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Question 1191549: Kevin has money in two savings accounts. One rate is 6% and the other is 12%. If he has $600 more in the 12% account and the total interest is $279, how much is invested in each savings account?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52905)   (Show Source):
You can put this solution on YOUR website! .
Kevin has money in two savings accounts. One rate is 6% and the other is 12%.
If he has $600 more in the 12% account and the total interest is $279,
how much is invested in each savings account?
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x = the amount saved at 12%;
y = the amount saved at 6%.
From the condition, you have these two equations
x - y = 600 dollars (1)
0.12x + 0.06y = 279 dollars (2)
To solve this system, express x = y + 600 from equation (1) and substitute it into equation (2). You will get
0.12(y+600) - 0.06y = 279
0.12y + 0.12*600 + 0.06y = 279
0.12y + 0.06y = 279 - 0.12*600
0.18y = 207
y = 207/0.18 = 1150.
Then from equation (1),
x = 600 + 1150 = 1750.
ANSWER. $1750 invested at 12% and $1150 invested at 6%.
CHECK. I will check equation (2). Its left side is 0.12*1750 + 0.06*1150 = 279 dollars. ! Correct !
Solved.
Answer by greenestamps(13215)  (Show Source):
You can put this solution on YOUR website!
There are numerous ways to set up this problem for solving using formal algebra; the other tutor has shown one.
If a formal algebraic solution is not required, you can get some good mental exercise by solving the problem informally, maybe something like this:
The "extra" $600 invested at 12% yields $72 interest.
The remaining amount, invested equally at 6% and 12%, gives an average rate of 9%; and it yields the remaining $207 interest.
$207 interest at 9% means the amount invested was $207/.09 = $2300.
That $2300 is equally divided between the two investments, so the amount invested in each was $1150.
Now remember the additional $600 invested at 12% interest to find that....
ANSWER: $1150 at 6%; $1150+$600 = $1750 at 12%
CHECK: .06(1150)+.12(1750) = 69+210 = 279
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