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Difficulty: HardLinear Equations in One Variable

A container originally holds a liquid mixture consisting of substance A and substance B, where substance A constitutes 25\frac{2}{5} of the total volume. After 1515 liters of substance B are added to the container and 33 liters of substance A evaporate, the volume of substance A in the container becomes 14\frac{1}{4} of the new total liquid volume. What was the original total volume, in liters, of the liquid mixture in the container?

  1. A
    32
  2. 40Answer
  3. C
    45
  4. D
    50
  5. E
    52

Answer

40 liters
Letting VV represent the original volume of the liquid mixture in liters, the initial amount of substance A is 25V\frac{2}{5}V. After adding 15 liters of substance B and losing 3 liters of substance A to evaporation, the updated volume of substance A is 25V−3\frac{2}{5}V - 3, and the updated total volume is V+15−3=V+12V + 15 - 3 = V + 12. Setting up the relationship 25V−3=14(V+12)\frac{2}{5}V - 3 = \frac{1}{4}(V + 12) and expanding the right side gives 25V−3=14V+3\frac{2}{5}V - 3 = \frac{1}{4}V + 3. Subtracting 14V\frac{1}{4}V from both sides yields 320V=6\frac{3}{20}V = 6, which solves to V=40V = 40 liters.

Step-by-Step Solution

1
Define the unknown variable and express initial quantities algebraically.
Let VV be the original total volume of the liquid mixture in liters. The original volume of substance A is 25V\frac{2}{5}V.
Establishing a variable for the initial total volume allows all changes to be modeled in terms of one variable.
2
Express the modified quantities after additions and evaporation.
New volume of substance A =25V−3= \frac{2}{5}V - 3. New total volume =V+15−3=V+12= V + 15 - 3 = V + 12.
Adding 15 liters of substance B increases the total volume by 15, and losing 3 liters of substance A decreases both substance A and the total volume by 3.
3
Set up the linear equation based on the given ratio condition.
\frac{2}{5}V - 3 = \frac{1}{4}(V + 12)
Substance A forms one-fourth of the updated total liquid volume.
4
Expand and solve the linear equation for VV.
\frac{2}{5}V - 3 = \frac{1}{4}V + 3 \implies \frac{2}{5}V - \frac{1}{4}V = 6 \implies \frac{8 - 5}{20}V = 6 \implies \frac{3}{20}V = 6 \implies V = 40.
Clearing terms and subtracting 14V\frac{1}{4}V from 25V\frac{2}{5}V gives 320V=6\frac{3}{20}V = 6, which yields V=40V = 40.

Key Concept

Linear Equations in One Variable
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Topics

Linear Equations in One Variable

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