Question 1113471
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If *[tex \Large P] is invested in two accounts, one at interest rate *[tex \Large r_1] and the other at interest rate *[tex \Large r_2] and the total interest earned is *[tex \Large I] after 1 year at simple annual interest, and the amount invested in the first account is *[tex \Large P_1] while the amount invested in the second account is *[tex \Large P_2],  the following relationships hold:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_1\ +\ P_2\ =\ P]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_1r_1\ +\ P_2r_2\ =\ I]


Note that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_2\ =\ P\ -\ P_1]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_1r_1\ +\ (P\ -\ P_1)r_2\ =\ I]


For your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_1\,\cdot\,0.08\ +\ (11000\ -\ P_1)\,\cdot\,0.06\ =\ 842]


Solve for *[tex \LARGE P_1] and then calculate *[tex \LARGE 11000\ -\ P_1]


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 \ \ \ \ \ \ \ \ \ \  

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