Question

will give medal
Mr. Brownwood invests a certain amount of money at 9% interest and $1,800 more than that amount in another account at 11% interest. At the end of one year, he earned a total of $818 in interest. How much money was invested in each account?

A) $3,500 at 9%; $4,300 at 11%
B) $3,400 at 9%; $3,200 at 11%
C) $3,100 at 9%; $4,900 at 11%

Answer 1

The equation that you need to solve this is:
0.09a + 0.11(a + 1800) = 818
Where a is the amount invested a 9%.

Answer 2

so would it be c ?

Answer 3

Yes, C is correct.

Answer 4

k thanks :)

Answer 5

You're welcome :)