1. Word Problems and Their Types
We will apply the Kaplan Approach to the different types of word problems.

2. Distance = Rate x Time, or D = R x T
Motion Problems
Virtually, all motion problems involve the formula
Distance = Rate x Time, or D = R x T
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

3. Kaplan Approach Step 1: Analyze the question Step 2: Identify the task
Step 3: Approach Strategically
Step 4: Confirm your answer

4. Analyze the question What type of math is it going to require?
Alegbra and Simultaneous equations

5. Identify the task What is the question we are answering?
If Garrett overtakes Scott in 2 hours, how many miles will Garrett have covered?
What form will the answer be in?
Distance or Miles

6. Approach Strategically
What approach will we be using? Textbook, Picking Numbers, Backsolving, or Strategic Guessing
Textbook

7. Textbook Approach: Step 1 We are looking for Garret’s Distance
Define your variables knowing Distance=rate x time
SCOTT
SR= Scott's rate
SD= Scott’s Distance
ST= Scott’s Time
GARRETT
GR=Garrett’s rate
GD= Garrett’s Distance
GT= Garrett’s Time

8. Step 2 We are looking for Garret’s Distance
Define your equations and variables
SCOTT
SR= SR
SD= SD
ST= 2.5 hrs
GARRETT
GR=One less than twice Scott’s= 2SR-1
GD= GD
GT= 2 hrs

9. Step 3 We are looking for Garret’s Distance
Set up your equations knowing D=RxT
SCOTT
SD= SR x 2.5
GARRETT
GD= (2SR-1) X (2)
We know that Garrett overtakes Scott which means Garrett’s Distance is equal to Scott’s Distance
2.5SR=(2)(2SR-1)
2.5SR=4SR-2
SR= 4/3

10. Step 4 We are looking for Garret’s Distance
Answer the question
GARRETT
GD= (2SR-1) X (2)
GD=(2)(2(4/3)-1)
GD= 2((8/3)-1)
GD= 2(5/3)
GD= 10/3 or 3 1/3 Miles

11. Confirm your answer
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

12. Practice Problems
Complete the following problems in small groups or individually and submit via Edmodo.
If this is not completed in class, then it is your homework.

13. Work Problems
The formula for work problems is Work = Rate x Time, or W = R x T.
The amount of work done is usually one 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

14. 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

15. 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

16. 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.
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

17. Answers
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
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 Setting these expressions equal to each other gives
4r - 2 = r x 2 ½
Solving this equation for r gives r = 4/3. Hence, Garrett will have traveled
D = 4r - 2 = 4(4/3) - 2 = 3 1/3 miles
The answer is (B).

18. Answers
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) = 1/t 1/60 = 1/t t = 60
Hence, working alone, Bobby can do the job in 1 hour. The answer is (C).

19. Answers
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 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:
10%(50) + 30%x = 20%(50 + x)Multiply this equation by 100 to clear the percent symbol and then solving for x yields x = 50. The answer is (D).

20. Answers
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 = 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) = D D = D = D = 13
Hence, there are 13 dimes, and the answer is (D).
Dimes
Quarters
Total
Number D
20 - D 20
Value
10D
25(20 - D)
305

21. Answers
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 x + 30 = 2(x + 10) x + 30 = 2x x = 10
Hence, Steve is 10 years old, and the answer is (C).
Age now
Age in 10 years
Steve x
x + 10
John
x + 20
x + 30