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Randall invests $7200 in two different accounts. The first account paid 11 %, the second account paid 13 % in interest. At the end of the first year he had earned $832 in interest. How much was in each account?

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Answer:

  • 11% -- $5200
  • 13% -- $2000

Explanation:

Let x represent the amount invested in the higher-earning account. Then the investment in the other account is (7200-x), and the total interest is ...

0.13x +0.11(7200 -x) = 832

0.02x = 832 -792 . . . . . . eliminate parentheses, collect terms, subtract 792

x = 40/0.02 = 2000 . . . . divide by the coefficient of x

$2000 was invested at 13%; $5200 was invested at 11%.

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