Ramon invested $2,400 into two accounts. One account paid 3% interest and the other paid 6% interest. He earned 5% interest on the total investment. How much money did he put in each account?
Accepted Solution
A:
Answer:In the account that paid 3% Ramon put [tex]\$800[/tex]
In the account that paid 6% Ramon put [tex]\$1,600[/tex]
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
[tex]P(rt)=Pa(rat)+Pb(rbt)[/tex]
in this problem we have
[tex]t=t\ years\\ P=\$2,400\\ Pa=\$x\\ Pb=\$(2,400-x)\\r=0.05\\ra=0.03\\rb=0.06[/tex]
substitute[tex]2,400(0.05t)=x(0.03t)+(2,400-x)(0.06t)[/tex]
solver for xSimplify[tex]2,400(0.05)=x(0.03)+(2,400-x)(0.06)[/tex]
[tex]120=0.03x+144-0.06x[/tex]
[tex]0.03x=24[/tex]
[tex]x=\$800[/tex]
thereforeIn the account that paid 3% Ramon put [tex]\$800[/tex]
In the account that paid 6% Ramon put [tex]\$2,400-\$800=\$1,600[/tex]