Asked by Carol
Bob invested $30,000, part at 10% and part at 1%. If the total interest at the end of the year is $1,560, how much did he invest at 10%?
Let x be the amount invested at 10%. Then, 30,000-x is invested at 1%
0.1 x + 0.01 (30,000 - x) = 1560
Solve that algebraic equation for x.
First step:
0.09 x = 1560 - 300 = 1260
x = ?
Let x be the amount invested at 10%. Then, 30,000-x is invested at 1%
0.1 x + 0.01 (30,000 - x) = 1560
Solve that algebraic equation for x.
First step:
0.09 x = 1560 - 300 = 1260
x = ?
Answers
Answered by
Bob
14000
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