Can you solve this??

Question

Call me stupid or dumb or whatever but I cant remember how to solve a problem like this....

Morgan invested $5000for one year, part at 9% annual interest and the rest at12% annual interest. The interest from the investment at 9% was $198 more than the investment at 12%. How much money did she invest at 12%?

Thanks for the help but all i need is a easy formula

Answer 1

Let x = amount at 9%, then 5000 - x = amount at 12%
(9% times x) = (12% times (5000-x)) + 198
.09x = .12(5000-x) + 198
Solve for x.
.09x = 600 - .12x + 198
.09x = -.12x + 798
Add .12x to each side
.21x = 798
Divide both sides by .21
x = 798/.21 = 3800
x stood for the amount at 9% so
5000 - 3800 = 1200 = amount at 12%

Answer 2

Use the formula

A=Pe^(rt)

where P is the initial amount, r is the annual interest rate, and t is the time in years. Subtract off P to get just the interest.

NOTE: I assume continuous compounding of interest; others may assume annual (for greatest simplicity), or monthly, or whatever.

Let X be the amount invested at 9%. Then $5000-X was invested at 12%. t=1.

You have this equation, based on the given info:

X[e^(0.09)-1] = (5000-X)[e^(0.12)-1] + 198.

Expandn everything:

X[e^(0.09)+e^(0.12)-2] = 5000[e^(0.12)-1] + 198.

This gives you

0.2217X = 835.4843

or

X = $3768.54. This is the amount invested at 9%, and $1231.46 was invested at 12%.

Answer 3

Let's say she invested x dollars at 12%. That means she invested the rest, or 5000-x dollars, at 9%. The resulting equation would look like this:

.12x=.09(5000-x)-198
.12x=450-.09x-198
.21x=252
x=1200

She invested $1200 at 12%, and $3800 at 9%. Check:
12% of $1200 is $144
9% of $3800 is $342, exactly $198 more

Answer 4

$1200

Answer 5

I'm not going to solve it for you, but I'll give a hint, you should be able to work out the details.

(5000-x)*1.09-x*1.12 = 198

Solve for x