SOLUTION: A woman has a total of $10,000 to invest. She invests part of the money in an account that pays 8% per year and the rest in an account that pays 11% per year. If the interest earne
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-> SOLUTION: A woman has a total of $10,000 to invest. She invests part of the money in an account that pays 8% per year and the rest in an account that pays 11% per year. If the interest earne
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Question 1160573: A woman has a total of $10,000 to invest. She invests part of the money in an account that pays 8% per year and the rest in an account that pays 11% per year. If the interest earned in the first year is $890, how much did she invest in each account? Answer by greenestamps(13200) (Show Source):
x amount at 8%; 10000-x at 11%; total interest 890:
Solve using basic algebra; I leave that to you.
A much faster and easier informal solution method....
(1) $10,000 all at 8% would yield $800 interest; all at 11% would yield $1100 interest.
(2) The actual interest amount, $890, is 3/10 of the way from $800 to $1100. ($800 to $1100 is $300; $800 to $890 is $90; 90/300 = 3/10).
(3) That means 3/10 of the $10,000 was invested at the higher rate.
ANSWER: 3/10 of $10,000, or $3000, at 11%; the rest at 8%.
