SOLUTION: A man has $250,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $23,900 and the amount invested at 8% is twice that inv

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A man has $250,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $23,900 and the amount invested at 8% is twice that inv      Log On


   

Question 1203745: A man has $250,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $23,900 and the amount invested at 8% is twice that invested at 12%.
(a) How much is invested in each property?
12%property $
10%property $
8%property $

(b) What is the annual income from each property?
12%property $
10%property $
8%property $

Found 2 solutions by mananth, MathTherapy:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
A man has $250,000 invested in three properties.
One earns 12%, Let investment in this one be $ x
second at 10% Let investment in this one be $y
and the amount invested at 8% is twice that invested at 12%.
Third one 8%. So amount invested in this one is $ 2x

$250,000 invested in three properties.
x+y+2x =250000
3x+y = 250 000--------------------------------(1)
His annual income from the properties is $23,900
Interest equation
12%x +10%y + 8%(2x)= 23 900
0.12x+0.10y +0.16x =23900
0.28x + 0.10y - 23900----------------------------(2)
Divide (1) by 10
0.30x+0.1y =25000
0.28x + 0.10y - 23900 (from (2) Subtract
0.02x =1100
x = 1100/0.02
x =55 000
2x = 110 000
3x+y =250 000
3*55000+y = 250 000
165 000 +y =250 000
y =85000
12%property $ 55 000
10%property $ 85 000
8%property $ 110 000
Check
12%*55000+10%*85000+8%*110000 =23,900
you can now calculate the annual income from each property?
12%property $
10%property $
8%property $

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
A man has $250,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $23,900 and the amount invested at 8% is twice that invested at 12%.
(a) How much is invested in each property?
12%property			$ 
10%property			$ 
8%property			$ 


(b) What is the annual income from each property?
12%property			$ 
10%property			$ 
8%property			$ 

Let amount invested in the property earning 12%, be T
Then amount invested in the property earning 8% = 2T
So, amount invested in the property earning 10% = 250,000 - (T + 2T) = 250,000 - 3T

Income from property earning 12%: .12T
Income from property earning 8%: .08(2T) = .16T
Income from property earning 10%: .1(250,000 - 3T) = 25,000 - .3T

Since total income from the 3 investments is $23,900, we get: .12T + .16T + 25,000 - .3T = 23,900
                                                                       .12T + .16T - .3T = 23,900 - 25.000
                                                                                  - .02T = - 1,100 
    Amount invested in property earning 12%, or 

                         Amount invested in property earning 8%: 2T = 2(55,000) = $110,000

  Amount invested in property earning 10%: 250,000 - (55,000 + 110,000) = 250,000 - 165,000 = $85,000

Income from property earning 12%, property earning 8%, and property earning 10% are: .12T, .16T, and 25,000 - .3T,
respectively. Use those facts, along with T being $55,000, to find the income from each of the 3 properties.