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 $7348 dollars. How much did he invest at each rate?

Answer 1

Answer: At 6% = $66,800

               At 10% = $34,400

Step-by-step explanation:

Let S be his savings. He divided S into x that goes at 6% and y at 10%

As he put twiced as much at the lower-yielding one, x = 2y

Simple interest is given by: Amount x interest x time

This way I₁ = x . 0.06 . 1 = 0.06x

               I₂ = y. 0.10 . 1 = 0.10y

As x = 2y, I₁ = = 0.06 . 2y = 0.12y

As we know that the annual interest was $7348

I₁ + I₂ = 7348

0.12y + 0.10y = 7348

0.22y = 7348

y = 7348/0.22

y = 33,400

As x = 2y, x = 2.33,400 = 66,800

This was he invested $33,400 and $66,800