larry mitchell invested part of his $27,000 advance at 2% annual simple interest and the rest at 6% annual simple interest. If h
is total yearly interest from both accounts was $860, find the amount invested at each rate
2 answers:
X=$4,000 INVESTED5%.
28,000-4,000=$24,000 INVESTED 3%.
Step-by-step explanation:
PROOF:
.05*4,000+.03*24,000=920
200+720=920
920=920
Answer:
- $19000 at 2%, $8000 at 6%
Step-by-step explanation:
Let x is the amount at 2%, then 27000 - x is the amount at 6%.
<u>The total interest is:</u>
- 0.02x + 0.06(27000 - x) = 860
- 0.02x - 0.06x + 1620 = 860
- 0.04x = 1620 - 860
- 0.04x = 760
- x = 760/0.04
- x = 19000
<u>The amount at 6% is:</u>
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