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Last year, Mauricio had US$30.000 to invest. He invested some of it in an account that paid 5% simple interest per year, and he invested the rest in an account that paid 9% simple interest per year. After one year, he received a total of US$2380 in interest. How much did he invest in each account?

Question

Last year, Mauricio had US30.000toinvest.Heinvestedsomeofitinanaccountthatpaid530.000 to invest. He invested some of it in an account that paid 5% simple interest per year, and he invested the rest in an account that paid 9% simple interest per year. After one year, he received a total of US2380 in interest. How much did he invest in each account?

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Solution

Let's denote the amount Mauricio invested at 5% as x x and the amount he invested at 9% as y y .

We know the following:

  1. The total amount invested is $30,000. x+y=30,000 x + y = 30,000

  2. The total interest received after one year is $2,380. 0.05x+0.09y=2,380 0.05x + 0.09y = 2,380

We now have a system of linear equations: x+y=30,000 x + y = 30,000 0.05x+0.09y=2,380 0.05x + 0.09y = 2,380

First, solve the first equation for y y : y=30,000−x y = 30,000 - x

Next, substitute y y in the second equation: 0.05x+0.09(30,000−x)=2,380 0.05x + 0.09(30,000 - x) = 2,380

Distribute the 0.09: 0.05x+2,700−0.09x=2,380 0.05x + 2,700 - 0.09x = 2,380

Combine like terms: −0.04x+2,700=2,380 -0.04x + 2,700 = 2,380

Subtract 2,700 from both sides: −0.04x=−320 -0.04x = -320

Divide both sides by -0.04: x=8,000 x = 8,000

Now, substitute x x back into the equation y=30,000−x y = 30,000 - x : y=30,000−8,000 y = 30,000 - 8,000 y=22,000 y = 22,000

Therefore, Mauricio invested 8,000intheaccountthatpaid58,000 in the account that paid 5% interest and 22,000 in the account that paid 9% interest.

This problem has been solved

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