Question

$5400 is invested, part of it at 12% and part of it at 6%. For a certain year, the total yield is $504.00. How much was invested at each rate?

Answer 1

Let x represent amount invested at 12% and y represent amount invested at 6%.

We have been given that $5400 is invested, part of it at 12% and part of it at 6%.

We can represent this information in an equation as:

`x+y=5400...(1)`

`y=5400-x...(1)`

We are also told that for a certain year, the total yield is $504.00. This means that amount of interest for one year was $504.00. We can represent this information in an equation as:

`0.12x+0.06y=504...(2)`

Upon substituting equation (1) in equation (2), we will get:

`0.12x+0.06(5400-x)=504`

`0.12x+324-0.06x=504`

`0.06x+324=504`

`0.06x+324-324=504-324`

`0.06x=180`

`(0.06x)/(0.06)=(180)/(0.06)`

`x=3000`

Therefore, $3000 was invested at 12%.

Upon substituting
`x=3000` in equation (1), we will get:

`y=5400-3000=2400`

Therefore, $2400 was invested at 6%.