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  • Three identical inlet pipes were opened at 6:00 AM to fill up an empty container. At 8:30 AM, 2 identical pipes (having different rate from existing pipe’s rate) are also opened. At 9:30 AM, it still required 1 more hour for all these pipes to fill up the remaining 30 m3 of water. If each of the 2 pipes which were added at 8:30 AM can individually fill up the empty container in 40 hours, then the capacity of the container is ____________ m3.

in Numerical Ability by (195 points)
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2 Answers

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120 m3
by (21 points)
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Let y be the rate per hour of the initial 3 taps.

The 2 identical new taps can individually fill the tank in 40 hours.

Let the capacity of tank be x m^3.

Therefore, the rate per hour of new taps = x/40.


2*x/40 + 3*y = 30

Since all the taps when kept open for 1 hour fill 30 m^3 water, then the 3 initial taps will fill (x-60) m^3 in 2.5 hours.


3*2.5*y = x - 60

On solving the above two linear equations, we get

x = 120 m^3 and y = 8 m^3/hr.
by (417 points)

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