A trust account manager has $220,000 to be invested. The investment choices have current yields of 8%, 7%, and 10%. Suppose that the investment goal is to earn interest of $16,300, and risk factors make it prudent to invest some money in all three investments. What is the maximum amount that will be invested at 7%, and in this case how much will be invested at the other rates?
A)$220,000 at 7%, $0 at 8%, $0 at 10%
B)$180,000 at 7%, $10,000 at 8%, $30,000 at 10%
C)$190,000 at 7%, $0 at 8%, $30,000 at 10%
D)$30,000 at 7%, $0 at 8%, $190,000 at 10%
E)$190,000 at 7%, $20,000 at 8%, $10,000 at 10%

Question

A trust account manager has $220,000 to be invested. The investment choices have current yields of 8%, 7%, and 10%. Suppose that the investment goal is to earn interest of $16,300, and risk factors make it prudent to invest some money in all three investments. What is the maximum amount that will be invested at 7%, and in this case how much will be invested at the other rates? ​
A)$220,000 at 7%, $0 at 8%, $0 at 10%
B)$180,000 at 7%, $10,000 at 8%, $30,000 at 10%
C)$190,000 at 7%, $0 at 8%, $30,000 at 10%
D)$30,000 at 7%, $0 at 8%, $190,000 at 10%
E)$190,000 at 7%, $20,000 at 8%, $10,000 at 10%

Quizplus · Accepted Answer

Let x be the amount invested at 8%, y be the amount invested at 7%, and z be the amount invested at 10%. Then we have the following system of equations:

x + y + z = 220,000 (total amount invested)
0.08x + 0.07y + 0.10z = 16,300 (total interest earned)

We want to maximize the amount invested at 7%, which is y. To do this, we can use substitution to eliminate x and z:

z = 220,000 - x - y
0.08x + 0.07y + 0.10(220,000 - x - y) = 16,300
0.02x - 0.03y = 3,700
2x - 3y = 370,000

Solving this system with the equation x + y + z = 220,000, we get:

x = 0, y = 190,000, z = 30,000

Therefore, the maximum amount invested at 7% is $190,000, and the amounts invested at 8% and 10% are $0 and $30,000, respectively. Therefore, the best choice is C.

A Trust Account Manager Has $220,000 to Be Invested