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Tim
Lv 4
Tim asked in Science & MathematicsMathematics · 1 decade ago

I need help with a mixture problem?

A jar contains 6 liters of a 25% alcohol solution. How much should be poured out and replaced with 100% alcohol to make a 33% solution? Last time I posted this I got two different answers( .72L and .48L) and my answer was .64L. Which one is right? or are all of them wrong?

3 Answers

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  • Don E Knows
    Lv 6
    1 decade ago
    Favorite Answer

    You are correct.

    I solve these with the formula : (High - Mix / High - Low) * Total = Low amount needed

    So this becomes : (100 - 33 / 100 - 25) * 6 = 5.36 Low amount needed.

    So if you pour out 6.00 - 5.36 = 0.64, then you fill this with the 100% (High), you get the correct concentration of 33%

    Proof :

    5.36(1/4) + 0.64(1) = 6(1/3)

    1.34 + 0.64 = 2.00

  • hcbiochem
    Lv 7
    1 decade ago

    I think that your answer of 0.64 L is correct. The way I solved this problem was:

    6L ( 33%) = x(25) + (6-x)(100)

    solving for x gave me 5.36 L, so the volume of 100% ethanol you need is 0.64 L

  • M3
    Lv 7
    1 decade ago

    first work out the PROPORTION of 25% & 100% solutions reqd. to give 33%

    the easiest way is like this:

    25% .....+8% .... 33% .........+67%........ 100%

    67/75 of 25% soln. & 8/75 of 100%

    so 6*[8/75] L of 25% soln. = 0.64 L must be poured out

    you were right & the others were wrong !

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