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
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10 months ago
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