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24 January, 07:25

A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400 L solution that is 62% acid?

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  1. 24 January, 08:35
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    Let x be the quantity of solution one in mixture.

    so solution 2 would be (400-x) L

    according to ques.

    80% of x + 30% of (400-x) = 62% of 400

    i. e. 80*x/100 + 30 * (400-x) / 100 = 62*400/100

    => 80x - 30x = 62*400 - 30*400

    => 50x = 32*400

    x=256 L

    therefore 256L of solution one is required
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