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There are 3 containers : A, B and C which contain water, milk and acid respectively in equal quantities. 10% of the content of A is taken out and poured into B. Then the same amount from B is transferred to C, from which again the same amount is transferred to A. What is the proportion of milk in container A at the end of the process?
- a)9/10
- b)1/11
- c)1/121
- d)2/13
Correct answer is option 'C'. Can you explain this answer?
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There are 3 containers : A, B and C which contain water, milk and acid...
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There are 3 containers : A, B and C which contain water, milk and acid...
Solution:
Let's assume that each container initially contains x units of liquid.
After taking out 10% from A, the amount of liquid in A becomes 0.9x and the amount of liquid in B becomes 0.1x + x = 1.1x.
When the same amount of liquid from B is transferred to C, the amount of liquid in B becomes 1x and the amount of liquid in C becomes 0.1x + x = 1.1x.
When the same amount of liquid from C is transferred to A, the amount of liquid in C becomes 1x and the amount of liquid in A becomes 0.9x + 0.1x = x.
Now, let's calculate the amount of milk in each container.
Since each container initially contains water, milk, and acid in equal quantities, the proportion of milk in each container is 1/3.
The amount of milk in container A after the first transfer is (1/3) * 0.9x = 0.3x.
The amount of milk in container B after the second transfer is (1/3) * 1x = 0.33x.
The amount of milk in container C after the third transfer is (1/3) * 1.1x = 0.37x.
Finally, when the same amount of liquid from C is transferred back to A, the amount of milk in A becomes (1/3) * (0.9x + 0.1x/3) = 0.03x.
Therefore, the proportion of milk in container A at the end of the process is (0.03x/x) = 1/33 which is option (C).
Hence, the correct answer is option (C).
Let's assume that each container initially contains x units of liquid.
After taking out 10% from A, the amount of liquid in A becomes 0.9x and the amount of liquid in B becomes 0.1x + x = 1.1x.
When the same amount of liquid from B is transferred to C, the amount of liquid in B becomes 1x and the amount of liquid in C becomes 0.1x + x = 1.1x.
When the same amount of liquid from C is transferred to A, the amount of liquid in C becomes 1x and the amount of liquid in A becomes 0.9x + 0.1x = x.
Now, let's calculate the amount of milk in each container.
Since each container initially contains water, milk, and acid in equal quantities, the proportion of milk in each container is 1/3.
The amount of milk in container A after the first transfer is (1/3) * 0.9x = 0.3x.
The amount of milk in container B after the second transfer is (1/3) * 1x = 0.33x.
The amount of milk in container C after the third transfer is (1/3) * 1.1x = 0.37x.
Finally, when the same amount of liquid from C is transferred back to A, the amount of milk in A becomes (1/3) * (0.9x + 0.1x/3) = 0.03x.
Therefore, the proportion of milk in container A at the end of the process is (0.03x/x) = 1/33 which is option (C).
Hence, the correct answer is option (C).
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There are 3 containers : A, B and C which contain water, milk and acid respectively in equal quantities. 10% of the content of A is taken out and poured into B. Then the same amount from B is transferred to C, from which again the same amount is transferred to A. What is the proportion of milk in container A at the end of the process?a)9/10b)1/11c)1/121d)2/13Correct answer is option 'C'. Can you explain this answer?
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