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Math Help - Mixture word problem

  1. #1
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    Mixture word problem

    How many liters of water must be added to 35 liters of 80% hydrochloric acid solution to reduce its strength to 75%?
    Ans: 6.53 liters

    solution:
    x (0%water) +35 (80% HCL) = (35+x)(75% strength HCL)
    35*(0.8) = (35+x)0.75
    x = 2.3333 liters
    incorrect
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  2. #2
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    Re: Mixture word problem

    Hello, MathMinors!


    How many liters of water must be added to 35 liters of 80% HCl solution
    to reduce its strength to 75%?

    Ans: 6.53 liters . . . . Wrong!

    We concerned with the amount of water to be added.

    The amount of water is our concern.


    We start with 35 liters of soluton which is 20% water,
    . . It contains:. 0.20 \times 35 \:=\:7 liters of water.

    We add x liters of water.

    The mixture contains:. x + 7 liters of water. .[1]


    We know that the mixture will contain 35 + x liters of which 25% is water.
    . . The mixture contains:. 0.25(35+x) liters of water. .[2]


    We just described the final amount of water in two ways.

    There is our equation . . . \boxed{x + 7 \:=\:0.25(35 + x)}


    Solve for x\!:\;\;x + 7 \:=\:8.75 + 0.25x \quad\Rightarrow\quad 0.75x \:=\:1.75

    . . . . . . . . . . x \:=\:\frac{1.75}{0.75} \:=\:\frac{175}{75} \:=\:\frac{7}{3}


    \text{Therefore, we must add }2\tfrac{1}{3}\text{ liters of water.}

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  3. #3
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    Re: Mixture word problem

    Using a concentration line of 0%, 75%, 80%
    80 -75 = parts H2O =5 , 75-0 = parts of 80% to make 80 parts of 75%
    parts of water required for 35 parts 80% = 35/75 *5 = 2.333 parts water.

    Water solutions of HCL exceeding 40% do not exist at normal temp and pressure
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