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:

$\frac{5}{100}x+\frac{6}{100}(6000-x)=337.5$

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