Question 80008: I just need to know how to right this into a math problem. Here it is:
John now has two investments that produce a $150 income each month. If $1,000 more is invested at 9% than at 10% per year, how much was invested at each percent? There were 12 months in that year.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! John now has two investments that produce a $150 income each month. If $1,000 more is invested at 9% than at 10% per year, how much was invested at each percent?
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Amt invested at 9% = x; amt of monthly interest on this is (0.09/12) x dollars
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Amt invested at 10% = x+1000 ; amt of monthly int. on this is (0.10/12)(1000+x)
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EQUATION:
int + int = 150 dollars
0.0075x + 8.333 + 0.008333x=150
0.0158333x=141.667
x=$8947.41(amt. invested at 9%)
10000-8947.41=$1052.59 (amt. invested at 10%)
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Cheers,
Stan H.
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