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 $3080 dollars. How much did he invest at each rate?

Answer 1

Explanation:

Let x be the amount invested in the 6% account.

Since he invested twice as much in the lower-yielding account, the amount invested in the 10% account is 2x.

We know that the annual interest is $3080 dollars, and the interest is calculated on the invested amount at each rate.

The interest in the 6% account is (6/100) * x = 3/50 * x

The interest in the 10% account is (10/100) * 2x = 1/10 * 2x

The total annual interest is 3/50x + 1/102x = 3/50x + 1/5x = (3+1)/5x = 4/5x

Therefore, we can write the equation:

4/5*x = 3080

To find x, we can divide both sides by 4/5:

x = 3080*5/4

x = $2310

So, the man invested $2310 in the 6% account and $2310*2 = $4620 in the 10% account.