Maria invested $2,400 into two accounts. one account paid 4% interest and the other paid 6% interest. she earned 5.5% interest on the total investment. how much money did she put in each account?

Step-by-step explanation:

Let x represent the amount invested at the higher rate (6%). Then the amount invested at the lower rate is (2400-x) and the total interest earned is ...

  6%·x + 4%·(2400-x) = 5.5%·2400

Dividing by % and rearranging, we have ...

  x(6 -4) = 2400(5.5 -4)

  x = 2400·(5.5 -4)/(6 -4) = 2400(1.5/2) = 2400·0.75

  x = 1800 . . . . . . . . amount invested at 6%

  2400-x = 600 . . . amount invested at 4%

Maria put $1800 in the 6% account and $600 in the 4% account.

Comment on the solution

You will note that the proportion of the investment that went to the higher interest rate account is (5.5-4)/(6-4). This is the ratio of the mixed interest rate less the lower rate to the difference of account rates. This will be the generic solution to mixture problems, so is worthy of note for that reason.

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step-by-step explanation:

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