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MalachiKeti
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20% of a milk and water solution with 80% water 20 Dec 2024, 02:32
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C
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45% (medium)

Question Stats:

75% (02:33) correct 25% (02:40) wrong based on 85 sessions

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20% of a milk and water solution with 80% water is replaced with water. The same process is repeated two more times. What is the concentration of water in final solution?

A. 70.12%
B. 85.2%
C. 89.76%
D. 78.12%
E. 96%
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C
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skry209
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20% of a milk and water solution with 80% water 21 Dec 2024, 07:52
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% of milk left in the final solution = 20(1-0.2)^3 = 10.24%
% of water left in the final solution = 100 - 10.24 = 89.76%
Option C
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parthwagh
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20% of a milk and water solution with 80% water 20 Dec 2024, 10:21
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MalachiKeti
20% of a milk and water solution with 80% water is replaced with water. The same process is repeated two more times. What is the concentration of water in final solution?

A. 70.12%
B. 85.2%
C. 89.76%
D. 78.12%
E. 96%
We are tasked with finding the final concentration of water in the solution after replacing 20% of it with water three times. Here's a step-by-step explanation:
Initial Information:
  1. The solution initially contains 80% water and 20% milk.
  2. 20% of the solution is replaced with water.
Step 1: Calculate the effect of the first replacement.
  • Initial water content = 80%80\%80%.
  • After removing 20% of the solution, the remaining solution is 80% of the original, and it retains the same water-to-milk ratio:
    • Water retained = 80%×80%=64%80\% \times 80\% = 64\%80%×80%=64% of the original volume.
  • The 20% that was removed is replaced with pure water, which adds 20%20\%20% water to the solution.
New water concentration = 64%+20%=84%64\% + 20\% = 84\%64%+20%=84%.
Step 2: Calculate the effect of the second replacement.
  • The solution now contains 84% water.
  • Again, 20% of the solution is removed and replaced with pure water:
    • Water retained = 80%×84%=67.2%80\% \times 84\% = 67.2\%80%×84%=67.2%.
    • Water added = 20%20\%20% (from the replacement with pure water).
New water concentration = 67.2%+20%=87.2%67.2\% + 20\% = 87.2\%67.2%+20%=87.2%.
Step 3: Calculate the effect of the third replacement.
  • The solution now contains 87.2% water.
  • Again, 20% of the solution is removed and replaced with pure water:
    • Water retained = 80%×87.2%=69.76%80\% \times 87.2\% = 69.76\%80%×87.2%=69.76%.
    • Water added = 20%20\%20% (from the replacement with pure water).
New water concentration = 69.76%+20%=89.76%69.76\% + 20\% = 89.76\%69.76%+20%=89.76%.
Final Answer:
The concentration of water in the final solution is 89.76%, which corresponds to Option C.
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