Jump to a New ChapterIntroductionThe Discipline of DisciplineSAT StrategiesThe SAT Personal TrainerMeet the Writing SectionBeat the EssayBeat Improving SentencesBeat Identifying Sentence ErrorsBeat Improving ParagraphsMeet the Critical Reading sectionBeat Sentence CompletionsReading Passages: The Long and Short of ItThe Long of ItThe Short of ItSAT VocabularyMeet the Math SectionBeat Multiple-Choice and Grid-InsNumbers and OperationsAlgebraGeometryData, Statistics, and Probability
 19.1 To Algebra or Not to Algebra? 19.2 A Very Short Algebra Glossary 19.3 Substitution Questions 19.4 Solving Equations 19.5 Algebra, ABSOLUTE Value, and Exponents 19.6 Beat the System (of Equations) 19.7 Inequalities 19.8 Binomials and Quadratic Equations

 19.9 Variation 19.10 How Do Functions Function? 19.11 Evaluating Functions 19.12 Compound Functions 19.13 Domain and Range 19.14 Functions As Models 19.15 Defeating Word Problems 19.16 The Most Common Word Problems
The Most Common Word Problems
Word problems come in all shapes and sizes. But each and every year, the SAT includes certain particular varieties. We’ve got the skinny on ’em.
Rates
A rate is a ratio of related qualities that have different units. For example, speed is a rate that relates the two quantities of distance and time. Here is the general rate formula:
No matter the specifics, the key to a rate word problem is in correctly placing the given information in the three categories: A, r, and B. Then, you can substitute the values into the rate formula. We look at the three most common types of rate: speed, work, and price.
Speed
In the case of speed, time is quantity A and distance is quantity B. For example, if you traveled for 4 hours at 25 miles per hour, then
Usually, the new SAT won’t simply give you one of the quantities and the rate and ask you to plug it into the rate formula. Since rate questions are always in the form of word problems, the information that you’ll need to solve the problem is often given in the befuddling complicated manner you’ve grown to know and hate.
Here’s an example:
 Jim rollerskates 6 miles per hour. One morning, Jim starts rollerskating and doesn’t stop until he has gone 60 miles. How many hours did he spend rollerskating?
This question provides more information than simply the speed and one of the quantities. You get unnecessary facts such as how Jim is traveling (by rollerskates) and when he started (in the morning). Ignore them and focus on the facts you need to solve the problem.
• Quantity A: x hours rollerskating
• Rate: 6 miles per hour
• Quantity B: 60 miles
Here’s a more difficult rate problem:
 At a cycling race, the cyclist from California can cycle 528,000 feet per hour. If the race is 480 miles long, how long will it take her to finish the race? (1 mile = 5280 feet)
You should immediately pick out the given rate of 528,000 feet per hour and notice that the total distance traveled is 480 miles. You should also notice that the question presents a units problem: The given rate is in feet per hour, while the total distance traveled is given in miles.
Sometimes a question gives you inconsistent units, as in this example. Always read over the problem carefully and don’t forget to adjust the units—the SAT makes sure that the answer you would come to if you had forgotten to correct for units appears among the answer choices.
For the cycling question, since the question tells you that there are 5,280 feet in a mile, you can find the rate for miles per hour:
Now you can plug the information into the rate formula:
• Time: x hours cycling
• Rate: 100 miles per hour
• Distance: 480 miles
Work
Work sucks. You’re there from 9 to 5 and, at best, you get two weeks off per year, and you’ve got a boss constantly checking up on you. Work word problems on the SAT are a breeze in comparison. On work word problems, you’ll usually find the first quantity measured in time (t), the second quantity measured in work done (w), and the rate measured in work done per time (r). For example, if you knitted for 8 hours and produced 2 sweaters per hour, then
Here’s a sample work problem. It’s one of the harder rate word problems you might come across on the SAT:
 Four workers can dig a 40-foot well in 4 days. How long would it take for 8 workers to dig a 60-foot well? Assume that these 8 workers work at the same pace as the 4 workers.
First, examine what that problem says: 4 workers can dig a 40-foot well in 4 days. You know how much total work was done and how many people did it, you just don’t know the rate at which the workers worked. You need that rate, since the 8 workers digging the 60-foot wells are working at the same rate. Since , you can get the rate by dividing 40 by 4, which equals 10. The workers together dig at a pace of 10 feet per day.
Now for that group of 8 workers digging a 60-foot well. The total work (w) done by the 8 workers is 60 feet, and they work at a rate (r) of 10 feet per day per 4 workers. Can you use this information to answer the question? Oh yeah. The rate of 10 feet per day per 4 workers converts to 20 feet per day per 8 workers, which is the size of the new crew. Now you can use the rate formula:
• Time: x days of work
• Rate: 20 feet per day per 8 workers
• Work Done (in this case, distance dug): 60 feet
This last problem required a little bit of creativity—but nothing you can’t handle. Just remember the classic rate formula and use it wisely.
Price
In rate questions dealing with price, you’ll usually find the first quantity measured in numbers of items, the second measured in price, and the rate in price per item. If you have 8 basketballs, and you know that each basketball costs \$25,
Exponential Growth and Decay
Exponential growth and decay problems are like percent change problems on steroids: You must perform a percent change over and over again. You can use exponents on these repeated percent change questions. Here’s an example:
 If a population of 100 grows by 5% per year, how large will the population be in 50 years?
You could do two things to solve this problem. You could multiply each successive generation by 5% fifty times to get the final answer, or you could use this formula:
Final Amount = Original Amount × (1 + Growth Rate)(number of changes)
The formula is probably the better bet. So, to solve this problem,
final amount =
Exponential decay only slightly modifies the formula:
Final Amount = Original Amount × (1 – Growth Rate)(number of changes)
Exponential decay is often used to model population decreases as well as the decay of physical mass.
We’ll work through a few example problems to get a feel for both exponential growth and decay problems.
A Simple Exponential Growth Problem
 A population of bacteria grows by 35% every hour. If the population begins with 100 specimens, how many are there after 6 hours?
You’ve got an original population of 100, a growth rate of .35 every hour, and 6 hours. To solve the problem, you just need to plug the appropriate values into the formula for exponential growth.
final amount =
A Simple Exponential Decay Problem
 A fully inflated beach ball loses 6% of its air every day. If the beach ball originally contains 4000 cubic centimeters of air, how many cubic centimeters does it hold after 10 days?
Since the beach ball loses air, you know this is an exponential decay problem. The decay rate is .06, the original amount is 4000 cubic centimeters of air, and the time is 10 days. Plugging the information into the formula,
final amount =
A More Annoying Exponential Growth Problem
 A bank offers a 4.7% interest rate on all savings accounts, per month. If 1000 dollars is initially put into a savings account, how much money will the account hold two years later?
This problem is a bit tricky because the interest rate is per month, while the time period is given in years. You need to make the units match up. In the two-year time period given by the question, there will be 2 × 12 = 24 months
final amount =
Here’s another compounding problem:
 Ben puts \$2000 into a savings account that pays 5% interest compounded annually. Justin puts \$2500 into a different savings account that pays 4% annually. After 15 years, whose account will have more money in it if no more money is added or subtracted from the principal?
Ben’s account will have \$2000 × 1.0515 ≈ \$4157.85 in it after 15 years. Justin’s account will have \$2500 × 1.0415 ≈ \$4502.36 in it. Justin’s account will still have more money in it than Ben’s after 15 years. Notice, however, that Ben’s account is gaining on Justin’s account.
And with that, you’ve covered everything you need to know to rock SAT algebra. Geometry’s next.
 Jump to a New ChapterIntroductionThe Discipline of DisciplineSAT StrategiesThe SAT Personal TrainerMeet the Writing SectionBeat the EssayBeat Improving SentencesBeat Identifying Sentence ErrorsBeat Improving ParagraphsMeet the Critical Reading sectionBeat Sentence CompletionsReading Passages: The Long and Short of ItThe Long of ItThe Short of ItSAT VocabularyMeet the Math SectionBeat Multiple-Choice and Grid-InsNumbers and OperationsAlgebraGeometryData, Statistics, and Probability
Test Prep Centers
 New SAT Test Center Mini SAT
SAT Vocab Novels
 Rave New World S.C.A.M. Sacked Busted Head Over Heels
SAT Power Tactics
 Algebra Data Analysis, Statistics & Probability Geometry Numbers & Operations Reading Passages Sentence Completions Writing Multiple-Choice Questions The Essay Test-Taking Strategies Vocabulary Builder
SparkCollege
 College Admissions Financial Aid College Life