Question

A man has $210,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $20,200 and the amount invested at 8% is twice that invested at 12%.(a) How much is invested in each property?12%=10%=08%=(b) What is the annual income from each property?12%=10%=08%=

Answer 1

A)

first property= 40,000

2nd property=90,000

3rd property= 80,000

B)

annual income from first property= 4800

annual income from 2nd property= 9000

annual income from third property = 6400

Step-by-step explanation:

So, we can form two equation.

first is, total amount invested in three properties is 210,000

A+B+C=210,000 eq (1)

A is the first property

B is the 2nd property

C is the 3rd property

2nd is, annual income from the property is 20,200

12%*A+10%*B+8%*C= 20,200 eq (2)

As, we have given

the amount invested at 8% is twice that invested at 12%.

it means C is the twice of A

so, the only way we can make A equal to C by multiplying 2 with A, otherwise both won't be equal.

so, 2A=C eq(3)

substituting eq 3 with eq (1) and eq (2)

then we get

A+B+2A=210,000

3A+B=210,000 eq(4)

And,

0.12A+0.10B+0.08*2*A=20,200

0.28A+0.10B=20,200 eq(5)

Finally,

3A+B=210,000

0.28A+0.10B=20,200

multiplying equation (4) with 0.10 then

0.3A+0.10B=21,000

multiplying -1 with equation (5)

-0.28A-0.10B=-20,200

Add them, we get -0.02A=-800

A=40,000

so, putting the value of A in eq (3)

then 2*40,000=C

C=80,000

putting the value of A and B in equation (1)

We get

B=90,000

2nd part of the question

• 12%*A= 12%*40,000

so, income from A is 4800

• 10%*B=10%*90,00

so, income from B is 9000

• 8%*C=8%*80,000

so, income from C is 6400