Question 1206863: You invested $24 comma 00024,000 in two accounts paying 5 %5% and 7 %7% annual interest, respectively. If the total interest earned for the year was $1400, how much was invested at each rate?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website! .
You invested $24 comma 00024,000 in two accounts paying 5 %5% and 7 %7% annual interest, respectively.
If the total interest earned for the year was $1400, how much was invested at each rate?
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Let x dollars be the amount invested at 7%.
then the amount invested at 5% is (24000-x) dollars.
Write equation for the total annual interest
0.07x + 0.05*(24000-x) = 1400 dollars.
Simplify and find x
0.07x + 0.05*24000 - 0.05x = 1400
0.07x - 0.05x = 1400 - 0.05*24000
0.02x = 200
x = 200/0.02 = 20000/2 = 10000.
ANSWER. $10000 invested at 7%, the rest, or 24000-10000 = 14000 dollars invested at 5%.
CHECK. 0.07*10000 + 0.05*14000 = 1400 dollars, the combined annual interest. ! correct !
Solved.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
The response from the other tutor shows a typical formal algebraic solution.
The discussion below shows a less formal process for solving any 2-part mixture problems like this.
The idea behind this method is that the ratio between the amounts invested in the two accounts is exactly determined by where the overall percentage return lies between the percentages of the two accounts.
$24,000 all invested at 5% would earn $1200 interest; all invested at 7% would earn $1680 interest.
Picture the three interest amounts -- $1200, $1400, and $1680 -- on a number line and observe/calculate that $1400 is (1400-1200)/(1680-1200) = 200/480 = 5/12 of the way from $1200 to $1680.
That means 5/12 of the total was invested at the higher rate.
5/12 of $24,000 is $10,000. So $10,000 was invested at 7% and the other $14,000 at 5%.
ANSWER: $10,000 at 7%; $14,000 at 5%
CHECK: .07(10000)+.05(14000) = 700+700 = 1400
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