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Question 525995: A person invests $50,000 for one year; some is invested at 7%, some at 8%, and the remainder at 12%. The combined interest earned at the end of the year from these investments was $4,770. the amount invested at 8% is $4000 more then the amount invested at 7% and 12% combined. find the amount of money invested at each rate
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! 7%--------------x
8%--------------(x+y+4000)
12%-------------y
x+x+y+4000+y= 50000
2x+2y+4000=50000
2x+2y=46000
/2
x+y=23000..............(1)
7x+8(x+y+4000)+12y= 4770 *100 ( the % is converted by multiplying by 100)
7x+8x+8y+32000+12y=477000
15x+20y=477000-32000
15x+20y=445000
/5
3x+4y=89000................(2)
1 x + 1 y = 23000 .............1
3 x + 4 y = 89000 .............2
Eliminate y
multiply (1)by -4
Multiply (2) by 1
-4 x -4 y = -92000
3 x 4 y = 89000
Add the two equations
-1 x = -3000
/ -1
x = 3,000.00
plug value of x in (1)
1 x + 1 y = 23000
3000 + 1 y = 23000
1 y = 23000 -3000
7%--------------x= $3,000
8%--------------(x+y+4000)= $27000
12%-------------y= 20000
m.ananth@hotmail.ca
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