Question

$6300 is invested, part of it at 11% and part of it at 7%. For a certain year, the total yield is $565.00. How much was invested at each rate?

Answer 1

Welcome to Open Study.

Answer 2

Try to find the the interest formula.

Answer 3

Exactly.

Answer 4

But I already have the principle, I'm looking for the amount at each rate

Answer 5

@aum @zepdrix @satellite73

Answer 6

Assume "x" dollars is invested at 11%.
Then (6300-x) dollars is invested at 7%

Interest = P*r*t
where P is the Prinicipal
r is the rate of interest in decimal
t is the number of years invested.

Interest for "x" dollars invested at 11% for 1 year is: x * 0.11 * 1 = 0.11x

Interest for "(6300-x)" dollars invested at 7% for 1 year is: (6300-x) * 0.07 * 1 = 441 - 0.07x

Total interest = 0.11x + 441 - 0.07x = 565
solve for x.

Answer 7

thanks for all your work but it's not looking for interest. I really appreciate it.

Answer 8

Once you solve for x, that is the amount invested at the 11% rate.
(6300-x) is the amount invested at the 7% rate.

Answer 9

Total interest = 0.11x + 441 - 0.07x = 565
(0.11x - 0.07x) = 565 - 441
0.04x = 124
x = 124 / 0.04 = 3100
6300-x = 6300-3100 = 3200

Therefore, $3,100 was invested at the 11% rate and $3,200 was invested at the 7% rate.