Question

Ms. Wilson invested $30,000 in two accounts, one yielding 8% interest and the

other yielding 9%. If she received a total of $2,590 in interest at the end of the
year, how much did she invest in each account?
The amount invested at 8% was $
The amount invested at 9% was $

Answer 1

The amount invested at 8% was $11,000

The amount invested at 9% was $19,000

Explanation:

Let the variable x represent the amount in $ invested at 8% and let y be the amount in $ invested at 9%

Total amount invested:
x + y = 30000 [1]

8% = 8/100 = 0.08

9% = 9/100 = 0.09

Interest at 8% on $x = 0.08x

Interest at 9% on $x = 0.09x

Total Interest :

0.08x + 0.09y = 2590 [2]

Using equations [1] and [2] we can solve for x and y

We have

x + y = 30000 [1]
0.08x + 0.09y = 2590 [2]

Multiply equation 1 by 0.08 to get
0.08x + 0.08y = 0.08(30,000)

0.08x + 0.08y = 2,400 [3]

Subtract [3] from [2] :

0.08x + 0.09y = 2590
-
0.08x + 0.08y = 2400
----------------------------------
0x + 0.01y = 190

Divide both sides by 0.01
0.01y/0.01 = 190/0.01

y = = 19,000

Use [1] to get value of x

x + y = 30,000

x + 19,000 = 30,000

x = 30,000 - 19,000

x = 11,000