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10 February, 06:44

How many ml of a 20% acid mixture and a 80% acid mixture should be mixed to get 120ml of a 35% acid mixture?

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  1. 10 February, 06:56
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    Volume of Mixture A = 90 ml

    Volume of Mixture B = 30 ml

    Step-by-step explanation:

    Let say, Mixture A + Mixture B = Mixture C

    Volume of Mixture A is x

    Volume of Mixture B is y

    So, Volume of Mixture C is x+y = 120 ml

    Now, Acid contain in Mixture A is 20% = 0.2x

    Acid contain in Mixture B is 80% = 0.8y

    Also, Acid contain in Mixture C is 35% = (0.35) (x+y) = 0.35*120=42

    Now, we know that,

    Acid contain of Mixture A + Acid contain of Mixture B=Acid contain of Mixture C

    ∴ 0.2x+0.8y=42

    ∴ 2x+8y=420

    We get two linear equations

    2x+8y=420 and x+y = 120

    Solving above equation ...

    ∴ x=120-y

    Replacing x value in 2x+8y=420

    ∴ 2 (120-y) + 8y=420

    ∴ 240-2y+8y=420

    ∴ 6y=180

    ∴ y=30

    Replacing y value in any equation

    ∴ x=120-y=120-30=90

    ∴ x=90

    Thus,

    Volume of Mixture A is x=90 ml

    Volume of Mixture B is y=30 ml
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