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 100 mins. In how much time the 3rd pipe alone can empty the tank? A) 100 B) 200 C) 300 D) 400

1 Answer

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

Related questions

Description : Two tap can fill a dumper in 7 hrs and 8 hrs respectively. The taps are opened simultaneously and it is found that due to leakage in the bottom it took 16 mins more to fill the dumper. When the dumper is full, in what time will the leak empty it? A) 56hrs B) 56hrs C) 46hrs D) 36hrs

Last Answer : A Work done by 2 taps in 1 hr = (1/7 + 1/8) = 15/56 Time taken by these taps to fill the dumper = 56/15 hrs =3 hrs 44 mins Due to leakage, time taken = 3 hrs +44mins+16 mins =4hrs Work done by (2 ... the leak in 1 hr = (15/56 - ¼) =15 - 4/56 =1/56 Leak will empty the full cistern in 56 hrs.

Description : 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

Last 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

Description : A tank can be filled by a tap in 8 hrs while it can be emptied by another tap in 18 hrs. if both the taps are opened simultaneously, then after how much time will the cistern get filled? A) 14hrs 24mins B) 14 hrs 12 mins C) 14hrs 56 mins D) 14hrs 30mins

Last Answer : A Net part filled in 1hr = (1/8 – 1/18) = 9 – 4/72 = 5/72 Therefore the tank will be filled in 72/5 hrs i.e., 14 hrs 24mins

Description : 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

Last 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.

Description : Two taps can fill a bunker in 40 and 48 mins respectively and a waste tap can empty 6 gallons per minute. All the 3 taps working together can fill the bunker in 30 mind. The capacity of the bunker is. A) 480 B) 480 C) 680 D) 780 

Last Answer : A Work done by the waste tap in 1 min = 1/30 – (1/40 + 1/48) = 8 – (6+5)/240 = 8-11/240 = -3/240 = -1/80 (negative sign means emptying) Volume of 1/80 part = 60 gallons Volume of whole = (6*80) = 480 gallons

Description : Two pipe P & Q fill a tab in 5 hrs and 20 hrs respectively. If both the pipes are open then due to the leakage, it took 30mins more to fill the tab. If the tab is full, how long will it take for the leakage alone to empty the tank? A) 28 B) 24 C) 36 D) 12

Last Answer : C Part filled by(P+Q) in 1 hr = 1/5 + 1/20 = ¼ So P & Q together can fill the tank in 4 hrs. Work done by the leak in 1 hr = 1 / 4 - 2/9 Leak will empty the tank in 36 hrs.

Description : Taps P, Q and R can fill a tank in 3, 4 and 5 hours respectively. If all the taps are opened together and after 30 minutes taps Q and R are turned off, find the total time in which the tank is full. A) 2(3/8)hrs B) 1(1/7)hrs C) 2(13/40)hrs D) 3(13/43)hrs

Last Answer : C In 1 hr P, Q, R = 1/3+1/4+1/5 = 20+15+12/60 = 47/60 Filled in 30m = 47/120 Remaining = 1-47/120 =73/120 Tap P = 3*73/120= 219/120 Total = 219/120+1/2 =219+60 /120= 279/120= 2 13/40 hrs

Description : A water cistern is two-fifth full. Tap A can fill a cistern in 20 mins and tap B can empty it in 12mins. If both the taps are open, how long will take to empty or fill the cistern completey? A) 10 mins B) 12mins C) 14mins D) 16 mins

Last Answer : B Clearly tap B is faster than the Tap A and so, the cistern will be emptied Part to be emptied = 2/5 Part emptied by (A+B) in 1 min = (1/12 – 1/20) = 5-3/60 = 2/60 =>1/30 Therefore 1/30 : 2/5 :: 1;x x= 1*2/5 *30/1 = 12mins so the tank will be emptied in 12mins.

Description : 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

Last 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

Description : One tap can fill a cistern four times as fast as another tap. If together the two taps can fill the cistern in 48 mins then the slower tap alone will be able to fill the cisterns. A) 1hr B) 2hr C) 3hr D) 4hr

Last Answer : D Let the slower tap alone fill the cistern in x mins Then faster tap will fill it in x/4 mins Therefore 1/x +4/x = 1/48 5/x = 1/48 X = 48*5 x = 240 mins x = 4 hrs.

Description : A tank has two faucet which fill it in 15 mins and 18 mins respectively. There is also a waste faucet in the tank. When all the 3 are opened the empty tank is full in 25 mins. How long the waste faucet take to empty the full tank? A) 450/17 mins B) 450/28 mins C) 450/37 mins D) 450/22 mins

Last Answer : C Work done by the waste faucet in 1 minute =>1/25 – (1/15 + 1/18) =>1/25 – (6+5/90) = 1/25 – 11/90 => - 37/450 Here negative sign means emptying Therefore waste faucet will empty the full tank in 450/37 mins.

Description : 3 faucet A,B,C can fill a cistern from empty to full in 45mins, 35mins, and 25 mins respectively. When the cistern is empty all the 3 faucets are opened. A,B,c discharge chemical solutions A,B,C respectively. What is ... the liquid in the tank after 5 mins? A) 63/143 B) 72/151 c) 67/173 D) 81/167

Last Answer : A Part filled by (A+B+C) in 5 mins = 5(1/45 +1/35 + 1/25) = 5 (35+45+63/1575) = 143/315 Part filled by c in 5 mins = 5/25 = 1/5 Required ratio = 1/5 * 315/143 =63/143.

Description : Two taps M and N can alone fill a tank in 80 minutes and 120 minutes respectively. But due to a leakage of tank, it took 12 more minutes to fill the tank. In how many hours, the leak can alone empty the full tank? A) 240 B) 230 C) 248 D) 256

Last Answer : A A and B can fill tank in (1/80 + 1/120) = 1/48 so 48minutes But it took 12 more minutes, this means the tank got full in 48+12= 60 minutes So (1/80 + 1/120 – 1/x) = 1/60 1/X=1/80+1/120-1/60 Solve, x = 240

Description : Two faucet A and B can fill a tank in 20 hours and 40 hours respectively. If they are opened simultaneously. Sometimes later, tap B was closed, then it takes total 14 hours to fill up the whole tank. After how many hours B was closed? A) 4 hours B) 15.2 hours C) 12 hours D) 17.6 hours

Last Answer :  A Let x is the time when B is closed X(1/20+1/40)+14/20=1 X=4 hours

Description : Two taps A and B are opened together to fill a tank. Both the taps fill the tank in time x If B separately took 25 minutes more time than x to fill the tank and B took 49 minutes more time than x ... the tank, then find out the value of x? A) 48 minutes B) 24 minutes C) 54 minutes D) 35 minutes

Last Answer : D Time is taken to fill the tank by both taps x = √a*b x = √25*49 => 5*7 => 35

Description : Two faucet P and Q can fill a tank in 10 min. and 20 min. respectively. A water faucet R can empty the tank in 10 min. First P and Q are opened. After 3 min, R is also opened. In how much time, the tank is full? A) 17m B) 15m C) 11m D) 19m

Last Answer :  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/ ... (1/20). Now,(1/20) part is filled in one minute. (11/20) part is filled in = (20*(11/20))=11minutes.

Description : A tank is 1/4th full. Two pipes which fill the tank in 30minutes and 40 minutes respectively are opened simultaneously. After 10 minutes, a third pipe which empties the full tank in 60 minutes is also opened. In how many minutes the tank will be full? A) 14 B) 12 C) 15 D) 17

Last Answer : A Since 1/4th is already filled, 3/4th is to filled now. (1/30 + 1/40)*(10+x) – (1/60)*x = ¾ 7/120(10+x)-x/60=3/4 7x-2x/120=3/4-7/12 5x/120=1/6 30x=120 X=4 So total 10+4=14minutes

Description : One tap can fill a tank thrice as fast as another tap. If together the two taps can fill the tank in 12 minutes, then the slower tap alone will be able to fill the tank in A) 30min B) 33min C) 32min D) 35min

Last Answer : C (1/x)+(1/3x)=(1/12) 3+1/3x = 4/3x = 1/12 3x = 48 x=32mins

Description : A tank is filled by 3 taps K, L and M with uniform flow. The second tap L takes 3 times the time taken by K to fill the tank, while M takes two times the time taken by L to fill the tank. If all ... hours, find the time required by tap K alone to fill the tank. A) 10hrs B) 21hrs C) 14hrs D) 22hrs

Last Answer : B 1/x + 1/ (3x) + (1/6x) = 1/14 9/6x= 1/14 6x/9 = 14 6x=126 X=126/6 X=21

Description : 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

Last 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

Description : A large cistern can be filled by 2 pipes P & Q in 120 mins and 80 mins respectively. How many mins will it take to fill the cistern from empty state if Q is used for half the time and P & Q fil it together for the other half? A) 2hrs B) 1hr C) 3hrs D) 4hrs

Last Answer : B Part filled by (P+Q) in 1 min = 1/120 + 1/80 =2+3/240 = 5/240 = 1/48 Suppose the cistern is filled in x mins then, x/2 (1/48 + 1/80) = 1 x/2(5+3/240)=1 8x/480 = 1 x=480/8 x= 60 mins or 1 hr

Description : A tank can be filled by an inlet tap at the rate of 8 litres per minute. A leak in the bottom of a tank can empty the full tank in 16 hours. When the tank is full, the inlet is opened and due to ... hours. How many litres does the tank hold? A) 8000 litre B) 9560 litre C) 8525 litre D) 9600 litre

Last Answer : D Part emptied by the leak in 1 hour = 1/16 part filled by (leak & inlet open) in 1 hour = 1/80 Part filled by the inlet tap in 1 hour = 1/16 – 1/80 = 1/20 Inlet tap fills the tank in = 20 hours Inlet tap fills water at the rate of 8 litres a minute. Capacity of tank = 20 * 60 * 8= 9600 litre

Description : Two tube A and B can fill a cistern in 48 mins and 64 mins. If both the tubes are opened simultaneously, after how much time B should be closed so that the cistern is full in 36 minutes? A) 20 mins B) 16mins C) 22mins D) 14mins

Last Answer : B Let B be closed after ‘x’ mins. Then part filled by (A+B) in x mins + part filled by A in (36 – x) mins = 1 Therefore x(1/48 + 1/64) + (36 – x)* 1/48 = 1 7x/192 + 36 – x/48 = 1 7x + 4(36-x)/192 = 1 7x+144 – 4x = 192 3x = 48 =>x = 16 Here B must be closed after 16 mins.

Description : Faucet A and B can fill a tank in 15 and 9 hrs respectively. Faucet C can empty it in 45 h. The tank is half full. All the three faucets are in operation simultaneously. After how much time the tank will be full ? A) 1(7/15)hrs B) 2(1/11)hrs C) 3(3/14)hrs D) 3(3/11)hrs

Last Answer : C In 1 hr = 1/15+1/9 – 1/45 =>3+5-1/45 = 7/45 ½ tank filled by 3 Faucets = 45/7*1/2 => 45/14 =>3(3/14)

Description : Pipes P, Q and R which fill the tank together in 12 hours are opened for 2hours after which pipe R was closed. Find the number of hours taken by pipe R to fill the tank if the remaining tank is filled in 14 hours. A) 16 B) 14 C) 20 D) 42

Last Answer : D 1/P + 1/Q + 1/R = 1/12 Now given that first all open for 2hours, then R closed and P+Q completes in 14 hours, so (1/P + 1/Q + 1/R) *2 + (1/P + 1/Q)*14 = 1 Put 1/P + 1/Q = 1/12 – 1/R (1/12 – 1/R + 1/R) *2 + (1/12 – 1/R)*14 = 1 1/6+ 14/12 – 14/R = 1 Solve, R = 42

Description : Two filling tap can fill a dumper in 14 and 28 min. respectively and when the waste tap is open, they can together fill it in 35 min. The waste pipe can empty the full dumper in – A) 152/9 hrs. B) 137/7 hrs. C) 60/9 hrs. D) 140/11 hrs.

Last Answer :  D 1/14 + 1/28 – 1/35 =>10 + 5 – 4/140 =>11/140 =>140/11 => hence answer is 140/11 hours.

Description : A cistern is filled in 15 hours by three pipes P, Q and R. The pipe R is thrice as fast as Q and Q is thrice as fast as P. How much time will pipe P alone take to fill the tank? A) 120 hrs B) 195 hrs C) 135 hrs D) Cannot be determined 

Last Answer : B Suppose pipe P alone takes x hours to fill the tank. Then, pipes Q and R will take x/3 and x/9hours respectively to fill the tank. 1/x + 3/x + 9/x = 1/15 13/x = 1/15 => x = 195 hrs.

Description : Two taps can fill a tank in 4 hours and 5 hours .If two taps are operate simultaneously, In how much time will the tank be filled ? A) 4hrs 18min B) 10min 12min C) 2hrs 13min D) 12hrs 10min

Last Answer : C 1/(A+B) =1/4 + 1/5 = (5+4)/20 = 9/20 =>20/9=2hrs 13min to filled the tank

Description : Tap 1 can fill a tank in 10 hrs , tap 2 in 20 hrs and tap 3 in 60 hrs. if all the taps are open, in how many hour will the tank be filled? A) 3 hrs B) 5 hrs C) 7hrs D) 6 hrs

Last Answer : D Net part filled in 1 hr = 1/10 + 1/20 + 1/60 = 6 + 3 + 1/60 = 10/60 = 1/6 Therefore all the 3 taps together fill tank in 6 hrs

Description : Three taps P,Q,R can fill a bunker in 3 hrs. After working at it together for 1 hr, R is closed and P,Q can fill the remaining part in 3 hrs. The number of hrs taken by R alone to fill the bunker. A) 9hrs B) 7hrs C) 11hrs D) 5hrs

Last Answer : A  Part filled in 1 hr = 1/3  Remaining part = 1 – 1/3 = 2/3  (P + Q)’s 3 hours work = 2/3  (P+Q)’s 1 hour work = 2/9  R’s 1 hr work = [(P+Q+R)’s 1 hr work – (P+Q)’s 1 hr work]  = 1/3 – 2/9 = 3-2/9 = 1/9  R alone can fill the tank in 9 hours.

Description : Two faucet can separately fill a cistern 30minutes and 45 minutes respectively and when the waste faucet is open, they can together fill it in 54 minutes. The waste faucet can empty the full cistern in? A) 27 min B) 23 min C) 23 min D) 29 min 

Last Answer : A 1/30 + 1/45 - 1/x = 1/54 1/x=1/30+1/45-1/54 1/x=9+6-5/270 1/x=10/270 1/x=1/27 x = 27

Description : Two pumps M and N can fill a cistern in 10m and 15m respectively. If both the pumps are opened simultaneously, after how much time should N be closed so that the cistern is full in 8 minutes ? A) 5min B) 6min C) 3min D) 7min

Last Answer :  C X(1/10+1/15) +(8-x)1/10 = 1 5x/30+8-x/10 =1 5x+24-3x/30 =1 2x+24 = 30 2x=6 X=6/2 = 3

Description : Two pumps M and N can separately fill a dumper in 36 minutes and 45 minutes respectively. Both the pumps are opened together but 12 minutes after the start the pump M is turned off. How much time will it take to fill the dumper? A) 9 min B) 10 min C) 30 min D) 32 min

Last Answer : C 12/36 + x/45 = 1 (12*5)/180+4x/180=1 60+4x=180 4x=120 X=120/4=30min

Description : two pipes X and Y can fill a cistern in 18 hrs and 24 hrs respectively. If both the pumps are opened simultaneously. How much time will be taken to fill the cistern? A) 10hrs 50mints B) 10 hrs 5mints C) 10hrs 17mints D) 10hrs 44mints

Last Answer : C Part filled by x in 1hour = 1/18 Part filled by y in 1 hour = 1/24 Part filled by (x+y) in 1 hour = 1/18 + 1/24 =>4+3/72 =>7/72 The cistern will be full in 72/7 hrs = 10 hrs 17mins.

Description : A pipe can fill the tank in 12 hours.Because of a leak in the tank it took 16(1/2) hours to fill the tank.If the tank is full,how much time will the leak take to empty it? A) 18hrs51mins B) 18hrs 20min C) 18hrs 40min D) 18hrs 45min E) None of these

Last Answer : A =>1/12-2/33 = 11-4/132. ==>7/132 =>132/7 hrs =>132/7 hrs=>1131 mins =>18hrs 51min

Description : A dumper is filled in 15hrs by 3 tubes P,Q,and R. the tube R is thrice as fast as Q and Q is thrice as fast as P. How much time will tube Q alone take to fill the tank? A) 195hrs B) 190hrs C) 185hrs D) 180hrs

Last Answer : A Suppose tube P alone takes x hrs to fill the tank Then tubes Q and R will take x/3 and x/9 hrs respectively to fill the dumper. Therefore 1/x + 3/x+ 9/x = 1/15 13x = 1/15 X = 195 hrs

Description : Faucet A basically used as inlet pipe and Faucet B is used as outlet pipe. Faucet A and B both are opened simultaneously, all the time. When Faucet A fills the tank and Faucet B empty the tank, it will take thrice the ... efficiency of Faucet A and Faucet B respectively? a) 1:2 b) 2:1 c) 1:3 d) 3:1

Last Answer : B Efficiency when both pipes used to fill = A + B And efficiency when Faucet A is used to fill and Faucet B is used to empty the tank = A-B So (A+B)/(A-B)=3/1 A/B+1/(A/B-1)=3/1 A/B+1=3(A/B-1) 3A/B-A/B=4 2A/B=4 A/B=2 Thus, the ratio of efficiency of Faucet A and B =2:1

Description : A tank can be filled by a tap in 6 hours and emptied by an outlet pipe in 8 hours. How long will it take to fill the cistern if both the taps are opened together? 1. 24 hours 2. 22 hours 3. 21 hours 4. 18 hours 5. 20 hours

Last Answer : 1;

Description : Two pipes, Xand Y can fill a tank in 36 and 30 minutes respectively. Both are opened together, but at the end of 3minutes, X is turned off. In how many more minutes will Y fill the cistern? A) 7(24/5) B) 7(1/2) C) 24(1/3) D) 8(1/4) 

Last Answer : C X can fill cistern in 36 minutes. X fills cistern in 1 minute = 1/36 Y can fill cistern in 30 minutes. Y fills cistern in 1 minute = 1/30 part. Xand Y together can fill cistern in 1 minute, = {(1/ ... part. 147/180 part cistern must be filled by Y in, [(147/180)/(1/30)] = 24(1/3) minutes.

Description : A Bunker has a leak which would empty the completely filled bunker in 20 hours. If the bunker is full of water and a tap is opened which admits 4 litres of water per minutes in the bunker, how many litres does the bunker holds? A) 2400 B) 4500 C) 4800 D) 7200

Last Answer : C Leak emptied the bunker per hour = 100/20 = 5% of water per hour; Quantity of water emptied per hour = 4*60 = 240; Thus, 5% = 240 liter; Hence, capacity of water, 100% = 240*100/5= 4800 liter.

Description : A tap can fill a cistern in 11hours, but due to a leakage it took 13hours to fill the cistern. If the cistern is full, in what time will the cistern become empty due to leakage ? A) 73.30hrs B) 77.50hrs C) 71.5 hrs D) 72.30hrs E) None of these

Last Answer : C 1/11-1/13 = 13-11/143 =>2/143 =>1/71.5 =>71.5hours

Description : A tap can fill a bunker in 4 hours. After half the bunker is filled, three more similar tap are opened. What is the total time taken to fill the bunker completely? A) 3 hours B) 2.5 hours C) 5 hours D) 4.2hours

Last Answer : B In One hour tap can fill = 1/4 Time is taken to fill half of the bunker = 1/2 * 4= 2hours Part filled by four taps in one hour = (4*1/4) = 1 Required Remaining Part = 1/2 Total time = 2+ 1 /2=2.5 hrs

Description : Two faucet can fill a tank in 12hours and 16 hours .While a third faucet empties the full tank in 24 hours.If all the three faucet are operate simultaneously, In how much time will the tank be filled ? A) 4 hours 12mins B) 4hours 48min C) 9hours 36mins D) 5hours 48min

Last Answer : C  1/12+1/16 – 1/24 = (4+3-2)/48  =>5/48  Time taken to fill the tank = 48/5 = 9hours 36min

Description : A leak in the bottom of a bunker can empty the full bunker in 4 hrs. An inlet pump fills water at the rate of 3 litres a min. when the bunker is full, the inlet is opened and due to the leak, ... hrs. How many litres does the bunker field? A) 1160 litrs B) 1610 litres C) 2160 litres D) 2610 litres

Last Answer :  C Work done by the inlet in 1 hr = (1/4 – 1/6) = (3-2/12) = 1/12 Work done by the inlet in 1 min = 1/12 * 1/60 = 1/720 Volume of 1/720 part = 3 litres Volume of whole = (720 * 3 )=2160 litres.

Description : Three pipes P, Q and R can fill a tank from empty to full in 40 minutes, 15 minutes, and 30 minutes respectively. When the tank is empty, all the three pipes are opened. P, Q and R discharge chemical solutions X, Y and Z respectively. ... 5 minutes? a) 8 / 15 b) 7 / 15 c) 8 / 17 d) 6 / 13 e) 8 / 13

Last Answer : a Part of the tank filled by pipe P in 1 minute = 1 / 40 Part of the tank filled by pipe Q in 1 minute = 1 / 15 Part of the tank filled by pipe R in 1 minute = 1/ 30 Here we have to find the proportion of ... together in 5 minute = 5 1/8 = 5/ 8 Required proportion = (1/3) / ( 5/8) = 8 / 15

Description :  3 faucet P,Q & R can fill a cistern in 12 hrs. After working together for 4hrs, R is closed P and Q can fill the remaining part in 14 hrs. The number of hrs taken by R alone to fill the cistern is. A) 18 hrs B) 20 hrs C) 28hrs D) 30hrs

Last Answer : C  Part filled in 4 hrs = 4/12 = 1/3 Remaining part = 2/3 (P+Q)’s 14 hrs work = 2/3 (P+Q)’s 1 hr work = 2/ 3 *14 = 1/21 R’s 1 hr work = (P+Q+R)’s 1 hr work – (P+Q)’s 1 hr work = 1/12 – 1/21 = 7-4/84 = 3/84 = 1/28 R alone can fill the tank in 28 hrs.

Description : A dumper is normally filled in 18 hours but takes 6 hours longer to fill because of a leak in the bottom of the dumper. If the dumper is full the leak will empty it in how many hours? A) 76 hours B) 78 hours C) 72 hours D) 74 hours

Last Answer : C Work done by leak in 1 hr=(1/18-1/24) =>4-3/72 =>1/72 Leak will empty the dumper in 72 hours

Description : Two faucet A and Bcan fill a cistern in 6 hours and 2 hours respectively. If they are opened on alternate hours and if faucet A is opened first, in how many hours will the cistern be full? A) 4 hours B) 5 hours C) 7 hours D) 6 hours

Last Answer : A Faucet A can fill = 1/6 faucet B can fill = 1/2 For every two hour, 1/6 + 1/2 = 1+3/6 =>4/6=2/3 Part filled Total filled in 3 hours = 2/3+1/6= 5/6 In next hour it will be fille full. So total time will be 4 hours.

Description : A cistern is filled by three faucets with uniform flow. The first two faucets operating simultaneously fill the cistern in the same during which the cistern is filled by the third faucet alone. The second faucet fills the cistern ... by the first faucet is? A) 6 hrs B) 10 hrs C) 15 hrs D) 30 hrs

Last Answer : C Suppose, first faucet alone takes x hours to fill the cistern. Then, second and third faucets will take (x - 5) and (x - 9) hours respectively to fill the cistern. 1/x + 1/(x - 5) = 1/(x - 9) (2x - 5)(x - 9) = x(x - 5) x2 - 18x + 45 = 0 (x- 15)(x - 3) = 0 => x = 15

Description : A tube can fill a tank completely in 18 hours. After half the tank is filled , one more similar tube is opened. What is the total time taken to fill the tank completely ? A) 14hrs 20min B) 13hrs 30min C) 13hrs 10min D) 14hrs 30min

Last Answer : C A tube can fill the half tank in 9hrs Now another similar tube opened 1/18+1/18 = 2/18 = 1/9 Remaining half tank filled in 4.5hrs Total time = 9+4.5 =13.5 = 13hrs 30min