Date: 4-11-12
Topic: 7-6 Percent
Problems
Objective
Essential Question: How do you solve problems using
percents?
To solve problems involving percents.
Whenever a price is changed, you can find the percent of cent
of decrease by using the following formula:
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑖𝑐𝑒
=
100
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑝𝑟𝑖𝑐𝑒
To find the change in price, you calculate the difference
between the original price and the new price.
Example 1
Find the change in price.
a. The original price of the car Jasmine wants was
$10,000. It is now on sale for $8,999.
The price decreased by $10,000 - $8999, or $1001.
b. Jerry originally paid $600 per month to rent his
apartment. It now costs him $650.
The price increased by $650 - $600, or $50.
Summary
1
Exercise 1
The original price for a pair of jeans was $25.00. The new
price is $30.00. What is the change in price?
The original price for a sweater was $60.00. The new price is
$45.00. What is the change in price?
Example 2
To attract business, the manager of a musical instruments store
decreased the price of an alto saxophone from $500 to $440.
What was the percent of decrease?
Step 1
The problem asks for the percent of decrease.
Step 2
Let x = the percent of decrease.
ℎ
Step 3
Step 4
100
100
=
=
ℎ
0
00
00 = 0000
= 12
Step 5
= 0 12
0
= 0 12
00
2
Exercise 2
The Gold’s home was assessed this at a value of $162,000.
Last year, it had been assessed at $150,000.What was the
percent of increase?
Step 1
Step 2
Step 3
Step 4
Step 5
3
Example 3
Ricardo paid $27 for membership in the Video Club. This was
an increase of 8% from last year. What was the price of
membership last year?
Step 1
The problem asks for the original price.
Step 2
Let x = the original price. Then 27  x = the change in price.
ℎ
Step 3
100
100
= 2 00
Step 4
10
ℎ
=
=
2
100
= 2 00
=2
Step 5
00
2 00
2 = 2 00
2 00 = 2 00
4
Exercise 2
The number of students at Westwood High School with a
driver’s license is now 558. This is 24% more than last year.
How many students had a driver’s license last year?
Step 1
Step 2
Step 3
Step 4
Step 5
5
Definition
Amount invested  Annual interest rate = Annual simple interest
Example 4
Sheila invests part of $6000 at 6% interest and the rest at 11%
interest. Her total annual income from these investments is $460.
How much is invested at 6% and how much at 11%?
Step 1
The problem asks for the amounts invested at 6% and at 11%.
Step 2
Let x = the amount invested at 6%. Then 6000  x = the amount
invested at 11%.
Amount invested  Rate =
At 6%
0.06
x
At 11%
6000 - x
0.11
Step 3
00
0 11( 000
Step 4
100[0 0
0 11( 000
000
=
Step 5
00
0.11(6000  x)
0
Total interest
)] = 100(
)=
000
11 =
000
11( 000
=
)=
Interest
0.06x
0)
Multiply both
sides by 100
20 000
000
000
0 11
2000 =
0
6
Example 4
Craig invested $4000 in bank certificates and in bonds. The
certificates pay 5.5% interest, and the bonds pay 11% interest.
His interest income is $352 this year. How much money was
invested in bank certificates?
Step 1
Step 2
Amount invested  Rate =
Interest
At 5.5%
At 11%
Step 3
Step 4
Step 5
7