Complete question :
Larry Mitchell invested part of his $32000 advance at 4% annual simple interest and the rest at 9% annual simple interest. If his total yearly interest from both accounts was 2430, find the amount invested at each rate. The amount invested at is $ nothing. The amount invested at is $ nothing.
Answer:
Investment at 4% = $9000
Investment at 9% = $23,000
Explanation:
Given that :
Total investment = 32000
Let;
amount invested at 4% = a
amount invested at 9% = 32000 - a
Total interest earned = 2430
Simple interest = principal * rate * time
Investment at 4%:
(a * 0.04) + [(32000 - a) * 0.09] = 2430
0.04a + 2880 - 0.09a= 2430
-0.05a + 2880 = 2430
-0.05a = 2430 - 2880
-0.05a = - 450
a = 9000
Hence,
Investment at 4% = $9000
Investment at 9% = $32,000 - $9000 = $23,000