Question

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?​

Answer 1

Answer:

In the account that paid 3% Ramon put $\ 800$

In the account that paid 6% Ramon put $\ 1,600$

Step-by-step explanation:

we know that

The simple interest formula is equal to

$I=P(rt)$

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

$P(rt)=Pa(rat)+Pb(rbt)$

in this problem we have

$t=t\ years\\ P=\ 2,400\\ Pa=\ x\\ Pb=\ (2,400-x)\\r=0.05\\ra=0.03\\rb=0.06$

substitute

$2,400(0.05t)=x(0.03t)+(2,400-x)(0.06t)$

solver for x

Simplify

$2,400(0.05)=x(0.03)+(2,400-x)(0.06)$

$120=0.03x+144-0.06x$

$0.03x=24$

$x=\ 800$

therefore

In the account that paid 3% Ramon put $\ 800$

In the account that paid 6% Ramon put $\ 2,400-\ 800=\ 1,600$