tonebeeze wrote:

If x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year, the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $56. If $2,000 is the total amount invested, how much is invested at 8 percent?

a. $280

b. $800

c. $892

d. $1108

e. $1200

We are given that x = the number of dollars invested at 10% and y = the number of dollars invested at 8%. We are also given that the total amount invested is $2000, and thus we have x + y = 2000 or x = 2000 - y. Finally, we are given that the annual income from the 10% investment exceeds the annual income from the 8% investment by $56, so we can create the following equation to determine the value of y:

0.10x = 0.08y + 56

0.10(2000 - y) = 0.08y + 56

200 - 0.10y = 0.08y + 56

144 = 0.18y

14400 = 18y

y = 14400/18 = 800

Answer: B

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