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Interest Problems

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 Problems

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  1. Interest Problems

  2. Interest Formula • Principal • Rate • Time = Interest • Principal – money invested • % Rate – change to a decimal • Time – years

  3. 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?

  4. 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?

  5. Investment • What is the total amount invested? • $30,000

  6. Investment • Solve, and round your answer to the nearest cent. • x = $27,333.33 (invested at 7%)

  7. 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?

  8. 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?

  9. 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?

  10. Mixture Problems

  11. Mixture Problems • quantity • percentage (strength) = the PART described by the percentage in the units described by the quantity

  12. Mixture Problems • 45 lbs. of 80% iodine solution • 45 • 0.80 = x • 36 lbs. of iodine

  13. 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

  14. 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?

  15. 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?

  16. Section 2.6 • pp. 67-69

  17. Problem 1 • Prt = I • (148)(.06)(1) = I • $8.88 = I

  18. Problems 2-3 • Quantity • Strength • = 50(.03) • = 1.5 gallons • 12  30 = Strength • 0.4 = Strength • 40% Strength

  19. Problem 4 • 120 + 150 = x • x = $270 P • r • t = I 3000 .04 1 120 2500 .06 1 150

  20. Problem 5 1.1 + 0.42 = x x = 1.52 gallons

  21. 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)

  22. Problem 9 0.02x + 0.11(3 – x) = 0.24 1 kg of 2% HCl solution 2 kg of 11% HCl solution

  23. 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)

  24. 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

  25. 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

  26. Problem 15 2 + x = 0.08(100 + x) x  6.5 mL of pure acetic acid

  27. Problem 16 9 = 0.001(300 + x) x = 8700 liters of water

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