Question

You invested a total of $13,000 at 4 1/2% and 5% simple interest. During one year, the two accounts earned $615. How much did you invest in each account?

A. $5,000 at 4 1/2%, $8,000 at 5%
B. $8,000 at 4 1/2%, $5,000 at 5%
C. $6,500 at 4 1/2%, $6,500 at 5%
D. $7,000 at 4 1/2%, $6,000 at 5%

Answer 1

After setting up a system of equations and using substitution to solve them, it is determined that $7,000 was invested at 4 1/2% and $6,000 at 5% simple interest to earn a total of $615 in one year.

Step-by-step explanation:

The question relates to determining how much was invested in each of two accounts with different simple interest rates, given the total amount invested and the total interest earned in one year. To solve this problem, we need to set up a system of equations based on the information given:

• Let x be the amount invested at 4 1/2% interest.
• Let y be the amount invested at 5% interest.

The total amount invested is $13,000, so we have the equation:

x + y = 13,000

The total interest earned from both accounts is $615, so we also have the equation:

0.045x + 0.05y = 615

Using these two equations, we can solve for x and y. We can either use substitution or elimination. Let's use substitution:

1. From the first equation, y = 13,000 - x.
2. Substitute y in the second equation by (13,000 - x):
   
   0.045x + 0.05(13,000 - x) = 615
3. Expand and solve for x:
   
   0.045x + 650 - 0.05x = 615
   
   -0.005x = -35
   
   x = 7,000
4. Substitute x into y = 13,000 - x:
   
   y = 13,000 - 7,000
   
   y = 6,000

So you have invested $7,000 at 4 1/2% and $6,000 at 5%. Therefore the correct answer is D: $7,000 at 4 1/2%, $6,000 at 5%.