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Math Help - A word problem...confusing.

  1. #1
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    A word problem...confusing.

    (First, if this is in the wrong section, im sorry!)

    Alright so here is the word problem.

    A bar of metal contains 10% Silver.
    Another bar of metal contains 15% Silver
    How many kg of each is required to make a 10kg bar of 12% Silver?

    The only problem, is that the teacher stated you can only use ONE variable in your equation.

    I am entirely lost on this one...If i can use 2 variables, I can do that...but one?
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  2. #2
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    Hello Aric

    Welcome to Math Help Forum!
    Quote Originally Posted by Aric View Post
    (First, if this is in the wrong section, im sorry!)

    Alright so here is the word problem.

    A bar of metal contains 10% Silver.
    Another bar of metal contains 15% Silver
    How many kg of each is required to make a 10kg bar of 12% Silver?

    The only problem, is that the teacher stated you can only use ONE variable in your equation.

    I am entirely lost on this one...If i can use 2 variables, I can do that...but one?
    I know this doesn't really answer your question, but let me show you a simple method that doesn't need any variables at all. It just uses ratios.

    The difference between 10 and 12 is 2, and the difference between 12 and 15 is 3. Write these two numbers as a ratio 2:3.

    Then swap them around, that's 3:2, to give you the relative quantities of the two metal-mixtures.

    So for the first mixture, 1 kg contains 10% = 100 g of silver. So take 3 kg of this, which will contain 300 g of silver.

    The second mixture is 15% silver; so for every 1 kg of this, there will be 150 g of silver. Take 2 kg of this, and you'll also have 300 g of silver.

    So, when we add these two quantities together we get 5 kq altogether containing 300 + 300 = 600 g of silver. This is the required 12%.

    To get 10 kg altogether, we obviously double the quantities and take 6 kg of the first, and 4 kg of the second.

    (It will work with any other number - provided it's between 10% and 15% of course. If you want 11% of silver, the ratio is 1:4 this time. Swap them around and take 4 kg of the first (that will contain 400 g of silver) for every 1 kg of the second (that's 150 g of silver). Total 550 g in 5 kg = 11%. Works every time!)

    Grandad
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  3. #3
    Senior Member pacman's Avatar
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    "A bar of metal contains 10% Silver. Another bar of metal contains 15% Silver. How many kg of each is required to make a 10kg bar of 12% Silver?"

    Let,

    x = weight of metal bar,

    (10 - x) = weight of another metal bar,

    10 = weight of the mixture

    Computation:

    (Wt Mixture)(% Ag) = (Wt Metal)(% Ag) + (Wt Mixture - Wt Metal)(% Ag)

    (10 kg)(12%) = (x kg)(10%) + (10 kg - x kg)(15%),

    10(0.12) = 0.10x + (10 - x)(0.15),

    1.2 = 0.10x + 1.5 - 0.15x,

    now solve for x:





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  4. #4
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    Ah! both these answers were very helpful! Thanks to the both of you!
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