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Question 1117137: Part of $6,000 was invested at 10% interest and the rest at 12%. If the annual income from these investments was $680, how much was invested at each rate?
$______ was invested at 10%, while $_______ was invested at 12%
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
The $6000 all invested at 10% would produce $600 interest income; all invested at 12% would produce $720 income.
The actual income, $680, is 2/3 of the way from $600 to $720; that means 2/3 of the money must be invested at the higher rate.
2/3 of $6000 is $4000.
Answer: $4000 at 12%; $2000 at 10%.
Here is a nice way to display this method for finding the answer to a problem like this, where there are investments at two different rates.
In this display, the numbers in the first column are the amounts of interest if all the money were invested at each of the two rates, and the number in the middle column is the actual amount of interest.
The numbers in the third column are the differences, calculated diagonally, between the numbers in the first and second columns.
When the numbers are displayed this way, the numbers in the third column show the ratio in which the money must be split between the two rates.
In your example, those numbers show that twice as much money should be invested at the second (higher) rate -- leading again to the answer of $4000 at 12% and $2000 at 10%.
An algebraic solution is of course possible; but it requires far more work....
let x = amount invested at 10%
then 6000-x = amount invested at 12%
The total interest income is $680:
The amount that should be invested at 10% is x=$2000; the amount that should be invested at 12% is $6000-x = $4000.
Same answer, of course; but a lot more work....
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