Math Study Guide for the SAT Exam

Problem Solving

The SAT uses a variety of questions to test your math literacy. Can you use math concepts to solve work- and life-related problems? These questions deal with things like ratios, proportions, and percentages in ways you will use in real life. In addition to knowing what math terms mean, you’ll need to be able to use them appropriately to solve problems. If any of these procedures are difficult for you, doing practice questions can help you improve your fluency.

Properties of Operations

To be accurate in math, it is necessary to abide by some standard rules and properties. You’ve already learned about the order of operations. Here are some other commonly needed properties.

Rules for Calculation

There are three basic properties for adding and multiplying numbers: the associative, commutative, and distributive properties. Understanding these properties makes it easy for you to manipulate expressions and equations. It’s important to know that these properties do not apply to either subtraction nor division.

The associative property simply says that the grouping of numbers in addition or multiplication will not affect the result of operations. For example:

\[(3 \times 5) \times 4 = 3 \times (5 \times 4)\]

More generally, this can be represented as:

\[m + (n + o) = (m + n) + o \quad \text{ or } \quad m \times (n \times o) = (m \times n) \times o\]

The commutative property states that the elements in the operation can be moved around without affecting the result. This is represented by this equation:

\[m + n + o = o + n + m \quad\text{ or } \quad m \times n \times o = m \times n \times o\]

Here is an example:

\[3 \times 5 \times 4 = 4 \times 5 \times 3\]

When we refer to the distributive property of numbers, we mean performing multiplication distributed over addition:

\[A(2B + C) = 2AB + AC\]

For example:

\[3(x + y) = 3x + 3y\]

Ratios, Rates, Proportional Relationships, and Units

As you study for the SAT Math section, be sure you both understand and know how to apply the following concepts. This will save you a great deal of time that can be devoted to actually working the problems out instead of trying to jog your memory for specific definitions.

Ratio

Ratios are used to compare one quantity to another quantity. Say for instance you are taking a math exam that has a total of \(36\) questions: \(19\) questions on algebra and \(17\) questions on data analysis. When we compare the quantity of one part to another part or one part to the total, the comparison or relationship is called the ratio.

If we want to show the ratio of the number of algebra questions to data analysis questions, we write it as \(19\text{:}17\) (read as “\(19\) to \(17\)”). This can also be written as a fraction, \(\frac{19}{17}\).

To find the ratio of the number of questions on data analysis to the whole math exam, we write it as \(17\text{:}36\) or \(\frac{17}{36}\).

Proportion

Ratios and proportions are related math concepts. Equal ratios are said to be in proportion, or proportionate, to each other. For instance, the ratios \(2\text{:}4\) and \(10\text{:}20\) are proportionate.

Proportions are useful when you don’t know the value of a variable in a ratio. For instance, suppose you want to know how much you will be paid if you work \(12\) hours in a week. If you are paid \(\$8\) an hour, then you can set up the following proportion:

\[\frac{8}{1} = \frac{x}{12}\]

where \(x\) is the unknown payment. To solve a proportion like this, we cross-multiply:

\[8 \times 12 = 1 \times x\] \[x = 96\]

Rate and Unit Rate

The concept of rate will appear in questions that involve distance traveled over time (e.g., miles per hour, meters per second), work done per unit of time, cost per unit, area, density, and similar topics. Almost all of these rate questions require the ability to manipulate units and convert them if necessary.

Rate is generally a special kind of ratio expressing one term or quantity (measured in one unit) in comparison to another term or quantity (measured in another unit). For instance, we say that speed (\(s\)), which is a rate, is the measure of distance (\(d\)) over a measure of time (\(t\)). This is represented as a simple formula:

\[s = \frac{d}{t}\]

If you know speed and time but not distance, you can rearrange the formula as:

\[d = s \cdot t\]

For example, a car travels \(45\) miles in \(1.5\) hours. This is a rate showing the quantity measured as \(45\) miles over another quantity measured, \(1.5\) hours. To know the speed at which the car traveled, simply divide the distance by the time: \(45 \div 1.5 = 30\) miles per hour.

“Miles per hour” is a unit rate, meaning the rate of exactly one unit of a quantity in relation to another quantity. For “miles per hour”, we are measuring the number of miles in relation to exactly one hour.

Percentages

Percentage is an extension of ratios in which the whole is always \(100\%\). For instance, if you have two parts of a whole that has a total of five parts, the ratio is written as \(2\text{:}5\) or \(\frac{2}{5}\). As a percentage, though, it is expressed as \(40\%\) (\(2\text{:}5 = 40{:}100\)).

A percentage, represented by the percent symbol (\(\%\)), is found using this formula:

\[\% = \frac{X}{Z}\times 100\]

where \(X\) is the part and \(Z\) is the whole.

From the example above on the ratio of data analysis questions to the whole math exam expressed as \(17:58\), we may also say that the data analysis questions make up \(29.31\%\) of the entire math exam:

\[\% = \frac{17}{58} \times 100\% = 29.31\%\]

Percent Change

For the SAT, you will also need to know how to solve questions involving percent change (decrease or increase). You will generally be given both the final amount (\(F\)) and the initial amount (\(I\)). Percent change (\(\Delta\%\)) is expressed with this formula:

\[\Delta\% = \frac{F-I}{I} \times 100\]

Sometimes, you will be given the percentage change and need to find either the final or initial amount. As with previous formulas you’ve learned, you will simply need to rearrange it to get your answer. Let’s try a sample problem.

The total sales for a brand of soda at a retail store were \(\$2\text{,}000\) for the month of February. If the sales increased by \(3.4\%\) in March, what were the total sales of that soda in March?

Solution

Let’s rearrange our initial formula and input the given values:

\[\Delta\% = \frac{F – I}{I} \times 100\%\] \[\frac {3.4 \times 2\text{,}000}{100} = F – 2\text{,}000\] \[\frac {3.4 \times 2\text{,}000}{100} + 2\text{,}000 = F\] \[F = 2\text{,}068\]

So, after a \(3.4\%\) increase, the soda sales in March were \(\$2\text{,}068\).

Simple and Compound Interest

Interest is the amount paid or earned for the use of money, such as in money borrowed or invested. So, for instance, say you borrow \(\$10,000\) from a bank. The bank will expect to be repaid the initial \(\$10,000\) plus an additional percentage, which is the interest rate.

Here are important terms to understand:

• principal—the amount borrowed or invested

• interest rate—the ratio of the amount paid for borrowing money to the principal (customarily presented as a percentage per year).

• time—the time or duration that the principal amount is borrowed or invested

• simple interest—the amount paid for the use of money based on an annual interest rate

• compound interest—the amount paid for the use of money when interest is calculated on both the principal and any interest already earned, causing the total to grow at an increasing (exponential) rate over time

Solving Problems with Simple and Compound Interest

On the SAT, you will be expected to know how to calculate both simple and compound interest. As the name suggests, simple interest is the easier of the two. It is calculated with this formula:

\[I = p \cdot r \cdot t\]

where \(p\) is the principal, \(r\) is the annual interest rate expressed in decimal form, and \(t\) is time, normally expressed in years (though it can be other time periods).

Let’s take our initial example of borrowing \(\$10\text{,}000\) from the bank. If the interest rate the bank charges is \(10\%\), the total interest you will pay after one year is:

\[10\text{,}000 \cdot 0.10 \cdot 1 = $1\text{,}000.00\]

If the loan’s duration was two years, the interest charged would be \(\$2\text{,}000\).

Questions involving compound interest are a bit more complicated. They basically entail these steps:

1) Solve for the interest in the first period (e.g., one year, one quarter).

2) Add the computed interest in the first period to the principal.

3) Use this total to solve for the interest in the second period.

4) Add the computed value to the previous total.

5) Repeat the procedure as many times as the number of time periods desired (e.g., five times for the interest in five years, compounded annually).

6) Add all interest for the different periods to get the total interest of the amount borrowed compounded annually.

Using the same example as above, the interest for the first year will be the same at \(\$1\text{,}000\).

For the second year, however, it will be:

\[(10\text{,}000 + 1\text{,}000) \cdot 0.10 \cdot (1) = 1\text{,}100\]

Adding the interests for two years, the total compound interest charged would be \(\$2\text{,}100\), for a total of \(\$12\text{,}100\). Note that this compound interest amount is bigger than the simple interest on the same principle for the same period of time (\(\$2\text{,}000\)).

When the period of time (loan duration or investment term) involves bigger numbers, doing all those steps will be far too time-consuming. In such cases, it is more practical to use this formula:

\[A = p \cdot (1 + r)^t\]

where \(A\) is the future value to be paid, \(p\) is the principal, \(r\) is the annual interest rate in decimal form, and \(t\) is time, or the period in years.

Let’s use the example of \(\$10\text{,}000\) with a \(10\%\) interest rate over two years:

\[A = 10\text{,}000 \cdot (1 + 0.10)^2 = 10\text{,}000 \cdot (1.10)^2 = 10\text{,}000 \cdot 1.21 = 12\text{,}100\]

Now, you may get a question that states “\(10\%\) per year with monthly compounding.” This does not mean \(10\%\) per month. Rather, it means, a monthly interest of \(0.83\%\) (\(10\%\) divided by \(12\)), compounded monthly.

The formula for this concept varies a bit from the one you just learned:

\[A = P \cdot (1 + \frac{r}{n})^{nt}\]

Here, the \(n\) is the number of times the interest compounds each year.

So, if we take our same principle (\(\$10\text{,}000\)) but use monthly compounding interest, the interest for two years is:

\[A = 10\text{,}000 \cdot (1 +\frac{0.10}{12})^{12 \cdot 2} = 12\text{,}203.91\] \[I = 12\text{,}203.91 – 10\text{,}000 = 2\text{,}203.91\]

Multi-Step Problems

Just completing a single procedure is not enough to answer most questions on this exam. The SAT is really a test of reasoning as much as math, and a problem often involves several procedures along the way. When addressing a multi-step problem, these are some practical first steps:

• Determine what math topics are included. Is the question asking you to use algebra or trigonometry?

• Look for keywords that tell you what operations are necessary.

• If dealing with a word problem, assign variables to quantities and determine what is the math problem that needs solved.

• Develop a plan to solve the problem, breaking it into smaller, easier-to-solve problems.

Let’s do two word problems that involve multiple steps.

Problem #1

Ratios that are equal are said to be proportionate to each other, such as \(\frac{5}{7}\) and \(\frac{25}{35}\).

A bowl contains a mixture of one cup of cornstarch and four cups of flour. A second bowl contains the same ingredients in the same proportion, but instead has \(1.5\) cups of cornstarch. If half a cup of oats is added to the second bowl, what is the ratio of oats to flour in the second bowl?

Solution

This is a two-step problem. First, we must determine the amount of flour in the second bowl using the same ratio in the first bowl (which is \(1:4\)). We represent the amount of flour with the variable \(x\). Ratios are in proportion if they are equal, so:

\[\frac{1}{4} = \frac{1.5}{x}\]

Now, we cross-multiply, which gives us:

\[x = 6\]

Now, the second step is determining the ratio of oats to flour. Half a cup of oats was added to the second bowl, so the ratio of oats to flour is \(0.5\) to \(6\), or \(1:12\).

Problem #2

Andy usually buys large Grade A eggs from her supplier at \(\$2.50\) per dozen. She chanced upon a farm where the eggs were sold at \(70\%\) of this price and bought \(10\) dozens. How much more would she have paid for the \(10\) dozen eggs if she had bought from her regular supplier?

Solution

First, calculate the price of eggs at the farm, which is \(70\%\) of \(\$2.50\), remembering to convert the percentage to a decimal:

\[0.7 \times 2.50 = \$1.75\]

Now, you need to know the price Andy paid for the whole purchase. She paid \(\$1.75\) per dozen and bought \(10\) dozen, so the amount she paid was:

\[1.75 \times 10 = $17.50\]

Now, calculate the price for \(10\) dozen eggs from her regular supplier:

\[2.50 \times 10 = \$25.00\]

Finally, getting the difference of the two total amounts, we get Andy’s savings:

\[25 – 17.50 = \$7.50\]

Use Unit Conversion

Familiarity with the unit conversion method, or dimensional analysis, comes in very handy when dealing with rate questions and when double-checking answers.

Dimensional Analysis

Dimensional analysis, also called the factor-label method, is a systematic way to convert one unit into another. This method is especially helpful in rate problems and whenever you need to be certain that your units make sense throughout a calculation.

The idea is simple: You multiply the quantity you’re given by conversion factors, which are fractions that equal \(1\). These conversion factors cancel out the units you don’t want and replace them with the units you do want.

Here is a clear step-by-step way to apply this method:

1) Write the value with its unit.—Start by writing the given number as a fraction, even if the denominator is just \(1\). For example:

\[50 \text{mi} = \frac{50 \, \text{mi}}{1}\]

2) Decide what unit you want to end with.—Always keep the final goal in mind. Are you converting miles to feet or hours to minutes?

3) Multiply by the conversion factors.—A conversion factor is a fraction where the numerator and denominator are equal amounts written in different units, such as:

\(\frac{60 \, \text{min}}{1 \, \text{hr}} \,\text{ or }\,\frac{1 \, \text{ft}}{12 \, \text{in}}\) Since these fractions equal \(1\), multiplying by them doesn’t change the value, only the unit.

4) Arrange each conversion factor so units cancel. Place the unwanted unit opposite the one you have. For example, to convert miles to feet:

\(\frac{50 \, \text{mi}}{1} \times \frac{5\text{,}280 \, \text{ft}}{1 \, \text{mi}}\) ​ So, the miles (\(\text{mi}\)) cancel out, leaving feet.

5) Continue multiplying until only the desired unit remains.

6) Do the arithmetic to get the final answer.

This method works for nearly any type of conversion, distance, time, rate, volume, and more, and helps you verify that your answer is dimensionally correct. Here is a table of common distance and volume conversions:

Distance Conversions	Volume Conversions
\(1 \text{ km} = 1\text{,}000 \text{ m}\)	\(1 \text{ kL} = 1\text{,}000 \text{ L}\)
\(1 \text{ m} = 100 \text{ cm}\)	\(1 \text{ L} = 100 \text{ cL}\)
\(1 \text{ cm} = 10 \text{ mm}\)	\(1 \text{ cL} = 10 \text{ mL}\)
\(1 \text{ mi} = 1\text{,}760 \text{ yd}\)	\(1 \text{ gal} = 4 \text{ qt}\)
\(1 \text{ yd} = 3 \text{ ft}\)	\(1 \text{ qt} = 4 \text{ c}\)
\(1 \text{ ft} = 12 \text{ in}\)	\(1 \text{ c} = 8 \text{ fl oz}\)

Let’s do a problem that involves basic time conversions.

Alia has an annual basic salary rate of \(\$53\text{,}760\). If she works eight hours a day, four days a week, and four weeks a month, what is her hourly rate?

Solution

We want to convert dollars per year into dollars per hour. Start with what we know:

\[\frac{\$53\text{,}760}{1 \, \text{yr}}\]

Now use the factor-label method step-by-step.

First, we will convert years to months. There are \(12\) months in \(1\) year, so:

\[\frac{\$53\text{,}760}{1 \, \text{yr}} \times \frac{1 \, \text{yr}}{12 \, \text{mo}}=\frac{\$53\text{,}760}{12 \, \text{mo}}\]

The “years” canceled out, leaving dollars per month.

Next, we will convert months to weeks. Alia works \(4\) weeks per month:

\[\frac{\$53\text{,}760}{12 \, \text{mo}} \times \frac{1 \, \text{mo}}{4 \, \text{wk}}\]

The months canceled out. Then, we convert weeks to days. She works \(4\) days per week:

\[\frac{1 \, \text{mo}}{4 \, \text{wk}} \times \frac{1 \, \text{wk}}{4 \, \text{d}}\]

The weeks canceled out.

Lastly, we will convert days to hours. She works \(8\) hours per day:

\[\frac{1 \, \text{wk}}{4 \, \text{d}} \times \frac{1 \, \text{d}}{8 \, \text{hr}}\]

The days canceled out and we’re left with the unit we wanted all along, hours.

Now, multiply everything together:

\[\frac{\$53\text{,}760}{1 \, \text{yr}} \times \frac{1}{12} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{8} = \frac{\$35}{\text{hr}}\]

So, Alia’s hourly wage is \(\$35\) per hour.

This example shows how writing the units carefully helps ensure that everything cancels correctly, leaving only the unit you want.