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29 Jul 2024, 08:32
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Pipe A, working alone at its constant rate, starts filling an empty pool and then stops. Subsequently, Pipe B, working alone at its constant rate, completes the task and fills the remaining part of the pool. If the entire filling process took 7 hours, and Pipe A pumped in three times as much water as Pipe B, how many hours will it take Pipe B, working alone at its constant rate, to fill the entire pool on its own?
(1) After 4 hours, half of the pool was filled.
(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.
(1) After 4 hours, half of the pool was filled.
(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.
29 Jul 2024, 08:32
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Official Solution:
Pipe A, working alone at its constant rate, starts filling an empty pool and then stops. Subsequently, Pipe B, working alone at its constant rate, completes the task and fills the remaining part of the pool. If the entire filling process took 7 hours, and Pipe A pumped in three times as much water as Pipe B, how many hours will it take Pipe B, working alone at its constant rate, to fill the entire pool on its own?
Pipe A pumping in three times as much water as Pipe B implies that Pipe A pumped in 75% of the pool's capacity, while Pipe B pumped in the remaining 25% of the pool's capacity.
Assuming Pipe A takes \(x\) hours to fill the entire pool and Pipe B takes \(y\) hours to fill the entire pool, Pipe A would require \(\frac{3x}{4}\) hours to fill 75% of the pool, and Pipe B would require \(\frac{y}{4}\) hours to fill the remaining 25% of the pool. Hence, we are given that \(\frac{3x}{4} + \frac{y}{4} = 7\), which simplifies to \(3x + y = 28\) and we need to find the value of \(y\).
(1) After 4 hours, half of the pool was filled.
Given that initially Pipe A filled 75% of the pool, it must have also filled initial 50% of the pool by itself. Hence, this statement implies that Pipe A filled half of the pool in 4 hours. Thus, Pipe A would need 8 hours to fill the entire pool, making \(x\) equal to 8. Substituting \(x = 8\) into \(3x + y = 28\) gives \(y = 4\). Sufficient.
(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.
This implies that the combined rate of Pipe A and Pipe B is three times that of Pipe A alone. Thus, \(\frac{1}{x} + \frac{1}{y} = 3*\frac{1}{x}\). Simplifying this gives \(x = 2y\). Substituting \(x = 2y\) into \(3x + y = 28\) gives \(y = 4\). Sufficient.
Answer: D
Pipe A, working alone at its constant rate, starts filling an empty pool and then stops. Subsequently, Pipe B, working alone at its constant rate, completes the task and fills the remaining part of the pool. If the entire filling process took 7 hours, and Pipe A pumped in three times as much water as Pipe B, how many hours will it take Pipe B, working alone at its constant rate, to fill the entire pool on its own?
Pipe A pumping in three times as much water as Pipe B implies that Pipe A pumped in 75% of the pool's capacity, while Pipe B pumped in the remaining 25% of the pool's capacity.
Assuming Pipe A takes \(x\) hours to fill the entire pool and Pipe B takes \(y\) hours to fill the entire pool, Pipe A would require \(\frac{3x}{4}\) hours to fill 75% of the pool, and Pipe B would require \(\frac{y}{4}\) hours to fill the remaining 25% of the pool. Hence, we are given that \(\frac{3x}{4} + \frac{y}{4} = 7\), which simplifies to \(3x + y = 28\) and we need to find the value of \(y\).
(1) After 4 hours, half of the pool was filled.
Given that initially Pipe A filled 75% of the pool, it must have also filled initial 50% of the pool by itself. Hence, this statement implies that Pipe A filled half of the pool in 4 hours. Thus, Pipe A would need 8 hours to fill the entire pool, making \(x\) equal to 8. Substituting \(x = 8\) into \(3x + y = 28\) gives \(y = 4\). Sufficient.
(2) Working together at their respective constant rates, Pipes A and B can fill the empty pool in one-third of the time it takes Pipe A, working alone at its constant rate.
This implies that the combined rate of Pipe A and Pipe B is three times that of Pipe A alone. Thus, \(\frac{1}{x} + \frac{1}{y} = 3*\frac{1}{x}\). Simplifying this gives \(x = 2y\). Substituting \(x = 2y\) into \(3x + y = 28\) gives \(y = 4\). Sufficient.
Answer: D
