Last year Kareem had $20,000

Question

to invest. He invested some of it in an account that paid
8% simple interest per year, and he invested the rest of it in an account that paid
7% simple interest per year. After one year, he received a total of $1540
in interest. How much did he invest in each account?

Answer 1

Let \( x \) be the amount Kareem invested at 8% interest. Then, the amount he invested at 7% interest is \( 20000 - x \).

The interest earned from the investment at 8% is given by: \[ 0.08x \]

The interest earned from the investment at 7% is given by: \[ 0.07(20000 - x) \]

According to the problem, the total interest earned from both investments after one year is $1540. Therefore, we can set up the equation: \[ 0.08x + 0.07(20000 - x) = 1540 \]

Now, let's simplify and solve for \( x \): \[ 0.08x + 1400 - 0.07x = 1540 \] \[ (0.08x - 0.07x) + 1400 = 1540 \] \[ 0.01x + 1400 = 1540 \] \[ 0.01x = 1540 - 1400 \] \[ 0.01x = 140 \] \[ x = \frac{140}{0.01} = 14000 \]

So, Kareem invested \( x = 14000 \) dollars at 8% interest. To find the amount invested at 7%, we calculate: \[ 20000 - x = 20000 - 14000 = 6000 \]

Thus, Kareem invested:

• $14,000 at 8% interest
• $6,000 at 7% interest

To verify, we can check the total interest:

• Interest from the 8% account: \( 0.08 \times 14000 = 1120 \)
• Interest from the 7% account: \( 0.07 \times 6000 = 420 \)
• Total interest: \( 1120 + 420 = 1540 \), which matches the given information.

Therefore, the solution is correct.