You invested 19000 in two accounts paying 3 and 4% annual interest respectively. If the total earned for the year was $740, how much was invested at each rate

Answer 1

Hi Stefan, let n represent the amount invested at 3%. Then the amount invested at 4% would be (19000 - n).

Equation to solve:

.03 * n + .04 * (19000 - n) = 740

.03n + .04 * 19000 - .04n = 740

.03n + 760 - .04n = 740

Combine terms:

-.01n = -20

n = 2000

The amount of money invested at 3% was $2000.

The amount of money invested at 4% was $19000 - $2000, or 17000

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