Part filled by (A + B) in 1 minute = (1/60 + 1/40) = 1/24
Suppose the tank is filled in x minutes.
Then, x/2(1/24 + 1/40) = 1
(x/2) * (1/15) = 1 => x = 30 min.
Given A alone can fill the tank of capacity 240 lit in 16 hrs.
=> A can fill in 1 hr = 240/16 = 15 lit
=> B alone can fill the tank of capacity 240 lit in 12 hrs.
=> B can fill in 1 hr = 240/12 = 20 lit
Now, (A + B) in 1 hr = 15 + 20 = 35 lit
But they are opened for 2 hrs
=> 2 x 35 = 70 lit rae filled
Remaining water to be filled in tank of 240 lit = 240 - 70 = 170 lit.
Let the slower pipe alone fill the tank in x minutes.
Then,faster pipe will fill it in x/3 minutes.
Capacity of the tank =(12 x 13.5) liters =162 liters.
Capacity of each bucket =9 liters
Number of buckets needed = 162/9 =18.
Volume of water collected in the tank in 1 hour
? (0.3 × 0.2 × 20km × 1000mts) = 1200 m cubic
If after t hours, the water is at height of 12m,
1200t=200×150×12
? t = 300 Hours.
Given taps X and Y can fill the tank in 30 and 40 minutes respectively. Therefore,
part filled by tap X in 1 minute = 1/30
part filled by tap Y in 1 minute = 1/40
Tap Z can empty the tank in 60 minutes. Therefore,
part emptied by tap Z in 1 minute = 1/60
Net part filled by Pipes X,Y,Z together in 1 minute = [1/30 +1/40 - 1/60] = 5/120 = 1/24
i.e., the tank can be filled in 24 minutes.
Let filling capacity of the pump be 'F'
Then, the filling capacity of the pump will be (F + 10)
From the given data,
Let the pipe B be closed after 'K' minutes.
30/16 - K/24 = 1 => K/24 = 30/16 - 1 = 14/16
=> K = 14/16 x 24 = 21 min.
Let the efficiencies of filling pipes is 4p and 5p respectively.
Efficiency of pipe which empty the tank = 2/3 x 9p/2 = 3p
Total work = 3p x 36 = 108p
Time to fill the tank by both the pipes = 108p/9p = 12 min.
Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the taps in 1 hour = 4 x 1/6 = 2/3
Remaining part = 1 - 1/2 = 1/2
2/3 : 1/2 :: 1 : p
p = 1/2 x 1 x 3/2 = 3/4 hrs. i.e., 45 min
So, total time taken = 3 hrs 45 min.
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