You invested $5000 between two accounts paying 4% and 9% annual interest, respectively. If the total interest earned for the
year was $350, how much was invested at each rate?
$
was invested at 4% and
was invested at 9%.

Question

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calculista calculista · Answer 1 · 2019-07-03T18:50:23+02:00

Answer:

Part 1) The amount invested at 4% was $2,000

Part 2) The amount invested at 9% was $3,000

Step-by-step explanation:

we know that

The simple interest formula is equal to

[tex]I=P(rt)[/tex]

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

Let

x------> the amount invested at 4%

5,000-x ----> the amount invested at 9%

in this problem we have

[tex]t=1\ year\\ P=\$5,000\\I=\$350\\r1=0.04\\r2=0.09[/tex]

substitute in the formula above

[tex]350=x(0.04*1)+(5,000-x)(0.09*1)[/tex]

[tex]350=0.04x+450-0.09x[/tex]

[tex]0.05x=450-350[/tex]

[tex]0.05x=100[/tex]

[tex]x=2,000[/tex]

so

[tex]5,000-x=5,000-2,000=3,000[/tex]

therefore

The amount invested at 4% was $2,000

The amount invested at 9% was $3,000