Given:
John invest $5,000 in one account and $8,000 in an account paying 4% more in interest.
Total interest in one year = $1230
To find:
The rates of the investments.
Solution:
Let x% be the rate of interest on investment $5000.
Then, (x+4)% is the rate of interest on investment $8000.
Total interest is $1230. So,
[tex]5000\times \dfrac{x}{100}+8000\times \dfrac{x+4}{100}=1230[/tex]
[tex]50x+80(x+4)=1230[/tex]
[tex]50x+80x+320=1230[/tex]
[tex]130x=1230-320[/tex]
[tex]130x=910[/tex]
Divide both sides by 130.
[tex]x=\dfrac{910}{130}[/tex]
[tex]x=7[/tex]
Now,
[tex]x\%=7\%[/tex]
[tex](x+4)\%=(7+4)\%=11\%[/tex]
Therefore, $5,000 is invested at 7% and $8,000 is invested at 11%.
