Easy Solutions to Investment and Mixture Problems
Learn how to solve investment and mixture problems using simple interest formulas and mixture calculation methods with detailed examples.
Easy Solutions to Investment and Mixture Problems
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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?
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 • What is the total amount invested? • $30,000
Investment • Solve, and round your answer to the nearest cent. • x = $27,333.33 (invested at 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?
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?
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 Problems • quantity • percentage (strength) = the PART described by the percentage in the units described by the quantity
Mixture Problems • 45 lbs. of 80% iodine solution • 45 • 0.80 = x • 36 lbs. of iodine
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
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?
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?
Section 2.6 • pp. 67-69
Problem 1 • Prt = I • (148)(.06)(1) = I • $8.88 = I
Problems 2-3 • Quantity • Strength • = 50(.03) • = 1.5 gallons • 12 30 = Strength • 0.4 = Strength • 40% Strength
Problem 4 • 120 + 150 = x • x = $270 P • r • t = I 3000 .04 1 120 2500 .06 1 150
Problem 5 1.1 + 0.42 = x x = 1.52 gallons
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 0.02x + 0.11(3 – x) = 0.24 1 kg of 2% HCl solution 2 kg of 11% HCl solution
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 • .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 • 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 2 + x = 0.08(100 + x) x 6.5 mL of pure acetic acid
Problem 16 9 = 0.001(300 + x) x = 8700 liters of water