Question

Anita invested $6000, some at 5% and the rest at 6%. If the yearly income is $337.50, what is the amount invested at each rate?

Answer 1

Let the amount invested with 5% interest be x

Therefore, the amount invested with 6% interest will be (6000-x)

It is given that the total interest earned yearly is $337.5. Thus, the equation of interest will be:

`(5)/(100)x+(6)/(100)(6000-x)=337.5`

Multiplying both sides by 100 we get:

`5x+6(6000-x)=33750`

`5x+36000-6x=33750`

Subtracting both sides by 36000 we get:

`-x=-2250`

`\therefore x=2250`

Thus, the amount invested at 5% is $2250

Therefore, the amount invested at 6% will be $(6000-2250)=$3750