Question 1153755: Joey Boy has some money invested at 8% and more invested at 12% the second amount is $25,000 more than the first and the annual simple interest on the second amount exceeds the interest on the first amount by $5000 how much does he have invested at each rate
Found 2 solutions by mananth, josgarithmetic:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! Joey Boy has some money invested at 8% ------x
and more invested at 12% the second amount is x+ $25,000
interest on first amount = 8%x
Interest on second amount = 12%*(x+25000)
interest on second amount - interest on the first amount= $5000
12%*(x+25000) - 8%*x =5000
Multiply the equation by 100
12(x+25000) -8x =500000
12x+ 300000-8x =500000
4x=200000
x= 50000
amount invested at 8% = $50,000
amount invested at 12% = $50,000 +$25000 = $75,000
Check
8% of 50 000 = $ 4000
12% 0f 75,000 = $9000
Difference =$5,000
Answer by josgarithmetic(39630)  (Show Source):
You can put this solution on YOUR website! x, amount at 8%
x+25000, amount at 12%
0.08(x), interest on first amount, the 8%
0.12(x+25000), interest on second amount, the 12%
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annual simple interest on the second amount exceeds the interest on the first amount by $5000
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Solve this equation.
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