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

is this rsm? lol okay.

x is the amount of money invested at 5%
y is the amount of money invested at 6%
the total amount of money invested is $6000, or x+y
when it says the yearly income is $337.50, it means that the 5% of x plus the 6% of y equals $337.50.

now write the two equations:

x+y =6000

0.05x + 0.06y = 337.50

now, find y for the first equations, which is y= 6000 - x. use this equation to substitute for the second equation.

0.05x + 0.06(6000 -x) = 337.50

blah, blah, blah, solve for x... you should get x = 2250, and y = 3750.

I hate money problems like this, hope it helps.

Answer 2

Anita invested $3750 at 6%.

Anita invested $2250 at 5%.

Explanation:

Let x be amount of money invested at 5% and y be the amount of money invested at 6%.

We have been given that Anita invested $6000, some at 5% and the rest at 6%. We can represent this information in an equation as:

`x+y=6000...(1)`

The income from the money invested at 5% will be equal to 5% of x (0.05x). The income from the money invested at 6% will be equal to 6% of y (0.06y).

We are also told that the yearly income is $337.50. We can represent this information in an equation as:

`0.05x+0.06y=337.50...(2)`

We will use substitution method to solve our system of equations.

From equation (1) we will get,

`x=6000-y`

Substituting this value in equation (2) we will get,

`0.05(6000-y)+0.06y=337.50`

`300-0.05y+0.06y=337.50`

`300+0.01y=337.50`

`300-300+0.01y=337.50-300`

`0.01y=37.50`

`(0.01y)/(0.01)=(37.50)/(0.01)`

`y=3750`

Therefore, Anita invested $3750 at 6%.

Upon substituting
`y=3750` in equation (1) we will get,

`x+3750=6000`

`x+3750-3750=6000-3750`

`x=2250`

Therefore, Anita invested $2250 at 5%.