Question

Things did not go quite as planned. You invested $8000, part of it in stock that paid 12% annual interest. However, the rest of the money suffered a 5% loss. If the total annual income from both investments was $620, how much was invested at each rate

Answer 1

Answer: $ 6000 and $ 2000

Explanation: Let X represent the investment at 12%.

If x was invested at 12%

Then 8000-x suffered a 5% loss.

0.12x -0.05(8000-x) = 620

0.12x - 400 + 0.05x = 620

0.17x = 1020

Solve for x

x = 1020/ 0.17

x = $6000

Hence $6000 was invested at 12% and $2000 was the loss at 5%

Answer 2

Answer:

$6,000 was invested at 12% interest rate, while $2,000 was invested at a loss of 5%.

Explanation:

Let P represents the the amount invested at 12%, this implies that the amount money that suffered 5% loss is $8,000-P. We can therefore have the following:

620 = 0.12P - 0.05(8000 - P)

620 = 0.12P - 400 + 0.05P

620 + 400 = 0.12P + 0.05P

1,020 = 0.17P

P = 1020/0.17

P = $6,000

Amount invested on the one that suffered loss = $8,000 - $6,000 = $2,000.

Therefore, $6,000 was invested at 12% interest rate, while $2,000 was invested at a loss of 5%.