Pre-Algebra 8-7 More Applications of Percents Learn to compute simple interest.

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Pre-Algebra 8-7 More Applications of Percents Learn to compute simple interest.

Pre-Algebra 8-7 More Applications of Percents When you borrow money from a bank, you pay interest for the use of the bank’s money. When you deposit money into a savings account, you are paid interest. Simple interest is one type of fee paid for the use of money. I = P  r  t Simple Interest Principal is the amount of money borrowed or invested Rate of interest is the percent charged or earned Time that the money is borrowed or invested (in years)

Pre-Algebra 8-7 More Applications of Percents To buy a car, Sarah borrowed $15,000 for 3 years at an annual simple interest rate of 9%. How much interest will she pay if she pays the entire loan off at the end of the third year? What is the total amount that she will repay? Example 1: Finding Interest and Total Payment on a Loan First, find the interest she will pay. I = P  r  tUse the formula. I = 15,000  0.09  3 Substitute. Use 0.09 for 9%. I = 4050 Solve for I.

Pre-Algebra 8-7 More Applications of Percents Example 1 Continued Sarah will pay $4050 in interest. A = P + Iamount = principal + interest A = 15, Substitute. A = 19,050 Solve for A. You can find the total amount A to be repaid on a loan by adding the principal P to the interest I. Sarah will repay a total of $19,050 on her loan.

Pre-Algebra 8-7 More Applications of Percents Example 2: Determining the Amount of Investment Time I = P  r  t Use the formula. 450 = 6,000  0.03  t Substitute values into the equation. 2.5 = t Solve for t. Melissa invested $6000 in a bond at a yearly rate of 3%. She earned $450 in interest. How long was the money invested? 450 = 180 t The money was invested for 2.5 years, or 2 years and 6 months.

Pre-Algebra 8-7 More Applications of Percents When you borrow money from a bank, you pay interest for the use of the bank’s money. When you deposit money into a savings account, you are paid interest. Simple interest is one type of fee paid for the use of money. I = P  r  t Simple Interest Principal is the amount of money borrowed or invested Rate of interest is the percent charged or earned Time that the money is borrowed or invested (in years)

Pre-Algebra 8-7 More Applications of Percents To buy a laptop computer, Elaine borrowed $2,000 for 3 years at an annual simple interest rate of 5%. How much interest will she pay if she pays the entire loan off at the end of the third year? What is the total amount that she will repay? Try This: Example 1 First, find the interest she will pay. I = P  r  tUse the formula. I = 2,000  0.05  3 Substitute. Use 0.05 for 5%. I = 300 Solve for I.

Pre-Algebra 8-7 More Applications of Percents Try This: Example 1 Continued Elaine will pay $300 in interest. P + I = Aprincipal + interest = amount = A Substitute = A Solve for A. You can find the total amount A to be repaid on a loan by adding the principal P to the interest I. Elaine will repay a total of $2300 on her loan.

Pre-Algebra 8-7 More Applications of Percents Try This: Example 2 I = P  r  t Use the formula. 200 = 4,000  0.02  t Substitute values into the equation. 2.5 = t Solve for t. TJ invested $4000 in a bond at a yearly rate of 2%. He earned $200 in interest. How long was the money invested? 200 = 80t The money was invested for 2.5 years, or 2 years and 6 months.

Pre-Algebra 8-7 More Applications of Percents I = P  r  t Use the formula. I = 1000   18 Substitute. Use for 3.25%. I = 585 Solve for I. Now you can find the total. Example 3: Computing Total Savings Bryan’s parents deposited $1000 into a savings account as a college fund when he was born. How much will Bryan have in this account after 18 years at a yearly simple interest rate of 3.25%?

Pre-Algebra 8-7 More Applications of Percents P + I = AUse the formula = A 1585 = A Bryan will have $1585 in the account after 18 years. Example 3 Continued

Pre-Algebra 8-7 More Applications of Percents Mr. Middleswarth borrowed $8000 for 4 years to make home improvements. If he repaid a total of $10,320, at what interest rate did he borrow the money? Example 4: Finding the Rate of Interest P + I = A Use the formula I = 10,320 I = 10,320 – 8000 = 2320 Find the amount of interest. He paid $2320 in interest. Use the amount of interest to find the interest rate.

Pre-Algebra 8-7 More Applications of Percents Example 4 Continued 2320 = 32,000  rMultiply. I = P  r  t Use the formula = 8000  r  4 Substitute ,000 = r Divide both sides by 32, = r Mr. Middleswarth borrowed the money at an annual rate of 7.25%

Pre-Algebra 8-7 More Applications of Percents When you borrow money from a bank, you pay interest for the use of the bank’s money. When you deposit money into a savings account, you are paid interest. Simple interest is one type of fee paid for the use of money. I = P  r  t Simple Interest Principal is the amount of money borrowed or invested Rate of interest is the percent charged or earned Time that the money is borrowed or invested (in years)

Pre-Algebra 8-7 More Applications of Percents I = P  r  t Use the formula. I = 1000   50 Substitute. Use for 7.5%. I = 3750 Solve for I. Now you can find the total. Try This: Example 3 Bertha deposited $1000 into a retirement account when she was 18. How much will Bertha have in this account after 50 years at a yearly simple interest rate of 7.5%?

Pre-Algebra 8-7 More Applications of Percents P + I = A Use the formula = A 4750 = A Bertha will have $4750 in the account after 50 years. Try This: Example 3 Continued

Pre-Algebra 8-7 More Applications of Percents Mr. Mogi borrowed $9000 for 10 years to make home improvements. If he repaid a total of $20,000 at what interest rate did he borrow the money? Try This: Example 4 P + I = A Use the formula I = 20,000 I = 20,000 – 9000 = 11,000 Find the amount of interest. He paid $11,000 in interest. Use the amount of interest to find the interest rate.

Pre-Algebra 8-7 More Applications of Percents Try This: Example 4 Continued 11,000 = 90,000  r Multiply. I = P  r  t Use the formula. 11,000 = 9000  r  10 Substitute. 11,000 90,000 = r Divide both sides by 90, = r Mr. Mogi borrowed the money at an annual rate of about 12.2%.