lesson 6-7 pages 298-302 using percent equations lesson check 6-6
DESCRIPTION
Percent equation Discount Simple interestTRANSCRIPT
Lesson 6-7 Pages 298-302
Using Percent Equations
Lesson Check 6-6
What you will learn!1. How to solve percent problems using percent equations.
2. How to solve real-life problems involving discount and interest.
Percent Percent equationequationDiscountDiscountSimple interestSimple interest
What you really need to know!
In a percent equation, the percent is written as a decimal.
What you really need to know!
% of BASE = PART
The word “ of ” means multiply!
What you really need to know!
The number after the word “of” is the base!
The number near the word “is” is the part!
Example 1:
Find 38% of 22.% x BASE = PART38% x 22 = PART0.38 x 22 = PART8.36 = PART
Example 2:
19 is what percent of 25?
% x BASE = PART% x 25 = 19% = 19 ÷ 25% = 0.76 76%
Example 3:
84 is 16% of what number?
% x BASE = PART16% x BASE = 840.16 x BASE = 84BASE = 84 ÷ 0.16BASE = 525
Example 4:
The regular price of a ring is $495. It is on sale at a 20% discount. What is the sale price of the ring?20% x $495 = $99$495 - $99 = $396
Example 5:
Suppose you invest $2,000 at an annual interest rate of 4.5%. How long will it take to earn $495?
I = prt
I = prtI is interest
p is principal
r is annual interest rate
t is time in years
Example 5:
Suppose you invest $2,000 at an annual interest rate of 4.5%. How long will it take to earn $495?
495 = 2,000 • 0.045 • t5.5 years = t
Page 300
Guided Practice
#’s 4-11
Pages 298-300 with someone at home and
study examples!
Read:
Homework: Pages 301-302
#’s 12-40 even
#’s 44-61
Lesson Check 6-7
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Lesson 6-7
Lesson Check 6-7