Swimming Pools in club A and club B contain 500 and 200 liters of wate

1.	Swimming Pools in club A and club B contain 500 and 200 liters of water,respectively. Water is pumped out of swimming pool A at the rate of Qliters per minute and water is being added to swimming pool B at the rateof G liters per minute. , How many hours will elapse before the two tankscontain equal amounts of water?
Answer» Given: Water in swimming POOL of club A = 500 litres Water in swimming pool of club B = 200 litres Rate at which water is pumped out from A = Q litres/min Rate at which water is pumped out from B = G litres/min To find: Number of HOURS at which volume of both the pools become equal. Solution: LET 't' be the number minutes taken so that the volume of the two pools become equal. Volume taken out from pool A after 't' minutes: $Q \times t$ So, volume left in pool A after 't' minutes = 500- $Q\times t$ Volume PUT to pool B after 't' minutes: $G \times t$ So, volume of pool B after 't' minutes = 200 + $G \times t$ Now, let US put both the volumes equal: $500-Qt=200+Gt\\\Rightarrow 300 = Gt+Qt\\\Rightarrow 300 = (G+Q)t\\\Rightarrow t = \dfrac{300}{G+Q}\ minutes\\\Rightarrow t = \dfrac{300}{60\times (G+Q)}\ hours\\\Rightarrow t = \dfrac{5}{G+Q}\ hours$ So, after $\frac{5}{G+Q}\ hours$ the volume of the two pools will be equal.

Swimming Pools in club A and club B contain 500 and 200 liters of water,respectively. Water is pumped out of swimming pool A at the rate of Qliters per minute and water is being added to swimming pool B at the rateof G liters per minute. , How many hours will elapse before the two tankscontain equal amounts of water?

Answer»

Given:

Water in swimming POOL of club A = 500 litres

Water in swimming pool of club B = 200 litres

Rate at which water is pumped out from A = Q litres/min

Rate at which water is pumped out from B = G litres/min

To find:

Number of HOURS at which volume of both the pools become equal.

Solution:

LET 't' be the number minutes taken so that the volume of the two pools become equal.

Volume taken out from pool A after 't' minutes: $Q \times t$

So, volume left in pool A after 't' minutes = 500- $Q\times t$

Volume PUT to pool B after 't' minutes: $G \times t$

So, volume of pool B after 't' minutes = 200 + $G \times t$

Now, let US put both the volumes equal:

$500-Qt=200+Gt\\\Rightarrow 300 = Gt+Qt\\\Rightarrow 300 = (G+Q)t\\\Rightarrow t = \dfrac{300}{G+Q}\ minutes\\\Rightarrow t = \dfrac{300}{60\times (G+Q)}\ hours\\\Rightarrow t = \dfrac{5}{G+Q}\ hours$

So, after $\frac{5}{G+Q}\ hours$ the volume of the two pools will be equal.

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