Question

A sum of $3000 was invested, part at 6% and the rest at 9%. How much was invested at each rate if the interest on each investment is the same?

Answer 1

Amount invested at 6% yield`= 1800`
Amount invested at 9% yield`= 1200`

Explanation:

Amount invested at 6% yield`= x`
Let the yield be `=y`
Then `x xx 6/100 = y`
`(6x)/100 = y` -----------------------(1)

Amount invested at 9% yield `= 3000 - x`
Let the Yield be `= y` [ since yield from both investment is the same)
Then
`(3000-x)9/100=y` -----------------(2)

Substitute `y = (6x)/100` in equation (2)
`(3000-x)9/100=(6x)/100` [multiplying both sides by 100

`(3000-x)9=(6x)`
`27000-9x -6x=0`
`-15x=-27000`
`x=27000/15 = 1800`
Amount invested at 6% yield`= 1800`
Amount invested at 9% yield`= 3000-1800=1200`