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

Answer:

Anita invested $3750 at 6%.

Anita invested $2250 at 5%.

Step-by-step 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,

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

$300-0.05y+0.06y=337.50$

$300+0.01y=337.50$

$300-300+0.01y=337.50-300$

$0.01y=37.50$

$\frac{0.01y}{0.01}=\frac{37.50}{0.01}$

$y=3750$

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%.