INVESTMENT and MIXTURE PROBLEMS

Sample Problems

INVESTMENT PROBLEMS

1. Carlos invested a sum of money at 7%. He invested a second sum, $200 more than the first sum, at 8%. The annual income from the two investments is $346. Find the amount he invested at each rate.

The formula we use is

PR = I

(Principal) X (Rate) = Interest

P	R	I
x	.07	.07x
x + 200	.08	.08(x + 200)

Step 1 - Write the equation	.07x + .08(x + 200) = 346

Step 2 - Multiply both sides of the equation by 100 to get rid of the decimals	100[.07x + .08(x + 200)] = 100(346) 100(.07x) + 100(.08)(x + 200) = 100(346) 7x + 8(x + 200) = 34600

Step 3 - Use the distributive property to multiply 8(x + 200)	7x + 8x + 1600 = 34600

Step 4 - Combine like terms	15x + 1600 = 34600

Step 5 - Subtract 1600 from both sides	15x = 33000

Step 6 - Divide both sides of the equation by 15 to get the solution for x	x = 2200

Therefore the solution to the problem is:

$2,200 was invested at 7% and ($2,200 + $200) = $2,400 was invested at 8%

2. Danny invested $11,000. Part of his money is invested in bonds which yield 8% and the remainder is invested in bonds which yield 10%. His total annual income from these bonds is $1,020. Find the amount he has invested in each kind of bond.

P	R	I
x	.08	.08x
11000 ‑ x	.10	.10(11000 ‑ x)

Step 1 - Write the equation	.08x + .10(11000 ‑ x) = 1020

Step 2 - Multiply both sides of the equation by 100 to get rid of the decimals	100[.08x + .10(11000 ‑ x)] = 100(1020) 100(.08x) + 100(.10)(11000 ‑ x) = 100(1020) 8x + 10(11000 ‑ x) = 102000

Step 3 - Use the distributive property to multiply 10(11000 ‑ x)	8x +110000 ‑ 10x = 102000

Step 4 - Combine like terms	-2x + 110000 = 102000

Step 5 - Subtract 110000 from both sides	-2x = -8000

Step 6 - Divide both sides of the equation by -2 to get the solution for x	x = 4000

Therefore the solution to the problem is:

$4,000 was invested at 8% and ($11000 ‑ $4000) = $7,000 was invested at 10%

MIXTURE PROBLEMS

3. A chemist has one solution that is 14% salt and another solution which is 18% salt. How many ounces of each must be used to produce 60 ounces that is 15% salt?

The formula we use is

A X % = T

(Amount) X % = Total amount of each substance

A

%

T

x

.14

.14x

60 ‑ x

.18

.18(60 ‑ x)

60

.15

.15(60)

Step 1 - Write the equation

.14x + .18(60 ‑ x) = .15(60)

Step 2 - Multiply both sides of the equation by 100 to get rid
              of the decimals

100[.14x + .18(60 ‑ x)] = 100(.15)(60)
100(.14x) + 100(.18)(60 ‑ x) = 15(60)
14x + 18(60 ‑ x) = 900

Step 3 - Use the distributive property to multiply

18(60 ‑ x)

14x + 1080 ‑ 18x = 900

Step 4 - Combine like terms

-4x + 1080 = 900

Step 5 - Subtract 900 from both sides

-4x = -180

Step 6 - Divide both sides of the equation by -2 to get the
              solution for x

x = 45

Therefore the solution to the problem is:

45 ounces of the 14% solution and (60 ‑ 45) = 15 ounces of the 18% solution

4.   How many pounds of pure salt must be added to 60 pounds of a 8% solution of salt and water to increase it to a 20% solution?

Note: When a substance is pure we write 100% for the percentage of its content.

A

%

T

60

.08

.08(60)

x

1.00

1.00x

(60 + x)

.20

.20(60 + x)

Step 1 - Write the equation

.08(60) + 1.00x = .20(60 + x)

Step 2 - Multiply both sides of the equation by 100 to get rid
              of the decimals

100[.08(60) + 1.00x] = 100[(.20)(60 + x)]
100(.08)(60) + 100(1.00x) = 100(.20)(60 + x)
8(60) + 100x = 20(60 + x)

Step 3 - Use the distributive property to multiply

20(60 + x)

480 + 100x = 1200 + 20x

Step 4 - Subtract 20x from both sides

480 + 80x = 1200

Step 5 - Subtract 480 from both sides

              Note: Steps 4 and 5 can be combined into one step

80x = 720

Step 6 - Divide both sides of the equation by 80 to get the
              solution for x

x = 9

Therefore the solution to the problem is that 9 pounds of pure salt must be added to 60 pounds of a 8% solution of salt and water to increase it to a 20% solution.