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Math Help - Mixture/Solution Problem 2

  1. #1
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    Mixture/Solution Problem 2

    Hi All,
    One term I'm not getting in the below..

    Q: 5 litres of water is added to a certain quantity of pure milk costing $3 per litre. If, by selling the mixture at the same price as before a profit of 20% is made, what is the amount of pure milk in the mixture?

    A: Ok so we have 3x as our cost which should be equal to 1.2(3x) (or 120% of 3x given profit). I would imagine the expression therefore to be 3x + 5 = 1.2(3x). However the solution is 3x + 15 = 1.2(3x). Why 15? (I realize this is 3*5 litres of water but not sure why!).

    As usual thanks in advance,
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  2. #2
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    Hello, dumluck!

    Q: 5 litres of water is added to a certain quantity of pure milk costing $3 per litre.
    If, by selling the mixture at the same price as before, a profit of 20% is made,
    what is the amount of pure milk in the mixture?

    A: Ok so we have 3x as our cost which should be equal to 1.2(3x)
    . . (or 120% of 3x given profit).

    I would imagine the expression therefore to be 3x + 5 = 1.2(3x). . No

    We have \,x litres of milk.
    We add 5 litres of water.

    We have: x + 5 liters of mixture, which will sold at $3 per litre.

    Its value is: . 3(x+5) dollars.

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, dumluck!


    We have \,x litres of milk.
    We add 5 litres of water.

    We have: x + 5 liters of mixture, which will sold at $3 per litre.

    Its value is: . 3(x+5) dollars.

    That makes a lot more sense, thanks for that.

    Paul
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