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  1. #1
    An Urch Guru Pundit Swami Sage
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    mixtures

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    Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?

    1/5
    1/4
    1/2
    3/4
    4/5

  2. #2
    An Urch Guru Pundit Swami Sage DWarrior's Avatar
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    If I understand the problem correctly, C.

    Original solution was all 50% acid, then x % of that was removed and x % of 30% acid added.

    So: (1-x)50+x30=40 solve for x

    Or: the average of 50 and 30 is 40, so you need a 50/50 split.

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    TestMagic Fan! coyote's Avatar
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    IMO the answer is 1/2.

    Method 1: the ratio of the two parts will be (50-40) : (40-30)

    Method 2: Let x be original quantity.
    Let y be the replaced fraction .

    0.5x - 0.5xy + 0.3xy = 0.4x

    y=1/2

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    An Urch Guru Pundit Swami Sage DWarrior's Avatar
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    Quote Originally Posted by coyote View Post
    Method 1: the ratio of the two parts will be (50-40) : (40-30)
    Enlighten me please, this method seems much faster.

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    TestMagic Fan! coyote's Avatar
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    Quote Originally Posted by DWarrior View Post
    Enlighten me please, this method seems much faster.
    This is a standard problem applicable to mixtures as well as some problems on averages.

    Code:
    A                 :           B
    a                             b
                            
                      c
        
    (b~c)          :        (a~c)
    What I've shown above is that if A and B are the items whose certain parameter is given as 'a' and 'b' respectively and same parameter of resultant mixture C is given to be 'c',
    Then the ratio in which A and B have been mixed is (c~b) : (c~a)
    ['~' sign stands for difference not subtraction]

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    Nice, thanks

  7. #7
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    Ans is 1/2.

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    An Urch Guru Pundit Swami Sage
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    OA is C.

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