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 $3520. How much did

he invest at each rate?

Short answer

For the given investment problem the man invested $32,000 at a rate of 6% and $16,000at a rate of 10%.

Step-by-step solution

Step 1. Given information.

x = the amount invested in the first account.

y = the amount invested in the second account.

Step 2. Write down the concept.

x = 2y

x – 2y = 0 …….(1)

Also the total sum of interest earned would be,

0.06x + 0.1y = 3520 ……….(2)

Step 3. Determining the values

Solving equation (1) and (2) we have,

20$\times$ equation (1) gives 1.2x + 2y = 70400

Now adding this to equation (2), we get

2.2x = 70400

x = 32000

Substituting this in (1)

2y = 32000

Y = 16000