C
Part filled in 3 min. = 3*((1/10)+(1/20))
= 3 * (2+1/20)
=>3(3/20) = 9/20
Remaining part=(1-(9/20))=(11/20).
Net part filled in 1min. when P,Q and R are opened=(1/10)+(1/20)-(1/10)=(1/20).
Now,(1/20) part is filled in one minute.
(11/20) part is filled in = (20*(11/20))=11minutes.
Two faucet p and q can fill a tank in 10 minutes and 20 minutes.If both faucet are opened simultaneously , after how much time should q be closed so that the tank is full in 9 minutes ? A) 2 min B) 9 min C) 4 min D) 7min Answer : A p fill the tank in 1 minute (10×2 = 20) = 2 units q fill the tank in 1 minute (20×1 = 20) = 1units For 9 min(p) = 9×2 = 18 units Remaining = 20 – 18 = 2 units Time for q be closed so that the tank is full in 9 minutes = 2/1 = 2 min...
Two pumps P & Q can fill a tank in 12 mins and 15 mins respectively while a third pipe R can empty the full tank in 6 mins. P & Q kept open for 5 mins in the beginning and then R is also opened. In what time is tank emptied? A) 30mins B) 35mins C) 40mins D) 45 mins Answer : D Part filled in 5 min = 5(1/12 +1/15) = 5*9/60 = ¾ Part emptied in 1 min when all the pumps are opened, = 1/6 – (1/12 + 1/15) = 1/6 – 3/20 = 1/60 Now 1/60 is part emptied in 1 min Therefore ¾ part will be emptied in 60 * ¾ = 45 mins...
Two taps P and Q can separately fill a tank in 120mins and 150 mins respectively. There is a 3rd tap in the bottom of the tank to empty it. If all of the 3 taps are simultaneously opened, then the tank is full ... . In how much time the 3rd pipe alone can empty the tank? A) 100 B) 200 C) 300 D) 400 Answer : B Work done by the 3 tap in 1 min = 1/100 – ( 1/120 + 1/150) = 1/100 – (5+4/600) =1/100 – 9/600 =6-9/600 = - 3/600 = - 1/200(negative signs mean empting) Therefore the 3rd tap alone can empty the tank in 200 mins...
Two taps P and Q can fill a cistern in 24 min and 32 min respectively. Both the taps are opened together for a certain time but due to some obstruction the flow of water was restricted to 7/4 of full flow in tap P and ... How long was it before the full flow. A) 8 min B) 3 min C) 5.6 min D) 4.5 min Answer : D Let the obstruction remain for X min. Hence, Part of cistern filled in X min + part of cistern filled in 6 min = full cistern [(7X/4*24)+(5X/3*32)]+[(6/24)+(6/32)] = 1 (12X/96)+(7/16) = 1 12x/96=9/16 Thus, X = 4.5 min....
Three pumps, M, N and O are opened to fill a tank such that M and N can fill the tank alone in 18 min. and 23 min. respectively and O can empty it in 15 min. After 3 minutes the emptying pipe is closed. In how many minutes the tank will be full in this way? A) 20 B) 25 C) 18 D) 12 Answer : D Let the tank full in x minutes, then M and N opened for x minutes and O for 3 minutes. (1/18 + 1/23)*x – (1/15)*3 = 1 (23+18/414)X=1+1/5 Solve, x = 12...
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