1. Solving Word Problems: The Value of Money and Percents
Section 2.7
Solving Word Problems: The Value of Money and Percents

2. Example
A business executive rented a car. The Supreme Car Rental Agency charged $39 per day and $0.28 per mile. The executive rented the car for two days and the total rental cost was computed to be $176. How many miles did the executive drive the rented car? 1. Understand the problem. Let m = the number of miles driven 2. Write an equation. per day cost + mileage cost = total cost 39(2) m = 176

3. Example (cont)
3. Solve and state the answer. 39(2) m = m = m = 98 m = 350 The executive drove 350 miles. 4. Check. 39(2) (350) = = = 176

4. Example
A sofa was marked down with the following sign: “The price of this sofa has been reduced by 23%. You can save $138 if you buy now.” What was the original price of the sofa? 1. Understand the problem. Let s = the original price of the sofa. Let 0.23s = the amount of the price reduction, which is $ Write an equation and solve. 0.23s = 138 s = 600 The original price of the sofa was $600.

5. Example
Kylie received a pay raise this year. The raise was 3% of last year’s salary. This year she will receive $35,535. What was her salary last year before the raise?
1. Understand the problem.
Let x = Kylie’s salary last year
0.03 = the amount of the raise
2. Write an equation and solve.
last year’s salary + the amount of raise = new salary
x x = 35,535
1.03x = 35,535
x = 34,500
The check is left to you. 5

6. Solving Simple Interest Problems
Simple interest = principal ∙ rate ∙ time
I = P ∙ R ∙ T
The interest rate is assumed to be per year unless otherwise stated. The time T must be in years. 6

7. Example
Find the interest on a loan of $2500 that is borrowed at a simple interest rate of 9% for 3 months.
First change 3 months to years.
3 months = 3/12 = ¼ year
I = P ∙ R ∙ T
I = 2500 ∙ 0.09 ∙ ¼
= 225 ∙ ¼
We divide 225 by 4.
= 56.25
The interest for 3 months is $56.25. 7

8. Example
When Bob got out of math class, he had to make a long distance call. He had exactly enough dimes and quarters to make a phone call that would cost $2.55. He had one fewer quarter than he had dimes. How many coins of each type did he have? Let d = the number of dimes Let d – 1 = the number of quarters Each dime is worth 0.10 Each quarter is worth 0.25 The value of the dimes + value of quarters = d (d – 1) = 2.55

9. Example (cont)
0.10d (d – 1) = d d – 0.25 = d – 0.25 = d = 2.80 d = 8 Bob had 8 dimes and (8 – 1) or 7 quarters. The check is left to you.