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Question 1124263: Things did not go quite as planned. You invested $20,000, part of it in a stock that paid 12% annual interest. However, the rest of the money suffered a 5% loss. If the total annual income from both investments was $2026, how much was invested at each rate?
Found 2 solutions by Theo, solver91311:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! total amount of money that was invested was 20,000.
one part of it earned 12%.
the other part lost 5%.
x = the part that earned 12%.
y = the part that lost 5%.
the total income from both investments was 2026.
your equation is:
.12 * x - .05 * y = 2026
also, x + y = 20,000
from the second equation, solve for y to get y = 20,000 - x
in the first equation replace y with 20,000 - x to get:
.12 * x - .05 * (20,000 - x) = 2026
simplify to get:
.12 * x - .05 * 20,000 + .05 * x = 2026
combine like terms to get:
.17 * x - 1000 = 2026
add 1000 to both sides of this equation to get:
.17 * x = 3026
solve for x to get x = 3026 / .17 = 17,800.
since x + y = 20,000, and x = 17,800, then y must be equal to 2,200.
he earned 12% on 17,800 and lost 5% on 2,200.
his net return was 2136 - 110 = 2026
that agrees with the problem statement, so the solution looks good.
the solution is he invested 17,800 that earned 12% and 2,200 that lost 5%.
Answer by solver91311(24713)  (Show Source):
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