Question

You invested $27,000 in two accounts paying 8% and 9% annual interest, respectively. If the total interest eamed for the year was $2320, how much was invested at each rate?

Answer 1

$11000 was invested at 8%

$16000 was invested at 9%

Explanation:

We will need a system of equations to determine the amount invested at each rate.

• Let x represent the amount invested at 8%.
• Let y represent the amount invested at 9%.
• Let 0.08x represent the interest earned at 8%.
• Let 0.09y represent the interest earned at 9%.

We know that:

amount invested at 8% + amount invested at 9% = 27000

Thus, x + y = 27000 and 0.08x + 0.09y = 2320.

We further know that:

interest earned at 8% + interest earned at 9% = 2320

Thus, 0.08x + 0.09y = 2320.

We can solve using substitution by isolating x in x + y = 27000. Then, we'll plug in the result for x in 0.08x + 0.09y = 2320 to find y:

Isolating x in x + y = 27000:

(x + y = 27000) - y

x = -y + 27000

Substituting x = -y + 27000 for x in 0.08x + 0.09y = 2320 to find y:

0.08(-y + 27000) + 0.09y = 2320

-0.08y + 2160 + 0.09y = 2320

0.01y + 2160 = 2320

0.01y = 160

16000 = y

Now we plug in 16000 for y in x + y = 27000 to find x:

x + 16000 = 27000

x = 11000

Thus, you invested $11000 at 8% and $16000 at 9%.

Check work:

We can check our work by plugging in 11000 for x and 16000 for y in both equations in our system. If we get the same answer on both sides, our answers are correct.

Checking solutions in x + y = 27000:

11000 + 16000 = 27000

27000 = 27000

Checking solutions in 0.08x + 0.09y = 2320:

0.08(11000) + 0.09(16000) = 2320

880 + 1440 = 2320

2320 = 2320

Thus, our answers are correct.