SOLUTION: Mr. Wilson invested money in two accounts. His total investment was $20,000. If one account pays 6% interest and the other pays 8% interest, how much does he have in each account i

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Question 1094914: Mr. Wilson invested money in two accounts. His total investment was $20,000. If one account pays 6% interest and the other pays 8% interest, how much does he have in each account if he earned a total of $1,380 in interest in 1 year?
Found 3 solutions by ikleyn, greenestamps, MathTherapy:
Answer by ikleyn(52909) About Me  (Show Source):
You can put this solution on YOUR website!
.
It is a standard problem on investment.

If you want to learn on how to solve it on your own, read the lesson
    - Using systems of equations to solve problems on investment
in this site.

Consider it as a sample/template/prototype.  Read it attentively.
Then solve your problem by substituting your data.
In this way you will learn the method.

Good luck and happy learning  !!



Answer by greenestamps(13215) About Me  (Show Source):
You can put this solution on YOUR website!

Try this alternative method and see if it "works" for you. If you understand it, it will get you to the answer to problems like this much faster, and with much less work, than the traditional algebraic approach.

If all the $20,000 were invested at 6%, the interest would have been $1200.
If all the $20,000 were invested at 8%, the interest would have been $1600.
The actual amount of interest is, of course, somewhere between those two figures.

More specifically, WHERE the actual amount of interest lies between those two figures exactly determines the ratio in which the investment must be split between the two accounts.

Just as a preliminary calculation, we can note that the $1380 is a bit closer to $1200 than it is to $1600, so we know a bit more needs to be invested at 6% than at 8%.

Now for the details.... We find exactly how far the $1380 is from $1200 and from $1600; the ratio of those distances is the ratio in which the investment must be split. So
1600-1380+=+220
1380-1200+=+180

The ratio of those differences is 220:180, or 22:18, or 11:9.

Since the ratio is 11:9 and the total amount of the investment is $20,000, the amounts in the two accounts need to be $11,000 and $9,000.

And since we have already determined that the larger amount has to be invested at the 6% rate, the final answer is

$11,000 at 6% and $9000 at 8%.

With all the words of explanation, it looks like a long process. But without the words, this is the whole process:

.06%2A20000+=+1200
.08%2A20000+=+1600
1600-1380+=+220
1380-1200+=+180
220%3A180+=+11%3A9

11/20 of the $20,000 = $11,000 at 6%; 9/20 of the $20,000 = $9000 at 8%.

Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!
Mr. Wilson invested money in two accounts. His total investment was $20,000. If one account pays 6% interest and the other pays 8% interest, how much does he have in each account if he earned a total of $1,380 in interest in 1 year?
Let the 6%-investment amount be S
Then the 8% investment amount = 20,000 - S
We then get: .06S + .08(20,000 - S) = 1,380
.06S + 1,600 - .08S = 1,380
.06S - .08S = 1,380 - 1,600
- .02S = - 220
S, or 6%-investment amount =
Amount invested at 8%:
That's all!! Nothing COMPLEX and/or CONFUSING!
P.S. I could've also used 2 variables: 1 for the 6% and another for the 8%.