xn67
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Ms. Garza invested $50,000 in three different accounts. She invested three times as
much money in an account that paid 8% interest than an account that paid 10% interest.
The third account earned 12% interest. If she earned a total of $5160 in interest in a year,
how much did she invest in each account?

Respuesta :

calculista

Answer:

The amount invested in the account that paid 8%  was $18,000

The amount invested in the account that paid 10%  was $6,000

The amount invested in the account that paid 12%  was $26,000

Step-by-step explanation:

Let

3x -----> the amount invested in the account that paid 8%

x -----> the amount invested in the account that paid 10%

$50,000-4x ----> the amount invested in the account that paid 12%

in this problem we have

[tex]8\%=8/100=0.08\\10\%=10/100=0.10\\12\%=12/100=0.12[/tex]

we know that

[tex]3x(0.08)+x(0.10)+(50,000-4x)(0.12)=5,160[/tex]

Solve for x

[tex]0.24x+0.10x+6,000-0.48x=5,160[/tex]

[tex]0.48x-0.24x-0.10x=6,000-5,160[/tex]

[tex]0.14x=840[/tex]

[tex]x=\$6,000[/tex]

[tex]3x=\$18,000[/tex]

[tex]\$50,000-4x=\$26,000[/tex]

therefore

The amount invested in the account that paid 8%  was $18,000

The amount invested in the account that paid 10%  was $6,000

The amount invested in the account that paid 12%  was $26,000

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