Maths Concepts for cracking SAT Quantitative Section

Here’s a look at all the concepts that you need to know for cracking SAT Quantitative Section


Arithmetic is the fundamental building block of math. The other three subject areas tested in SAT Math are all pretty much unthinkable without arithmetic. You’ll certainly need to know your arithmetic to power through algebra, geometry, and data analysis problems, but the Math section also includes some pure arithmetic problems as well. So, it makes sense to start Math  with a discussion of numbers and the typical things we do with them.

  • Common Math Symbols

You may remember these from way back when, but in case you need a quick refresher, here’s a list of some of the most commonly used math symbols you should know for the SAT. We’ll discuss some of them in this arithmetic section and others later.





Less than

The quantity to the left of the symbol is less than the quantity to the right.


Greater than

The quantity to the left of the symbol is greater than the quantity to the right.

Less than or equal to

The quantity to the left of the symbol is less than or equal to the quantity to the right.

Greater than or equal to

The quantity to the left of the symbol is greater than or equal to the quantity to the right.


Square root

A number which when multiplied by itself equals the value under the square root symbol.

| x |

Absolute value

The positive distance a number enclosed between two vertical bars is from 0.



The product of all the numbers up to and including a given number.



In geometry, two lines separated by this symbol have the same slope (go in exactly the same direction).



In geometry, two lines separated by this symbol meet at right angles.



A measure of the size of an angle. There are 360 degrees in a circle.



The ratio of the circumference of any circle to its diameter; approximately equal to 3.14.

  • Number Terms

The test makers assume that you know your numbers. Make sure you do by comparing your knowledge to our definitions below.




Whole numbers

The set of counting numbers, including zero

0, 1, 2, 3

Natural numbers

The set of whole positive numbers except zero

1, 2, 3, 4


The set of all positive and negative whole numbers, including zero, not including fractions and decimals. Integers in a sequence, such as those in the example to the right, are called consecutive integers.

–3, –2, –1, 0, 1, 2, 3

Rational numbers

The set of all numbers that can be expressed as integers in fractions—that is, any number that can be expressed in the form , where m and n are integers


Irrational numbers

The set of all numbers that cannot be expressed as integers in a fraction

π, , 1.010100001000110000

Real numbers

Every number on the number line, including all rational and irrational numbers

Every number you can think of

  • Even and Odd Numbers
  • An even number is an integer that is divisible by 2 with no remainder, including zero.

              Even numbers: –10, –4, 0, 4, 10

  • An odd number is an integer that leaves a remainder of 1 when divided by 2.

             Odd numbers: –9, –3, –1, 1, 3, 9

Even and odd numbers act differently when they are added, subtracted, multiplied, and divided. The following chart shows the rules for addition, subtraction, and multiplication (multiplication and division are the same in terms of even and odd).




even + even = even

even – even = even

even × even = even

even + odd = odd

even – odd = odd

even × odd = even

odd + odd = even

odd – odd = even

odd × odd = odd

Zero, as we’ve mentioned, is even, but it has its own special properties when used in calculations. Anything multiplied by 0 is 0, and 0 divided by anything is 0. However, anything divided by 0 is undefined.

  • Positive and Negative Numbers
  • A positive number is greater than 0. Examples include , 15, and 83.4.
  • A negative number is less than 0. Examples include –0.2, –1, and –100.
  • One tip-off is the negative sign (–) that precedes negative numbers.
  • Zero is neither positive nor negative.
  • On a number line, positive numbers appear to the right of zero, and negative numbers appear to the left:

                                        –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5

Positive and negative numbers act differently when you add, subtract, multiply, or divide them. Adding a negative number is the same as subtracting a positive number:

5 + (–3) = 2, just as 5 – 3 = 2

Subtracting a negative number is the same as adding a positive number:

7 – (–2) = 9, just as 7 + 2 = 9

To determine the sign of a number that results from multiplication or division of positive and negative numbers, memorize the following rules.



positive × positive = positive

positive ÷ positive = positive

positive × negative = negative

positive ÷ negative = negative

negative × negative = positive

negative ÷ negative = positive

Here’s a helpful trick when dealing with a series of multiplied or divided positive and negative numbers: If there’s an even number of negative numbers in the series, the outcome will be positive. If there’s an odd number, the outcome will be negative.

When negative signs and parentheses collide, it can get pretty ugly. However, the principle is simple: A negative sign outside parentheses is distributed across the parentheses. Take this question:

3 + 4 – (3 + 1 – 8) = ?

We first work out the parentheses, which gives us:

3 + 4 – (4 – 8)

This can be simplified to:

3 + 4 – (– 4)

As discussed earlier, subtracting a negative number is the same as adding a positive number, so our equation further simplifies to:

3 + 4 + 4 = 11

An awareness of the properties of positive and negative numbers is particularly helpful when comparing values in Quantitative Comparison questions.

  • Remainders

A remainder is the integer left over after one number has been divided by another. Take, for example, 92 ÷ 6. Performing the division we see that 6 goes into 92 a total of 15 times, but 6 × 15 = 90, so there’s 2 left over. In other words, the remainder is 2.

  • Divisibility

Integer x is said to be divisible by integer y when x divided by y yields a remainder of zero. The SAT sometimes tests whether you can determine if one number is divisible by another. You could take the time to do the division by hand to see if the result is a whole number, or you could simply memorize the shortcuts in the table below.

Divisibility Rules


All whole numbers are divisible by 1.


A number is divisible by 2 if it’s even.


A number is divisible by 3 if the sum of its digits is divisible by 3. This means you add up all the digits of the original number. If that total is divisible by 3, then so is the number. For example, to see whether 83,503 is divisible by 3, we calculate 8 + 3 + 5 + 0 + 3 = 19. 19 is not divisible by 3, so neither is 83,503. NOTE: (Short cut) If you want to reduce the addition, strike out the multiples of 3 from the number. For example, in the number 9726311, 9, 72, 63 are all multiples of 3, so no need to add them. Simply add the remaining digits: 1+1 =2. Since, 2 is not divisible by 3, the number 9726311 also is not divisible by 3.


A number is divisible by 4 if its last two digits, taken as a single number, are divisible by 4. For example, 179,316 is divisible by 4 because 16 is divisible by 4.


A number is divisible by 5 if its last digit is 0 or 5. Examples include 0, 430, and –20.


A number is divisible by 6 if it’s divisible by both 2 and 3. For example, 663 is not divisible by 6 because it’s not divisible by 2. But 570 is divisible by 6 because it’s divisible by both 2 and 3 (5 + 7 + 0 = 12, and 12 is divisible by 3).


7 may be a lucky number in general, but it’s unlucky when it comes to divisibility. Although a divisibility rule for 7 does exist, it’s much harder than dividing the original number by 7 and seeing if the result is an integer. So if the SAT happens to throw a “divisible by 7” question at you, you’ll just have to do the math i.e. divide the number by 7 and find the remainder.


A number is divisible by 8 if its last three digits, taken as a single number, are divisible by 8. For example, 179,128 is divisible by 8 because 128 is divisible by 8.


A number is divisible by 9 if the sum of its digits is divisible by 9. This means you add up all the digits of the original number. If that total is divisible by 9, then so is the number. For example, to see whether 531 is divisible by 9, we calculate 5 + 3 + 1 = 9. Since 9 is divisible by 9, 531 is as well. NOTE: The shortcut applicable to the divisibility test of 3 could be applied for the divisibility test of 9 as well, the only difference being that the multiples have to be 9 and the sum of the remaining digits has to be divisible by 9.


A number is divisible by 10 if the units digit is a 0. For example, 0, 490, and –20 are all divisible by 10.


This one’s a bit involved but worth knowing. (Even if it doesn’t come up on the test, you can still impress your friends at parties.) Here’s how to tell if a number is divisible by 11: Add every other digit starting with the leftmost digit and write their sum. Then add all the numbers that you didn’t add in the first step and write their sum. If the difference between the two sums is divisible by 11, then so is the original number. For example, to test whether 803,715 is divisible by 11, we first add 8 + 3 + 1 = 12. To do this, we just started with the leftmost digit and added alternating digits. Now we add the numbers that we didn’t add in the first step: 0 + 7 + 5 = 12. Finally, we take the difference between these two sums: 12 – 12 = 0. Zero is divisible by all numbers, including 11, so 803,715 is divisible by 11.


A number is divisible by 12 if it’s divisible by both 3 and 4. For example, 663 is not divisible by 12 because it’s not divisible by 4. 162,480 is divisible by 12 because it’s divisible by both 4 (the last two digits, 80, are divisible by 4) and 3 (1 + 6 + 2 + 4 + 8 + 0 = 21, and 21 is divisible by 3).

  • Factors

A factor is an integer that divides into another integer evenly, with no remainder. In other words, if is an integer, then b is a factor of a. For example, 1, 2, 4, 7, 14, and 28 are all factors of 28, because they go into 28 without having anything left over. Likewise, 3 is not a factor of 28 since dividing 28 by 3 yields a remainder of 1. The number 1 is a factor of every number.

Some SAT problems may require you to determine the factors of a number. To do this, write down all the factors of the given number in pairs, beginning with 1 and the number you’re factoring. For example, to factor 24:

  • 1 and 24 (1 × 24 = 24)

  • 2 and 12 (2 × 12 = 24)

  • 3 and 8 (3 × 8 = 24)

  • 4 and 6 (4 × 6 = 24)

Five doesn’t go into 24, so you’d move on to 6. But we’ve already included 6 as part of the 4 × 6 equation, and there’s no need to repeat. If you find yourself beginning to repeat numbers, then the factorization’s complete. The factors of 24 are therefore 1, 2, 3, 4, 6, 8, 12, and 24.

  • Prime Numbers

They are the only numbers whose sole factors are 1 and themselves. More precisely, a prime number is a number that has exactly two positive factors, 1 and itself. For example, 3, 5, and 13 are all prime, because each is only divisible by 1 and itself. In contrast, 6 is not prime, because, in addition to being divisible by 1 and itself, 6 is also divisible by 2 and 3. Here are a couple of points about primes that are worth memorizing:

  • All prime numbers are positive. This is because every negative number has –1 as a factor in addition to 1 and itself.

  • The number 1 is not prime. Prime numbers must have two positive factors, and 1 has only one positive factor, itself. It is co-prime.

  • The number 2 is prime. It is the only even prime number. All prime numbers besides 2 are odd.

Here’s a list of the prime numbers less than 100:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97

It wouldn’t hurt to memorize this list. In addition, you can determine whether a number is prime by using the divisibility rules listed earlier. If the number is divisible by anything other than 1 and itself, it’s not prime.

If a number under consideration is larger than the ones in the list above, or if you’ve gone and ignored our advice to memorize that list, here’s a quick way to figure out whether a number is prime:

  1. Estimate the square root of the number.

  2. Check all the prime numbers that fall below your estimate to see if they are factors of the number. If no prime below your estimate is a factor of the number, then the number is prime.

Let’s see how this works using the number 97.

  1. Estimate the square root of the number:

  2. Check all the prime numbers that fall below 10 to see if they are factors of 97: Is 97 divisible by 2? No, it does not end with an even number. Is 97 divisible by 2? No, it does not end with an even number. Is 97 divisible by 3? No, 9 + 7 = 16, and 16 is not divisible by 3. Is 97 divisible by 5? No, 97 does not end with 0 or 5. Is 97 divisible by 7? No, 97 ÷ 7 = 13, with a remainder of 6.

Therefore, 97 is prime.

  • Prime Factorization

A math problem may ask you to directly calculate the prime factorization of a number. Other problems, such as those involving greatest common factors or least common multiples (which we’ll discuss soon), are easier to solve if you know how to calculate the prime factorization. Either way, it’s good to know how to do it.

To find the prime factorization of a number, divide it and all its factors until every remaining integer is prime. The resulting group of prime numbers is the prime factorization of the original integer. Want to find the prime factorization of 36? We thought so:

36 = 2 × 18 = 2 × 2 × 9 = 2 × 2 × 3 × 3

That’s two prime 2s, and two prime 3s, for those of you keeping track at home.

It can be helpful to think of prime factorization in the form of a tree:

As you may already have noticed, there’s more than one way to find the prime factorization of a number. Instead of cutting 36 into 2 and 18, you could have factored it into 6 × 6, and then continued from there. As long as you don’t screw up the math, there’s no wrong path—you’ll always get the same result.

Let’s try one more example. The prime factorization of 220 could be found like so:

220 = 10 × 22

10 is not prime, so we replace it with 5 × 2:

10 × 22 = 2 × 5 × 22

22 is not prime, so we replace it with 2 × 11:

2 × 5 × 22 = 2 × 2 × 5 × 11

2, 5, and 11 are all prime, so we’re done. The prime factorization of 220 is thus 2 × 2 × 5 × 11.

  • Greatest Common Factor /Divisor

The greatest common factor (GCF) of two numbers is the largest number that is a factor of both numbers—that is, the GCF is the largest factor that both numbers have in common. For example, the GCF of 12 and 18 is 6, because 6 is the largest number that divides evenly into 12 and 18. Put another way, 6 is the largest number that is a factor of both 12 and 18.

To find the GCF of two numbers, you can use their prime factorizations. The GCF is the product of all the numbers that appear in both prime factorizations. In other words, the GCF is the overlap of the two factorizations.

For example, let’s calculate the GCF of 24 and 150. First, we figure out their prime factorizations:

24 = 2 × 2 × 2 × 3

150 = 2 × 3 × 5 × 5

Both factorizations contain 2 × 3. The overlap of the two factorizations is 2 and 3. The product of the overlap is the GCF. Therefore, the GCF of 24 and 150 is 2 × 3 = 6.

  • Multiples

A multiple can be thought of as the opposite of a factor: If is an integer, then x is a multiple of y. Less formally, a multiple is what you get when you multiply an integer by another integer. For example, 7, 14, 21, 28, 70, and 700 are all multiples of 7, because they each result from multiplying 7 by an integer. Similarly, the numbers 12, 20, and 96 are all multiples of 4 because 12 = 4 × 3, 20 = 4 × 5, and 96 = 4 × 24. Keep in mind that zero is a multiple of every number. Also, note that any integer, n, is a multiple of 1 and n, because 1 × n = n.

  • Least Common Multiple

The least common multiple (LCM) of two integers is the smallest number that is divisible by the two original integers. As with the GCF, you can use prime factorization as a shortcut to find the LCM. For example, to find the least common multiple of 10 and 15, we begin with their prime factorizations:

10 = 5 × 2

15 = 5 × 3

The LCM is equal to the product of each factor by the maximum number of times it appears in either number. Since 5 appears once in both factorizations, we need to include it once in our final product. The same goes for the 2 and the 3, since each of these numbers appears one time in each factorization. The LCM of 10 and 15, then, is 5 × 3 × 2 = 30. In other words, 30 is the smallest number that is divisible by both 10 and 15. Remember that the LCM is the least common multiple—you have to choose the smallest number that is a multiple of each original number. So, even though 60 is a multiple of both 10 and 15, 60 is not the LCM, because it’s not the smallest multiple of those two numbers.

This is a bit tricky, so let’s try it again with two more numbers. What’s the LCM of 60 and 100?

First, find the prime factorizations:

60 = 2 × 2 × 3 × 5

100 = 2 × 2 × 5 × 5

So, 2 occurs twice in each of these factorizations, so we’ll need to include two 2s in our final product. We have one 5 in our factorization of 60, but two 5s in our factorization of 100. Since we’re looking to include the maximum number of appearances of each factor, we’ll include two 5s in our product. There’s also one 3 in the first factorization, and no 3s in the second, so we have to add one 3 to the mix. This results in an LCM of 2 × 2 × 3 × 5 × 5 = 300.

LCM of two numbers, n1 and n2 could be found as: (n1 * n2) / GCF of n1 and n2.

  • Order of Operations

What if you see something like this on the test:

You basically have two choices. You can (a) run screaming from the testing site yelling “I’ll never, ever, EVER get into graduate school!!!” or (b) use PEMDAS.

PEMDAS is an acronym for the order in which mathematical operations should be performed as you move from left to right through an expression or equation. It stands for:

  • Parentheses

  • Exponents

  • Multiplication

  • Division

  • Addition

  • Subtraction

You may have had PEMDAS introduced to you as “Please Excuse My Dear Aunt Sally.” Come up with one of your own if you want. Just remember PEMDAS. It is also called the BODMAS  rule(Brackets Open Division Multiplication Addition Subtraction)


If an equation contains any or all of these PEMDAS elements, first carry out the math within the parentheses, then work out the exponents, then the multiplication, and the division. Addition and subtraction are actually a bit more complicated. When you have an equation to the point that it only contains addition and subtraction, perform each operation moving from left to right across the equation. Let’s see how this all plays out in the context of the example above:

First work out the math in the parentheses, following PEMDAS even within the parentheses. So here we focus on the second parentheses and do the multiplication before the subtraction:

Now taking care of the subtraction in both sets of parentheses:

Now work out the exponent (more on those later):

Then do the multiplication:

Then the division:

12 – 7 + 17

We’re left with just addition and subtraction, so we simply work from left to right:

5 + 17

And finally:


PEMDAS is the way to crunch down the most difficult-looking equations or expressions. Take it one step at a time, and you’ll do just fine.

  • Fractions

The number of questions on the Math section that involve fractions in some way or another is nothing short of stupefying. This means you must know fractions inside and out. Know how to compare them, reduce them, add them, and multiply them. Know how to divide them, subtract them, and convert them to mixed numbers. Know them. Love them like the SAT does. Make them your friend on the test, not your enemy.

To begin, here are the basics: A fraction is a part of a whole. It’s composed of two expressions, a numerator and a denominator. The numerator of a fraction is the quantity above the fraction bar, and the denominator is the quantity below the fraction bar. For example, in the fraction , 1 is the numerator and 2 is the denominator. The denominator tells us how many units there are in all, while the numerator tells us how many units out of that total are specified in a given instance. The general concept of fractions isn’t difficult, but things can get dicey when you have to do things with them. Hence, the following subtopics that you need to have under your belt.

  • Fraction Equivalencies

Fractions represent a part of a whole, so if you increase both the part and whole by the same multiple, you will not change the relationship between the part and the whole.

To determine if two fractions are equivalent, multiply the denominator and numerator of one fraction so that the denominators of the two fractions are equal (this is one place where knowing how to calculate LCM and GCF comes in handy). For example, because if you multiply the numerator and denominator of by 3, you get:. As long as you multiply or divide both the numerator and denominator of a fraction by the same nonzero number, you will not change the overall value of the fraction.

  • Reducing Fractions

Reducing fractions makes life simpler, and we all know life is complicated enough without crazy fractions weighing us down. Reducing takes unwieldy monsters like and makes them into smaller, friendlier critters. To reduce a fraction to its lowest terms, divide the numerator and denominator by their GCF. For example, for , the GCF of 450 and 600 is 150. So the fraction reduces down to , since and.

A fraction is in its simplest, totally reduced form when the GCF of its numerator and denominator is 1. There is no number but 1, for instance, that can divide into both 3 and 4, so is a fraction in its lowest form, reduced as far as it can go. The same goes for the fraction , but is a different story because 3 is a common factor of both the numerator and denominator. Dividing each by this common factor yields , the fraction in its most reduced form.

  • Adding, Subtracting, and Comparing Fractions

To add fractions with the same denominators, all you have to do is add up the numerators and keep the denominator the same:

Subtraction works similarly. If the denominators of the fractions are equal, just subtract one numerator from the other and keep the denominator the same:

Remember that fractions can be negative too:

Some questions require you to compare fractions. Again, this is relatively straightforward when the denominators are the same. The fraction with the greater numerator will be the larger fraction. For example, is greater than , while is greater than . (Be careful of those negative numbers! Since –5 is less negative than –13, –5 is greater than –13.)


Working with fractions with the same denominators is one thing, but working with fractions with different denominators is quite another. So we came up with an easy alternative: the Magic X. For adding, subtracting, and comparing fractions with different denominators, the Magic X is a lifesaver. Sure, you can go ahead and find the least common denominator, a typical way of tackling such problems, but we don’t call our trick the “Magic X” for nothing. Here’s how it works in each situation.

Adding. Consider the following equation:

You could try to find the common denominator by multiplying by 9 and by 7, but then you’d be working with some pretty big numbers. Keep things simple, and use the Magic X. The key is to multiply diagonally and up, which in this case means from the 9 to the 3 and also from the 7 to the 2:

In an addition problem, we add the products to get our numerator: 27 + 14 = 41. For the denominator, we simply multiply the two denominators to get:

The numerator is 41, and the denominator is 63, which results in a final answer of .


Subtracting. Same basic deal, except this time we subtract the products that we get when we multiply diagonally and up. See if you can feel the magic in this one:

Multiplying diagonally and up gives:

The problem asks us to subtract fractions, so this means we need to subtract these numbers to get our numerator: 24 – 25 = –1. Just like in the case of addition, we multiply across the denominators to get the denominator of our answer:

That’s it! The numerator is –1 and the denominator is 30, giving us an answer of . Not the prettiest number you’ll ever see, but it’ll do.

Comparing. The Magic X is so magical that it can also be used to compare two fractions, with just a slight modification: omitting the step where we multiply the denominators. Say you’re given the following Quantitative Comparison problem.

Column A

Column B

* *

Now, if you were a mere mortal with no magic at your fingertips, this would be quite a drag. But the Magic X makes it a pleasure. Again, begin by multiplying diagonally and up:

Now compare the numbers you get: 161 is larger than 150, so is greater than . Done.

Learn how to employ the Magic X in these three circumstances, and you’re likely to save yourself some time and effort.

  • Multiplying Fractions

Multiplying fractions is a breeze, whether the denominators are equal or not. The product of two fractions is merely the product of their numerators over the product of their denominators:

Want an example with numbers? You got one:

Canceling Out. You can make multiplying fractions even easier by canceling out. If the numerator and denominator of any of the fractions you need to multiply share a common factor, you can divide by the common factor to reduce both numerator and denominator before multiplying. For example, consider this fraction multiplication problem:

You could simply multiply the numerators and denominators and then reduce, but that would take some time. Canceling out provides a shortcut. We can cancel out the numerator 4 with the denominator 8 and the numerator 10 with the denominator 5, like this:

Then, canceling the 2s, you get:

Canceling out can dramatically cut the amount of time you need to spend working with big numbers. When dealing with fractions, whether they’re filled with numbers or variables, always be on the lookout for chances to cancel out.

  • Dividing Fractions

Multiplication and division are inverse operations. It makes sense, then, that to perform division with fractions, all you have to do is flip the second fraction and then multiply. Check it out:

Here’s a numerical example:

Compound Fractions. Compound fractions are nothing more than division problems in disguise. Here’s an example of a compound fraction:

It looks intimidating, sure, but it’s really only another way of writing , which now looks just like the previous example. Again, the rule is to invert and multiply. Take whichever fraction appears on the bottom of the compound fraction, or whichever fraction appears second if they’re written in a single line, and flip it over. Then multiply by the other fraction. In this case, we get . Now we can use our trusty canceling technique to reduce this to , or plain old 6. A far cry from the original!

  • Mixed Numbers

It concerns fractions mixed with integers. Specifically, a mixed number is an integer followed by a fraction, like . But operations such as addition, subtraction, multiplication, and division can’t be performed on mixed numbers, so you have to know how to convert them into standard fraction form. Since we already mentioned , it seems only right to convert it.

The method is easy: Multiply the integer (the 1) of the mixed number by the denominator of the fraction part, and add that product to the numerator: 1 × 3 + 2 = 5. This will be the numerator. Now, put that over the original denominator, 3, to finalize the converted fraction: .

Let’s try a more complicated example:

  • Decimals

A decimal is any number with a nonzero digit to the right of the decimal point. Like fractions, decimals are a way of writing parts of wholes. Some SAT questions ask you to identify specific digits in a decimal, so you need to know the names of these different digits. In this case, a picture is worth a thousand (that is, 1000.00) words:

Notice that all of the digits to the right of the decimal point have a th in their names.

In the number 839.401, for example, here are the values of the different digits.

Left of the decimal point

Right of the decimal point

Units: 9

Tenths: 4

Tens: 3

Hundredths: 0

Hundreds: 8

Thousandths: 1

  • Converting Fractions to Decimals

So, what if a problem contains fractions, but the answer choices are all decimals? In that case, you’ll have to convert whatever fractional answer you get to a decimal. A fraction is really just shorthand for division. For example, is exactly the same as 6 ÷ 15. Dividing this out on your scratch paper results in its decimal equivalent, .4.

  • Converting Decimals to Fractions

What comes around goes around. If we can convert fractions to decimals, it stands to reason that we can also convert decimals to fractions. Here’s how:

  1. Remove the decimal point and make the decimal number the numerator.

  2. Let the denominator be the number 1 followed by as many zeros as there are decimal places in the original decimal number.

  3. Reduce this fraction if possible.

Let’s see this in action. To convert .3875 into a fraction, first eliminate the decimal point and place 3875 as the numerator:

Since .3875 has four digits after the decimal point, put four zeros in the denominator following the number 1:

We can reduce this fraction by dividing the numerator and denominator by the GCF, which is 125, or, if it’s too difficult to find the GCF right off the bat, we can divide the numerator and denominator by common factors such as 5 until no more reduction is possible. Either way, our final answer in reduced form comes out to .

  • Ratios

Ratios look like fractions and are related to fractions, but they don’t quack like fractions. Whereas a fraction describes a part of a whole, a ratio compares one part to another part.

A ratio can be written in a variety of ways. Mathematically, it can appear as or as 3:1. In words, it would be written out as “the ratio of 3 to 1.” Each of these three forms of the ratio 3:1 means the same thing: that there are three of one thing for every one of another. For example, if you have three red alligators and one blue alligator, then your ratio of red alligators to blue alligators would be 3:1. For the SAT, you must remember that ratios compare parts to parts rather than parts to a whole. Why do you have to remember that? Because of questions like this:

For every 40 games a baseball team plays, it loses 12 games. What is the ratio of the team’s losses to wins?











The question says that the team loses 12 of every 40 games, but it asks you for the ratio of losses to wins, not losses to games. So the first thing you have to do is find out how many games the team wins per 40 games played: 40 – 12 = 28. So for every 12 losses, the team wins 28 games, for a ratio of 12:28. You can reduce this ratio by dividing both sides by 4 to get 3 losses for every 7 wins, or 3:7. Choice C is therefore correct. If you instead calculated the ratio of losses to games played (part to whole), you might have just reduced the ratio 12:40 to 3:10, and then selected choice A. For good measure, the test makers include 10:3 to entice anyone who went with 40:12 before reducing. There’s little doubt that on ratio problems, you’ll see an incorrect part: whole choice and possibly these other kinds of traps that try to trip you up.

  • Proportions

Just because you have a ratio of three red alligators to one blue alligator doesn’t mean that you can only have three red alligators and one blue one. It could also mean that you have six red and two blue alligators or that you have 240 red and 80 blue alligators. Ratios compare only relative magnitude. To know how many of each color alligator you actually have, in addition to knowing the ratio, you also need to know how many total alligators there are. This concept forms the basis of another kind of ratio problem you may see on the SAT, a problem that provides you with the ratio among items and the total number of items, and then asks you to determine the number of one particular item in the group. Sounds confusing, but as always, an example should clear things up:

Egbert has red, blue, and green marbles in the ratio of 5:4:3 and he has a total of 36 marbles. How many blue marbles does Egbert have?

First let’s clarify what this means. For each group of 5 red marbles, Egbert has a group of 4 blue marbles and a group of 3 green marbles. If he has one group of each, then he’d simply have 5 red, 4 blue, and 3 green marbles for a total of 12. But he doesn’t have 12—we’re told he has 36. The key to this kind of problem is determining how many groups of each item must be included to reach the total. We have to multiply the total we’d get from having one group of each item by a certain factor that would give us the total given in the problem. Here, as we just saw, having one group of each color marble would give Egbert 12 marbles total, but since he has 36 marbles, we have to multiply by a factor of 3 (since 36 ÷ 12 = 3). That means Egbert has 3 groups of red marbles with 5 marbles in each group, for a total of 3 × 5 = 15 red marbles. Multiplying the other marbles by our factor of 3 gives us 3 × 4 = 12 blue marbles, and 3 × 3 = 9 green marbles. Notice that the numbers work out, because 15 + 12 + 9 does add up to 36 marbles total. The answer to the question is therefore 12 blue marbles.

So here’s the general approach: Add up the numbers given in the ratio. Divide the total items given by this number to get the factor by which you need to multiply each group. Then find the item type you’re looking for and multiply its ratio number by the factor you determined. In the example above, that would look like this:

5 (red) + 4 (blue) + 3 (green) = 12

36 ÷ 12 = 3 (factor)

4 (blue ratio) × 3 (factor) = 12 (answer)

For the algebraic-minded among you, you can also let x equal the factor, and work the problem out this way:

5x + 4x + 3x = 36

12x = 36

x = 3

blue = (4)(3) = 12

  • Percents

Percents occur frequently in Data Interpretation questions but are also known to appear in Problem Solving and Quantitative Comparison questions as well. The basic concept behind percents is pretty simple: Percent means divide by 100. This is true whether you see the word percent or you see the percentage symbol, %. For example, 45% is the same as or .45.

Here’s one way percent may be tested:

4 is what percent of 20?

The first thing you have to know how to do is translate the question into an equation. It’s actually pretty straightforward as long as you see that “is” is the same as “equals,” and “what” is the same as “x.” So we can rewrite the problem as 4 equals x percent of 20, or:

Since a percent is actually a number out of 100, this means:

Now just work out the math:

Therefore, 4 is 20% of 20.

Percent problems can get tricky, because some seem to be phrased as if the person who wrote them doesn’t speak English. The SAT test makers do this purposefully because they think that verbal tricks are a good way to test your math skills. And who knows—they may even be right. Here’s an example of the kind of linguistic trickery we’re talking about:

What percent of 2 is 5?

Because the 2 is the smaller number and because it appears first in the question, your first instinct may be to calculate what percent 2 is of 5. But as long as you remember that “is” means “equals” and “what” means “x” you’ll be able to correctly translate the word problem into math:

So 5 is 250% of 2.

You may also be asked to figure out a percentage based on a specific occurrence. For example, if there are 200 cars at a car dealership, and 40 of those are used cars, then we can divide 40 by 200 to find the percentage of used cars at the dealership:. The general formula for this kind of calculation is:

Percent of a specific occurrence =


  • Converting Percents into Fractions or Decimals

Converting percents into fractions or decimals is an important SAT skill that may come into play in a variety of situations.

  • To convert from a percent to a fraction, take the percentage number and place it as a numerator over the denominator 100. If you have 88 percent of something, then you can quickly convert it into the fraction .

  • To convert from a percent to a decimal, you must take a decimal point and insert it into the percent number two spaces from the right: 79% equals .79, while 350% equals 3.5.

  • Percent Increase and Decrease

One of the most common ways the SAT tests percent is through the concept of percent increase and decrease. There are two main varieties: problems that give you one value and ask you to calculate another, and problems that give you two values and ask you to calculate the percent increase or decrease between them. Let’s have a look at both.

  • One Value Given

In this kind of problem, they give you a single number to start, throw some percentage increases or decreases at you, and then ask you to come up with a new number that reflects these changes. For example, if the price of a $10 shirt increases 10%, the new price is the original $10 plus 10% of the $10 original. If the price of a $10 shirt decreases 10%, the new price is the original $10 minus 10% of the $10 original.

One of the classic blunders test takers make on this type of question is to forget to carry out the necessary addition or subtraction after figuring out the percent increase or decrease. Perhaps their joy or relief at accomplishing the first part distracts them from finishing the problem. In the problem above, since 10% of $10 is $1, some might be tempted to choose $1 as the final answer, when in fact the answer to the percent increase question is $11, and the answer to the percent decrease question is $9.

Try the following example on your own. Beware of the kind of distracter  just discussed above.

A vintage bowling league shirt that cost $20 in 1990 cost 15% less in 1970. What was the price of the shirt in 1970?











First find the price decrease (remember that 15% = .15):

$20 × .15 = $3

Now, since the price of the shirt was less back in 1970, subtract $3 from the $20 1990 price to get the actual amount this classic would have set you back way back in 1970 (presumably before it achieved “vintage” status):

$20 – $3 = $17

If you finished only the first part of the question and looked at the choices, you might have seen $3 in choice A and forgotten to finish the problem. B is the choice that gets the point.

Try the next problem.

The original price of a banana in a store is $2.00. During a sale, the store reduces the price by 25% and Joe buys the banana. Joe then raises the price of the banana 10% from the price at which he bought it and sells it to Sam. How much does Sam pay for the banana?

This question asks you to determine the cumulative effect of two successive percent changes. The key to solving it is realizing that each percentage change is dependent on the last. You have to work out the effect of the first percentage change, come up with a value, and then use that value to determine the effect of the second percentage change.

We begin by finding 25% of the original price:

Now subtract that $.50 from the original price:

$2 – $.50 = $1.50

That’s Joe’s cost. Then increase $1.50 by 10%:

Sam buys the banana for $1.50 + $.15 = $1.65. A total rip-off, but still 35 cents less than the original price.

Some test takers, sensing a shortcut, are tempted to just combine the two percentage changes on double-percent problems. This is not a real shortcut.  Here, if we reasoned that the first percentage change lowered the price 25%, and the second raised the price 10%, meaning that the total change was a reduction of 15%, then we’d get:

Subtract that $.30 from the original price:

$2 – $.30 = $1.70 = WRONG!

If you see a double-percent problem on the SAT, it will include this sort of wrong answer as a trap.

  • Two Values Given

In the other kind of percent increase/decrease problem, they give you both a first value and a second value, and then ask for the percent by which the value changed from one to the other. If the value goes up, that’s a percent increase problem. If it goes down, then it’s a percent decrease problem. Luckily, we have a handy formula for both:

percent increase = OR


percent decrease =                 OR


To borrow some numbers from the banana example, Sam pays $1.65 for a banana that was originally priced at $2.00. The percent decrease in the banana’s price would look like this:

percent decrease =

So Sam comes out with a 17.5% discount from the original price, despite lining Joe’s pockets in the process.


A basic question of this type would simply provide the two numbers for you to plug into the percent decrease formula. A more difficult question might start with the original banana question above, first requiring you to calculate Sam’s price of $1.65 and then asking you to calculate the percent decrease from the original price on top of that. If you find yourself in the deep end of the SAT’s question pool, that’s what a complicated question might look like.

  • Common Fractions, Decimals, and Percents

Some fractions, decimals, and percents appear frequently on the SAT. Being able to quickly convert these into each other will save time on the exam, so it pays to memorize the following table.







* *(the little line above the 6 means that 6 repeats indefinitely, so 0.166 = .166666 . . .) *






* * *












* * *









  • Exponents


An exponent is a shorthand way of saying, “Multiply this number by itself this number of times.” In ab, a is multiplied by itself b times. Here’s a numerical example: 25 = 2 × 2 × 2 × 2 × 2. An exponent can also be referred to as a power: 25 is “two to the fifth power” or simply “two raised to five.” Before jumping into the exponent nitty-gritty, learn these five terms:

  • Base: The base refers to the 3 in 35. In other words, the base is the number multiplied by itself, however many times specified by the exponent.

  • Exponent: The exponent is the 5 in 35. The exponent tells how many times the base is to be multiplied by itself.

  • Squared: Saying that a number is squared is a common code word to indicate that it has an exponent of 2. In the expression 62, 6 has been squared.

  • Cubed: Saying that a number is cubed means it has an exponent of 3. In the expression 43, 4 has been cubed.

  • Power: The term power is another way to talk about a number being raised to an exponent. A number raised to the third power has an exponent of 3. So 6 raised to the third power is 63.

  • Common Exponents

It can be very helpful and a real time saver on the SAT if you can easily translate back and forth between a number and its exponential form. For instance, if you can easily see that 36 = 62, it can really come in handy when you’re dealing with binomials, quadratic equations, and a number of other algebraic topics we’ll cover later. Below are some lists of common exponents.



Powers of 2

We’ll start with the squares of the first ten integers:

Here are the first ten cubes:

Finally, the powers of 2 up to 10 are useful to know for various applications:

12 = 1

13 = 1

20 = 1

22 = 4

23 = 8

21 = 2

32 = 9

33 = 27

22 = 4

42 = 16

43 = 64

23 = 8

52 = 25

53 = 125

24 = 16

62 = 36

63 = 216

25 = 32

72 = 49

73 = 343

26 = 64

82 = 64

83 = 512

27 = 128

92 = 81

93 = 729

28 = 256

102 = 100

103 = 1000

29 = 512

210 = 1,024

  • Adding and Subtracting Exponents

The rule for adding and subtracting values with exponents is pretty simple : Just Don’t Do It.

This doesn’t mean that you won’t see such addition and subtraction problems; it just means that you can’t simplify them. For example, the expression 215 + 27 does not equal 222. The expression 215 + 27 is written as simply as possible, so don’t make the mistake of trying to simplify it further. If the problem is simple enough, then work out each exponent to find its value, then add the two numbers. For example, to add 33 + 42, work out the exponents to get (3 × 3 × 3) + (4 × 4) = 27 + 16 = 43.

However, if you’re dealing with algebraic expressions that have the same base variable and exponents, then you can add or subtract them. For example, 3×4 + 5×4 = 8×4. The base variables are both x, and the exponents are both 4, so we can add them. Just remember that expressions that have different bases or exponents cannot be added or subtracted.

  • Multiplying and Dividing Exponents with Equal Bases

Multiplying or dividing exponential numbers or terms that have the same base is so quick and easy it’s like a little math oasis. When multiplying, just add the exponents together. This is known as the Product Rule:

To divide two same-base exponential numbers or terms, subtract the exponents. This is known as the Quotient Rule:

  • Multiplying and Dividing Exponents with Unequal Bases

The same isn’t true if you need to multiply or divide two exponential numbers that don’t have the same base, such as, say, . When two exponents have different bases, you just have to do your work the old-fashioned way: Multiply the numbers out and multiply or divide the result accordingly: .

There is, however, one trick you should know. Sometimes when the bases aren’t the same, it’s still possible to simplify an expression or equation if one base can be expressed in terms of the other. For example:

25× 84

Even though 2 and 8 are different bases, 8 can be rewritten as a power of 2; namely, 8 = 23. This means that we can replace 8 with 23 in the original expression:

25 × (23) 4

Since the base is the same for both values, we can simplify this further, but first we’re going to need another rule to deal with the (23) 4 term. This is called . . .

  • Raising an Exponent to an Exponent

This one may sound like it comes from the Office of Redundancy Office, but it doesn’t. To raise one exponent to another exponent (also called taking the power of a power), simply multiply the exponents. This is known as the Power Rule:

Let’s use the Power Rule to simplify the expression that we were just working on:

25 × (23)9 = 25 × 23×9 = 25 × 227

Our “multiplication with equal bases rule” tells us to now add the exponents, which yields:

25 × 227 = 232

232 is a pretty huge number, and the SAT would never have you calculate out something this large. This means that you can leave it as 232, because that’s how it would appear in the answer choices.

To Recap: Multiply the exponents when raising one exponent to another, and add the exponents when multiplying two identical bases with exponents. The test makers expect lots of people to mix these operations up, and they’re usually not disappointed.

  • Fractions Raised to an Exponent

To raise a fraction to an exponent, raise both the numerator and denominator to that exponent:

  • Negative Numbers Raised to an Exponent

When you multiply a negative number by another negative number, you get a positive number, and when you multiply a negative number by a positive number, you get a negative number. Since exponents result in multiplication, a negative number raised to an exponent follows these rules:

  • A negative number raised to an even exponent will be positive. For example, (–2)4 = 16. Why? Because (–2)4 means –2 × –2 × –2 × –2. When you multiply the first two –2s together, you get positive 4 because you’re multiplying two negative numbers. When you multiply the +4 by the next –2, you get –8, since you’re multiplying a positive number by a negative number. Finally, you multiply the –8 by the last –2 and get +16, since you’re once again multiplying two negative numbers. The negatives cancel themselves out and vanish.

  • A negative number raised to an odd exponent will be negative. To see why, just look at the example above, but stop the process at –23, which equals –8.

  • Special Exponents

It’s helpful to know a few special types of exponents for the SAT.

  • Zero

Any base raised to the power of zero is equal to 1.

1230 = 1

0.87750 = 1

a million trillion gazillion0 = 1

You should also know that 0 raised to any positive power is 0. For example:

01 = 0

073 = 0

  • One

Any base raised to the power of 1 is equal to itself: 21 = 2, –671 = –67, and x1 = x. This fact is important to know when you have to multiply or divide exponential terms with the same base:

The number 1 raised to any power is 1:

12 = 1

14,000 = 1

  • Negative Exponents

Any number or term raised to a negative power is equal to the reciprocal of that base raised to the opposite power. An example will make it clearer:

Here’s a more complicated example:

Here’s an English translation of the rule: If you see a base raised to a negative exponent, put the base as the denominator under a numerator of 1 and then drop the negative from the exponent. From there, just simplify.

  • Fractional Exponents

Exponents can be fractions too. When a number or term is raised to a fractional power, it is called taking the root of that number or term. This expression can be converted into a more convenient form:

The symbol is known as the radical sign, and anything under the radical is called the radicand. But first let’s look at an example with real numbers:, because 4 × 4 × 4 = 64. Here we treated the 2 as an ordinary exponent and wrote the 3 outside the radical.

  • Roots and Radicals

Usually the test makers will ask you to simplify roots and radicals.

As with exponents, though, you’ll also need to know when such expressions can’t be simplified.

Square roots require you to find the number that, when multiplied by itself, equals the number under the radical sign. A few examples:= 5, because 5 × 5 = 25 = 10, because 10 × 10 = 100 = 1, because 1 × 1 = 1 = , because

Here’s another way to think about square roots:

if xn = y, then = x


  • If you take the square root of a variable, however, the answer could be positive or negative.

For example, if you solve x2 = 100 by taking the square root of both sides, x could be 10 or –10. Both values work because 10 × 10 = 100 and –10 × –10 = 100 (recall that a negative times a negative is a positive).

You may see cube and higher roots on the SAT. These are similar to square roots, but the number of times the final answer must be multiplied by itself will be three or more. You’ll always be able to determine the number of multiplications required from the little number outside the radical, as in this example:= 2, because 2 × 2 × 2 = 8

Here the little 3 indicates that the correct answer must be multiplied by itself a total of three times to equal 8.

A few more examples:= 3, because 3 × 3 × 3 = 27 = 5, because 5 × 5 × 5 × 5 = 625 = 1, because 1 × 1 × 1 × 1 = 1

  • Simplifying Roots

Roots can only be simplified when you’re multiplying or dividing them. Equations that add or subtract roots cannot be simplified. That is, you can’t add or subtract roots. You have to work out each root separately and then perform the operation. For example, to solve , do not add the 9 and 4 together to get . Instead,.

You can multiply or divide the numbers under the radical sign as long as the roots are of the same degree—that is, both square roots, both cube roots, etc. You cannot multiply, for example, a square root by a cube root. Here’s the rule in general form:

Here are some examples with actual numbers. We can simplify the expressions below because every term in them is a square root. To simplify multiplication or division of square roots, combine everything under a single radical sign.

You can also use this rule in reverse. That is, a single number under a radical sign can be split into two numbers whose product is the original number. For example:

The reason we chose to split 200 into 100 × 2 is because it’s easy to take the square root of 100, since the result is an integer, 10. The goal in simplifying radicals is to get as much as possible out from under the radical sign. When splitting up square roots this way, try to think of the largest perfect square that divides evenly into the original number. Here’s another example:

It’s important to remember that as you’ve seen earlier, you can’t add or subtract roots. You have to work out each root separately and then add (or subtract). For example, to solve , you cannot add 25 + 9 and put 34 under a radical sign. Instead, = 5 + 3 = 8.


  • Absolute Value

The absolute value of a number is the distance that number is from zero, and it’s indicated with vertical bars, like this: |8|. Absolute values are always positive or zero—never negative. So, the absolute value of a positive number is that number: |8| = 8. The absolute value of a negative number is the number without the negative sign: |–12| = 12. Here are some other examples:

|5| = 5

| –4.234 | = 4.234 = =

|0| = 0

It is also possible to have expressions within absolute value bars:

3 – 2 + |3 – 7|

Think of absolute value bars as parentheses. Do what’s inside them first, then tackle the rest of the problem. You can’t just make that –7 positive because it’s sitting between absolute value bars. You have to work out the math first:

3 – 2 + | –4 |

Now you can get rid of the bars and the negative sign from that 4.

3 – 2 + 4 = 5



See if you can notice the difference between most of the math problems we’ve considered so far, and these:

Eek! There are letters in there! As you no doubt know, the letters are called variables, and they make possible the wonderful world of algebra.

Since the variables represent unspecified quantities, algebra brings arithmetic into the world of the unknown. Of course, much of algebra deals with making that unknown known; that is, solving equations so that variables can be replaced by good old-fashioned numbers. We’ll get to all that soon enough, but first a little vocabulary is in order.

Algebra Terms

There are six main terms that describe the world of algebra. The SAT won’t ask you to define them but will nonetheless give you questions that require you to work with them.

  1. Constant. A numerical quantity that does not change.

  1. Variable. An unknown quantity written as a letter. A variable can be represented by any letter in the English alphabet; x or y are common on the SAT, but you’ll see others as well. Variables may be associated with specific things, like x number of apples or y dollars. Other times, variables have no specific association, but you’ll need to manipulate them to show that you understand basic algebraic principles.

  1. Coefficient. A coefficient is a number that appears next to a variable and tells how many of that variable there are. For example, in the term 4x, 4 is the coefficient and tells us there are four xs. In the term 3×3, 3 is the coefficient and tells us there are three x3s.

  1. Term. The product of a constant and a variable. Or, a quantity separated from other quantities by addition or subtraction. For example, in the equation 3×3 + 2×2 – 7x + 4 = x – 1, the side to the left of the equal sign contains four terms (3×3, 2×2, –7x, 4), while the right side contains two terms (x, –1). The constants, 4 and –1, are considered terms because they are coefficients of variables raised to the zero power: 4 = 4×0. So technically, every algebraic term is the product of a constant and a variable raised to some power.

  1. Expression. Any combination of terms. An expression can be as simple as a single constant term, like 5, or as complicated as the sum or difference of many terms, each of which is a combination of constants and variables, such as . Expressions don’t include an equal sign—this is what differentiates expressions from equations. If you’re given the value of every variable in the expression, then you can calculate the numerical value of the expression. Lacking those values, expressions can’t be solved, although they can often be simplified.

  1. Equation. Two expressions linked by an equal sign. Much of the algebra on the SAT consists of solving equations.

Inputs and Outputs: Simple Substitutions

We mentioned above that we can calculate the numerical value of an expression if we’re given the value of every variable in it. In this case, the expression itself is like a machine that takes an input (the variable value) and outputs a solution. One of the simplest kinds of algebraic problems on the SAT operates like this, as in the following example:

If x = 2, what is the value of ?

There’s nothing to do but simply input 2 into the expression in place of x and do the math:

So here, an input of 2 yields an output of 6.

You may see something like this in the beginning of the Math section, but things will most likely get more complicated after that. For example, the inputs themselves may be a bit more complex than a single number; in fact, an input may itself contain variables:

If 2y + 8x = 11, what is the value of 3(2y + 8x)?

You might see this equation bubbling over with variables and panic. Don’t. Since the expression 2y + 8x appears in both parts of the question, and we’re told this expression equals 11, we can simply substitute that figure in place of the expression on the right to get 3(11) = 33.

A question may also involve multiple substitutions. For instance:

z = , y = 3x, and x = 2, then what is the value of z?

To approach this problem, you just have to input 2 for x to find y, and then input those values into the equation for z. Substituting 2 for x into y = 3x gives y = 3(2) = 6. Inputting x= 2 and y = 6 into the equation for z gives:

  • Simplifying Algebraic Expressions

Before we move on to solving more complicated equations, we need to cover a few simplification tools that allow us to change algebraic expressions into simpler but equivalent forms.

  • Distributing

The rule of distribution states:

The a in this expression can be any kind of term, meaning it could be a variable, a constant, or a combination of the two. When you distribute a factor into an expression within parentheses, multiply each term inside the parentheses by the factor outside the parentheses. 4(x + 2), for example, would become 4x + (4)(2), or 4x + 8. Let’s try a harder one: 3y(y2 – 6). Distributing the 3y term across the terms in the parentheses yields:

Seems logical enough. But the true value of distributing becomes clear when you see a distributable expression in an equation. We’ll see an example of this later in the section on linear equations.

  • Factoring

Factoring an expression is the opposite of distributing. 4×3 – 8×2 is one mean-looking expression, right? Or so it seems, until you realize that both terms share the greatest common factor 4×2, which you can factor out:

By distributing and factoring, you can group or ungroup quantities in an equation to make your calculations simpler, depending on what the other terms in the equation look like. Sometimes distributing will help; other times, factoring will be the way to go. Here are a few more examples of both techniques:

  • Combining Like Terms

After factoring and distributing, you can take additional steps to simplify expressions or equations. Combining like terms is one of the simplest techniques you can use. It involves adding or subtracting the coefficients of variables that are raised to the same power. For example, by combining like terms, the expression

can be simplified by adding the coefficients of the variable x3 (–1 and 3) together and the coefficients of x2 (1 and 4) together to get:

Variables that have different exponential values are not like terms and can’t be combined. Two terms that do not share a variable are also not like terms and cannot be combined regardless of their exponential value. For example, you can’t combine:

You can, however, factor the first expression to get x2(x2 + 1), which you should do if it helps you answer the question.

  • Linear Equations with One Variable

Simplifying is nice, and helpful to boot, but solving is really where it’s at. To solve an equation, you have to isolate the variable you’re solving for. That is, you have to “manipulate” the equation until you get the variable alone on one side of the equal sign. By definition, the variable is then equal to everything on the other side of the equal sign. You can’t manipulate an equation the way you used to manipulate your little brother or sister. When manipulating equations, there are rules. Here’s the first and most fundamental.

  • Whatever you do to one side of an equation, you must do to the other side.

If you divide one side of an equation by 3, divide the other side by 3. If you take the square root of one side of an equation, take the square root of the other.

sBy treating the two sides of the equation in the same way, you don’t change what the equation means. You change the form of the equation into something easier to work with—that’s the point of manipulating it—but the equation remains true since both sides stay equal.

Take, for instance, the equation 3x + 2 = 5. You can do anything you want to it as long as you do the same thing to both sides. Here, since we’re trying to get the variable x alone on the left, the thing to do is subtract 2 from that side of the equation. But we can only do that if we subtract 2 from the other side as well:

Now we can just divide both sides by 3 to get x = 1, and we’re done.

You should use the simplification techniques you learned above (distributing, factoring, and combining like terms) to help you solve equations. For example:

That seems fairly nasty, since there aren’t any like terms to combine. But wait a sec . . . what if you distribute that 3y on the left side of the equation? That would give:

Shiver our timbers! Now we can subtract 3y3 from both sides to get

and then simply divide both sides by 18 to get y = 2.

  • Reverse PEMDAS

Many equations include a combination of elements you learned about in our arithmetic discussion. Remember PEMDAS, the acronym you learned to help you remember the order of operations? Well, what do you get if you do PEMDAS in reverse? SADMEP, of course. Or, using the old corny mnemonic device, Sally Aunt Dear My Excuse Please. Wait, scratch that—mnemonics don’t work in reverse.

Why do we want to reverse our trusty order of operations, anyway? The idea is to undo everything that has been done to the variable so that it will be isolated in the end. So you should first subtract or add any extra terms on the same side as the variable. Then divide and multiply anything on the same side as the variable. Next, raise both sides of the equation to a power or take their roots according to any exponent attached to the variable. Finally, work out anything inside parentheses. In other words, do the order of operations backward: SADMEP!

We’ll need to demonstrate with an example. Here’s a little monstrosity:

In this equation, poor little x is being square rooted, multiplied by 2, added to 3, and encased in parentheses—all in the numerator of a fraction. That’s hardly what we’d call “alone time.” You’ve got to get him out of there! Undo all of these operations to liberate x and solve the equation.

Let SADMEP be your guide: First, subtract 2 from both sides of the equation:

There’s no addition or division possible at this point, but we can multiply both sides by 2 to get rid of the fraction:

Now divide both sides by 3 (see you later, parentheses!):

Now we’re in position to subtract 3 from each side:

Divide both sides by 2:

Finally, square each side to get rid of the square root:

Success! You’ve freed poor x from all of those bullying operations.

  • Variables in the Denominator

Remember, the key to solving equations is to isolate the variable, but how to do this depends on where the variable is located. A variable in the numerator of a fraction is usually pretty easy to isolate. But if the variable is in the denominator, things get more complicated. See what you can make of this one:

Following SADMEP, start by subtracting the 3:

But now you have to get the x out of the denominator, and the only way to do that is to multiply both sides of the equation by that denominator, x + 2:

Divide both sides by 4:

Subtract 2 from each side:

  • Equations with Absolute Value

To solve an equation in which the variable is within absolute value bars, you have to follow a two-step process:

  1. Isolate the expression within the absolute value bars.

  2. Divide the equation in two.

Divide the equation in two? What is this, a magic trick? Kind of. Watch:

If |x + 3| = 5, then x =

Since both 5 and –5 within absolute value bars equal 5, the expression inside the bars can equal 5 or –5 and the equation will work out. That’s why we have to work through both scenarios. So we’re actually dealing with two equations:

x + 3 = 5

x + 3 = –5

For a complete solution, we need to solve both. In the first equation, x = 2. In the second equation, x = –8. So the solutions to the equation |x + 3| = 5 are x = {–8, 2}. Both work. Substitute them back into the equation if you have any doubts and to reinforce why we need to solve two equations to get a full answer to the question.

  • Equations with Exponents and Radicals

Absolute value equations aren’t the only ones with more than one possible answer. Exponents and radicals can also have devilish effects on algebraic equations that are similar to those caused by absolute value. Consider the equation x2 = 25. Seems pretty simple, right? Just take the square root of both sides, and you end up with x = 5. But remember the rule of multiplying negative numbers? When two negative numbers are multiplied together the result is positive. In other words, –5 squared also results in 25: –5 ×–5 = 25. This means that whenever you have to take the square root to simplify a variable brought to the second power, the result will be two solutions, one positive and one negative: . (The only exception is if x = 0.) You’ll see what we mean by working through this question:

If 2×2 = 72, what is the value of x?

To solve this problem, we first divide both sides by 2 to get x2 = 36. Now we need to take the square root of both sides: .

Recognizing when algebraic equations can have more than one possible answer is especially important in Quantitative Comparison questions. Sometimes it may appear that the relationship between columns A and B leans in a definite direction, until you notice that one of the columns can actually have more than one possible value. Keep your eyes peeled for this common nuance.

  • Linear Equations with Two or More Variables

Some SAT questions contain two variables. Lucky for you, those questions also contain two equations, and you can use the two equations in conjunction to solve for the variables. These two equations together are called a system of equations or simultaneous equations.

There are two ways to solve simultaneous equations. The first method involves substitution, and the second involves adding or subtracting one equation from the other. Let’s look at both techniques.

  • Solving by Substitution

You’ve already seen examples of substitution in the Input/Output discussion earlier in the chapter, so you should be familiar with the mechanics of plugging values or variables from one place into another place to find what you’re looking for. In these next examples, both pieces of the puzzle are equations. Our method will be to find the value of one variable in one equation and then plug that into a second equation to solve for a different variable. Here’s an example:

If x – 4 = y – 3 and 2y = 6, what is x?

You’ve got two equations, and you have to find x. The first equation contains both x and y. The second equation contains only y. To solve for x, you first have to solve for y in the second equation and substitute that value into the first equation. If 2y = 6, then dividing both sides by 2, y = 3. Now, substitute 3 for y in the equation x – 4 = y – 3:

That’s all there is to it. But here’s one that’s more likely to give you trouble:

If 3x = y + 5 and 2y – 2 = 12k, what is x in terms of k?

Notice anything interesting? There are three variables in this one. To solve for x in terms of k, we have to first get x and k into the same equation. To make this happen, we can solve for y in terms of k in the second equation and then substitute that value into the first equation to solve for x:

Then substitute 6k + 1 for y in the equation 3x = y + 5:

This is our answer, since x is now expressed in terms of k. Note that you could also solve this problem by solving for y in terms of x in the first equation and substituting that expression in for y in the second equation. Either way works.

  • Solving by Adding or Subtracting

The amazing thing about simultaneous equations is that you can actually add or subtract the entire equations from each other. Here’s an example:

If 6x + 2y = 11 and 5x + y = 10, what is x + y?

Look what happens if we subtract the second equation from the first:

To add or subtract simultaneous equations, you need to know what variable or expression you want to solve for, and then add or subtract accordingly. We made the example above purposely easy to show how the method works. But you won’t always be given two equations that you can immediately add or subtract from each other to isolate the exact variable or expression you seek, as evidenced by this next example:

If 2x + 3y = –6 and –4x + 16y = 13, what is the value of y?

We’re asked to solve for y, which means we’ve got to get rid of x. But one equation has 2x and the other has –4x, which means the x terms won’t disappear by simply adding or subtracting them.

Golden Rule of Algebra comes to the rescue: If you do the same thing to both sides of an equation, you don’t change the meaning of the equation. That means that in this case, we could multiply both sides of 2x + 3y = –6 by 2, which would give us 2(2x +3y) = 2(–6). Using the trusty distributive law, and multiplying out the second part, gives us 4x +6y = –12. Now we’re in a position to get rid of those pesky x terms by adding this new (but equivalent) form of equation 1 to equation 2:

On the SAT, you will almost always be able to manipulate one of the two equations in a pair of simultaneous equations so that they can be added and subtracted to isolate the variable or expression you want. If you can’t see how to do this, or for questions with easy numbers, go ahead and solve by using substitution instead. As you practice with these types of problems, you’ll get a sense for which method works best for you.

  • Binomial and Quadratic Equations

A binomial is an expression containing two terms. The terms (x + 5) and (x – 6) are both binomials. A quadratic expression takes the form ax2 + bx + c, where. Quadratics closely resemble the products formed when binomials are multiplied. That’s why we treat these topics together. We’ll start off with binomials, and work our way to quadratics.

  • Multiplying Binomials

The acronym will help you remember how to multiply binomials. This acronym is FOIL, and it stands for First + Outer + Inner + Last. The acronym describes the order in which we multiply the terms of two binomials to get the correct product.

First, multiply the first (F) terms of each binomial:

x × x = x2

Next, multiply the outer (O) terms of the binomials:

x × 3 = 3x

Then, multiply the inner (I) terms:

1 × x = x

And then multiply the last (L) terms:

1 × 3 = 3

Add all these terms together:

Finally, combine like terms, and you get:

Here are a few more examples of multiplied binomials to test your FOILing faculties:

Note that the last one doesn’t form a quadratic equation, and that none of the terms can be combined. That’s okay. When presented with binomials, follow FOIL wherever it leads.

  • Working with Quadratic Equations

A quadratic equation will always have a variable raised to the power of 2, like this:

x2 = 10x – 25

Your job will be to solve for the given variable, as we’ve done with other algebraic equations throughout this section. The basic approach, however, is significantly different from what you’ve done so far. Instead of isolating the variable on the left, you’ll want to get everything on the left side of the quadratic equation, leaving 0 on the right. In the example above, that means moving the 10x and –25 to the left side of the equation:

x2 – 10x + 25 = 0

Now it’s looking like most of the products of binomials we saw in the previous section, except instead of being just a quadratic expression, it’s a quadratic equation because it’s set equal to 0.

To solve the equation, we need to factor it. Factoring a quadratic equation means rewriting it as a product of two terms in parentheses, like this:

x2 – 10x + 25 = (x – 5)(x – 5)

How did we know to factor the equation into these binomials? Here’s the secret: Factoring quadratic equations on the SAT always fits the following pattern:

(x ± m)(x ± n)

Essentially, we perform FOIL in reverse. When we approach a quadratic like x2 – 10x + 25, the two numbers we’re looking for as our m and n terms need to multiply to give the last number in the equation. The last number in the equation is 25, so we need to find two numbers whose product is 25. Some pairs of numbers that work are:

1 and 25

–1 and –25

5 and 5

–5 and –5

Further, the sum of the two numbers needs to give the middle number in the equation. Be very careful that you don’t ignore the sign of the middle number:

x2 – 10x + 25

Since you’re subtracting 10x, the middle number is –10. That means the m and n numbers we seek not only need to multiply to 25 but also need to add to –10. Going back to our factor list above, –5 and –5 is the only pair that works. Substituting this into the pattern gives:

(x – 5)(x – 5)

And since we originally set the equation equal to 0, we now have:

(x – 5)(x – 5) = 0

For the product of two terms to equal 0, that means that either one could be 0. Here both terms are (x – 5), so x – 5 must equal 0.

x – 5 = 0

x = 5

The final answer is x = 5.

In this example, m and n are equal, which is why we end up with only one answer. But that’s usually not the case. Let’s look at another example, using different numbers:

x2 = –10x – 21

To solve for x, first move everything to the left to set the equation equal to 0:

x2 + 10x + 21 = 0

Now we need to figure out what numbers fit our pattern:

(x ± m)(x ± n)

We know from our equation that m × n needs to equal 21, and m + n needs to equal 10. So, which numbers work? Let’s look at m × n = 21 first:

1 and 21

–1 and –21

3 and 7

–3 and –7

We can eliminate (1 and 21) and (–1 and –21), since neither of these pairs add up to 10. The third pair, 3 and 7, adds to 10, so we can stop right there and plug these numbers into our pattern:

(x + 3)(x + 7) = 0

If you need to double-check your factoring, just FOIL the resulting binomials, which should bring you right back to the original quadratic. Since we now have the product of two different binomials sets equal to 0, one of the two terms needs to be 0. So, either (x + 3) = 0 or (x + 7) = 0, which means x could be equal to –3 or –7. There’s no way to determine for sure, since both values work.

  • Quadratic Factoring Patterns

There are three patterns of quadratics that commonly appear on the SAT. Learn them now, and you’ll work faster on test day.

Pattern 1: x2 + 2xy + y2 = (x + y)(x + y) = (x + y)2

Example: x2 + 6xy + 9 = (x + 3)2

Pattern 2: x2 – 2 xy + y 2 = (x – y )( x – y) = (x – y)2

Example: x2 – 10xy + 25 = (x – 5)2

Pattern 3: (x + y)(x – y) = x2 – y2

Example: (x + 4)(x – 4) = x2 – 16

You may be wondering why we wrote the last pattern with the factored form first. We wrote it this way because this is the way it often appears on the SAT: You’ll be given an expression that fits the pattern on the left of the equal sign [(x + y)(x – y)], and you’ll need to recognize its equivalent form of (x2 – y2).

Here’s an example of how you can use this pattern on test day:

What is the value of?

This looks horrific, and, well, it is if you attempt to perform a full FOIL treatment on it. Who wants to multiply 72 by 85 and deal with all those radical signs? Not us. Not you. Not anyone. Luckily, quadratic pattern 3 helps us avoid all that work. That pattern states that whenever we have the sum of two values multiplied by the difference of those same values, the whole messy expression is equal to x2 – y2. Here, if we let x =85 and y =, then x2 – y2 =. If this doesn’t look any better to you, then you’re not realizing that the squared symbol (the little 2) and the square root symbol (the) cancel each other out. This gives the much simpler expression 85 – 72, which equals 13.

This is an excellent example of how changing your math mindset, something we implored you to do in the previous chapter, will help you on the SAT. In the difficult-looking problem we just tackled, your first instinct may have been to hack your way through the numbers. However, if you instead suspected that the SAT test makers probably wouldn’t present a problem like this if there wasn’t a more elegant solution, then you might have searched for an easy way in. Quadratic factoring pattern 3 does the trick.

  • Inequalities

An inequality is like an equation, but instead of relating equal quantities, it specifies exactly how two quantities are not equal. There are four types of inequalities:

  1. x > y x is greater than y.

  2. x < y x is less than y.

  3. x ≥ y x is greater than or equal to y.

  4. x ≤ y x is less than or equal to y.

So, for example, x + 3 ≤ 2x may be read as “x + 3 is less than or equal to 2x.” Similarly, y > 0 is another way of saying “y is greater than 0.” Inequalities may also be written in compound form, such as 4 < y – 7 < 3y – 10. This is really just two separate inequalities: 4 < y – 7 and y – 7 < 3y – 10. Another way to think about this compound inequality is that y – 7, the expression stuck in the middle, is between 4 and 3y – 10.

  • Solving Inequalities

Solving inequalities is a lot like solving equations: Get the variable on one side of the inequality and all the numbers on the other, using the algebraic rules you’ve already learned. There is, however, one exception to this, and it’s a crucial exception. This exception is crucial, so we’ll repeat it:

  • The Inequality Exception

Multiplying or dividing both sides of an inequality by a negative number requires you to flip the direction of the inequality.

Let’s try some examples.

Solve for x in the inequality .

First we knock that 3 away from the x by adding 3 to both sides:

That 2 in the denominator is quite annoying, but by now you should know the fix for that—just multiply both sides by 2 and it will disappear from the left side of the inequality:

We can simplify the right side with our handy distributive law, multiplying the 2 by both terms in the parentheses to get a final answer of:

We’re done. The x stands alone, and we know that it’s less than the expression 4y + 6. Now try this one:

Solve for x in the inequality .

Here are the steps, all at once:

Notice that in this example the inequality had to be flipped, since both sides had to be divided by –2 to isolate the variable in the end.

To help remember that multiplication or division by a negative number reverses the direction of the inequality, remember that if x > y, then –x < –y. Just as 5 is greater than 4, –5 is less than –4. The larger the number, the smaller it becomes when you make it negative. That’s why multiplying or dividing inequalities by negatives requires switching the direction of the inequality sign.

  • Inequalities with Two Variables

Another type of inequality problem involves two variables. For these, you’ll be given a range of values for each of the two variables. For example:

–8 ≤ a ≤ 0

5 ≤ b ≤ 25

This is just another way of saying that a is between –8 and 0, inclusive, and that b is between 5 and 25, inclusive. Inclusive means that we include the values at each end, which is what’s meant by the greater than or equal to and less than or equal to signs. If the test makers didn’t want –8 and 0 to be possible values for a, for example, they would have to use simple greater-than and less-than signs (> and <). That scenario corresponds to the word exclusive, which means you should exclude the values at each end.

Once the range of the two variables has been established, the problem will then ask you to determine the range of values for some expression involving the two variables. For example, you could be asked for the range of values of a – b. This is really just asking for the smallest and largest possible values of a – b.

One good way to tackle these problems is to whip up a handy Inequality Table, a table with columns for each of the two variables and one column for the expression whose range you’re trying to determine. First write in the largest and smallest values for a and b from the original inequalities. In this example, the extreme values for a are –8 and 0, and for b are 5 and 25. Write these values in the table so that each combination of a and b is represented. There will be four combinations total: the smallest value of a with the smallest value of b; the smallest value of a with the largest value of b; the largest value of a with the smallest value of b; and the largest value of a with the largest value of b:



a – b









Now simply evaluate the expression you’re asked about for each of the four combinations in the table—in this case, a – b:



a – b



–8 – 5 = –13



–8 – 25 = –33



0 – 5 = –5



0 – 25 = –25

The Inequality Table shows us that the smallest possible value of a – b is –33 and the largest is –5. Writing this as a compound inequality gives our final answer:

–33 ≤ a – b ≤ –5

  • Inequality Ranges

The previous question demonstrated how inequalities can be used to express the range of values that a variable can take. There are a few ways that inequality problems may involve ranges. We consider three scenarios below.

  • Operations on Ranges

Ranges can be added, subtracted, or multiplied. Consider the following:

If 4 < x < 7, what is the range of 2x + 3?

To solve this problem, manipulate the range like an inequality until you have a solution. Begin with the original range:

4 < x < 7

Since the range we’re ultimately looking for contains 2x, we need to turn the x in the inequality above into that. We can do this by multiplying the whole inequality by 2:

8 < 2x < 14

Since we’re doing the same thing to all parts of the inequality, this manipulation doesn’t change its value or meaning. In other words, we’re simply invoking the Golden Rule: Do unto one part what we do unto the other part. But we’re not there yet, because the range we seek is 2x + 3, not plain old 2x. No problem: Just add 3 to the inequality across the board, and you have the final answer:

11 < 2x + 3 < 17

Always remember the crucial rule about multiplying inequalities: If you multiply or divide a range by a negative number, you must flip the greater-than or less-than signs. For example, if you multiply the range 2 < x < 8 by –1, the new range will be –2 > –x > –8.

  • Absolute Value and Single Ranges

Absolute values do the same thing to inequalities that they do to equations. You have to split the inequality into two parts, one reflecting the positive value of the inequality and one reflecting the negative value. You’ll see an example just below. If the absolute value is less than a given quantity, then the solution will be a single range with a lower and an upper bound. An example of a single range would be the numbers between –5 and 5, as seen in the following number line:

A single range question will look something like this:

Solve for x in the inequality |2x – 4| ≤ 6.

First, split the inequality into two. In keeping with the rule for negative numbers, you’ll have to flip around the inequality sign when you write out the inequality for the negative scenario:

2x – 4 ≤ 6

2x – 4 ≥ –6

Solve the first:

2x – 4 ≤ 6

2x ≤ 10

x ≤ 5

Then solve the second:

2x – 4 ≥ –6

2x ≥ –2

x ≥ –1

So x is greater than or equal to –1 and less than or equal to 5. In other words, x lies between those two values. So you can write out the value of x in a single range,.

  • Absolute Value and Disjointed Ranges

You won’t always find that the value of the variable lies between two numbers. Instead, you may find that the solution is actually two separate ranges: one whose lower bound is negative infinity and whose upper bound is a real number, and one whose lower bound is a real number and whose upper bound is infinity. Yeah, words make it sound confusing. A number line will make it clearer. An example of a disjointed range would be all the numbers smaller than –5 and larger than 5, as shown below:

On the SAT, disjointed ranges come up in problems in which the absolute value is greater than a given quantity, such as the following:

Solve for x in the inequality |3x + 4| > 16.

You know the drill. Split ’er up, then solve each inequality:

3x + 4 > 16

3x + 4 < –16

Again, notice that we have to switch the inequality sign in the second case because of the negative-number rule.

Solving the first:

3x + 4 > 16

3x >12

x > 4

And the second:

3x + 4 < –16

3x < –20

Notice that x is greater than the positive number and smaller than the negative number. In other words, the possible values of x don’t lie between the two numbers; they lie outside the two numbers. So you need two separate ranges to show the possible values of x: –∞ < x < – and 4 < x < ∞. There are two distinct ranges for the possible values of x in this case, which is why the ranges are called disjointed. It doesn’t mean they can bend their fingers back all the way—that’s double-jointed.

  • Made-up Symbols

As if there aren’t enough real math symbols in the world, the SAT test makers occasionally feel the need to make up their own. You may see a made-up symbol problem on the SAT involving little graphics you’ve never seen before in a math context. Sure, they look weird, but they often involve variables and equations, which is why we thought we’d cover them here in the algebra section. Anyway, there’s a silver lining to this weirdness: Made-up symbol problems always give you exact instructions of what to do. Follow the instructions precisely, and you’ll do just fine. Let’s see how this works with an example:

If x Ω y = 5x + 2y – 10, what is 3 Ω 4?

Some test takers will see a problem like this and think, “Ω is one of those symbols that only math geniuses learned. I never learned this symbol. I’m not a math genius. AAAAGGGHHHH!!!!”

Relax: The fact is,  no one learned what this symbol means, because the test makers made it up. Fortunately, the test makers have also made up a definition for it—that is, they tell us exactly what do to when we see a Ω. All it requires is some simple calculating.

Look at the first part of the problem again: x Ω y = 5x + 2y – 10. All that’s really saying is that whenever we have two numbers separated by a Ω, we need to take 5 times the first number and 2 times the second number, add them together, and subtract 10. That’s it—these are the only instructions we need to follow.

For the expression 3 Ω 4, then, x is 3, which we multiply by 5 to get 15. Similarly, y is 4, which we’re told to multiply by 2, which gives us 8. Adding these together gives us 23 and subtracting 10 brings us to 13 and a quick and easy point. Quick and easy, that is, if you don’t panic and just follow the directions to a T.

Try one more example to get the hang of this symbol business:

If j@k =, what is 5@–1?

Forget the actual variables for a moment and focus on the instructions given: Whenever we have two numbers separated by an @, we need to divide the second number by 2, multiply the first number by 13 and subtract the two numbers. So, substituting 5 for j and –1 for k:

The test makers may keep the choices as fractions, or they may decide to write them in decimal form, in which case –65.5 will be correct.

  • Geometry-Algebra Hybrids: Coordinate Geometry

Coordinate geometry combines the graphical figures found in geometry with the variables used in algebra.

  • The Coordinate Plane

It’s the space in which coordinate geometry exists.

Every point on a coordinate plane can be mapped by using two perpendicular number lines. The horizontal x-axis defines the space from left to right. The vertical y-axis defines the space up and down. And the two meet at a point called the origin.

Every point on the plane has two coordinates. Because it’s the center of the plane, the origin gets the coordinates (0,0). The coordinates of all other points indicate how far they are from the origin. These coordinates are written in the form (x, y). The x-coordinate is the point’s location along the x-axis (its distance either to the left or right of the origin). If a point is to the right of the origin, its x-coordinate is positive. If a point is to the left of the origin, its x-coordinate is negative. If a point is anywhere on the y-axis, its x-coordinate is 0.

The y-coordinate of a point is its location along the y-axis (either up or down from the origin). If a point is above the origin, its y-coordinate is positive, and if a point is below the origin, its y-coordinate is negative. If a point is anywhere on the x-axis, its y-coordinate is 0.

So the point labeled (2, 2) is 2 units to the right and 2 units above the origin. The point labeled (–7, –5) is 7 units to the left and 5 units below the origin.

  • Quadrants

Each graph has four quadrants, or sections. The upper-right quadrant is quadrant I, and the numbering continues counterclockwise. At the exact center of the graph, the origin is not in any of the four quadrants. Similarly, the x-axis and y-axis are not in any quadrant, either.

You may be asked where a particular point exists. Although you could graph the point on a sketched graph, then use the diagram above, a faster method is to memorize the chart below:
















This chart makes coordinate-location questions a breeze. Check it out:

In which quadrant is coordinate pair (–8.4, –2) located?

Sure, you could draw the coordinate plane and plot the pair. Or you could simply note that both the x-coordinate and the y-coordinate are negative, which means that (–8.4, –2) lives in quadrant III.

  • Distance on the Coordinate Plane

You may come across a SAT math question that asks you to find the distance between two points on the coordinate plane or to find the midpoint between two points. There are two methods for finding distance and a formula for finding midpoints.

  • Finding Distance Using the Distance Formula

If you know the coordinates of any two points—we’ll call them (x1, y1) and (x2, y2)—you can find their distance from each other with the aptly named distance formula:

Let’s say you were suddenly overcome by the desire to calculate the distance between the points (4,–3) and (–3, 8). Just plug the coordinates into the formula:

  • Finding Distance Using Right Triangles

You can also solve distance questions using right triangles. Consider this one:

What is the distance between the points (2, 4) and (5, 0)?

Plot the two points on a graph and connect them with a straight line:

The distance between these two points is the length of the dark line connecting them. To calculate this length, let the dark line be the longest side of a right triangle. Then draw two perpendicular lines to construct the short sides, as indicated by the slashed lines in the diagram.

Since the horizontal side went from an x-coordinate of 2 to an x-coordinate of 5, its length is 5 – 2 = 3.

Since the vertical side went from a y-coordinate of 0 to a y-coordinate of 4, its length is 4 – 0 = 4.

These two sides fit the {3, 4, 5} right-triangle pattern, and so the long side must be 5. And that’s the distance between the two points.

So which method for calculating distance should you use? If you’re given the two points, there’s no reason why you shouldn’t be able to plug them into the distance formula and get the answer without bothering to sketch out a picture. The reason we also teach you the right-triangle method is because, hey, you may just like it better, but more important, a question may present you with a diagram to start, in which case the triangle method may be easier.

  • Finding Midpoints

To find the midpoint between points (x1, y1) and (x2, y2) in the coordinate plane, use this formula:

In other words, the x- and y-coordinates of the midpoint are simply the averages of the x- and y-coordinates of the endpoints. Applying the formula, the midpoint of the line in the coordinate plane connected by the points (6, 0) and (3, 7) is:

  • Slope

A line’s slope is a measurement of how steeply that line climbs or falls as it moves from left to right. The slopes of some lines are positive; the slopes of others are negative. Whether a line has a positive or negative slope is easy to tell just by looking at a graph of the line. If the line slopes uphill as you trace it from left to right, the slope is positive. If a line slopes downhill as you trace it from left to right, the slope is negative. Uphill = positive. Downhill = negative.

You can get a sense of the magnitude of the slope of a line by looking at the line’s steepness. The steeper the line, the greater the slope; the flatter the line, the smaller the slope. Note that an extremely positive slope is larger than a moderately positive slope, while an extremely negative slope is smaller than a moderately negative slope.

Check out the lines below and try to determine whether the slope of each line is negative or positive and which has the greatest slope:

Lines a and b have positive slopes, and lines c and d have negative slopes. In terms of slope magnitude, line a > b > c > d.

Slopes You Should Know by Sight

There are certain easy-to-recognize slopes that it pays to recognize by sight:

  • A horizontal line has a slope of zero.

  • A vertical line has an undefined slope.

  • A line that makes a 45º angle with a horizontal line has a slope of either 1 or –1, depending on whether it’s going up or down from left to right.

Of the four lines pictured below, which has a slope of 0, which has a slope of 1, which has a slope of –1, and which has an undefined slope?

Line a has slope 0 because it’s horizontal. Line b has slope –1 because it slopes downward at 45º as you move from left to right. Line c has slope 1 because it slopes upward at 45º as you move from left to right. The slope of line d is undefined because it is vertical.


Calculating Slope

If (x1, y1) and (x2, y2) are two of the points on a line, the slope of that line can be calculated using the following formula:

Rise is how far the line goes up between two points. Run is how far the line goes to the right between two points. The formula is just shorthand for the difference between the two y-coordinates divided by the difference between the two x-coordinates. Let’s work out an example:

Calculate the slope of the line passing through (–4, 1) and (3, 4).

The difference in y-coordinates, y2 – y1, is 4 – 1 = 3. The difference in x-coordinates, x2 – x1, is 3 – (–4) = 7. The slope is simply the ratio of these two differences, .

Slopes of Parallel and Perpendicular Lines

The slopes of parallel and perpendicular lines exhibit the following relationships:

  • The slopes of parallel lines are always the same. If one line has a slope of m, any line parallel to it will also have a slope of m.

  • The slopes of perpendicular lines are always the opposite reciprocals of each other. A line with slope m is perpendicular to a line with a slope of  

In the figure below, lines q and r both have a slope of 2, so they are parallel. Line s is perpendicular to both lines q and r, so it has a slope of .

  • Finding the Equation of a Line

  • The Point-Slope Formula

The equation of a line passing through two points in the coordinate plane can be expressed by the point-slope formula:

(y – y1) = m(x – x1)

You already know how to calculate the slope, m. And x and y are just variables. You don’t have to do anything special with them in the formula other than write them down, unchanged.

As for x1 and y1, these are just the x- and y-coordinates of one of the points given. You can use whichever point you like; just make sure you use the same point for both the x1 and y1 values.

Let’s use the formula to work through an example, using the same points we used in the previous question:

What is the equation of the line passing through (–4, 1) and (3, 4)?

We already calculated. Now we have to decide which of the two points to use in the equation. Again, either point will do. Maybe (3, 4) in this case would be a little easier, since it doesn’t have a negative sign. Letting x1 = 3 and y1 = 4 gives:

We’ve actually written the equation of the line passing between the two points; all that’s left is to simplify it. First, eliminate the fraction by multiplying both sides by 7:

7(y – 4) = 7(x – 3)

7y – 28 = 3(x – 3)

Then distribute the 3 on the right side:

7y – 28 = 3x – 9

Finally, move the 3x to the left and the –28 to the right to isolate the variables:

–3x + 7y = 19

This is the final equation of the line passing through (–4, 1) and (3, 4), written in simplest form.

x-intercept and y-intercept

The x- and y-intercepts are the points at which the graph intersects the x-axis and the y-axis, respectively. A useful way to think about this is that at the x- intercept, y = 0, and at the y-intercept, x = 0.

Graphically, this is what the x- and y-intercepts look like for the equation we derived above:

To calculate the precise values of these intercepts, recall the equation of this line, which we figured out earlier: –3x + 7y = 19. Since x = 0 at the y-intercept, you can calculate it simply by setting x = 0 in the equation, like this:

0 + 7y = 19

y =

So is where the graph intersects the y-axis. Since the x-coordinate at this point is 0, the coordinate pair for the y-intercept is. Following the same mysterious naming conventions that cause slope to be named m, the point where the graph intersects the y-axis is called b.

For the x-intercept, let y = 0 in the equation –3x + 7y = 19. This gives:

–3x + 0 = 19

x =

So is where the graph intersects the x-axis. Since the y-coordinate at this point is 0, the coordinate pair for the x-intercept is .

  • The Slope-Intercept Equation

The second method for calculating the equation of a line designated by two points in the coordinate plane is the slope-intercept equation. It looks like this:

y = mx + c

  • x and y are variables, which you don’t need to change.

  • m is the slope, calculated by .

  • c is the y-intercept, the point where the graph intersects the y-axis.

For the line we’ve been working with, the slope, m, is . The y-intercept, b, is. Plugging these values into the slope-intercept equation gives:

This is the equation in slope-intercept format. You could simplify it by multiplying both sides by 7, then moving the variables to the left and the numbers to the right. As you may or may not expect, this gives exactly the same result as the point-slope format above:

–3x + 7y = 19

This actually makes a lot of sense: The equation for the line should be the same, regardless of the method you use to calculate it.

  • Which Equation Rules?

You might be wondering when you should use each of the two equations, (y – y1) = m(x – x1) and

y = mx + c. Although slope-intercept is more commonly taught (most of us have at least heard of y = mx + c), the point-slope format is actually more useful when it comes to certain problems on the SAT. Both equations require the slope, so they’re similar in that respect. The slope-intercept formula requires a very specific point, the y-intercept, but the point-slope formula can be calculated with any point. Since it’s less restrictive, you may therefore find more uses for point-slope. But if they flat-out give you the slope and the y-intercept, then y = mx +c may be the way to go. The bottom line: Know both, and decide which to use depending on the question’s parameters.

  • Word Problems

Word problems are nothing more than algebra problems, complicated by one small inconvenience: You have to come up with your own equations. Word problems present particular scenarios; the trick is to translate the information into math. Once you have your equations, they usually aren’t hard to solve.

Pretty much any math concept is fair game for a word problem, but there are certain types that show up with some regularity. We’ll work through a few examples here.

  • Using Simultaneous Equations

In a sack of 50 marbles, there are 20 more red marbles than blue marbles. All of the marbles in the sack are either red or blue. How many blue marbles are in the sack?

To get equations, we’re going to need to round ourselves up some variables. We don’t need a variable for total marbles, since that’s given: 50. However, we do need variables for the red and blue marbles, since those are unknown quantities. We’ll use r for red and b for blue.

Now for some translating. According to the 1st statement:

r = b + 20

Since both r and b are unknown, we can’t solve the problem from this equation alone, so we’ll need to squeeze another equation out of this scenario. The only other facts are that there are 50 marbles total, and only red and blue ones in the sack. So it must also be true that:

r + b = 50

We have two different equations, each containing two variables, which means that we have enough information to solve the problem. And if you remember from earlier in the chapter, we even have a choice of how to solve such simultaneous equations. We’ll go with substitution. Since the first equation tells us that r = b + 20, we can go ahead and substitute b + 20 for r in the second equation, giving us:

b + 20 + b = 50; 2b + 20 = 50; 2b = 30; b = 15


Now perhaps the CAT(Computer Adaptive Test) software would present you with the problem in the form we just solved. Or, perhaps, if you answered one or two more questions correctly before this one, that CAT would up the ante and make the question a tad harder by requiring an additional step. For example, instead of merely asking you to find the number of blue marbles in the sack, the question could ask you to find the ratio of blue to red marbles. No problem; it requires a few more steps, but it’s not so difficult.

First, use either equation to find the number of red marbles. We’ll just plug 15 for b into r = b + 20, to get 35 reds. Then set up the ratio as 15:35 and simplify to 3:7.

  • Ratios

And speaking of ratios, here’s another kind of word problem based on this concept.

In a certain class, the ratio of boys to girls is 4 to 5. If the class has 12 boys, what is the total number of boys and girls in the class?

Again, the first step is to translate the English into algebra. Since ratios are part-to-part fractions, the ratio of boys to girls in the class is really just the fraction . A common careless mistake is to write the fraction upside down—that is, as . A great way to avoid making this mistake is to remember . In other words, whatever quantity appears after the word ‘of’ in the problem goes in the numerator, and whatever quantity appears after the word ‘to’ goes in the denominator.

So, the ratio of boys to girls is 4 to 5, which, in math terms, can be written like this:

We’re also told that the class has 12 boys. Substituting this into our equation gives:

We’ll now use the Magic X to cross-multiply diagonally and up:

5 × 12 = girls × 4

60 = girls × 4= girls

15 = girls

Think 15 is the answer? Think again. The question didn’t ask for the number of girls, but for the total number of boys and girls in the class. Easy: Total = 12 + 15 = 27, and we’re done.

  • Distance

Distance problems require only one fairly basic formula:

Rate(speed) × Time = Distance

If you bike at a rate of 10 miles per hour, and you bike for 2 hours, you’re going to cover 10 × 2 = 20 miles. What gets confusing is the various ways they state these problems, but rest assured they always give you enough information to set up an equation and solve for the variable you seek. Here’s an example:

Jim roller-skates 6 miles per hour. One morning, Jim roller-skates continuously for 60 miles. How many hours did Jim spend roller-skating?

Ignore the nonessentials and focus on the facts you need to solve the problem. Begin with the trusty distance formula:

rate × time = distance

Then fill in the values you know:

6 miles per hour × time = 60 miles

Divide both sides by 6 miles per hour to calculate the time spent as 10 hours.

This next distance problem is a bit more difficult because it requires one extra step. Instead of being given two variables and simply setting up an equation to solve for the third, you need to first calculate one of the variables before getting to that point. Try it out:

A traveler begins driving from California and heads east across the United States. If she drives at a rate of 528,000 feet per hour, and drives 4.8 hours without stopping, how many miles has she traveled? (1 mile = 5,280 feet)

The reason you need to go the extra mile (so to speak) is because the rate is given in feet per hour but the distance traveled is given in miles. The units must jibe for the problem to work, so a conversion is necessary. Luckily, the numbers are easy to work with: The driver drives 528,000 feet per hour, and 5,280 feet equal one mile. So to change her rate into miles per hour, simply divide:= 100 miles per hour

Now plug the values from the problem, including this new one, into the formula:

100 miles per hour × 4.8 hours = distance

The distance traveled is 480 miles.

  • Work

Work word problems are very similar to distance word problems, except the final outcome is units of something produced instead of distance covered. The other two variables are analogous to those in distance problems: the rate at which the units are produced, and the time someone spends producing them. If you knit 2 sweaters per hour, and knit for 8 hours, then you’ll knit 2 × 8 = 16 sweaters. This generalizes to the same basic formula we saw earlier:

rate × time = units produced

Since you’ve seen this basic mechanism in action in the previous section, we’ll jump right to a difficult work word problem:

Four workers can dig a 40-foot well in 4 days. How long would it take for 8 workers working at the same rate to dig a 60-foot well?

Sure, it’s complicated, but the three parameters are the same: rate, time worked, and amount of work completed. We know the latter: 60 feet dug. The middle amount, time, is what we’re looking for, so we’ll just leave that as our unknown variable for now. The question then hinges on the rate, which we need to calculate ourselves. Well, since 4 workers dig 40 feet in 4 days, we can divide by 4 to determine that those 4 workers can dig 10 feet in one day. The 8 workers that we’re interested in dig at the same rate, and since they represent twice as many people, they’ll be able to dig twice as many feet in a day: 20. Now we have the variables to plug into our formula:

20 feet per day × time = 60 units produced (in this case, feet dug)

Divide both sides by 20 to get a time of 3 days of work for the 8 workers to dig the 60-foot well.

Many other word problems follow the same basic form as the examples above. For example, in a price word problem, the price of an item times the number of items bought will equal the total cost. The particulars are slightly different, but the concept is basically the same. Use the information given in the problem to get two of the values, and you’ll always be able to calculate the third. And you’ll never have a word problem nightmare again.


  • Lines and Angles

An angle is composed of two lines and is a measure of the spread between them. The point where the two lines meet is called the vertex. On the SAT, angles are measured in degrees. Here’s an example:

The ° after the 45 is a degree symbol. This angle measures 45 degrees.


  • Acute, Obtuse, and Right Angles

Angles that measure less than 90° are called acute angles. Angles that measure more than 90° are called obtuse angles. A special angle that appears quite often on SAT is called a right angle. Right angles always measure exactly 90° and are indicated by a little square where the angle measure would normally be, like this:

  • Straight Angles

Multiple angles that meet at a single point on a line are called straight angles. The sum of the angles meeting at a single point on a straight line is always equal to 180°. You may see something like this, asking you to solve for n:

The two angles (one marked by 40° and the one marked by n°) meet at the same point on the line. Since the sum of the angles on the line must be 180°, you can plug 40° and n° into a formula like this:

40° + n° = 180°

Subtracting 40° from both sides gives n° = 140°, or n = 140.

Note: Angles measuring more than 180° but less than 360° are called reflex angles.

  • Vertical Angles

When two lines intersect, the angles that lie opposite each other, called vertical angles, are always equal.

Angles and are vertical angles and are therefore equal. Angles and are also vertical, equal angles.

  • Parallel Lines

A more complicated version of this figure that you may see on the SAT involves two parallel lines intersected by a third line, called a transversal. Parallel means that the lines run in exactly the same direction and never intersect. Here’s an example:

line 1 is parallel to line 2

Don’t assume lines are parallel just because they look like they are—the question will always tell you if two lines are meant to be parallel. Even though this figure contains many angles, it turns out that it only has two kinds of angles: big angles (obtuse) and little angles (acute). All the big angles are equal to each other, and all the little angles are equal to each other. Furthermore, any big angle + any little angle = 180°. This is true for any figure with two parallel lines intersected by a third line. In the following figure, we label every angle to show you what we mean:

As described above, the eight angles created by these two intersections have special relationships to each other:

  • Angles 1, 4, 5, and 8 are equal to one another. Angle 1 is vertical to angle 4, and angle 5 is vertical to angle 8.

  • Angles 2, 3, 6, and 7 are equal to one another. Angle 2 is vertical to angle 3, and angle 6 is vertical to angle 7.

  • The sum of any two adjacent angles, such as 1 and 2 or 7 and 8, equals 180º because they form a straight angle lying on a line.

  • The sum of any big angle + any little angle = 180°, since the big and little angles in this figure combine into straight lines and all the big angles are equal and all the little angles are equal. So, a big and little angle don’t need to be next to each other to add to 180°; any big plus any little will add to 180°. For example, since angles 1 and 2 sum to 180º, and since angles 2 and 7 are equal, the sum of angles 1 and 7 also equals 180º.

By using these rules, you can figure out the degrees of angles that may seem unrelated. For example:

If line 1 is parallel to line 2, what is the value of y in the figure above?

Again, this figure has only two kinds of angles: big angles (obtuse) and little angles (acute). We know that 55° is a little angle, so y must be a big angle. Since big angle + little angle = 180°, you can write:

y° + 55° = 180°

Solving for y:

y° = 180° – 55°

y = 125

  • Perpendicular Lines

Two lines that meet at a right angle are called perpendicular lines. If you’re told two lines are perpendicular, just think 90°.

  • Polygon Basics

A polygon is a two-dimensional figure with three or more straight sides. Polygons are named according to the number of sides they have.

Number of Sides























All polygons, no matter how many sides they possess, share certain characteristics:

  • The sum of the interior angles of a polygon with n sides is (n – 2)180°. For instance, the sum of the interior angles of an octagon is (8 – 2)180°= 6(180°) = 1080°.

  • A polygon with equal sides and equal interior angles is a regular polygon.

  • The sum of the exterior angles of any polygon is 360°.

  • The perimeter of a polygon is the sum of the lengths of its sides.

  • The area of a polygon is the measure of the area of the region enclosed by the polygon. Each polygon tested on the SAT has its own unique area formula, which we’ll cover below.

For the most part, the polygons tested in SAT math include triangles and quadrilaterals.

  • Triangles

Of all the geometric shapes, triangles are among the most commonly tested on the SAT. But since the SAT tends to test the same triangles over and over, you just need to master a few rules and a few diagrams. We’ll look at some special triangles shortly, but first we’ll explain four very special rules.

  • The Four Rules of Triangles

Commit these four rules to memory.

1. The Rule of Interior Angles. The sum of the interior angles of a triangle always equals 180°. Interior angles are those on the inside. Whenever you’re given two angles of a triangle, you can use this formula to calculate the third angle. For example:

What is the value of a in the figure above?

Since the sum of the angles of a triangle equals 180°, you can set up an equation:

a° + 100° + 60° = 180°

Isolating the variable and solving for a gives a = 20.

2. The Rule of Exterior Angles. An exterior angle of a triangle is the angle formed by extending one of the sides of the triangle past a vertex (or the intersection of two sides of a figure). In the figure below, d is the exterior angle.

Since, together, d and c form a straight angle, they add up to 180°: d + c = 180°. According to the first rule of triangles, the three angles of a triangle always add up to 180°, so a + b + c = 180°. Since d + c = 180° and a + b + c = 180°, d must be equal to a + b (the remote interior angles). This generalizes to all triangles as the following rules: The exterior angle of a triangle plus the interior angle with which it shares a vertex is always 180°. The exterior angle is also equal to the sum of the measures of the remote interior angles.

3. The Rule of the Sides. The length of any side of a triangle must be greater than the difference and less than the sum of the other two sides. In other words:

difference of other two sides < one side < sum of other two sides

Although this rule won’t allow you to determine a precise length of the missing side, it will allow you to determine a range of values for the missing side, which is exactly what the test makers would ask for in such a problem. Here’s an example:

What is the range of values for x in the triangle above?

Since the difference of 10 and 4 is 6, and the sum of 10 and 4 is 14, we can determine the range of values of x:

6 < x < 14

Keep in mind that x must be inside this range. That is, x could not be 6 or 14.


4. The Rule of Proportion. In every triangle, the longest side is opposite the largest angle and the shortest side is opposite the smallest angle. Take a look at the following figure and try to guess which angle is largest.

In this figure, side a is clearly the longest side and is the largest angle. Meanwhile, side c is the shortest side and is the smallest angle. So c < b < a and . This proportionality of side lengths and angles holds true for all triangles.


Use this rule to solve the question below:


What is one possible value of x if ?











According to the rule of proportion, the longest side of a triangle is opposite the largest angle, and the shortest side of a triangle is opposite the smallest angle. The question tells us that angle C < angle A < angle B. So, the largest angle in triangle ABC is angle B, which is opposite the side of length 8. We know too that the smallest angle is angle C, since angle C < angle A. This means that the third side, with a length of x, measures between 6 and 8 units in length. The only choice that fits this criterion is 7, choice C.

  • Isosceles Triangles

An isosceles triangle has two equal sides and two equal angles, like this:

The tick marks indicate that sides a and b are equal, and the curved lines inside the triangle indicate that angle A equals angle B. Notice that the two equal angles are the ones opposite the two equal sides. Let’s see how we might put this knowledge to use on the test. Check out this next triangle:

With two equal sides, this is an isosceles triangle. Even though you’re not explicitly told that it has two equal angles, any triangle with two equal sides must also have two equal angles. This means that x must be 70°, because angles opposite equal sides are equal. Knowing that the sum of the angles of any triangle is 180°, we can also calculate y: y + 70 + 70 = 180, or y = 40.

  • Equilateral Triangles

An equilateral triangle has three equal sides and three equal angles, like this:

The tick marks tip us off that the three sides are equal. We can precisely calculate the angles because the 180° of an equilateral triangle broken into three equal angles yields 60° for each. As soon as you’re given one side of an equilateral triangle, you’ll immediately know the other two sides, because all three have the same measure. If you ever see a triangle with three equal sides, you’ll immediately know its angles measure 60°. Conversely, if you see that a triangle’s three interior angles all measure 60°, then you’ll know its sides must all be equal.

  • Right Triangles

A right triangle is any triangle that contains a right angle. The side opposite the right angle is called the hypotenuse. The other two sides are called legs. The angles opposite the legs of a right triangle add up to 90°. That makes sense, because the right angle itself is 90° and every triangle contains 180° total, so the other two angles must combine for the other 90°. Right triangles are so special that they even have their own rule, known as the Pythagorean theorem.

  • The Pythagorean Theorem

It’s one of the most famous theorems in all of math, and it is tested with regularity on the SAT, to boot. Here it is:

In a right triangle, a2 + b2 = c2, where c is the length of the hypotenuse and a and b are the lengths of the two legs.

And here’s a simple application:

What is the value of b in the triangle above?

The little square in the lower left corner lets you know that this is a right triangle, so you’re clear to use the Pythagorean theorem. Substituting the known lengths into the formula gives:

32 + b2 = 52

9 + b2 = 25

b2 = 16

b = 4

This is therefore the world-famous 3-4-5 right triangle. Thanks to the Pythagorean theorem, if you know the measures of any two sides of a right triangle, you can always find the third.

  • Pythagorean Triples.

Because right triangles obey the Pythagorean theorem, only a specific few have side lengths that are all integers. For example, a right triangle with legs of length 3 and 5 has a hypotenuse of length . Positive integers that obey the Pythagorean theorem are called Pythagorean triples, and these are the ones you’re likely to see on your test as the lengths of the sides of right triangles. Here are some common ones:

{3, 4, 5}

{5, 12, 13}

{7, 24, 25}

{8, 15, 17}

In addition to these Pythagorean triples, you should also watch out for their multiples. For example, {6, 8, 10} is a Pythagorean triple, since it is a multiple of {3, 4, 5}, derived from simply doubling each value. This knowledge can significantly shorten your work in a problem like this:

What is the value of z in the triangle above?

Sure, you could calculate it out using the Pythagorean theorem, but who wants to square 120 and 130 and then work the results into the formula?

But armed with our Pythagorean triples, it’s no problem for us. The hypotenuse is 130, and one of the legs is 120. The ratio between these sides is 130:120, or 13:12. This exactly matches the {5, 12, 13} Pythagorean triple. So we’re missing the 5 part of the triple for the other leg. However, since the sides of the triangle in the question are 10 times longer than those in the {5, 12, 13} triple, the missing side must be 5 × 10 = 50.

  • 30-60-90 Right Triangles

Right-angle triangles include two extra-special ones that appear with astounding frequency on the SAT. They are 30-60-90 right triangles and 45-45-90 right triangles. When you see one of these, instead of working out the Pythagorean theorem, you’ll be able to apply standard ratios that exist between the length of the sides of these triangles.

The name derives from the fact that these triangles have angles of 30°, 60°, and 90°. So, what’s so special about that? This: The side lengths of 30-60-90 triangles always follow a specific pattern. If the short leg opposite the 30° angle has length x, then the hypotenuse has length 2x, and the long leg, opposite the 60° angle, has length . Therefore:

The sides of every 30-60-90 triangle will follow the ratio 1::2

Thanks to this constant ratio, if you know the length of just one side of the triangle, you’ll immediately be able to calculate the lengths of the other two. If, for example, you know that the side opposite the 30º angle is 2 meters, then by using the ratio you can determine that the hypotenuse is 4 meters, and the leg opposite the 60º angle is meters.

  • 45-45-90 Right Triangles

A 45-45-90 right triangle is a triangle with two angles of 45° and one right angle. It’s sometimes called an isosceles right triangle, since it’s both isosceles and right. Like the 30-60-90 triangle, the lengths of the sides of a 45-45-90 triangle also follow a specific pattern. If the legs are of length x (the legs will always be equal), then the hypotenuse has length .

The sides of every 45-45-90 triangle will follow the ration of

This ratio will help you when faced with triangles like this:

This right triangle has two equal sides, which means the two angles other than the right angle must be 45° each. So we have a 45-45-90 right triangle, which means we can employ the ratio. But instead of being 1 and 1, the lengths of the legs are 5 and 5. Since the lengths of the sides in the triangle above are five times the lengths in the right triangle, the hypotenuse must be , or.

  • Area of a Triangle

The formula for the area of a triangle is:

area =             or

Keep in mind that the base and height of a triangle are not just any two sides of a triangle. The base and height must be perpendicular, which means they must meet at a right angle.

Let’s try an example. What’s the area of this triangle?

Note that the area is not , because those two sides do not meet at a right angle. To calculate the area, you must first determine b. You’ll probably notice that this is the 3-4-5 right triangle you saw earlier, so b = 4. Now you have two perpendicular sides, so you can correctly calculate the area as follows:

We move now to quadrilaterals, which are four-sided figures. The first two we cover you’re no doubt familiar with, no matter how long you’ve been out of high school. The other two you may have forgotten about long ago.

  • Rectangles

A rectangle is a quadrilateral in which the opposite sides are parallel and the interior angles are all right angles. The opposite sides of a rectangle are equal, as indicated in the figure below:

  • Area of a Rectangle

The formula for the area of a rectangle is:

area = base × height or simply a = bh

Since the base is the length of the rectangle and the height is the width, just multiply the length by the width to get the area of a rectangle i.e. area = length  x width

  • Diagonals of a Rectangle

The two diagonals of a rectangle are always equal to each other, and either diagonal through the rectangle cuts the rectangle into two equal right triangles. In the figure below, the diagonal BD cuts rectangle ABCD into congruent right triangles BAD and BCD. Congruent means that those triangles are exactly identical.

Since the diagonal of the rectangle forms right triangles that include the diagonal and two sides of the rectangle, if you know two of these values, you can always calculate the third with the Pythagorean theorem. For example, if you know the side lengths of the rectangle, you can calculate the length of the diagonal. If you know the diagonal and one side length, you can calculate the other side length.

  • Squares

They, along with circles, are perhaps the most symmetrical shapes in the universe—nothing to sneeze at. A square is so symmetrical because its angles are all 90º and all four of its sides are equal in length. Like a rectangle, a square’s opposite sides are parallel and it contains four right angles. But squares one-up rectangles by virtue of their equal sides.

  • Area of a Square

The formula for the area of a square is:

area = s2

In this formula, s is the length of a side. Since the sides of a square are all equal, all you need is one side to figure out a square’s area.

  • Diagonals of a Square

The square has two more special qualities:

  • Diagonals bisect each other at right angles and are equal in length.

  • Diagonals bisect the vertex angles to create 45º angles. (This means that one diagonal will cut the square into two 45-45-90 triangles, while two diagonals break the square into four 45-45-90 triangles.)

Because a diagonal drawn into the square forms two congruent 45-45-90 triangles, if you know the length of one side of the square, you can always calculate the length of the diagonal:

Since d is the hypotenuse of the 45-45-90 triangle that has legs of length 5, according to the ratio, you know that. Similarly, if you know only the length of the diagonal, you can use the same ratio to work backward to calculate the length of the sides.

  • Parallelograms

A parallelogram is a quadrilateral whose opposite sides are parallel. That means that rectangles and squares qualify as parallelograms, but so do four-sided figures that don’t contain right angles.

In a parallelogram, opposite sides are equal in length. That means that in the figure above, BC = AD and AB = DC. Opposite angles are equal: angleABC = angleADC and angleBAD = angleBCD. Adjacent angles are supplementary, which means they add up to 180°. Here, an example is angleABC + angleBCD = 180°.

  • Area of a Parallelogram

The area of a parallelogram is given by the formula:

area = bh

In this formula, b is the length of the base, and h is the height. As shown in the figure below, the height of a parallelogram is represented by a perpendicular line dropped from one side of the figure to the side designated as the base.

  • Diagonals of a Parallelogram

  • The diagonals of a parallelogram bisect (split) each other: BE = ED and AE = EC

  • One diagonal splits a parallelogram into two congruent triangles: ∆ABD = ∆BCD

  • Two diagonals split a parallelogram into two pairs of congruent triangles: ∆AEB = ∆DEC and ∆BEC = ∆AED

  • Trapezoids

A trapezoid is a quadrilateral with one pair of parallel sides and one pair of nonparallel sides. Here’s an example:

In this trapezoid, AB is parallel to CD (shown by the arrow marks), whereas AC and BD are not parallel.

  • Area of a Trapezoid

The formula for the area of a trapezoid is a bit more complex than the area formulas you’ve seen for the other quadrilaterals in this section.

In this formula, s1 and s2 are the lengths of the parallel sides (also called the bases of the trapezoid), and h is the height. In a trapezoid, the height is the perpendicular distance from one base to the other.

If you come across a trapezoid question on the SAT, you may need to use your knowledge of triangles to solve it. Here’s an example of what we mean:

Find the area of the figure above.


First of all, we’re not told that the figure is a trapezoid, but we can infer as much from the information given. Since both the line labeled 6 and the line labeled 10 form right angles with the line connecting them, the 6 and 10 lines must be parallel. Meanwhile, the other two lines (the left and right sides of the figure) cannot be parallel because one connects to the bottom line at a right angle, while the other connects with that line at a 45° angle. So we can deduce that the figure is a trapezoid, which means the trapezoid area formula is in play.

The bases of the trapezoid are the parallel sides, and we’re told their lengths are 6 and 10. So far so good, but to find the area, we also need to find the trapezoid’s height, which isn’t given.

To do that, split the trapezoid into a rectangle and a 45-45-90 triangle by drawing in the height.

Once you’ve drawn in the height, you can split the base that’s equal to 10 into two parts: The base of the rectangle is 6, and the leg of the triangle is 4. Since the triangle is 45-45-90, the two legs must be equal. This leg, though, is also the height of the trapezoid. So the height of the trapezoid is 4. Now you can plug the numbers into the formula:

Another way to find the area of the trapezoid is to find the areas of the triangle and the rectangle, then add them together:

area = (4 × 4) + (6 × 4) (16) + 24

8 + 24 = 32

  • Circles

A circle is the collection of points equidistant from a given point, called the center. Circles are named after their center points. All circles contain 360°. The distance from the center to any point on the circle is called the radius (r). Radius is the most important measurement in a circle, because if you know a circle’s radius, you can figure out all its other characteristics, such as its area, diameter, and circumference. The diameter (d) of a circle stretches between endpoints on the circle, passing through the center. A chord also extends from endpoint to endpoint on the circle, but it does not necessarily pass through the center. In the following figure, point C is the center of the circle, r is the radius, and AB is a chord.

  • Tangent Lines

Tangents are lines that intersect a circle at only one point. Just like everything else in geometry, tangent lines are defined by certain fixed rules.

Here’s the first: A radius whose endpoint is the intersection point of the tangent line and the circle is always perpendicular to the tangent line, as shown in the following figure:

And the second rule: Every point in space outside the circle can extend exactly two tangent lines to the circle. The distances from the origin of the two tangents to the points of tangency are always equal. In the figure below, XY = XZ.

  • Central Angles and Inscribed Angles

An angle whose vertex is the center of the circle is called a central angle.

The degree of the circle (the slice of pie) cut by a central angle is equal to the measure of the angle. If a central angle is 25º, then it cuts a 25º arc in the circle.

An inscribed angle is an angle formed by two chords originating from a single point.

An inscribed angle will always cut out an arc in the circle that is twice the size of the degree of the inscribed angle. For example, if an inscribed angle is 40º, it will cut an arc of 80º in the circle.

If an inscribed angle and a central angle cut out the same arc in a circle, the central angle will be twice as large as the inscribed angle.

Circumference of a Circle

The circumference is the perimeter of the circle—that is, the total distance around the circle. The formula for circumference of a circle is:

circumference = 2πr

In this formula, r is the radius. Since a circle’s diameter is always twice its radius, the formula can also be written c = πd, where d is the diameter. Let’s find the circumference of the circle below:

Plugging the radius into the formula, c = 2πr = 2π (3) = 6π.

  • Arc Length

An arc is a part of a circle’s circumference. An arc contains two endpoints and all the points on the circle between the endpoints. By picking any two points on a circle, two arcs are created: a major arc, which is by definition the longer arc, and a minor arc, the shorter one.

Since the degree of an arc is defined by the central or inscribed angle that intercepts the arc’s endpoints, you can calculate the arc length as long as you know the circle’s radius and the measure of either the central or inscribed angle.

The arc length formula is:

arc length =

In this formula, n is the measure of the degree of the arc, and r is the radius. This makes sense, if you think about it: There are 360° in a circle, so the degree of an arc divided by 360 gives us the fraction of the total circumference that arc represents. Multiplying that by the total circumference (2pr) gives us the length of the arc.

Here’s the sort of arc length question you might see on the test:

Circle D has radius 9. What is the length of arc AB?

To figure out the length of arc AB, we need to know the radius of the circle and the measure of angleC, the inscribed angle that intercepts the endpoints of arc AB. The question provides the radius of the circle, 9, but it throws us a little curveball by not providing the measure of angleC. Instead, the question puts angleC in a triangle and tells us the measures of the other two angles in the triangle. Like we said, only a little curveball: You can easily figure out the measure of angleC because, as you know by now, the three angles of a triangle add up to 180º:

anglec = 180° – (50° + 70°)

anglec = 180° – 120°

anglec = 60°

Since anglec is an inscribed angle, arc AB must be twice its measure, or 120º. Now we can plug these values into the formula for arc length:

  • Area of a Circle

If you know the radius of a circle, you can figure out its area. The formula for area is:

area = πr2

In this formula, r is the radius. So when you need to find the area of a circle, your real goal is to figure out the radius. In easier questions the radius will be given. In harder questions, they’ll give you the diameter or circumference and you’ll have to use the formulas for those to calculate the radius, which you’ll then plug into the area formula.

  • Area of a Sector

A sector of a circle is the area enclosed by a central angle and the circle itself. It’s shaped like a slice of pizza. The shaded region in the figure on the left below is a sector. The figure on the right is a slice of pepperoni pizza. See the resemblance?

The area of a sector is related to the area of a circle just as the length of an arc is related to the circumference. To find the area of a sector, find what fraction of 360° the sector makes up and multiply this fraction by the total area of the circle. In formula form:

area of a sector =

In this formula, n is the measure of the central angle that forms the boundary of the sector, and r is the radius. An example will help. Find the area of the sector in the figure below:

The sector is bounded by a 70° central angle in a circle whose radius is 6. Using the formula, the area of the sector is:


  • Mish-Mashes: Figures with Multiple Shapes

The trick in these problems is to understand and be able to manipulate the rules of each figure individually, while also recognizing which elements of the mish-mashes overlap. For example, the diameter of a circle may also be the side of a square, so if you use the rules of circles to calculate that length, you can then use that answer to determine something about the square, such as its area or perimeter.

Mish-mash problems often combine circles with other figures. Here’s an example:

What is the length of minor arc BE in circle A if the area of rectangle ABCD is 18?

To find the length of minor arc BE, you have to know two things: the radius of the circle and the measure of the central angle that intersects the circle at points B and E. Because ABCD is a rectangle, and rectangles only have right angles, angleBAD is 90°. In this question, they tell you as much by including the right-angle sign. But in a harder question, they’d leave the right-angle sign out and expect you to deduce that angleBAD is 90° on your own. And since that angle also happens to be the central angle of circle A intercepting the arc in question, we can determine that arc BE measures 90°.

Finding the radius requires a bit of creative visualization as well, but it’s not so hard. The key is to realize that the radius of the circle is equal to the width of the rectangle. So let’s work backward from the rectangle to give us what we need to know about the circle. The area of the rectangle is 18, and its length is 6. Since the area of a rectangle is simply its length multiplied by its width, we can divide 18 by 6 to get a width of 3. As we’ve seen, this rectangle width doubles as the circle’s radius, so we’re in business: radius = 3. All we have to do is plug in the values we found into the arc length formula, and we’re done.

That covers the two-dimensional figures you should know, but there are also some three-dimensional figures you may be asked about as well. We’ll finish up this geometry section with a look at those.

  • Rectangular Solids

A rectangular solid is a prism with a rectangular base and edges that are perpendicular to its base. In English, it looks a lot like a cardboard box.

A rectangular solid has three important dimensions: length (l), width (w), and height (h). If you know these three measurements, you can find the solid’s volume, surface area, and diagonal length.

  • Volume of a Rectangular Solid

The formula for the volume of a rectangular solid builds on the formula for the area of a rectangle. As discussed earlier, the area of a rectangle is equal to its length times its width. The formula for the volume of a rectangular solid adds the third dimension, height, to get:

volume = lwh

Here’s a good old-fashioned example:

What is the volume of the figure presented below?

The length is 3x, the width is x, and the height is 2x. Just plug the values into the volume formula and you’re good to go: v = (3x)(x)(2x) = 6×3.


  • Surface Area of a Rectangular Solid

The surface area of a solid is the area of its outermost skin. In the case of rectangular solids, imagine a cardboard box all closed up. The surface of that closed box is made of six rectangles: The sum of the areas of the six rectangles is the surface area of the box. To make things even easier, the six rectangles come in three congruent pairs. We’ve marked the congruent pairs by shades of gray in the image below: One pair is clear, one pair is light gray, and one pair is dark gray.

Two faces have areas of l × w, two faces have areas of l × h, and two faces have areas of w × h. The surface area of the entire solid is the sum of the areas of the congruent pairs:

Surface area = 2lw + 2lh + 2wh.

Let’s try the formula out on the same solid we saw above. Find the surface area of this:

Again, the length is 3x, the width is x, and the height is 2x. Plugging into the formula, we get:

surface area = 2(3x)(x) + 2(3x)(2x) + 2(x)(2x)

=6×2 + 12×2 + 4×2

= 22×2

  • Dividing Rectangular Solids

If you’re doing really well on the Math section, the CAT program will begin scrounging for the toughest, most esoteric problems it can find to throw your way. One such problem may describe a solid, give you all of its measurements, and then tell you that the box has been cut in half, like so:

A number of possible questions could be created from this scenario. For example, you may be asked to find the combined surface area of the two new boxes. Or maybe a Quantitative Comparison question would ask you to compare the volume of the original solid with that of the two new ones. Actually, the volume remains unchanged, but the surface area increases because two new sides (shaded in the diagram) emerge when the box is cut in half. You may need to employ a bit of reasoning along with the formulas you’re learning to answer a difficult question like this, but it helps to know this general rule: Whenever a solid is cut into smaller pieces, its surface area increases, but its volume is unchanged.

  • Diagonal Length of a Rectangular Solid

The diagonal of a rectangular solid, d, is the line segment whose endpoints are opposite corners of the solid. Every rectangular solid has four diagonals, each with the same length, that connect each pair of opposite vertices. Here’s one diagonal drawn in:

It’s possible that a question will test to see if you can find the length of a diagonal. Here’s the formula:

Again, l is the length, w is the width, and h is the height. The formula is like a pumped-up version of the Pythagorean theorem. Check it out in action:

What is the length of diagonal AH in the rectangular solid below if AC = 5, GH = 6, and CG = 3?

The question gives the length, width, and height of the rectangular solid, so you can just plug those numbers into the formula:

The problem could be made more difficult if it forced you to first calculate some of the dimensions before plugging them into the formula.

  • Cubes

A cube is a three-dimensional square. The length, width, and height of a cube are equal, and each of its six faces is a square. Here’s what it looks like—pretty basic:

  • Volume of a Cube

The formula for finding the volume of a cube is essentially the same as the formula for the volume of a rectangular solid: We just need to multiply the length, width, and height. However, since a cube’s length, width, and height are all equal, the formula for the volume of a cube is even easier:

volume = s3

In this formula, s is the length of one edge of the cube.

  • Surface Area of a Cube

Since a cube is just a rectangular solid whose sides are all equal, the formula for finding the surface area of a cube is the same as the formula for finding the surface area of a rectangular solid, except with s substituted in for l, w, and h. This boils down to:

volume = 6s2

  • Diagonal Length of a Cube

The formula for the diagonal of a cube is also adapted from the formula for the diagonal length of a rectangular solid, with s substituted for l, w, and h. This yields , which simplifies to:

volume =

  • Right Circular Cylinders

A right circular cylinder looks like one of those cardboard things that toilet paper comes on, except it isn’t hollow. It has two congruent circular bases and looks like this:

The height, h, is the length of the line segment whose endpoints are the centers of the circular bases. The radius, r, is the radius of its base. For the SAT, all you need to know about a right circular cylinder is how to calculate its volume and area.

  • Volume of a Right Circular Cylinder

The volume of this kind of solid is the product of the area of its base and its height. Because a right circular cylinder has a circular base, its volume is equal to the area of the circular base times the height or:

volume = πr2h

Find the volume of the cylinder below:

This cylinder has a radius of 4 and a height of 6. Using the volume formula, its volume = π(4)2(6) = 96π.


  • Area of a Right Circular Cylinder


Total area of a right circular cylinder              = Area of the base +Area of the lid + Curved surface area

Thus, total area of a right circular cylinder     = πr2+ πr2  + 2πrh = 2 πr(r+h)

Data Analysis

You have to know about statistical concepts such as mean, median, mode, probability, and every other data analysis topic tested on the SAT.

  • Mean

On the SAT, mean and arithmetic mean both represent the concept that you may recognize by its more common name, average. No matter what we call it, the calculation is the same: Add up all the terms and divide by the number of terms. You’ve no doubt seen this in school: If you get scores of 90, 95, and 100 on three tests, then 95 is the average of the three test scores. In this basic example you can probably see at a glance that 95 is the average, but technically you can calculate it by taking the sum of the scores (90 + 95 + 100) and dividing it by the number of scores (3). The formula, in general terms, is:

Let’s try one out:

What’s the arithmetic mean of 3, –5, 7, and 0?

Solve by using the formula:

Some mean problems may be straightforward like the one above, but the more complicated ones may give you two values and ask you to solve for a third. For example, the test makers might give you the mean and the number of terms and ask you to solve for the sum of the terms. Your job will still be to plug the known values into the formula and solve from there. Here’s an example:

The average height of five people is 54 inches. One of the people leaves the group, and the average height of those remaining is 52 inches. How tall is the person who left?

In the first sentence, we’re given the number of people, five, and their average height, 54. We can use the mean formula to calculate the sum of the heights of these five people:

In the second sentence, we’re told that the average height of the remaining people is 52. Since one person left the group, four people remain. Plugging 4 and 52 into the mean formula gives:

The difference between the sum of the heights of the original five people and the sum of the heights of the remaining four must be equal to the height of the person who left. Subtract the second sum from the first to get the height of the person who left: 270 – 208 = 62 inches, the final answer.

  • Median

The median of a group of numbers is the middle term when the numbers are written in either ascending or descending order. That means that before you can calculate a median, you must first rewrite the terms of the group in ascending or descending order. For example, to calculate the median of 0.3, 7, 0, 9, and 10, you can’t choose 0 simply because it appears in the middle. You must first write the numbers in order: 0, 0.3, 7, 9, 10. Since 7 appears in the middle of this ordered list, 7 is the median.

If two numbers appear in the middle, which will happen whenever the total number of terms is even, take the mean of the two middle numbers to determine the median. For example, the median of 1, 2, 4, and 8 is 3, since 3 is the mean of 2 and 4.

One wrinkle you may come across in a median problem is a description of a list of consecutive numbers, instead of a list of the actual numbers, as in this example:

What is the median of all the integers between 210 and 260, inclusive?

You certainly don’t want to write out all the numbers from 210 to 260, and then try to find the one in the middle. It’s better to use the following formula:

In our example, the median would be:

Remember, inclusive means “including,” which is why we used 210 and 260. Had the question said exclusive, we would have used 211 and 259, divided by 2, and also gotten a median of 235.

  • Mode

The mode of a group of numbers is the number that occurs most frequently. If multiple numbers are tied for first place in the race to occur the most, then the group will have more than one mode. For example, the modes of the set of numbers {6, 6, 1, 3, 4, 1} are 6 and 1, since both 6 and 1 occur twice, and all the other numbers occur only once.

  • Range

The range of a group of numbers is the difference between the largest term and the smallest term. For example, the range of 10, –25, 3, 2, and 4 is 10 – (–25) = 35. One would need to travel 35 units on a number line to get from the smallest value, –25, to the largest value, 10.

  • Standard Deviation

Standard deviation is one of the most difficult statistical concepts, but thankfully you’ll only need a very general understanding of it for the SAT. The test makers won’t ask you to actually calculate standard deviation, as the formula for doing so is pretty difficult. You will, however, be expected to know that standard deviation is a measure of how spread out a group of numbers is. The more spread out a group of numbers, the larger its standard deviation. Let’s look at an example:

Which of the following groups of numbers has the greater standard deviation?

Group A: 11, 12, 13, 14, 15

Group B: 50, 51, 51, 52, 53

Even though the numbers in Group B are larger than those in Group A, they’re closer together thanks to the double occurrence of number 51. No such overlapping occurs in Group A. Group A exhibits a slightly greater spread and therefore has the greater standard deviation.

The 34-14-2 Rule

Standard deviation also lets you know how likely it is that a value will differ from the mean by a certain amount. In general, the farther a value is from the mean, the less likely it is to occur. The following graph, called a Normal Distribution, shows this in more detail:

This graph is the basis for the 34-14-2 Rule:

34%, 14%, and 2% represent the likelihood that a value will fall into each given region. For example, there is a 34% chance that a value will be between the mean and one standard deviation to the right or left of the mean. Similarly, there is a 14% chance that a value will be between one and two standard deviations to the right or left of the mean.

Here’s how to use the numbers. Say that the mean for some group of values is 10, and the standard deviation is 2. One standard deviation to the right of the mean will therefore be 10 + 2 = 12. The Normal Distribution graph states that there is a 34% chance that a value from the group will fall between 10 (the mean) and 12 (one standard deviation up from the mean). Based on this kind of analysis, you may be asked something like this:

What is the likelihood that a value within a group of values with a mean of 10 and a standard deviation of 2 equals 5?

If the mean is 10, then one standard deviation below the mean is 10 – 2 = 8, which creates a 34% chance that a value from the group falls between 10 and 8. A second standard deviation to the left would be 8 – 2 = 6, meaning that there’s a 14% chance that a particular value would fall between 8 and 6. More than two standard deviations to the left of the mean would be all values below 6. The graph tells us that these values have a 2% likelihood. The number 5 falls into this group, so 2% would be the answer.

  • Frequency Distribution

Say, for example, that fifteen college graduates were asked how many different jobs they had in their first five years following college, and the responses came back like so:

2, 4, 2, 1, 0, 1, 3, 2, 2, 5, 3, 4, 1, 2, 1

Not very pleasing to the eye, is it? One way to organize this information is to express it in the form of a frequency distribution: a chart that shows at a glance all of the answers given and the number of people who gave each answer. Frequency distributions typically designate x as the values (in this case, the answers given by those surveyed) and f as the frequency of each value. In the example above, that would look like this:

















We can see from the chart that one person surveyed had no jobs in the first five years after college, four people had one job, five people had two jobs, and so on. So what’s so great about this? Well, the best thing about it is that it allows us to quickly determine many of the other statistical features we’ve been discussing so far. For example, eyeing only the left-hand column tells us that the range of responses is 5 – 0 = 5. A quick scan of the right-hand column indicates that 5 is the largest frequency corresponding to any one answer, and it corresponds to the answer 2, which therefore qualifies 2 as the mode. The chart already lists the responses in ascending order, so the median will be the eighth value from the beginning—eighth because with fifteen values total, the eighth value is right in the middle with seven values below it and seven above it. The first value is 0, the next four values are 1, bringing us to the fifth value, and the next five values are 2, bringing us to the tenth value. The eighth value is therefore a 2, so 2 is the median of this group of values.

You may also be asked to calculate the mean from a frequency distribution. Recall the formula for mean:

Using the frequency distribution, we can quickly calculate the sum of the terms by finding the sum of each term multiplied by its frequency. The number of terms will be either given in the table (such as “total = 15” in our chart), or you can just add up the frequency numbers in the right-hand column to calculate the number of terms. In this case, the mean is:

If you learn the basics of frequency distributions, it should be a welcome sight if one of these appears on your test.

  • Probability

Probability is the measure of how often something is expected to occur, expressed as a fraction or decimal between 0 and 1. A probability of 0 means there’s no chance that the event under consideration will take place. A probability of 1 means it definitely will happen. Most probabilities tested on the SAT fall somewhere in between. We’ll use the common scenario of selecting colored marbles from a bag to illustrate the various kinds of probability questions you might see on your test.

  • Single Trials

The most basic kind of probability question involves a single selection from a given group of elements. Here’s an example:

In a bag containing 12 red, 13 white, and 15 black marbles, what is the probability of selecting a red marble on a single draw?

To tackle probability problems, use the following formula:

The number of favorable outcomes is math lingo for the number of ways you can get what the problem is asking you to get. Here, red marble is the favorable outcome, so the numerator of the fraction is 12, the number of red marbles in the bag. The total number of possible outcomes is the total number of possibilities, or, in our problem, the total number of marbles. Make sure to include all of the marbles, including those already counted as favorable outcomes. The total number of marbles is 12 + 13 + 15 = 40. Plugging 12 and 40 into the formula gives:

Simplifying this gives , the final answer.

  • Independent Events

If one event does not influence the occurrence or nonoccurrence of another event, the two events are independent. To find the probability of two independent events occurring, simply multiply their individual probabilities. For example, if there’s a 1 in 4 chance that Mary will be selected for a committee, and a 1 in 3 chance that Bill will be kicked out of college, and the events are independent (that is, Mary isn’t angling to join the committee with the purpose in mind of booting Bill), then there’s a chance that both Mary will be selected for the committee and Bill will be given his college walking papers.

We mention independent events at this point because this concept affects our next topic, multiple trials.

  • Multiple Trials

Frequently, probability questions on the SAT won’t be limited to a single draw, or trial, but will instead involve repeated draws. When a question involves drawing multiple times from the same group of entities, you need to distinguish between draws with replacement and draws without replacement. Let’s illustrate the difference using our marble example:

  1. 1 You select a marble, note its color, and put it back in the bag. You then select a marble again. This is called drawing with replacement.

  2. You select a marble and put it aside. Then you draw another marble from those remaining. This is called drawing without replacement.

The SAT will always make it clear which method is being used either by including the actual phrase with replacement or without replacement or by explicitly describing the method of selection in a way that makes it obvious which mechanism is in play. Let’s look at an example of each type.

  • Drawing with Replacement

A bag contains 12 red, 13 white, and 15 black marbles. What is the probability of selecting two black marbles in a row if the selection is made with replacement?

The number of black marbles, or favorable outcomes, is 15. The total number of marbles is 40. First, use the probability formula to find the probability of selecting a black marble on the first draw:

Since this problem involves drawing with replacement, we’ll need to put the black marble selected on the first draw back into the bag before selecting again. So the bag will still contain 15 black marbles out of 40 total for the second draw. The probability of drawing a black marble on the second draw is thus the same .

Now, even though the marbles are coming from the same bag, these two events—a black marble on the first draw and a black marble on the second—are independent; that is, what happens on one draw doesn’t affect what happens on the other. To get the probability of two black marbles in a row, we can therefore multiply the individual probabilities:

  • Drawing without Replacement

Now let’s see what happens when we don’t put the first marble back into the bag after selecting it:

A bag contains 12 red, 13 white, and 15 black marbles. What is the probability of selecting two black marbles in a row if the selection is made without replacement?

The probability of the first marble being black is , just as before. For the second draw, however, only 14 black marbles remain out of 39 total. (Remember, we took a black marble out of the bag and did not put it back.) This means that the probability of the second marble being black is . By assuming the first draw was favorable (a black marble selected), we adjusted the figures for our second probability. Since these figures are already adjusted to account for the first favorable outcome, the second drawing is independent from the first, so we can still multiply the individual probabilities to get the chances of selecting two black marbles in a row:

We can cancel the 3 and 39 before multiplying and also cancel a factor of 2 from the 8 and 14 to make our lives easier:

This can’t be reduced any further, so it’s the final answer.

  • The Probability of Something NOT Happening

If you’re told that the chance of snow tomorrow is 25%, it’s likely you recognize without much thought that the chance that it will not snow is 75%. Here’s the formula you used, whether you were aware of it or not:

The probability of an event NOT happening = 1 – The probability of that event

This formula can turn very hard probability questions into easier ones. Consider this next one:

A bag contains 12 red, 13 white, and 15 black marbles. What is the probability of selecting at least one red or one white marble in two draws if the selection is made with replacement?

This is harder than the previous problems, because it’s not altogether clear what must happen on any individual draw for a favorable outcome. For instance, the first draw might be black, and you still could have a favorable outcome if the second draw is red or white. Similarly, the second draw could be black, and you’d still have a favorable outcome if the first draw is red or white. And of course, a first and a second draw of red or white would also count as a favorable outcome. So how do we deal with this ambiguity?

Simple: Use the formula for “NOT happening.” It’s far easier in this case to calculate the probability of not getting at least one red or white marble in two draws because this is actually the same thing as drawing two black marbles, with replacement. We already calculated this earlier as . The probability of drawing at least one red or white in two draws is 1 minus the probability of that NOT happening, which is simply the probability of drawing two black marbles. The answer is therefore:

Still no piece of cake, but doable.

  • “Or” Questions

A difficult question may ask you for the probability of event A or event B occurring, which is different from the probability of A and B occurring. If you see one of these, use the formula:

Probability of A or B = Probability of A + Probability of B – Probability of A and B

For example, say the probability of Marcie passing a test is 70%, and the probability of Jerome passing the same test is 30%. The events in this problem are independent: Neither person passing the test influences whether the other person does so (unless of course they cheat from each other, which we’ll assume is not the case). So the final term of the expression, the probability of Marcie and Jerome passing, is equal to the product of the individual probabilities of those events, as we’ve seen all along. We’ll convert the percentages to fractions, since working with those may be easier. Then we’ll plug ‘em into the formula:

The probability of Marcie or Jerome passing the test is therefore equal to 79%.

In some cases, two events may be mutually exclusive, meaning that the probability of both occurring together is 0. For example, in choosing a single dog from a kennel, the chances of choosing a black Labrador and a white schnauzer are zero—you can’t have both. If these represented the A and B elements of the “or” formula, the final term would be 0, and you wouldn’t have to subtract anything.

  • Sequences

A sequence is a list of numbers that follows a particular pattern. If you get a sequence problem, you’ll probably be given at least one of the terms in the sequence, along with the rule that defines the pattern. You probably won’t have to figure out the pattern on your own; that’s more like the kind of thing you’d see on an IQ test.

However, what could make sequence problems tough is the notation. Each term in a sequence has the same variable, but each has a different subscript. This subscript indicates a particular term. For example:

a1 = the first term

a2 = the second term

a10 = the tenth term

a n = the nth term

an+1 = the term immediately after the nth term

For example, to indicate that the second term of a sequence is 5 and that the third term is 7, the test makers might write:

a2 = 5

a3 = 7

This subscript notation can also be used to indicate how each term relates to the others. For example:

an+1 = an – 3

This just means that each successive term is three less than the previous term. Here’s an example, using the notation we just discussed:

If a n = the nth term in a sequence, and a1 = 3 and an+1 = a n + 2, what is the value of a10?

Let’s use 1 as n to keep things simple:

a1 = a n = 3

So an+1 is the same as saying a1+1 or a2. This second term we’re told is equal to the first term, a n , plus 2, which means that the second term will be 3 + 2, or 5. So the notation, which looks intimidating, is really a shorthand way of saying that each successive term is two more than the previous term. Writing out this sequence from the first to the tenth terms gives 3, 5, 7, 9, 11, 13, 15, 17, 19, 21. The tenth term is 21, and so a10 = 21.

  • Arithmetic Sequences

In an arithmetic sequence, the difference between each term and the next is constant. This is the kind of sequence we saw in the previous example. In addition to understanding the notation and concepts for sequences, you should know the formula for arithmetic sequences:

a n = a1 + (n – 1)d


a n = the nth term

a1 = the first term

d = the difference between consecutive terms

This formula is useful if you need to determine the value of some very high term and don’t want to write down a long sequence of numbers. In our previous example, the first term (a1) is 3 and the difference between consecutive terms is 2. If you plug these numbers into the equation above looking for a10, you’ll get the same answer, 21, that we got earlier. In this example, it’s just as easy to write out the terms. But to determine the 100th term in that sequence, we’ll need to plug the numbers into the formula:

a100 = 3 + (100 – 1)2 = 3 + 99 × 2 = 201

  • Geometric Sequences and Exponential Growth

In a geometric sequence, the ratio between one term and the next is constant, not the actual difference between the terms. For example, in the sequence 3, 9, 27, 81, each successive term is three times greater than the preceding one, but the actual difference between the terms changes: 9 – 3 = 6, 27 – 9 = 18, and so on. Geometric sequences exhibit exponential growth, as opposed to the constant growth of arithmetic sequences. Here’s an example of the kind of geometric sequence that the test makers might toss at you:

g1 = 4

g n = 2gn – 1

Trying out some terms, this means that g2 = (2)g1, g3 = (2)g2, and so on. In other words, the first term is 4, and each successive term is twice the value of the preceding term. Writing out the first few terms lets us see that the ratio between terms is constant and thus confirms that this is a geometric sequence:

4, 8, 16, 32, . . .

The ratio between consecutive terms is always 2, even though the differences between the terms increase as you move to the right.

As with arithmetic sequences, you should learn the special formula for geometric sequences, just in case it’s not convenient to list out all of the terms up to the one you’re looking for:

g n = g1rn–1


g n = the nth term

g1 = the first term

r = the ratio between consecutive terms

Let’s use the formula to calculate the value of the tenth term in the geometric sequence defined by g1 = 4 and g n = 2gn – 1. We already know r = 2:

g10 = 4 × (2)10–1 = 4 × 29 = 4 × 512 = 2,048

  • Digit Counting

Here’s an interesting kind of problem that appears with some regularity on the SAT. In digit counting problems, you’re asked how many times a particular digit appears in a defined group of numbers, or how many numbers within such a group don’t contain a certain digit. An example will make this clearer:

How many three-digit positive even integers contain at least one digit that is a 7?











First make sure you understand the range of numbers under consideration. Positive three-digit numbers begin with 100 and go to 999. Moreover, we’re only interested in the even ones. The question is looking for how many numbers fit this description and contain at least one 7.

Well, we’re not going to go and count them all—that would take too long. But we will list a few examples that fit the criteria and then see if we can discern a pattern. Beginning with the 100s, the first number that has a 7 is 107, but since that’s not an even number, it doesn’t count. In fact, the next one, 117, doesn’t count either, nor does 127, 137, and so on. So we begin to see a pattern. The first even number we get to that has a 7 is 170, followed by 172, 174, 176, and 178. No number from 179 to 200 fits the bill, so in the entire 100s we have a total of 5 numbers that satisfy the question’s requirements.

Now we can generalize from what we’ve learned: The 200s will be no different, nor the 300s, nor the 400s, and so on. So if there are 5 cases in each of the nine groups of 100 numbers from 100 to 999, we can multiply 5 instances per each group of hundreds by nine groups of hundreds (100s, 200s, etc.) to get 45, choice A.

However, we’d be wrong. Naturally, A is a trap for people who forget that the 700s contain plenty of numbers that work, so we need to consider the numbers from 700 to 800 separately. The other eight groups of hundreds follow our pattern, so we’ll go with 5 × 8 = 40 to represent the instances that make the cut among those. But we also have to consider all of the numbers in the 700s, since every number in the 700s contains at least one 7. How many are even? There are 100 numbers between 700 and 799, and half are even, so we need to add 50 more cases to the 40 we’ve found already. The correct answer is 40 + 50 = 90, choice C.

Digit counting questions can be difficult, and if you see one, chances are the SAT software is throwing you more difficult questions because you’re doing pretty well. Understand the range, find a pattern and generalize it to as many cases as you can, and then check for special cases. If you’re careful, you’ll find them all.


You may see some problems on your exam in which a number is followed by an exclamation point, like this: 5!. An exclamation point used in math symbolizes a factorial. A factorial stands for the product of all the numbers up to and including the given number. So 5! = 5 × 4 × 3 × 2 × 1 = 120.

Some more examples:

3! = 3 × 2 × 1 = 6

4! = 4 × 3 × 2 × 1 = 24

55! = 55 × 54 × 53 × . . . × 3 × 2 × 1 = a really huge number you would never be expected to solve for

0! = 1

The proof of this last example is beyond the scope of what you need to know for the SAT. Just remember that 0! = 1 by definition.

The factorial of n also signifies the number of ways that the n elements of a group can be ordered. So, if you decide to ditch grad school and become a wedding planner instead, and need to figure out how many different ways six people can sit at a table with six chairs, 6! is the way to go: 6 × 5 × 4 × 3 × 2 × 1 = 720 possible seating arrangements. You’ll astound the other wedding planners with this quick calculation, never revealing the true source of your knowledge.

As you might guess from the name, factorials have many factors. Recall that a factor is a number that divides into another number with no remainder. Whenever you take the factorial of a number, the result will be divisible by all of the integers up to and including the original number. For example, 6! is divisible by 6, 5, 4, 3, 2, and 1, and all of those numbers are factors of 6!. This is all inherent in the definition of factorial, but it’s good to understand it in these terms too.

  • Simplifying Factorials

The test makers may ask you to work out a problem that involves factorials in fractions, and as you’ll soon see, this becomes downright necessary in permutation and combination problems. The trick is to cancel before calculating. As you’ve seen in earlier examples, canceling with fractions means dividing the numerator and denominator by the same number. A little cancellation makes complicated-looking factorial problems much easier to solve. Check it out:

What is ?

This expression looks like it might be a huge number. And, in fact, trying to calculate 98! or 100! would be near impossible without a computer or ultra-fancy calculator. Fortunately, we can simplify this equation significantly:

This works, because everything after and including the 98 cancels out in both the numerator and the denominator, leaving 100 × 99 in the numerator and 1 in the denominator. Here’s another way to think about this:

The 98! in the numerator cancels out with the 98! in the denominator, leaving only 100 × 99 = 9,900.

Truth be told, even factorial problems involving smaller numbers benefit from canceling out. For example:

Before you get stuck into multiplying out the factorials in the top and bottom parts of the fraction and then dividing the results, first cancel out what you can

Permutations and Combinations

Knowing how to calculate and simplify factorials is especially useful for problems involving permutations and combinations. These types of questions ask you to determine how many ways something can be done. For example, “In a race of eight horses, how many ways can the horses finish first, second, and third?” and “How many ways can two students be selected for a Grammar Jamboree out of a class of 20 students?” In this section, we’ll explain not only how to answer both of these questions but also the very important difference between them.

  • Permutations

In a permutation, order matters—that is, being first in a group is different from being second, third, or in any other position. The easiest way to tell that a question is a permutation is if it includes the word order or the word arrange. Even if it doesn’t contain these words, the question might describe some kind of ranking or race. Our horse race question above, for instance, is a permutation since finishing first is certainly different from finishing second or third in a horse race. If, for example, three of the horses are A, B, and C, then ABC is one possible finish, CAB is another, and BCA is a third. They all involve the same three horses, but switching them around yields additional arrangements that we need to add to our tally. In a combination problem, however, we’re not concerned with order, so BCA would be considered the same as CAB, and we wouldn’t count those twice. We’ll get to combinations in just a bit, but let’s continue with the permutation problem at hand.

Here’s the permutation formula:

In this case, nPr is the number of subgroups of size r that can be taken from within a set with n elements.

Here’s the horse race question again, and this time we’ll work through it using the formula:

In a race of eight horses, how many ways can the horses finish first, second, and third?

In this problem, n = 8, because that’s the total number of horses racing, and r = 3, because that’s the number of winners we’re interested in (first, second, and third place). Plugging into the formula gives:

Now we’ll solve, using our knowledge of factorials and canceling out:

Notice how the 5!s cancel out, leaving us with some basic multiplication to get our final answer.

  • Combinations

As we mentioned earlier, in a combination, order does not matter. For example, if you’re trying to buy three horses instead of ordering them first through third in a race, then it really doesn’t matter if you come away from the horse farm with horses ABC or horses CBA—they’re all the same horses.

You’ll be able to recognize a combination problem because it will involve selecting a small group from a larger group, with no regard to order. An example is the Grammar Jamboree problem introduced earlier:

How many ways can two students be selected for a Grammar Jamboree out of a class of 20 students?

Two students are to be selected from a class of 20, and no mention of order is made. This is common for situations involving teams: A team consisting of Jonathan and Gloria is the same as a team consisting of Gloria and Jonathan. Unless some specific mention is made of an ordering element, assume you’re dealing with a combination problem. Here’s the combination formula:

Here, unordered subgroups of size r are selected from a set of size n. Notice that this is the same as the permutation formula, except that it tacks on an extra r! term in the denominator. This means that we divide by a larger number in combination problems, resulting in a smaller number of final orderings. And that makes sense too: We’d expect fewer total orderings in combinations, since order doesn’t matter and we therefore count the shuffled orderings (ABC, CAB, BCA, etc.) as one.

Use the formula to solve our Grammar Jamboree problem:

Cancel the 18!s:

So, there are 190 possible two-person teams of jamboree-ers to choose from the class of 20.

  • Multiple Permutations and Combinations

If the CAT software might throw you a problem involving multiple permutations, combinations, or both. The key is to break the problem down into parts, solving each independently using the formulas discussed above. To obtain a final answer, multiply all of the individual results. Here’s an example:

How many ways can Suzanne order an ice cream cone if she is to select three different flavors out of fifteen available flavors and three different toppings out of five available toppings? The order in which the flavors are stacked is significant, but the order in which the toppings are added is not.

We admit it—this looks hard. However, it helps to think of this as two completely separate problems:

  1. Selecting ice cream flavors

  2. Selecting toppings

Since the order of flavors matters, this part of the problem is a permutation. Suzanne must select three different flavors out of a total of fifteen. Let’s use the P formula:

As per the problem, n = 15 and r = 3:

So Suzanne has 2,730 choices of flavors. But back to the problem: Since the order of toppings does not matter, this part is a combination. Suzanne must select three different toppings out of a total of five. Let’s use the C formula:

Now n = 5 and r = 3:

Cancel the 3!s, and restate 2! as 2 × 1:

Now we know that there are 10 possible combinations of toppings. The last step is to multiply the two results: 2,730 × 10 = 27,300. And there we go: Suzanne has an unbelievable 27,300 choices for her ice cream cone.

  • Groups

The basic idea is that some people or things belong to one group, others belong to another group, and still others belong to both groups or neither group. For example, at a certain country club, some members play golf, some play tennis, others play both sports, and still others prefer reading to playing. You’ll be given some of the specific numbers in such a problem and then asked to determine the missing values.

The approach you should use depends on whether the problem concerns two or three groups. We’ll cover the most effective techniques for both cases. You won’t see problems with more than three groups.

  • Problems with Two Groups

All you need for two-group problems is this formula:


group 1 + group 2 – both + neither = total


group 1 = the number of entities in one of the two groups

group 2 = the number of entities in the other group

both = the number of entities in both groups

neither = the number of entities in neither group

total = the total number of entities

You’ll most likely be given values for all of the parts of this formula except for one. You’ll then have to determine the value of the missing part. Let’s see how this works in the following example.

At a certain animal refuge, 180 animals have four legs, 240 are warm-blooded, and 85 both have four legs and are warm-blooded. If the animal refuge has 500 animals in total, how many animals at the refuge have neither four legs nor are warm-blooded?

We’ll let group 1 be the 180 animals that have four legs and group 2 be the 240 animals that are warm-blooded. Since we’re also given the number of animals that belong to both groups and the total number of animals, the missing value is the number of animals that belong to neither group. Not surprisingly, that’s what the question is after. Plugging the values into the formula gives:

180 + 240 – 85 + neither = 500

Solving for neither is a simple matter of solving this linear equation with one variable, something we discussed way back in the algebra section. Here goes:

180 + 240 – 85 + n = 500

335 + n = 500

n = 165

But things get a bit more difficult if they throw in an extra group.

  • Problems with Three Groups

The formula for three-group problems is really long and complicated (it involves nine unique terms!). So, we’ll skip it in favor of an easier approach: Venn diagrams. Remember these from junior high? Venn diagrams consist of intersecting circles, in which each circle represents the number of entities in a particular group. For example:

You’ll notice that the circles overlap. The upside-down triangular section in the middle, with the darker shading, represents the number of entities that belongs to all three groups. Sections in which only two circles overlap, indicated with the lighter shading, represent the number of entities that belongs to two overlapping groups. The outermost section of each circle, the part that doesn’t overlap with any of the other circles, represents the number of entities that belongs to each group alone. For example, in the swimming circle, the outermost section represents the number of swimmers who neither lift weights nor do aerobics.

The key to three-group problems is to work from the inside out. Begin with the entities that belong to all three groups, then address the entities that belong to two groups, and finally deal with the entities that belong to only one group.

Here’s an example of a three-group problem that conveniently makes use of the diagram above:

At the Get Fit Athletic Club, every member swims, lifts weights, does aerobics, or participates in some combination of these three activities. Sixty members swim, 75 lift weights, and 100 do aerobics. If 34 members both lift weights and swim, 25 members both do aerobics and swim, 44 members both use the weight room and do aerobics, and 10 participate in all three activities, how many members belong to the Get Fit Athletic Club?

We’ll start by filling in values, working from the inside out. Since 10 people belong to all three groups, we’ll write 10 in the middle section:

Next, we’ll fill in the values for people who belong to two groups. We’re told that 25 members participate in aerobics and in swimming, and it would be very tempting to write 25 in the section above the 10. Keep in mind, however, that 10 of these 25 people have already been accounted for: The 10 people who participate in all three activities are included among those who participate in aerobics and swimming. That leaves 25 – 10 = 15 people for the aerobics/swimming overlap section above the 10 in the middle. Similarly, 44 people do aerobics and use the weight room. Since 10 of these people have already been accounted for in the middle section, that leaves 44 – 10 = 34 people in the section that overlaps aerobics and weight room. Also, 34 people swim and use the weight room, so we’ll write 34 – 10 = 24 in the section overlapping those categories. That brings us to here:

Next we need to fill in values for people who belong to only one group. As with the previous step, we have to be very careful not to count anyone more than once. For example, we’re told that 60 members swim. That means that the total of all the numbers in the swimming circle must be 60. We already have 10 + 15 + 24 = 49 members in the swimming circle. That leaves 60 – 49 = 11 members for the outermost section of the swimming circle. Similarly, since 75 members use the weight room, that leaves 75 – 10 – 34 – 24 = 7 members for the outermost section of the weight room circle. Finally, since the total of the aerobics circle must be 100, the number in the remaining section must be 100 – 10 – 34 – 15 = 41. Filling these numbers in the appropriate sections of our Venn diagram yields the following:

The problem asks for the total number of club members, and we’re told specifically that every member participates in at least one of these activities. That means that no member exists outside of our three circles. We can therefore add all of the values in our diagram to arrive at the total number of club members. This gives us a final answer of 41 + 15 + 11 + 34 + 10 + 24 + 7 = 142.

We realize that this was a bit of work, but Venn diagrams are the only way to go on three-group problems. And the good news is that you’d have to be doing something right on the Math section for the CAT software to throw something this nasty your way, since the questions increase in difficulty the better you do.

General Math Strategies

In this section we provide some general strategies to get you thinking in the right direction regarding SAT math. Here’s a preview:

  • Change Your Math Mindset

  • Use Scratch Paper

  • Avoid Careless Mistakes

  • Change Your Math Mindset

Earlier we told you the good news that the SAT tests only basic math from junior high or early high school. However, since the concepts tested are basic and predictable, those wily test makers have to resort to certain tricks and traps to throw you off; otherwise, most test takers would ace the section. This fact has one very important ramification:

You need to change the way you’ve typically approached math questions in the past.

Think about the typical math tests you took in high school. Many were accompanied by three dreaded and imposing words: SHOW YOUR WORK. This mandate implies a slogging mentality: You’re taught to do a problem a certain way, and then required to spit back that exact method to get full credit.

SAT math, however, rewards cleverness to combat the traps the test makers set. It doesn’t matter whether you answer the question using a traditional or untraditional method. It doesn’t matter if you use algebra or don’t use algebra, draw a diagram or don’t draw a diagram, or simply get into the ballpark through approximation instead of calculating an answer precisely. All that matters is whether you answer the question correctly. Three elements of your new math mindset will be looking for shortcuts, approximating when possible, and keeping your eyes open for “common trap” and “left-field” answer choices. Let’s have a look at each one.

  • Shortcuts

As discussed above, your high school math experience may have instilled in you an instinct to jump into math problems with your sleeves rolled up, ready to slog away. And yes, sometimes that is the only way, or at least the only way you can see at the moment. Unless you perform math calculations lightning fast, however, you’ll probably need to sneak your way around at least some SAT Math problems to get to all 28 questions; that’s simply how the section is constructed. If you find yourself up against a real monster calculation that you think you need to work through to get the right answer, think again: Chances are the question is testing your math reasoning skills—that is, your ability to spot a more elegant solution that doesn’t require hacking through the math. Consider, for example, the following problem:

If x = 33.87, what is the value of ?

Is it really likely that they expect you to plug such an unwieldy number such as 33.87 in for all those x’s in the equation, especially given the fact that calculators aren’t allowed on the test? No, of course not, although that’s exactly what some people will attempt. Not you. Once you change your math mindset, you’d know instinctively that there must be some sort of shortcut here, and indeed there is.

If you multiply out the (x + 1) and (3x + 15) in the top part of the fraction (the “numerator”), you get 3×2 + 18x + 15, which cancels out the entire bottom part (the “denominator”), leaving the simplified value of the equation at (x – 2). Alternatively, you may have factored the bottom into (3x + 15)(x + 1) and then canceled out those terms from the numerator, again reducing the entire fraction to (x – 2). No matter which shortcut you employ, all that’s left is to substitute the given value for x into (x – 2) to get 33.87 – 2 = 31.87, and you’re done.

FOIL, factoring quadratic equations, and canceling are the concepts in play here, and if you need to brush up on them, don’t worry—we’ll get to these and plenty more bits of math minutiae soon enough in the following chapter. The point is simply to understand that many SAT math questions are written with shortcuts in mind, so begin right now to look for shortcuts as part of your new SAT math mindset.

  • Approximating (When Possible)

Another habit that may be ingrained in you from your math background is to “get the right answer.” Well duh, you may be saying; this is math, after all, so naturally you’ll want to solve the problems. Well, yes and no: yes in Problem Solving and Data Interpretation, but no in Quantitative Comparison questions where your job is not necessarily to solve the problems but rather to learn enough about quantities A and B to compare them. So especially in QCs—but also in the other question types—approximating values may save time. (Some DI questions even ask flat out for an approximate answer.)

Let’s see how we might use approximation on a sample QC question:

Column A

Column B

48% of 54

11% of 273


The quantity in Column A is greater.


The quantity in Column B is greater.


The two quantities are equal.


The relationship cannot be determined from the information given.


There’s no doubt that some test takers with an old-fashioned high school math mentality would wear down their pencils grinding out calculations to precisely determine the value of each quantity. Then they could say with complete confidence which column is greater, or if they’re the same. (Note that with only numbers and no variables in the question, the answer cannot be D. More on that in chapter 4.) Will this method get the right answer? Maybe, if they don’t botch the math—a not-so-unlikely prospect when dealing with awkward numbers like these. Even if this method does yield the right answer, it may take a good chunk of time.

Approximating is the way to go. Observe: 48% is pretty close to 50%, or one-half, so let’s work with that figure instead. Half of 54 is 27, so a little less than half of 54 (remember, the real figure is 48%, not the full 50%) must be a little less than 27, which is a fine approximation for Column A. Similarly, 11% is awfully close to 10%, an extremely manageable percentage. To take 10% of anything, we simply move the decimal point one place to the left. The value in Column B is therefore a little more than 27.3, since 11% of a number is larger than 10% of that same number. Since the quantity in Column A is less than 27, and the quantity in Column B is more than 27.3, Column B must be larger than Column A, which means that choice B is correct.

It would actually take a quick test taker less time to approximate the two values and settle on choice B than it took us to explain the method above. And, needless to say, it would take way less time (with less risk of careless mistakes, to boot) than it would take to actually do the math.

  • Common Traps and Left-Field Choices

One more element of your new math mindset concerns how you interact with the answer choices. The test makers prefer that you don’t stumble upon the right answer accidentally and therefore construct the choices accordingly. Let’s first discuss “common trap” and “left-field” choices individually, and then we’ll get to some examples.

Common Traps. Remember, the test makers often spice up what would otherwise be basic problems. That means that you should assume that they go out of their way to trap unwary test takers into selecting appealing wrong answer choices, sometimes called “distractors” since they’re meant to distract you from the correct choice. What might make a wrong choice appealing? Three main things:

  1. It repeats a number used in the problem itself.

  2. It represents a number you derive along the way to the right answer choice.

  3. It represents the answer that results from a common misunderstanding of the problem.

You’ll see examples in just a bit, but first let’s discuss another kind of answer choice you should keep on your radar.

Left-Field Choices. “Left-field” choices are just what they sound like—choices from way out in left field that simply make no sense in the context of the question. Say you get a complicated rate/time/distance problem in which you’re given a whole bunch of information and need to calculate how long it would take someone to drive from New York to Chicago. (Don’t worry—we’ll cover this kind of problem in chapter 2 along with every other essential math concept you need to know.) Say you forgot to divide by 100 at some point along the way and ended up with an answer of 1,500 hours. It seems ludicrous, but some people take the test with blinders on, and if they get 1,500, they get 1,500—period. So if that answer appeared among the choices, they’d choose it. This despite the fact that traveling even at a reasonably slow rate of 40 miles per hour, one could drive from New York to California twenty times in 1,500 hours. The answer just doesn’t make sense in the context of the question—it comes from left field. The test makers include some left-field choices to remind you that it’s not just a math test; it’s also a reasoning test, which means you can and should quickly eliminate choices that defy common sense.

Let’s now take a look at some traps and left-field choices in action. See what you can make of the following question:

Simone invests $10,000 in a bank account that pays 10% interest annually. If the interest is compounded quarterly, how much money will be in the account after two years assuming that no money is deducted from the account and no money other than interest and the initial investment is added to the account?











If you understand the formula for compound interest and can get the answer that way, that’s fine, although in this case we can eliminate choices to get there faster. Choice A repeats a number from the question, which makes it suspicious to begin with. Moreover, it contains shades of “left field” since it defies common sense. Does it sound reasonable that a bank account that receives interest will have the same amount it started with after two years, given that no deductions are made from it? No—it has to have more, so choice A bites the dust on this count as well. And speaking of left-field choices, we may as well cut E too: Does it seem logical that an account would more than double in two years at an interest rate of 10%? Any experience with an interest-bearing account should suggest that $21,000 is way out of the ballpark here, leaving only B, C, and D as contenders.

Ten percent of $10,000 is $1,000. If the problem were based on simple interest—which no doubt the test makers are hoping some people will think—then $11,000 would be in the account after one year, and $12,000 after two years. But neither of these takes into consideration that the interest is compounded quarterly. B and C are therefore traps, both written to tempt anyone who falls for this common misunderstanding. C, $12,000, is what results if you calculated simple instead of compound interest, while B, $11,000, represents a number on the way to that wrong answer. The correct answer is D, which is what we’d get if we plugged the numbers into the complicated equation for compound interest. In this case, we didn’t have to.

  • Intelligent Guessing

Your familiarity with distracters will help lead you to some quick and easy points, but that’s not the only use of this knowledge. Some questions are just downright tough, especially if you’re doing well and land yourself in the deep end of the question pool. Since in the computer-adaptive format you can’t move on to the next question until you answer the one in front of can never leave an answer blank. If you get stuck, you still have to pull the trigger on some choice or another. In those cases, eliminating even a few common traps or left-field choices will put the odds in your favor and allow you to guess intelligently.

  • Use Scratch Paper

On test day, you’ll receive at least three pieces of blank 8½ × 11 paper. Use them! Don’t try to solve equations in your head—you get no extra points, and the risk of error is high when you’re doing complex calculations or working with complicated strings of numbers. Some people even find it handy to jot down the letters A through E (or A through D for QC questions) on their scratch paper for each new question they face, allowing them to cross off choices they eliminate so they don’t get confused with which ones they chopped and which ones are still in contention. Try out this strategy to see whether it works for you.

If you use up your batch of paper, ask for more during the break between sections. The test center’s proctor will give you more, in batches of three sheets at a time. You’ll have to hand in your used scratch paper to the test proctor to get more, and you won’t be able to take the scratch paper with you when you leave the testing center.

  • Avoid Careless Mistakes

The bane of test takers at all levels is selecting the wrong answer to a question they know how to solve. Such mistakes are understandable, considering the pressure of the test and the timing restrictions which often force people to work faster than they’d like. But it doesn’t have to be this way. To avoid careless mistakes, follow these tips:

  • Slow down. Although rushing may allow you to answer more questions, a multitude of wrong answers, especially in the beginning of the test, could send your score plummeting. Think through the questions before jumping in to solve them. Taking the necessary time to select the proper approach will help you get off on the right foot before investing tons of time in a fruitless direction.

  • Read the question carefully. The test makers have a knack for asking strange or unexpected questions, the kind not usually asked in math class. Make sure that you answer the question asked rather than the question you think they might ask. If you have time, reread the question one last time before making your final selection to make sure you’re giving them what they want. For example, if they give you a question about boys and girls and ask for the number of boys, make sure you don’t accidentally go with the number of girls or the total number of boys and girls, things you very well may determine along the way. As we noted earlier, the test makers like to scatter such traps among the choices to catch the careless. Also pay attention to the units given in the problems: If they give you information in terms of minutes but ask for the answer in hours, you better take notice. In addition, if the test makers want you to round an answer, they’ll instruct you to do so, and when they’re looking for an approximate value (as is sometimes the case in DI questions), they’ll tell you that too. Listen for exactly what they want, and then give it to them.

  • Study your practice sets. Don’t gloss over careless errors in your practice problems. Study them! It’s one thing to simply not know how to do a problem, and quite another to think you aced it only to find out otherwise. Figure out where you went wrong. Were you rushing? Did you mix up numbers or fall for a common trap? Perhaps you did all the right math but then selected something other than what they asked for? Determine where your mistake lies and figure out what you need to do to avoid making that same mistake again.