Question

PLEASE HELP!!!!!!!

the annual interest rate on a $6000 investment exceeds the interest earned on a $4000 investment by $204. The $6000 is invested at a 0.4% higher rate of interest than the $4000. What is the interest rate of each investment?

Answer 1

6000(n%) = 4000(m%) + 204

n% = m% + .4

Answer 2

6000(m%+.4%) = 4000(m%) + 204

6000(m%) + 6000(.4%) = 4000(m%) + 204

6000(m%) -4000(m%) = 205 - 6000(.4%)

m%(6000-4000) = 205 - 6000(.4%)

m% = 205 - 6000(.004)
---------------
2000

Answer 3

m% = .0905

m%(100) = .0905(100)

m = 9.05%

n = 9.05% + .4% = 9.45%

dbl chk

Answer 4

i see a mistake, just a slight one tho;

i mistyped 204 into a 205 ....

Answer 5

m = 9%
n = 9.4% then ... sooo close :)

Answer 6

that checks out :)