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Question 1185811: A bank loaned out $14,000, part of it at the rate of 6% per year and the rest at 16% per year. If the interest received in one year totaled 1500$, how much was loaned at 6%
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
A standard "mixture" problem....
First a standard setup for solving the problem using algebra.
x dollars at 6%, plus (14000-x) dollars at 16%, yields $1500 interest.
Solve using basic algebra, although the numbers are a bit messy.
And now a quick and easy way to solve any 2-part mixture problem like this, if a formal algebraic solution is not required and your mental math skills are good.
$14,000 all invested at 6% would yield $840 interest; all at 16% would yield $2240 interest; the actual interest was $1500.
Picture the three interest amounts -- 840, 1500, and 2240 -- on a number line and observe/calculate that 1500 is 660/1400 of the way from 840 to 2240. That means 660/1400 of the total was loaned at the higher rate.
ANSWER: $6600 was loaned at 16%; the other $7400 at 6%.
CHECK: .16(6600)+.06(7400)=1056+444=1500
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