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11 January, 16:59

Find two numbers such that their product, sum and difference are in the ratio 5:4:1'

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  1. 11 January, 18:17
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    3 1/3 and 2

    Step-by-step explanation:

    Let the numbers be x and y and let z be the constant of proportion. Then

    xy = 5z ... (1)

    x + y = 4z ... (2)

    x - y = z ... (3) Adding (2) and (3)

    2x = 5z

    x = 5z/2 ... (4)

    Substituting for x in equation 1:

    5z/2 * y = 5z

    y = (5z * 2) / 5z = 2.

    Substituting for y in equations (2) and (3):

    x + 2 = 4z

    x - 2 = z Subtracting:

    4 = 3z

    z = 4/3.

    Substituting for z in equation (4)

    x = 5 (4/3) / 2

    x = 20/6 = 10/3 = 3 1/3.

    So the 2 numbers are 3 1/3 and 2.

    Checking the results:

    xy = 2 * 10/3 = 20/3 = 5 * 4/3 = 5z

    x + y = 10/3 + 2 = 16/3 = 4 * 4/3 = 4z.

    x - y = 10/3 - 2 = 4/3 = z.
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