Question 1100473: Yofi is investing $6,000 in two accounts, part of 4.5% interest and the balance at 6% interest. If the total annual interest earned from two accounts is 279. how much did yogi invest at each rate?
Found 2 solutions by richwmiller, greenestamps:
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! We know the total amount of money invested. $6000
x+y=6000
We know that the total interest for the year for the two accounts is $279
4.5/100*x+6/100*y= 279
x=6000-y
We substitute for x
4.5/100*(6000-y)+6/100*y= 279
We multiply out
270-4.5/100*y+6/100*y= 279
We combine like terms.
1.5*y/100=279-270
1.5*y/100=9
0.015*y=9
Isolate y
y=9/0.015
y=$600 at 6% earns $36 interest
Calculate x
x=6000-600
x=$5400 at 4.5% earns $243 interest
We check
4.5/100*5400+6/100*600=279
243+36=279
279=279
Since 279 = 279 is TRUE, all is well
It is good to be exposed to several ways to solve a problem. What seems easy to one person could seem hard to another and vice versa.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Here is a method which, if you understand it, will get you to the solution to this kind of "mixture" problem much faster than the traditional algebraic approach shown by the other tutor.
I'm not saying this is a better method, or that the traditional algebraic method is wrong. I'm only showing you an alternative method you can use if it makes sense to you.
The $6000 all invested at 4.5% would produce $270 interest; all invested at 6% would produce $360 interest. Then
The ratio 81:9 is equivalent to 9:1; that means the money must be split between the two accounts in the ratio 9:1 -- i.e., 9/10 of the total must be invested in one account and 1/10 in the other.
Since $279 is closer to $270 than to $360, most of the money must be invested in the 4.5% account.
So the amount invested in the 4.5% account is 9/10 of $6000, or $5400; the amount invested in the 6% account is 1/10 of $6000, or $600.
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