A man has $245,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $23,400 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?

for 8% investment = $110000

for 12% investment = $55000

for 10% investment = $80000

annual income for each property is

x amount for 8% investment = $8800

y amount for 12% investment = $6600

z amount for 10% investment = $8000

Explanation:

Given data

investment = $245000

rate 1 = 12%

rate 2 = 10%

rate 3 = 8%

annual income = $23400

rate = 8% twice that invested at 12%

to find out

invested in each property and the annual income from each property

solution

let us consider x amount for 8% investment

and consider y amount for 12% investment

and consider z amount for 10% investment

from question we say, 8% is twice that invested at 12%

x = 2y ... 1

and

x + y + z = 245000 ... 2

put 1 in equation 2

2y + y + z = 245000

z = 245000 - 3y ... 3

and we can say that

0.08x + 0.12y + 0.10z = 23400 ... 4

put equation 1 and 3 in 4

0.08 (2y) + 0.12y + 0.10 (245000 - 3y) = 23400

0.16y + 0.12y - 0.3y = - 1100

0.02y = 1100

y = $55000

so x = 2 (y) = 2 (5500) = $110000

and z = 245000 - 3 (55000) = $80000

so annual income for each property is

x amount for 8% investment = 0.08 * 110000 = $8800

y amount for 12% investment = 0.12 * 55000 = $6600

z amount for 10% investment = 0.10 * 80000 = $8000