Problem solving and business application

Question

Things did not go quite as planned. You invested $20,000, part of it in a stock that paid 12% annual interest. However, the rest of the money suffer a 5% loss. If the total annual income from both investments was $1720, how much was invested at each rate?

Answer 1

.12x -.05 (20-x) = 1.72

.12x +.05x = 1.72+1=2.72

x = 2.72/.17 = 16 = $16,000 invested at 12%

20-x =20-16 =4 = $4000 invested at 5%

4 times as much invested at 12%

check the answer.

1720/20000= 172/2000 = .086 = 8.6% average interest rate

12-8.6 = 3.4

8.6- -5 = 13.6

13.6/3.4 = 4 means 4 times more invested at 12% than at 5.%

.12(16000)-.05(4000) = 1,920 - 200 = 1,720 total interest

Answer 2

Let "x" be the amount you invest in the first stock. Let "y" be the amount you invest in the second stock.

Then x + y = 20000

Also, since the first stock earned 12%, then the amount you earned from it is 0.12x and the amount you lost from the second was 0,05y. We are told the total annual income was $1720 so 0.12x - 0.05y = 1720

x + y = 20000 or x = 20000 - y

0.12x - 0.05y = 1720 and substituting "20000 - y" in for "x" we get:

0.12(20000 - y) - 0.05y = 1720

2400 - 0.12y - 0.05y = 1720

-0.17y = -680

y = 4000

Since x = 20000 - y then x = 20000 - 4000 = 16000

So $16000 was invested in the first stock (that gained 12%) and $4000 was invested in the second (that lost 5%)