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2 June, 13:21

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?

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  1. 2 June, 15:27
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    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
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