Calculus Systems

Question

A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is $4290 dollars. How much did he invest at each rate?

Amount invested at 6% equals =

Amount invested at 10% equals =

Answer 1

Let x equal how much was invested at 10% interest.

Simple interest uses the formula i * invested amount, so the overall equation is:

.06(2x) + .1x = 4290

Then you get .22x = 4290.

x = 19500 

2x = 39000

$19,500 was invested at 10% interest

$39,000 was invested at 6% interest

Answer 2

.06(2x) + .1x = 4290

x = invested at 10%, 10%=.10

.12x + .10x = 4290

(.12+.10)x = 4290

.22x = 4290

x = 4290/.22 = $19,500 at 10%

twice as much at 6% = $39,000

grand total invested = 19500+39000 = $58,500

4290/58500 = about 7.3 % average return on both investments

7.3-6 = 1.3

10-7.3 = 2.7

2.7/1.3 = about 2 = 19550/39100

rough check on the answer, which looks perfect