SOLUTION: A total of $7000 is invested: part at 6% and the remainder at 10%. How much is invested at each rate if the annual interest is $470?
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Question 1153657: A total of $7000 is invested: part at 6% and the remainder at 10%. How much is invested at each rate if the annual interest is $470?
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
A total of $ is invested:
part at % and the remainder at %. How much is invested at each rate if the annual interest is $?
->invested at each % rate
then, other part is:
->invested at each % rate
check:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
A quick and easy way to solve this problem and a wide variety of similar "mixture" problems....
Consider the actual amount of interest and the amounts of interest if the whole amount were invested at each rate. $7000 at 6% gives $420 interest; $7000 at 10% gives $700 interest. Now consider those three amounts of interest on a number line.
$470 is 50/280 = 5/28 of the way from $420 to $700.
That means 5/28 of the total was invested at the higher rate.
5/28 of $7000 = 5*$250 = $1250.
ANSWER: $1250 at 10%; the other $5750 at 6%.
CHECK: .10(1250)+.06(5750) = 125+345 = 470
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