the biking regimen and the junior varsity athletes (the slower runners) incorporate usual training. After three weeks, the times of her varsity athletes on a 3-mile course decreased by an average of 1-minute, whereas her junior varsity athletes decreased their time on the same 3-mile course by approximately 30 seconds.

Which of the following is an appropriate conclusion?

A) The exercise bike regimen led to the reduction of the varsity runners’ time.
B) The exercise bike regimen would have helped the junior varsity team become faster.
C) No conclusion about cause and effect can be drawn because there might be fundamental differences between the way that varsity athletes respond to training in general and the way that junior varsity athletes respond.
D) No conclusion about cause and effect can be drawn because junior varsity athletes might have decreased their speed on the 3-mile course by more than 30 seconds had they completed the biking regimen.

Show Practice Question Answer and Explanation

When dealing with cause and effect in a study, or what the SAT calls a treatment, researchers need to ensure that they randomly select amongst the participants. Imagine that we wanted to test the effects on the immune system of a new caffeinated beverage. If researchers were to break our 100 subjects into under-40 and over-40, the results would not be reliable.

First off, young people are known to generally have stronger immune systems. Therefore, once we have randomly selected a group for a study, we need to further ensure that, once in the study, researchers randomly break the subjects into two groups. In this case, those who drink the newfangled beverage and those who must make do with a placebo, or beverage that is not caffeinated.

At this point, we are likely to have a group that is both representative of the overall population and will allow us to draw reliable conclusions about cause and effect.

Another scenario and this will help us segue to the practice question above, are treatments/trials in which the subjects are not randomly chosen. For instance, in the question about the runners, clearly, they are not representative of the population as a whole (I’m sure many people would never dare peel themselves off their couches to something as daft as run three miles).

Nonetheless, we can still determine cause and effect from a non-representative population (in this case runners) as long as those runners are randomly broken into two groups, exercise bike vs. usual one hour run. The problem with the study is the runner coach did not randomly assign runners but gave the slower runners one treatment. Therefore, the observed results cannot be attributed to the bike regimen; they could likely result from the fact that the two groups are fundamentally different. Think about it: a varsity runner is already the faster runner, one who is likely to improve faster at running a three-mile course than his or her junior varsity teammate. Therefore, the answer is C).

While D) might be true, and junior varsity subjects might have become faster had they been in the bike group, it doesn’t help us identify what was flawed about the treatment in the first place: the subjects were not randomly assigned.

### Subjects and Treatments: A Summary

Here are the key points regarding subjects and treatments (aka cause and effect questions) on SAT Problem Solving and Data Analysis:

1. Results from a study can only be generalized to the population at large if the group of subjects was randomly selected.
2. Once subjects have been selected, whether or not they were randomly selected, cause and effect can only be determined if the subjects were randomly assigned to the groups within the experiment/study/treatment.
3. There are three basic types of averages on the SAT that you should be pretty comfortable with at this point, and all of them start with the letter “m.” Those are the mean, the median, and the mode. In case those aren’t second nature, let’s define them, quickly.

## SAT Statistics (Mean, Median, and Mode)

The SAT Math test often asks you to do some statistics problems involving averages. Finding the mean is the most commonly used average and, as it so happens, the most commonly tested when it comes to SAT statistics. The formula is pretty simple:

{a+b+c+….}/n where n is the number of terms added in the numerator.  In the set of numbers {2,3,4,5}, 3.5 would be the mean, because 2+3+4+5=14, and $$14/4=3.5$$

If the numbers in a set are listed in order, the median is the middle number. In the set {1,5,130}, 5 is the median. In the set above, {2,3,4,5}, the median is 3.5, which is the mean of the middle two terms since there’s an odd number of them.

The mode is just the number that shows up the most often. It’s perfectly possible that there is no mode or that there are several modes. In the set {5,7,7,9,18,18}, both 7 and 18 are modes.

## What’s important to know about averages on the SAT?

Averages come up in algebra or word problems. You’ll usually have to find some value using the formula for a mean, but it may not be as simple as finding the average of a few numbers.  Instead, you’ll have to plug some numbers into the formula and then use a bit of algebra or logic to get at what’s missing.

For example, you might see a question like this:

If the arithmetic mean of x, 2x, and 6x is 126, what is the value of x?

To solve the question, you’ll need to plug it all into the formula and then do some variable manipulation.

$$frac {x+2x+6x}{3}=126$$

$${x+2x+6x}=378$$

$$9x = 378$$

$$x=42$$

Medians and modes, on the other hand, don’t show up all that often in Problem Solving and Data Analysis. Definitely be sure that you can remember which is which, but expect questions on means, most of the time. As for other types of statistical analysis, you may also be asked to solve some problems involving standard deviation.

Averages Practice Problem
If three sisters have an average (arithmetic mean) age of 24, and the youngest sister is 16, what is the sum of the ages of the two older sisters?

1. 28
2. 32
3. 56
4. 60
5. 72

If you’re careful to remember that the question is asking you for the sum of the sisters’ ages, you can solve this one pretty quickly. Keep in mind that we can’t find their individual ages, though. There’s not enough information for that. First we find the total combined age of the three, which must be 72, since $$24*3=72$$. Careful not to fall for the trap that is (E), we take the last step and subtract 16 from that total age to find the leftover sum, which is 56, or (C).

The phrase “weighted average” might be a little scary sounding, but it’s nothing to get freaked out over. Usually weighted averages on the SAT will use the basic formula for finding the mean (link to “SAT Math Types of Averages“). It’s pretty much the same skill.

### What is a “weighted average”?

Basically, weighted means uneven, here; the numbers that you’re looking at don’t carry the same importance. For example, if I’m trying to find the average number of fleas that my pets have, and each cat has 150 while each dog has 200, then those two numbers have equal “weight” only if I have the same number of cats as dogs. Let’s say I have 1 of each.

$$frac {150+200}{2}=175$$

That’s just a normal mean, so that’s no problem. Well, the fleas are a problem, I guess. And the fact that I’m counting fleas might have my family a little worried…anyway, the math is easy. But that’s a non-weighted average.

For a weighted average, I would have a different number of cats than dogs. Let’s say I had 3 cats and 2 dogs. (And they all have fleas…things are starting to get kinda gross. Sorry.)

In order to give them the appropriate weight, we’d have to multiply each piece appropriately and change the total (denominator) to reflect it.

$$frac {3(150)+2(200)}{5}=170$$

But if you expand that, you’ll see that it’s the same as the standard mean formula.

$$frac {150+150+150+200+200}{5}=170$$

Just make sure you divide by five (because I have five pets) not two (for two types of pets).

### Finding Average Rates

Average rates are a type of weighted average. Your SAT Problem Solving and Data Analysis section will include a problem or two about these, and you need to be sure not to fall for the common trap.

Maria’s drive to the supermarket takes her 20 minutes, during which she averages a speed of 21 miles per hour. She takes the same route home, but it only takes 15 minutes to cover the equal distance. What was Maria’s average speed while driving?

1. 15.5 mph
2. 21 mph
3. 24 mph
4. 24.5 mph
5. 28 mph

This is a tricky, multi-step problem, and you can’t plug in the answer choices to solve it, sadly.

Let’s first find all of our information, because the question has only given you part of it. You need to know the formula r=d/t (rate = distance/time), also expressed as d=rt (easily remembered as the “dirt” formula). We’re going to use it both ways.

Using that formula, let’s look at the first leg of her trip. She traveled for 1/3 of an hour at 21 mph, so she must have traveled 7 miles.

That’s $$21*0.333=7$$

Using that info, we can figure out the rate of her trip back home. Going 7 miles in 1/4 of an hour on the way home, she went an average of 28 mph.

That’s $$7/0.25=28$$

So now we need to find the total average. That’s not the average of the two numbers we have! Because each mile she traveled on the way there took more time than each mile on the way home, they have different weights!

✗ $$frac{21+28}{2}=24.5$$

Instead, you need to take the total of each piece—total time and total distance—to find the total, average rate.

✓ $$frac{14 text{ miles}}{.333 text{ hours} + .25 text{ hours}}=frac{14 text{ miles}}{.5833 text{ hours}}={24 text{ mph}}$$

### Average Rate Practice Question

Niles takes an interstate road trip over the course of two days. If he covers 610 miles in nine hours the first day and 300 miles in four hours on the second day, what is his average speed per hour?

A) 55 miles per hour
B) 65 miles per hour
C) 70 miles per hour
D) 75 miles per hour

To figure out the average speed of the entire trip, divide the total distance by the total number of hours. The handy equation D = rt, where D is total distance, r is rate, and t is time, will make this easier.

D = 910, r = ?, t = 9 + 4 = 13 hours.

910 = 13r, r = 70, answer C).

### Weighted Averages That You Won’t See on Your SAT

I’ve never seen an SAT Problem Solving and Data Analysis question that asks you to find an average based on percent weights (e.g. finding a final grade in a class where quizzes count for 70%, attendance for 20%, and participation for 10%). Finding that average is a little more complicated, so it’s nice that we don’t have to worry about it.

Simply put… if you’re finding the average of two sets of information that already are averages in their own right, as the number of fleas per cat and the number fleas per dog, you can’t just take the mean of those averages. You have to find the totals and then plug them into the formula. You should be excited about these kinds of problems, if for nothing more than having the opportunity to bust out your handy-dandy, brand-spanking’ new SAT calculator. 😛

## SAT Graphs: Tips and Tricks

Among the math skills that SAT Problem Solving and Data Analysis tests, reading data from tables or graphs is one of the more straightforward tasks. But there are a number of simple mistakes that might make you miss out on points if you’re not careful. The best way to avoid those totally avoidable slip-ups is to train yourself to follow a pattern.

1. Scanning Before Reading the Question
It’s tempting to jump right into the question, especially if you’re feeling clock pressure. But don’t do it! Usually, the question will make pretty much no sense if you don’t have the context that the figure gives you. You’ll end up reading the question, looking over the table/graph, rereading the question, and then finding the information you need. Why waste the time doing it twice? Scan the figure first.

You’ll want to read the headings, the axes, and the units of measurement, then make note of any missing information or obvious patterns.

After you’ve done that, go ahead to the question.

2. Add in Any Information From the Question
As is often true for other types of SAT math problems, the written question might have some info in it that the figure doesn’t include. Just like you would write in angle measurements, fill in any extra info; there’s no reason to try to keep it in your head.
3. Find the Areas of the Table That the Question Asks About
Your SAT might try to make the question more confusing by adding in a whole lot of superfluous information into the visual. Check what the question asks for, then circle the area in the table or graph that answers the question (or gives you info that will lead to the answer).

For the easiest SAT data questions, you’re already finished—they just want you to locate the info. But it might ask you to go another step.

4. Write the Math Out
If you’re asked about relationships between two things, look carefully at the relationships between 4-6 pieces of information (two xs and two ys), and write out the pattern. If you’re looking for some variable, write out the equation. If it’s not clear how to go about that, maybe you should try plugging answer choices in to see if they work.
After you’ve done whatever math you need to, check your units. It’s easy to make a mistake by using minutes instead of hours, and the SAT takes advantage of that in the incorrect answer choices.

A lot of SAT prep has to do with training yourself against hasty mistakes, and tables and graphs are classic places to make those slips. Follow the rules and you’ll be much safer.

That’s all for the SAT Problem Solving and Data Analysis section! We hope this breakdown was helpful for you. To read up on the other two sections of SAT Math, check out our posts on Heart of Algebra and Passport to Advanced Math.