Question

A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is $6600 dollars. How much did he invest at each rate?

Answer 1

Let x represent amount invested in the higher-yielding account.

We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be $2x$.

We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.

$I=Prt$, where,

I = Amount of interest,

P = Principal amount,

r = Annual interest rate in decimal form,

t = Time in years.

We are told that interest rates are 6% and 10%.

$6\%=\frac{6}{100}=0.06$

$10\%=\frac{10}{100}=0.10$

Amount of interest earned from lower-yielding account: $2x(0.06)=0.12x$.

Amount of interest earned from higher-yielding account: $x(0.10)=0.10x$.

$0.12x+0.10x=6600$

Let us solve for x.

$0.22x=6600$

$\frac{0.22x}{0.22}=\frac{6600}{0.22}$

$x=30,000$

Therefore, the man invested $30,000 at 10%.

Amount invested in the lower-yielding account would be $2x\Rightarrow 2(30,000)=60,000$.

Therefore, the man invested $60,000 at 6%.