You invested $5000 between two accounts paying 4% and 9% annual interest, respectively. If the total interest earned for theyear was $350, how much was invested at each rate?$was invested at 4% andwas invested at 9%.
Accepted Solution
A:
Answer:Part 1) The amount invested at 4% was $2,000Part 2) The amount invested at 9% was $3,000Step-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
Letx------> 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]
thereforeThe amount invested at 4% was $2,000The amount invested at 9% was $3,000