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maths

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a man has one kind of coffee at 'a' paise per kg and another at 'b' paise per kg.how much of each must he take to form a mixture of (a-b)kg, which he can sell at 'c' paise per kg without loss.

  • maths - ,

    I think you must mean a+b kg, since you're taking some of each.

    value of x kg of a is ax
    value of y kg of b is by

    value of mixture is thus ax+by
    to sell at c paise/kg,

    ax+by = (x+y)*c

    now, we only have one equation, so it's hard to pin down both x and y. However, we can let x=1 without loss of generality, because we can then figure y as a multiple of x.

    So, letting x=1,

    a+by = (y+1)c
    a+by-cy = c
    y(b-c) = c-a
    y = (c-a)/(b-c)

    This will be positive, since c is between a and b.

    So, if a=5 and b=10, and c=7, then we have x=1, y=2/3

    check: 1*5 + (2/3)*10 = 35/3 = (1 + 2/3)*7

    So, as long as y = 2x/3, the equation still holds.

  • maths - correction - ,

    Hmmm. I see I was too general. We need x=a and y=b

    a^2 + b^2 = (a+b)c

    so, if we let a=1, then
    1+b^2 = (1+b)c
    1+b^2 = c + bc
    b^2 - cb + (1-c) = 0
    b = 1/2 (c+√(c^2+4c-4))

    If we have a=a, instead of a=1, then

    b = a/2 (c+√(c^2+4c-4))

    check: if we have a=1 and b=1+√2 = 2.8, then we can sell a+b = 3.8kg at $2.

    1*1 + (1+√2)(1+√2) = 4+2√2
    (1 + 1+2√2)(2) = 4+2√2

    You cannot in general pick any random values for a and b and expect to make it fit any given c. You can pick a and b and figure c, or you can pick a and c and calculate b.

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