SOLUTION: What quantity of 60 per cent acid solution must be mixed with a 25 solution to produce 168 mL of a 50 per cent solution

Question 335412: What quantity of 60 per cent acid solution must be mixed with a 25 solution to produce 168 mL of a 50 per cent solution
Answer by nyc_function(2741) (Show Source): You can put this solution on YOUR website!
I'm assuming that your second solution is a 25% solution since there are no units given.
So the amount of 60% solution we'll call x, and the amount of 25% solution we'll call y, and the amount of 50% solution we'll call T since it's the total amount, therefore our first equation is
x + y = T so solving y in terms of x, we get y = T - x now to solve for the total amount of acid needed, we can use the equation
0.6x + 0.25y = 0.5T, but since we already know the total amount that T represents, T = 168mL so
0.6x + 0.25y = 0.5(168) so the total amount of acid is just 84mL, and since we already know what y is in terms of x, we can substitute
0.6x + 0.25(T - x) = 84 and now we just multiply out, and since we also know that T = 168
0.6x + 0.25(168) - 0.25x = 84 which reduces down to
(0.6 - 0.25)x + 42 = 84 then simplifying further gives
0.35x = 42 so dividing both sides we get
x = 120 and since T = 168 then y = 168 - 120 gives 48 so y = 48
now if we check our answers by plugging the values back in
0.6(120) + 0.25(48) = 0.5(168) we get
72 + 12 = 84 or
84 = 84 so that checks out.

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