Math, asked by spoken0281, 8 months ago

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 Q
liters per minute and water is being added to swimming pool B at the rate
of G liters per minute. , How many hours will elapse before the two tanks
contain equal amounts of water?​

Answers

Answered by isyllus
13

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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