Larry Invested part of his $12,000 advance at 6% annual simp. Interest and the rest at 3% annual simple interest.If his total yearly interest from both accounts

Question

Larry Invested part of his $12,000 advance at 6% annual simp. Interest and the rest at 3% annual simple interest.If his total yearly interest from both accounts was $540, find the amount invested at each rate.

Answer 1

For simple interest  I = Prt

 

If we invest part of the 12,000 at 6% interest then

one year of interest on that part would be I = .06x

with x representing the amount invested at 6%

 

The remainder of the money invested at 3% would

be represented by (12000-x) and the interest on

that amount for a year would be I = .03(12000-x)

 

We want these two interest amounts to total to $540

so add them together and set the sum equal to 540

 

.06x + .03(12000-x) = 540

.06x + 360 - .03x = 540

.03x=180

x = 180/.03 = $6000

 

We invested $6000 at 6%

Leaving $6000 invested at 3%

 

Check:  .06(6000)+.03(6000) = 540

            360 + 180 = 540

            540=540