A man has $290,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $27,600 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 $

Step-by-step explanation:

let amount invested at 12%=x

amount invested in 10 %=y

amount invested in 8 %=z

z=2x

x+y+z=290,000

x+y+2x=290,000

3x+y=290,000  

multiply by 5

15x+5y=1,450,000  ... (1)

12/100 x+10/100 y+8/100 z=27,600

12x+10 y+8z=2,760,000

6x+5y+4z=1,380,000

6x+5y+4(2x)=1,380,000

14x+5y=1,380,000   ... (2)

(1)-(2) gives

x=1,450,000-1,380,000=70,000

z=2(70,000)=140,000

70,000+y+140,000=290,000

y+210,000=290,000

y=290,000-210,000=80,000

invested in 12 %=$ 70,000

invested in 10 %=$ 80,000

invested in 8 %=$ 140,000

(b)

annual income from 12 %=12/100 ×70,000= $8,400

annual income from 10 % =10/100×80,000=$ 9,600

annual income from 8 %=8/100×140,000= $11,200

12 servings. 4 times 3= 12

i think it's either c or d

step-by-step explanation: the entire question is extremely odd. 04 obviously refers to the interest rate. honestly i will say the answer is c. because the (2 x 12) part would mean it's compounded semiannually. i'll say choose c.