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Question 677151: I'm not sure if I have this right. But, doing it as taught.
Mark invests $21,000 into two accounts with the simple interest rate of 4% and 6%. If he wants to earn $1020 annual interest. How much should he invest at each rate?
So far, I tabulated it and then came up with this equation and answer.
4%(x-21,000) + 6%(x) = 1020
I then multiplied times 10 to eliminate the percentages and then it looks like this.
4x - 84,000 + 6x = 10200
I combined like terms.
10x = 94200
Dividing by the coefficient, I got.
x = 9420
Then, I plugged x answer into x - 21,000 (9420 - 21,000 = 11580).
So, I believe my answer to be Mark must invest $9,420 at 6% and $11,580 at 4%.
Please be understanding of me, I'm painfully slow at this and I have no idea if I'm doing this right. I have no confidence at this point and I may forget the set up when exam time comes.
Answer by ReadingBoosters(3246)   (Show Source):
You can put this solution on YOUR website! Let's see...
Establish variables
x = $ at 4%
y = $ at 6%
...
Given equation
x+y = 21000
.04x + .06y = 1020
...
Substitute
.04(21000-y) + .06y = 1020
840 - .04y + .06y = 1020
840 + .02y = 1020
.02y = 1020 - 840
.02y = 180
 at 6%
...
x + 9000 = 21000
x = 21000 - 9000 =  at 4%
...
Check
.04(12000) + .06(9000) = 480+540 = 1020
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Delighted to help.
-Reading Boosters
Website: www.MyHomeworkAnswers.com
Wanting for others what we want for ourselves.
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