Question 1067918: $14,188 is invested, part at 10% and the rest at 6%. If the interest earned from the amount invested at 10% exceeds the interest earned from the amount invested at 6%6% by $1175.92, how much is invested at each rate? (Round to two decimal places if necessary.)
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! Solve by elimination:
Let x be the amount invested at 10% and y the amount invested at 0.06%
x+y = 14,188 multiply both sides of this equation times 0.06
0.06x+0.06y = 851.28 (equation 1)
and the problem tells us that
0.10x-0.06y = 1,175.92 (equation 2)
Add equations 1 and 2:
0.06x+0.06y = 851.28
+
0.10x-0.06y = 1,175.92
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0.16x-0y = 2,027.20
= 0.16x = 2,027.2
x = 12,670 this is the amount invested at 10%
14,188-12,670 = 1,518 this is the amount invested at 6%
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Check:
12,670(0.10) = 1,518(0.06)+1,175.92
1,267 = 1267 Correct.
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