SAT Math - Word Problems Review

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WORD PROBLEMS
Although exact steps for solving word problems cannot be given, the following guidelines will help:

(1) First, choose a variable to stand for the least unknown quantity, and then try to write the other unknown quantities in terms of that variable.

For example, suppose we are given that Sue's age is 5 years less than twice Jane's and the sum of their ages is 16. Then Jane's age would be the least unknown, and we let x = Jane's age. Expressing Sue's age in terms of x gives Sue's age = 2x - 5.

(2) Second, write an equation that involves the expressions in Step 1. Most (though not all) word problems pivot on the fact that two quantities in the problem are equal. Deciding which two quantities should be set equal is usually the hardest part in solving a word problem since it can require considerable ingenuity to discover which expressions are equal.

For the example above, we would get (2x - 5) + x = 16.

(3) Third, solve the equation in Step 2 and interpret the result.

For the example above, we would get by adding the x's: 3x - 5 = 16. Then adding 5 to both sides gives 3x = 21. Finally, dividing by 3 gives x = 7. Hence, Jane is 7 years old and Sue is 2x - 5 = 2(7) - 5 = 9 years old.

Motion Problems
Virtually, all motion problems involve the formula Distance = Rate x Time, or

Example: Scott starts jogging from point X to point Y. A half-hour later his friend Garrett who jogs 1 mile per hour slower than twice Scott's rate starts from the same point and follows the same path. If Garrett overtakes Scott in 2 hours, how many miles will Garrett have covered?
(A) 2 1/5 (B) 3 1/3 (C) 4 (D) 6 (E) 6 2/3

Following Guideline 1, we let r = Scott's rate. Then 2r - 1 = Garrett's rate. Turning to Guideline 2, we look for two quantities that are equal to each other. When Garrett overtakes Scott, they will have traveled the same distance. Now, from the formula D = R x T, Scott's distance is D = r x 2 1/2 and Garrett's distance is D = (2r - 1)2 = 4r - 2. Setting these expressions equal to each other gives

Solving this equation for r gives r = 4/3. Hence, Garrett will have traveled

The answer is (B).

Work Problems
The formula for work problems is Work = Rate x Time, or W = R x T. The amount of work done is usually 1 unit. Hence, the formula becomes 1 = R x T. Solving this for R gives R = 1/T.

Example: If Johnny can mow the lawn in 30 minutes and with the help of his brother, Bobby, they can mow the lawn 20 minutes, how long would take Bobby working alone to mow the lawn?
(A) 1/2 hour (B) 3/4 hour (C) 1 hour (D) 3/2 hours (E) 2 hours

Let r = 1/t be Bobby's rate. Now, the rate at which they work together is merely the sum of their rates:

Total Rate = Johnny's Rate + Bobby's Rate
1/20 = 1/30 + 1/t
1/20 - 1/30 = 1/t
(30 - 20)/(30)(20) = 1/t
1/60 = 1/t
t = 60

Hence, working alone, Bobby can do the job in 1 hour. The answer is (C).

Mixture Problems
The key to these problems is that the combined total of the concentrations in the two parts must be the same as the whole mixture.

Example: How many ounces of a solution that is 30 percent salt must be added to a 50-ounce solution that is 10 percent salt so that the resulting solution is 20 percent salt?
(A) 20 (B) 30 (C) 40 (D) 50 (E) 60

Let x be the ounces of the 30 percent solution. Then 30%x is the amount of salt in that solution. The final solution will be 50 + x ounces, and its concentration of salt will be 20%(50 + x). The original amount of salt in the solution is 10%(50). Now, the concentration of salt in the original solution plus the concentration of salt in the added solution must equal the concentration of salt in the resulting solution:

Multiply this equation by 100 to clear the percent symbol and then solving for x yields x = 50. The answer is (D).

Coin Problems
The key to these problems is to keep the quantity of coins distinct from the value of the coins. An example will illustrate.

Example: Laura has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?
(A) 3 (B) 7 (C) 10 (D) 13 (E) 16

Let D stand for the number of dimes, and let Q stand for the number of quarters. Since the total number of coins in 20, we get D + Q = 20, or Q = 20 - D. Now, each dime is worth 10 cents, so the value of the dimes is 10D. Similarly, the value of the quarters is 25Q = 25(20 - D). Summarizing this information in a table yields

Notice that the total value entry in the table was converted from $3.05 to 305 cents. Adding up the value of the dimes and the quarters yields the following equation:

10D + 25(20 - D) = 305
10D + 500 - 25D = 305
-15D = -195
D = 13

Hence, there are 13 dimes, and the answer is (D).

Age Problems
Typically, in these problems, we start by letting x be a person's current age and then the person's age a years ago will be x - a and the person's age a years in future will be x + a. An example will illustrate.

Example: John is 20 years older than Steve. In 10 years, Steve's age will be half that of John's. What is Steve's age?
(A) 2 (B) 8 (C) 10 (D) 20 (E) 25

Steve's age is the most unknown quantity. So we let x = Steve's age and then x + 20 is John's age. Ten years from now, Steve and John's ages will be x + 10 and x + 30, respectively. Summarizing this information in a table yields

Since "in 10 years, Steve's age will be half that of John's," we get

(x + 30)/2 = x + 10
x + 30 = 2(x + 10)
x + 30 = 2x + 20
x = 10

Hence, Steve is 10 years old, and the answer is (C).