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Purpose: To solve problems solving percents.

7.6 Percent Problems. Purpose: To solve problems solving percents. Homework: p. 317 -319 1-21 odd and problems 1-15 odd. Formula. Whenever a price is changed you can find the percent of increase or decrease by the following: Percent of change = change in price

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Purpose: To solve problems solving percents.

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  1. 7.6 Percent Problems • Purpose: To solve problems solving percents. • Homework: p. 317 -319 1-21 odd and problems 1-15 odd.

  2. Formula • Whenever a price is changed you can find the percent of increase or decrease by the following: • Percent of change = change in price • 1000 original price • To find the change in price, you calculate the difference between the original price and the new price.

  3. Examples • Alyce originally paid $6500 for a ca 3 years ago. The same car costs $7400 today. What is the change in price? • Find the change in price by subtracting $6500 from $7400. • $7400 – $6500 = $900 • The price increased $900

  4. Example #2 • The Bakers’ real estate taxes went from $1560 to $1683. What was the percent of increase? • Let x = the percent of increase. • % of Change = change in price • 100 original • x = 78 **Cross Products. • 100 1560 • 1560x = 7800; x = 5; 5% increase

  5. Example #3 • The number of registered voters in Fresno increased by 14% over last year. There are now 9633 registered voters. How many voters were there last year? • Let x = the # of voters last year. 9633 – x will = the change in # of voters. • 14 = 9633 – x ** Cross Products. • 100 x • 14x = 100(9633 – x) • 14x = 963300 – 100x 114x = 963300 • x = 8450; There were 8450 voters last year.

  6. Joe invested $2000 at 51/4% annual interest and $3000 at 8 ½% annual interest. What percent interest is he earning on his investment? • A + B = Total • .0525(2000) + .085(3000) = 5000x • 10000[.0525(2000) + .085(3000) = 5000x]10000 • 525(2000) + 850(3000) = 50,000,000x • 1,050,000 + 2,550,000 = 50,000,000x • 3,605,000 = 50,000,000x ; x = .072 ; 7.2%

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