SAT Study Guide/Part 2 - The Math Section/Word Problems

A typical SAT has several word problems, covering almost every math topic for which you are responsible. Expect to encounter word problems on all sorts of topics, such as: consecutive integers, fractions and percents, ratios and proportions, averages, circles, triangles, and other geometric figures. A few of these problems can be solved just with arithmetic, but most of them require basic algebra.

To solve word problems algebraically, you must treat algebra as a foreign language and learn to translate “word for word” from English into algebra, just as you would from English into French or Spanish or any other foreign language. When translating into algebra, you use some letter (often x) to represent the unknown quantity you are trying to determine. It is this translation process that causes difficulty for some students. Once the translation is completed, solving is easy using the techniques already reviewed.

Consider the pairs of typical SAT questions in Examples 1 and 2. The first ones in each pair (1A and 2A) would be considered easy, whereas the second ones (1B and 2B) would be considered harder.

Example 1A
What is 4% of 4% of 40,000?

Example 1B
In a lottery, 4% of the tickets printed can be redeemed for prizes, and 4% of those tickets have values in excess of $100. If the state prints 40,000 tickets, how many of them can be redeemed for more than $100?

Example 2A
If x + 7 = 2(x − 8), what is the value of x?

Example 2B
In 7 years, Erica will be twice as old as she was 8 years ago. How old is Erica now?

Once you translate the words into arithmetic expressions or algebraic equations, Examples 1A and 1B and 2A and 2B are clearly identical. The problem that many students have is doing the translation. It really isn’t very difficult, and you’ll learn how. First, though, look over the following English-to-algebra “dictionary.”

Let’s use this “dictionary” to translate some phrases and sentences.

In translating a statement, you first must decide what quantity the variable will represent. Often, this is obvious. Other times there is more than one possibility. Let’s translate and solve the two examples at the beginning of this section, and then look at a few new ones.

Example 1B
In a lottery, 4% of the tickets printed can be redeemed for prizes, and 4% of those tickets have values in excess of $100. If the state prints 40,000 tickets, how many of them can be redeemed for more than $100?

Solution
Let x be the number of tickets worth more than $100. Then,

x = 4% of 4% of 40,000 = 0.04 × 0.04 × 40,000 = 64,

which is also the solution to Example 1A.

Example 2B
In 7 years, Erica will be twice as old as she was 8 years ago. How old is Erica now?

Solution
Let x be Erica's age now; 8 years ago, she was x − 8, and 7 years from now, she will be x + 7. Then,

x + 7 = 2(x − 8),

and

x + 7 = 2(x − 8) ⇒ x + 7 = 2x − 16 ⇒ 7 = x − 16 ⇒ x = 23, which is also the solution to Example 2A.

Example 3
The product of 2 and 8 more than a certain number is 10 times that number. What is the number?

Solution
Let x represent the unknown number. Then,

2(8 + x) = 10x

and

2(8 + x) = 10x ⇒ 16 + 2x = 10x ⇒ 8x = 16 ⇒ x = 2

Example 4
If the sum of three consecutive integers is 20 more than the middle integer, what is the smallest of the three?

Solution
Let n represent the smallest of the three integers. Since the integers are consecutive, the middle number is n + 1, and the largest of the three numbers is n + 2. Then,

n + (n + 1) + (n + 2) = 20 + (n + 1)

and

n + (n + 1) + (n + 2) = 20 + (n + 1) ⇒ 3n + 3 = 21 + n ⇒ 2n + 3 = 21 ⇒ 2n = 18 ⇒ n = 9.

The three integers are 9, 10, and 11. Party like it's September 10, 2011.

Most algebraic word problems on the SAT are not very difficult. If, after studying this section, you still get stuck on a question, don’t despair. In each of Examples 3 and 4, if you had been given choices, you could have back-solved, and if the questions had been grid-ins, you could have used trial and error (effectively, back-solving by making up your own choices). Here’s how.

Alternative Solution to Example 3
Pick a starting number and test (using your calculator, if necessary).


 * Try 10: 8 + 10 = 18, and 2 × 18 = 36, but 10 × 10 = 100, which is much too big!
 * Try 5: 8 + 5 = 13, and 2 × 13 = 26, but 10 × 5 = 50, which is still too big.
 * Try 2: 8 + 2 = 10, and 2 × 10 = 20, and 10 × 2 = 20. That's it.

Alternative Solution to Example 4
You need three consecutive integers whose sum is 20 more than the middle one. Obviously, [1, 2, 3] and [5, 6, 7] are too small; neither one even adds up to 20.


 * Try [10, 11, 12]: 10 + 11 + 12 = 33, which is 22 more than 11. A bit too much.
 * Try [9, 10, 11]: 9 + 10 + 11 = 30, which is 20 more than 10.

Of course, if you can do the algebra, that’s usually the best way, to handle these problems. Ongrid-ins you might have to backsolve with several numbers before zooming in on the correct answer; also, if the correct answer was a fraction, such as $$\frac{23}{5}$$, you might never find it. In the rest of this section, the proper ways to set up and solve various word problems are stressed.

Helpful Hint
In all word problems on the SAT, remember to circle what you're looking for. Don't answer the wrong question!

Age Problems
In problems involving ages, remember that “years ago” means you need to subtract, and “years from now” means you need to add.

Example 5
In 1980, Judy was 3 times as old as Adam, but in 1984 she was only twice as old as he was. How old was Adam in 1990?


 * C) 4
 * L) 8
 * I) 12
 * M) 14
 * B) 16

Solution
Let x be Adam’s age in 1980, and fill in the table below.

Now translate: Judy’s age in 1984 was twice Adam’s age in 1984:

3x + 4 = 2(x + 4) 3x + 4 = 2(x + 4) ⇒ x + 4 = 8 ⇒ x = 4

Adam was 4 in 1980. However, 4 is not the answer to this question. Did you remember to circle what you’re looking for? The question could have asked for Adam’s age in 1980 (choice C) or 1984 (choice L) or Judy’s age in any year whatsoever (choice I is 1980, and choice B is 1984); but it didn’t. It asked for Adam’s age in 1990. Since he was 4 in 1980, then 10 years later, in 1990, he was 14 (M).