Two water containers

In the first container, there are 200 m3 of water and in the second 40 m3. The first container will flow down at a rate of 10 m3 water per hour. At the same time flows to the second rate of 5 m3 per hour. After how many hours there will be three times less water in the first container than in the second?

Correct result:

t =  16 h

Solution:


200 - 10t = (40+5t)/3

200 - 10•t = (40+5•t)/3

35t = 560

t = 16

Calculated by our simple equation calculator.



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