Three distinct pipes A, B, and C can transform an empty tank to full in 30 minutes, 20 minutes, and 10 minutes individually. Upon starting with an empty reservoir, all three valves are activated simultaneously, emitting chemical solutions P, Q, and R respectively. What is the fraction of solution R present in the tank fluid after 3 minutes?
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Answer: B. 6/11
Explanation
Calculating individual pipe contributions over three minutes gives 3 over 30, 3 over 20, and 3 over 10 of the tank. Summing these fractional fillings together results in a total filled amount equal to thirty-three over sixty of the tank. Dividing Pipe R's contribution of three-tenths by the total filled amount simplifies to the exact fraction six over eleven. Pipe filling rates represent the fraction of a total tank volume filled by that pipe in one unit of time. The total amount of fluid in a container is the sum of all individual contributions from active inflow pipes. Fractions of a mixture can be compared by finding a common denominator and simplifying the resulting ratio.
Option A represents an incorrect fraction resulting from an arithmetic error in simplifying the fluid ratio. Option C underestimates Pipe R's proportional contribution to the total fluid volume accumulated in the tank. Option D overestimates the share contributed by Pipe R relative to the combined output of all three pipes.
- Subject
- Mathematics
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- Mathematics
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- medium
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