Question

Larry Mitchell invested part of his $11,000 advance at 3% annual simple interest and the rest at 9% annual simple interest. If his total yearly interest from both accounts was $750​,

find the amount invested at each rate



The amount invested at 3% is $

​

The amount invested at 9% is $

Answer 1

• at 3%: $4000
• at 9%: $7000

Explanation:

The answer to this problem can be found by writing an equation for the total interest in terms of the amounts invested at each rate.

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setup

Let x represent the amount invested at 9% (the higher rate). Then (11000-x) is the amount invested at 3%, and the total interest earned is ...

0.09x +0.03(11000 -x) = 750

solution

0.06x = 420 . . . . . . . subtract 330 and simplify

x = 7000 . . . . . . . divide by 0.06 . . . amount invested at 9%

11000 -7000 = 4000 . . . amount invested at 3%

Larry invested $4000 at 3%, and $7000 at 9%.

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Additional comment

In "mixture" problems of this type, it generally works well to let the variable represent the amount of the highest-value contributor. That way, the equations end up with positive coefficients, tending to reduce errors.