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

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) + 0.28m = 176

Example (cont) 3. Solve and state the answer. 39(2) + 0.28m = 176 78 + 0.28m = 176 0.28m = 98 m = 350 The executive drove 350 miles. 4. Check. 39(2) + 0.28(350) = 176 78 + 98 = 176 176 = 176

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 $138 2. Write an equation and solve. 0.23s = 138 s = 600 The original price of the sofa was $600.

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 + 0.03x = 35,535 1.03x = 35,535 x = 34,500 The check is left to you. 5

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

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

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 = 2.55 0.10d + 0.25(d – 1) = 2.55

Example (cont) 0.10d + 0.25(d – 1) = 2.55 0.10d + 0.25d – 0.25 = 2.55 0.35d – 0.25 = 2.55 0.35d = 2.80 d = 8 Bob had 8 dimes and (8 – 1) or 7 quarters. The check is left to you.