Question

Oscar invests $20,000 in three investments earning 6% ,8% and 10%. He invests $9000 more in the 10% investment than in the 6% investment. How much does he have invested at each rate if he receives $1780 interest the first year?

Answer 1

This problem is mainly concerned with setting up an equation.

let's add the interest rates times their respective investment amounts:

$a = (9000 + x)0.10$

$b = x \times 0.06$

$c=(20000 - ((9000 + x) + x)) 0.08$

simplify:

$c = (11000 - 2x) \times 0.08$

sum of interest:

$1780 = a + b + c$

1780 = (9000+x) 0.10 + x 0.06 + (11000-2x)0.08

$1780 = 1780 + 0 \times x$

$0=0$

solution:

He does not have any invested in the 6%, while he has 9000 invested at 10% and 11,000 invested at 8%

to check:

$9000 \times 0.10 + 11000 \times 0.08$

$= 900 + 880 = 1780$

So this ended up being a trick problem because he does not actually have any invested in the 6% option.