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AR15.COM
2/22/2008 9:49:33 PM EDT
Hey guys does anyone know how to solve this problem?


A total of $12,000 is invested in two funds paying 9% and 11% simple interest. If the interest for the first year is $1,180, how much of the $12,000 is invested at 9%?


if anyone knows, any chance you could very briefly explain it, so i can get on the right track...
2/22/2008 9:56:00 PM EDT
[#1]
X + Y = 12000
.09X + .11y = 1180

two unknowns, two equations.  Easy./

Now to solve it.

X = 12000 - Y

.09(12000 - Y) + .11Y = 1180
1080 - .09Y + .11Y = 1180
.02Y = 100
Y= 5000

Check.
5000*.11 + 7000*.09 = 1180

2/22/2008 9:56:47 PM EDT
[#2]
$7000 at 9%, $5000 at 11%
2/22/2008 9:57:34 PM EDT
[#3]
I believe the answer is .9bar = 1.
2/22/2008 9:58:24 PM EDT
[#4]

Quoted:
X + Y = 12000
.09X + .11y = 1180

two unknowns, two equations.  Easy./


oh wow, now i feel like and idiot.

Haha, thanks. i was trying to make it way more complicated than it needed to be.
2/22/2008 9:58:48 PM EDT
[#5]
what's the name of the funds, I want to put some 401K money in those if they're paying 9% and 11%
2/22/2008 10:31:49 PM EDT
[#6]
12000 * x = 1180

12000 = 1180/x

12000/1180 = 1/x

10.1695 = 1/x

.098333 = x

That's the 41.5% mark between 9% and 11%.  We know the average of 9% & 11% is 10% and that yields $1200, which is too much, so the majority must be at 9%.  That would be the remainder of 41.5% from 100%, which is 58.5%.

12000(.585) = 7020
12000(.415) = 4980


$7020 @ 9% = 631.8

$4980 @ 11% = 547.8

631.8 + 547.8 = 1179.6 (round up to 1180)






2/22/2008 10:40:03 PM EDT
[#7]
Little intermediate algebra eh?