Story problem with percentages

Question

A man has 275000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $26000 and the amount invested at 8% is twice that invested at 12%. How much is invested in each property?

Answer 1

Hi Colby,

suppose a man invested $ x which earns 12%

                                   $ y which earns 10% 

Amount invested at 8% is twice at 12%, 

     he invested $ 2x which earns 8% .

Total investment = x+y+2x = 275000

                            3x +y =275000

                                  y = 275000-3x.......eq 1

Annual income=$ 26000

 

income from 12% interest for $x= 12x/100

income from 10 % interest for $y =10y/100 

income from 8% interest for $2x =16x/100

 

Total annual income = 26000=12x/100+10y/100+16x/100 

                                2600000=28x+10y........eq 2 

substituting value of y from eq 1 to eq 2, we get,

                                 2600000=28x [phone]x)

                                 2600000 = 28x [phone]x

                                  2600000 = -2x+2750000

                                       2x= 2750000-2600000

                                       2x= 150000

                                         x= 75000

                                        y = 275000-3(75000) =275000-225000=50000

investment $75000 earns 12%, $50000 earns 10%, $150000 earns 8%.