Question

You invested $6,000 between two accounts paying 5% and 6% annual interest, respectively. If the total interest earned for the year was $320, how much was invested at each rate?

a) $3,200 at 5% and $2,800 at 6%
b) $2,800 at 5% and $3,200 at 6%
c) $2,400 at 5% and $3,600 at 6%
d) $3,600 at 5% and $2,400 at 6%

Answer 1

To solve this problem, set up a system of equations where x represents the amount invested at 5% and (6000 - x) represents the amount invested at 6%. Simplify and solve the equation to find that $2400 was invested at 5% and $3600 was invested at 6%.

Step-by-step explanation:

To solve this problem, we can set up a system of equations. Let's say you invest x dollars at 5% and (6000 - x) dollars at 6%. The interest earned on the first account is 0.05x, and the interest earned on the second account is 0.06(6000 - x). The total interest earned is the sum of these two amounts, which is $320. We can write this as the equation:

0.05x + 0.06(6000 - x) = 320

Simplifying and solving the equation, we find that x = 2400. Therefore, $2400 was invested at 5% and $3600 was invested at 6%.