$17,200 is invested, part at 6% and the rest at 2%. If the interest earned from the amount invested at 6% exceeds the interest earned from the amount invested at 2% by $441.00, how much is invested at each rate? (Round to two decimal places if necessary.)

Solution:

Let x = amount invested at 6% and

let y = amount invested at 2%.

We can set up some equations that describe x and y:

"$17,200 is invested total" means

x + y = $17,200

 "The interest earned from the amount invested at 6% exceeds the interest earned from the amount invested at 2% by $441.00" means

0.06x = 0.02y + $441.00

Solve for x in the first equation to get x = 17,200 - y, then plug that into the second equation and solve for y:

0.05(17,200 - y) = 0.02y + 441.00

860 - 0.06y = 0.02y + 865.35

5.35 - 0.09y = 0.02y

154.80 = 0.11 y

1407 = y

So, $1407 was invested at 2%. Plug y = 1407 into the first equation and solve for x:

x + 1407 = 17,200

x = 15,793

So, $15,793 was invested at 6%.

the investment was $14,174.196

explanation:

a lump sum means that the father of the example made one deposit and leave it be.

we need to know the value of a capital that generated at 4.05% per year during 28 years the sum of $48,613.24

now clearing the formula to solve for capital we got:

and we reach the answer investment = $14,174.20