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Interest Problem

by tonebeeze » Mon May 16, 2011 3:40 pm
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

OA = B
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by pankajks2010 » Mon May 16, 2011 6:55 pm
as per the question:

10x/100-8y/100=56; 10x-8y=5600

Now, let the amount invested at 8% be a, then the amount invested at 10%=2000-a. Now substituting these values in the equation above;

10(2000-a)-8(a)=5600; a=800
Thus, B
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