Question

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?

Answer 1

40413.13, with the 13 being repetitive, so it could be rounded to .131 or .1

Answer 2

$3100 is invested at 9%

$4900 is invested at 11%

Explanation:

Let's take "x" be the amount invested at 9%.

(x + 1800) is invested in another account at 11%.

The interest amount earned by the two accounts is $818.

Here we can use the simple interest formula and find the amount invested in each account.

Simple interest (I) =
`(PNR)/(100)`, where P- is the principal , N is the number of years and R is the interest rate.

Simple interest =
`(x*1*9)/(100) + ((x+1800)*1*11)/(100)`

0.09x + 0.11(x+1800) = 818

Now we have to simplify and find the value of x .

Use the distributive property and simplify the second term.

0.09x + 0.11x + 198 = 818

0.2x + 198 = 818

0.2x =818 - 198

0.2x = 620

x = 620/0.2

x = 3100.

So $3100 is invested at 9%

x + 1800 = 3100 + 1800

= $4900

$4900 is invested at 11%