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$10,807 is invested, part at 12% and the rest at 5% . If the interest earned from the amount invested at 12% exceeds the interest earned from the amount invested at 5% by $580.63 , how much is invested at each rate?

Sagot :

Sqdancefanidn

Answer:

  • $6594 at 12%
  • 4213 at 5%

Step-by-step explanation:

You want the amounts of $10807 invested at 12% and 5% if the interest from the former exceeds that from the latter by $580.63.

Setup

We can let x represent the amount invested at 12%. Then (10807-x) will be the amount invested at 5%. Interest earned is the product of the interest rate and the invested amount, so the difference in interest earned is ...

  0.12x -0.05(10807 -x) = 580.63

Solution

  0.17x -540.35 = 580.63 . . . . . . . . . . simplify

  0.17x = 1120.98 . . . . . . . . . . . . . . add 540.35

  x = 6594 . . . . . . . . . . . . . . . . . divide by 0.17

The remaining amount is invested at 5%:

  10807 -6594 = 4213

$6594 is invested at 12%, and $4213 is invested at 5%.

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