Examkin

Question

Difficulty: Very hardMixture and Concentration

A container initially holds 100100 liters of an acid solution that is 40%40\% acid by volume. First, xx liters of the solution are drained and replaced with an equal volume of pure water. After the mixture is thoroughly stirred, xx liters of the new solution are drained and replaced with an equal volume of pure acid. If the final solution is 45.6%45.6\% acid by volume, what is the value of xx?

Answer: 20 liters

Answer

The value of xx is 2020 liters.
The solution requires tracking the amount of pure acid through two distinct replacement operations. Initially, there are 4040 liters of acid in 100100 liters of solution. Draining xx liters removes 0.4x0.4x liters of acid, leaving (40−0.4x)(40 - 0.4x) liters of acid after adding xx liters of pure water. In the second step, draining xx liters removes a fraction x100\frac{x}{100} of the remaining acid, leaving (40−0.4x)(1−x100)(40 - 0.4x)(1 - \frac{x}{100}) liters of acid. Adding xx liters of pure acid yields a total acid volume of (40−0.4x)(1−x100)+x=45.6(40 - 0.4x)(1 - \frac{x}{100}) + x = 45.6 liters. Expanding and solving the resulting quadratic equation x2+50x−1400=0x^2 + 50x - 1400 = 0 gives x=20x = 20 (since x>0x > 0).

Step-by-Step Solution

1
Determine initial volume of solute (pure acid).
Initial acid volume = 40%×100=4040\% \times 100 = 40 liters.
Establishing the starting amount of pure acid in the 100100-liter container.
2
Model the acid amount after the first replacement (with water).
Acid volume after first replacement = 40(1−x100)40\left(1 - \frac{x}{100}\right) liters.
Draining xx liters removes x100\frac{x}{100} of the total acid, and adding water adds zero acid.
3
Model the acid amount after the second replacement (with pure acid).
Final acid volume = 40(1−x100)(1−x100)+x40\left(1 - \frac{x}{100}\right)\left(1 - \frac{x}{100}\right) + x liters.
Draining xx liters of the new solution removes x100\frac{x}{100} of its acid, and replacing with pure acid adds xx liters of acid.
4
Set up and solve the quadratic equation given final acid volume of 45.645.6 liters.
x2+50x−1400=0  ⟹  (x+70)(x−20)=0  ⟹  x=20x^2 + 50x - 1400 = 0 \implies (x+70)(x-20) = 0 \implies x = 20.
Discarding the negative root (x=−70x = -70) because volume must be positive.

Key Concept

Multi-stage sequential mixture removal and replacement
Rate this question

Topics

Mixture and Concentration

More questions from this topic