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Question 918239: A man has $215,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $20,600 and the amount invested at 8% is twice that invested at 12%.
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! total invested 215000
total yield 20600
interest rates 8,10,12
2 times amount at 12% invested at 8%
x+y+z=215000
0.08*x+0.1*y+0.12*z=20600
x=2z
3 equations 3 unknowns
x+y+z=215000,
2z+y+1z=215000,
3z+y=215000
y=215000-3z
0.08*x+0.1*y+0.12*z=20600
0.08*2z+0.1*y+0.12*z=20600
0.16*z+0.1*y+0.12*z=20600
0.16*z+0.12z+0.1*y=20600
0.28*z+0.1*y=20600
2.8*z+1*y=206000
1*y=206000-2.8*z
3z+y=215000
1*y=206000-2.8*z
1*y=215000-3z
0=-9000--0.2*z
-9000=-0.2*z
-9000/-0.2=z
45000=z
z=45000
1*y=215000-3z
1*y=215000-3*45000
1*y=215000-135000
1*y=80000
x+y+z=215000
x+80000+45000=215000
x=215000-y-z
x=90000
x=90000 y=80000 z=45000
check
0.08*90000+0.1*80000+0.12*45000= 20600
7200+8000+5400= 20600
ok
codeint3r
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