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?

Question
Answer:
This problem is mainly concerned with setting up an equation.

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

[tex]a = (9000 + x)0.10[/tex]

[tex]b = x \times 0.06[/tex]

[tex]c=(20000 - ((9000 + x) + x)) 0.08[/tex]

simplify:

[tex]c = (11000 - 2x) \times 0.08[/tex]

sum of interest:

[tex]1780 = a + b + c[/tex]

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

[tex]1780 = 1780 + 0 \times x[/tex]

[tex]0=0[/tex]

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:

[tex]9000 \times 0.10 + 11000 \times 0.08[/tex]

[tex] = 900 + 880 = 1780[/tex]

So this ended up being a trick problem because he does not actually have any invested in the 6% option.
solved
general 10 months ago 5339