Transcript Beginning Algebra Early Graphing

Section 2.7
Solving Word
Problems: The
Value of Money
and Percents
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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
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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
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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.
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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.
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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.
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Example
Find the interest on a loan of $2500 that is borrowed at a
simple interest rate of 9% for 3 months.
3 months = 3/12 = ¼ year
First change 3 months
to years.
I=P ∙R∙T
I = 2500 ∙ 0.09 ∙ ¼
= 225 ∙ ¼
= 56.25
We divide 225 by 4.
The interest for 3 months is
$56.25.
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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
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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.
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