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Interest Problems. Interest Formula. Principal • Rate • Time = Interest Principal – money invested % Rate – change to a decimal Time – years. Example. James invested $500 in a 2% CD. If simple interest were used by the bank, how much interest would he have at the end of 4 years?.

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interest formula
Interest Formula
  • Principal • Rate • Time = Interest
  • Principal – money invested
  • % Rate – change to a decimal
  • Time – years
example
Example
  • James invested $500 in a 2% CD. If simple interest were used by the bank, how much interest would he have at the end of 4 years?
example1
Example
  • Mr. Hawkey invested $30,000 in two accounts. He made a total of $2,140 annual interest. How much did he invest in each account if one account pays 7% annual interest and the other pays 8.5% annual interest?
investment
Investment
  • What is the total amount invested?
  • $30,000
investment1
Investment
  • Solve, and round your answer to the nearest cent.
  • x = $27,333.33 (invested at 7%)
example2
Example
  • A merchant has 2 loans totaling $25,000. The interest rates are 6% and 7.5%. If the annual interest charge on the 6% loan is $520 more than the 7.5% loan, how much has he borrowed at each rate?
example3
Example
  • Susan wants to invest the $23,000 she earned in a year. Her bank has a savings account which earns 2.5% annually as well as a CD which earns 5.5%. Susan wants to put some money in each account, and she wants to earn at least $1,000 by the end of the year. How much should she put in each?
example4
Example
  • Mr. and Mrs. Bell received a $40,000 inheritance, so they decide to invest it in two different accounts and use the earned interest to go on a vacation. They put $30,000 in a 5% CD and $10,000 in a 3% money market account. If they need $3,000, how long will they need to leave the money in these accounts?
mixture problems1
Mixture Problems
  • quantity • percentage (strength) = the PART described by the percentage in the units described by the quantity
mixture problems2
Mixture Problems
  • 45 lbs. of 80% iodine solution
  • 45 • 0.80 = x
  • 36 lbs. of iodine
mixture problems3
Mixture Problems
  • What is the strength of a hydrochloric acid solution if there are 34 kilograms of water and 6 kilograms of acid?
  • 40 • x = 6
  • 15% strength
example5
Example
  • A pharmacist needs to make a facial cream that is 0.5% medicine and the rest lanolin. He has 80 grams of a 1.5% mixture in his lab. How much lanolin should he add to the mixture to make the desired strength?
example6
Example
  • The same pharmacist gets another order for 240 grams of 0.5% facial cream. From earlier orders, he has some 0.2% cream and some 0.8% cream. How much of each should he add together to get 240 g at the desired strength?
slide16
Section 2.6
  • pp. 67-69
problem 1
Problem 1
  • Prt = I
  • (148)(.06)(1) = I
  • $8.88 = I
problems 2 3
Problems 2-3
  • Quantity • Strength
  • = 50(.03)
  • = 1.5 gallons
  • 12  30 = Strength
  • 0.4 = Strength
  • 40% Strength
problem 4
Problem 4
  • 120 + 150 = x
  • x = $270

P • r • t = I

3000 .04 1 120

2500 .06 1 150

problem 5
Problem 5

1.1 + 0.42 = x

x = 1.52 gallons

problem 7
Problem 7
  • 0.085x + 0.1(15250 – x) = 1411.75
  • $7550 at 8.5%
  • $7700 at 10%

P •r • t = I

x .085 1 .085x

15250-x .1 1 .1(15250-x)

problem 9
Problem 9

0.02x + 0.11(3 – x) = 0.24

1 kg of 2% HCl solution

2 kg of 11% HCl solution

problem 10
Problem 10
  • 4000x + 8200(x+.015) = 1282
  • $4000 at 9.5%
  • $8200 at 11%

P •r • t = I

4000 x 1 4000x

8200 x+.015 1 8200(x+.015)

problem 13
Problem 13
  • .075(152000 – x) + .105x = 13350
  • $65,000 at 21%
  • $87,000 at 15%

P •r • t = I

152000-x .15 .5 .075(152000-x)

x .21 .5 .105x

problem 14
Problem 14
  • 102x + 180x = 70.50
  • x = 0.25
  • ¼ of a year or 3 months

P •r • t = I

1200 .085 x 102x

2000 .09 x 180x

problem 15
Problem 15

2 + x = 0.08(100 + x)

x  6.5 mL of pure acetic acid

problem 16
Problem 16

9 = 0.001(300 + x)

x = 8700 liters of water