Question

A total of $6000 was invested, part of it at 3% interest and the remainder at 8%. If the total yearly interest amounted to $380, how much was invested at each rate?

Answer 1

$ 2000 is invested at 3 % interest and $ 4000 is invested at 8 % interest

Solution:

Given that, total of $6000 was invested

Let "x" be the amount invested at 3 % interest

Then, (6000 - x) is the amount invested at 8 % interest

Given that,

The total yearly interest amounted to $380

Then, we can frame a equation as:

$x \times 3 \% + (6000 - x) \times 8 \% = 380$

Solve the above expression for "x"

$x \times \frac{3}{100} + (6000-x) \times \frac{8}{100} = 380\\\\0.03x + (6000-x) \times 0.08 = 380\\\\0.03x + 480 - 0.08x = 380\\\\-0.05x = -100\\\\\text{Divide both sides by -0.05 }\\\\x = 2000$

Thus, $ 2000 is invested at 3 % interest

(6000 - x) = 6000 - 2000 = 4000

$ 4000 is invested at 8 % interest