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Difficulty: HardAlgebraic Word Problems and Modeling

Container X contains a liquid solution that is 80%80\% alcohol by volume, and Container Y contains a liquid solution that is 25%25\% alcohol by volume. A chemist removes a specific volume of solution from Container X and mixes it with a solution from Container Y to produce 100100 liters of a new mixture that is 58%58\% alcohol by volume. Following this removal, 2020 liters of pure alcohol are added to the liquid remaining in Container X. If Container X initially held 120120 liters of solution, what is the concentration of alcohol, by volume, in Container X after the pure alcohol is added?

  1. A
    56.67%56.67\%
  2. B
    68%68\%
  3. C
    70%70\%
  4. D
    80%80\%
  5. 85%85\%Answer

Answer

85%
To find the final concentration, we first determine the volume of solution removed from Container X. Using the weighted average for the 100-liter mixture: 0.80Vx+0.25(100−Vx)=580.80 V_x + 0.25 (100 - V_x) = 58, which simplifies to 0.55Vx=330.55 V_x = 33, so Vx=60V_x = 60 liters. Container X originally held 120 liters, so removing 60 liters leaves 60 liters of solution containing 80%×60=4880\% \times 60 = 48 liters of alcohol. Adding 20 liters of pure alcohol increases the total alcohol to 48+20=6848 + 20 = 68 liters and the total volume to 60+20=8060 + 20 = 80 liters. The final concentration is 6880=85%\frac{68}{80} = 85\%.

Step-by-Step Solution

1
Find the volume of solution VxV_x removed from Container X to make the 100-liter mixture.
Vx=60V_x = 60 liters.
Let VxV_x be the volume from X and 100−Vx100 - V_x be the volume from Y. Setting up the alcohol concentration equation: 0.80Vx+0.25(100−Vx)=0.58(100)  ⟹  0.55Vx+25=58  ⟹  0.55Vx=33  ⟹  Vx=600.80 V_x + 0.25 (100 - V_x) = 0.58(100) \implies 0.55 V_x + 25 = 58 \implies 0.55 V_x = 33 \implies V_x = 60 liters.
2
Determine the remaining solution volume and alcohol volume in Container X after removing 60 liters.
Remaining solution = 6060 liters; Remaining alcohol = 4848 liters.
Container X initially had 120120 liters. Removing 6060 liters leaves 120−60=60120 - 60 = 60 liters. Since the mixture is homogeneous, the remaining liquid retains an 80%80\% alcohol concentration, yielding 0.80×60=480.80 \times 60 = 48 liters of alcohol.
3
Calculate the total alcohol volume and total solution volume in Container X after adding 20 liters of pure alcohol.
New alcohol volume = 6868 liters; New total solution volume = 8080 liters.
Adding 2020 liters of pure alcohol increases both the alcohol amount (48+20=6848 + 20 = 68 liters) and the total liquid volume (60+20=8060 + 20 = 80 liters).
4
Calculate the final concentration of alcohol in Container X.
Concentration = 6880=0.85=85%\frac{68}{80} = 0.85 = 85\%.
The final alcohol concentration is the ratio of final alcohol volume to final total liquid volume.

Key Concept

Algebraic Modeling of Multi-Step Mixture Solutions
Estimated Time:2m 30s
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Topics

Algebraic Word Problems and Modeling

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