Jump to a New ChapterIntroduction to the SAT IIContent and Format of the SAT II Math IICStrategies for SAT II Math IICMath IIC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
 5.1 Math IIC Algebra Strategies 5.2 Equation Solving 5.3 Writing Equations 5.4 Manipulating Equations 5.5 Systems of Equations

 5.6 Common Word Problems 5.7 Polynomials 5.8 Key Formulas 5.9 Review Questions 5.10 Explanations
Writing Equations
For some questions on the Math IIC test, you’ll need to translate the problem from a language you’re used to—English—into a more useful, albeit less familiar, language. We mean the language of math, of course, and one of your main test-taking responsibilities is to be able to write an equation based on the information given in a problem.
You’ll also be asked to find an expression for a certain quantity described in a word problem. The best way to learn how to do these things quickly and effectively is to practice. Here’s a sample problem:
 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 start with, you can write r + b = 50, where r is the number of red marbles and b the number of blue marbles in the sack. This equation tell us that the 50 marbles in the sack consist entirely of red marbles and blue marbles.
Now that you have an initial equation, you need to decipher what exactly the question is asking for. In this problem it is clear-cut: How many blue marbles are in the sack? You must therefore find the value of b.
Unfortunately, you need more information to do that. You can create a second equation based on the knowledge that there are 20 more red marbles than blue marbles. This part of the word problem can be written in the form of an equation as r = b + 20 (or b = r – 20).
Let’s list the two equations we have so far:
Using both of these equations, you can solve for b. After a little manipulation, which we’ll cover in the coming sections, you’ll find that b = 15 (and r = 35). Don’t worry about the solution for now—just focus on how we translated the word problem into equations that lead to the solution.
That problem was easy. Here’s a harder one:
 Stan sells oranges for c cents apiece. The minimum number of oranges that Stan will sell to an individual is r, but the first f oranges are free (f < r). Find an expression for the price in dollars of 35 oranges if 35 > r.
According to the problem, we need to find an expression (notice, not an equation) for the price in dollars of 35 oranges. The key to a problem like this one is working step by step. First, find out how many of the 35 oranges aren’t free of charge.
Next, find the price of those oranges.
But wait. Did you notice that the question asked for the price of 35 oranges in dollars? The writers of the Math IIC are a clever bunch, if a bit sneaky. They figure that a good number of test-takers will see only the word price and not notice what units are asked for. Be careful not to fall into their carefully laid trap.
We know there are 100 cents per dollar, so we can easily convert the price by dividing by 100.
Before we move to another problem, note that the variable r didn’t appear anywhere in the answer. Egad! It is yet another attempt (and a common one at that) by those devious test writers to lower your score. You may come across many problems, especially word problems, in which extraneous information is provided only to confuse you. Just because a variable or number appears in a problem doesn’t mean that it will be useful in finding the answer.
Here’s another problem:
 Gus needs to paint his house, which has a surface area of x square feet. The brand of paint he buys (at a cost of p dollars a can) comes in cans that cover y square feet each. Gus also needs to buy ten pairs of new jeans (he is uncoordinated and spills often). They cost d dollars a pair. If Gus makes these purchases, what is the difference (in dollars) between the cost of the paint and the cost of the jeans? Assume he doesn’t buy any excess paint—that is, the required amount is not a fraction of a can.
This word problem is long and complicated, but you need to carry out just four steps to solve it:
1. Gus must buy x /y cans of paint to cover his house.
2. This will cost him xp /y dollars.
3. The jeans Gus buys cost 10d dollars.
4. Thus, the difference, in dollars, between the cost of the paint and the cost of the jeans is xp /y – 10d.
For the rest of this chapter, we’ll constantly be converting word problems into equations. If you’re still uncomfortable doing this, don’t worry. You’ll get a lot more practice in the sections to come.
 Jump to a New ChapterIntroduction to the SAT IIContent and Format of the SAT II Math IICStrategies for SAT II Math IICMath IIC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
Test Prep Centers
SparkCollege
 College Admissions Financial Aid College Life