Question 1120427
<font face="Times New Roman" size="+2">


Bonds + Balanced + Aggressive = 6700.


Aggressive + 300 = Bonds + Balanced


So


2(Aggressive) + 300 = 6700


Aggressive = 3200


Since the interest rate on the Aggressive account was 12%, the amount of interest paid on the Aggressive account must be 0.12 X $3200 = $384


That means the amount paid on the other two accounts was $619 - $384 = $235.


Since the amount in the Aggressive account was $3200, the amount in the combination of the other two accounts must have been $6700 - $3200 = $3500.


So Bonds + Balanced = 3500.  Let *[tex \Large x] represent the amount invested in the Balanced account, then the amount invested in the Bonds account must have been *[tex \Large 3500\ -\ x].  Then the amount of interest earned on the Balanced account must have been *[tex \Large 0.08x] and the amount of interest earned on the Bonds account must have been *[tex \Large 0.05(3500\ -\ x)].  The sum of these two amounts is, as previously demonstrated, $235.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.08x\ +\ 0.05(3500\ -\ x)\ =\ 235]


Solve for *[tex \Large x] and then calculate *[tex \Large 3500\ -\ x]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>