A man has $195,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $18,700 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:

Given that a man has $195,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $18,700 and the amount invested at 8% is twice that invested at 12%.

12%10%8%total

Principalxyz195000

Interest0.12x0.1y0.08z18700

z=2x    

Thus we get three equations as

x+y+z = 195000:  0.12x+0.1y+0.08z = 18700 and z =2x

substitute z value in first 2 equations to get

3x+y = 195000:  0.28x+0.1y = 18700 or 28x+10y =1870000

                                                                 30x+10y =1950000

2x = 80000 or x = 40000

3(40000)+y = 195000: y = 75000

z = 80000

a) How much is invested in each property?

12% property $   ... 40000

10% property $  ... 75000

8% property $  ... 80000

b) Annual income =

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3