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kevincan
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Verynice problem.

Firstly using allegations we get two equations:

x-40/10 = a/b
x-55/25 = a/4b.

Making 2nd equation equal to a/b:
4x-220/25 = a/b.

Equations equation 1 and 2:
x = 80.

Thus concentration of the stronger acid = 80%

Again using allegation we get the ratios initially:
4:1
2nd pour:
1:1 or 4:4.

Which means 1 litre corresponds to 4x, thus x = 1/4litres.

Thus each glass = 250ml.

Totally 8 glasses poured eventually = 2litres of stronger acid.

So we have 1 litre of 30% acid and 2 litres of 80% acid.

30*1+80*2 / 3 = 190/3.

30ml is poured out, thus:

190/3*100 * (30) = 19ml.

30ml contains 19ml of acid solution.

====================================
kevincan
A 1-liter solution is 30% acid. One glass of a stronger acid solution is added, producing a mixture that is 40% acid. Three more glasses of the stronger solution are then added, producing a mixture that is 55% acid. Finally, four additional glasses of the stronger solution are added.

If 30 mL of the final mixture is removed, how many milliliters of acid does it contain?

A. 19 mL
B. 20 mL
C. 21 mL
D. 22 mL
E. 24 mL
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