Important Pipe And Cistern Problem Shortcut Tricks SSC CGL SSC CHSL Exam Type-1

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Pipe And Cistern Problem Shortcut Tricks for SSC CGL, SSC CHSL, FCI Exam Type-1

In Quantitive aptitude, Pipe And Cistern are one of the most important topics. It Plays a major role in any competitive Examination like SSC CGL, SSC CHSL, FCI etc. Generally, 1-2 questions are expected from Pipe And Cistern Chapter.

In Pipe And Cistern Type-1, I am going to discuss some important Pipe And Cistern Problem Shortcut Tricks which are frequently asked in SSC CGL, SSC CHSL, FCI etc Exam.

(Q1)Two pipes A and B can fill a tank in 20 minutes and 30 minutes respectively.If both pipes are opened together, the time taken to fill the tank is.

Solution: Fig.

pipe and cistern problem shortcut tricks

Rate of Filling the tank by A and B in 1 minute=3+2=5 unit

So, total time taken by Pipe (A+B)=60/5

                            =12 minutes Ans.

[Concept: First step-L.C.M=20,30=60 i.e total unit. now 60/20=3 unit per minutes tank filling by pipe A and 60/30=2 unit per minutes tank filling by pipe B.]

(Q2) A tap can empty a tank in 30 minutes. A second tap can empty it in 45 minutes. If both the taps operate simultaneously, how much time is needed to empty the tank?

Solution: Fig.

pipe and cistern problem shortcut tricks

Rate of emptying the tank by 1st and 2nd tap in 1 minute=(3+2)

                                =5 unit

So, total time taken by (1st+2nd)tap=90/5

                                =18 minutes Ans.

[Concept: First step-L.C.M=30,45=90 i.e total unit. now 90/30=3 unit per minutes tank emptying by 1st tap and 90/45=2 unit per minutes tank emptying by 2nd tap.]

(Q3)Three tapes A, B, C can fill an overhead tank in 4, 6 and 12 hours respectively. How long would the three taps take to fill the tank if all of them are opened together?

Solution: Fig.

pipe and cistern problem shortcut tricks

Rate of Filling the tank by A, B and C in 1 hour=(3+2+1)

                                =6 unit

So, total time taken by (A+B+C)=12/6

                              =2 hours Ans.

Common Type Of Pipe And Cistern Problem Shortcut Tricks That Are Frequently Asked In Various Competitive Exam like SSC CGL, SSC CHSL, FCI etc. Type-1.1  

(Q1)A water tank can filled by a tap in 30 minutes and another tap can fill it in 60 minutes. If both the taps are kept open for 5 minutes and then the first tap is closed, how much time 2nd tap will take to fill the remaining tank.

Solution: Fig.

pipe and cistern problem shortcut tricks

Rate of Filling the tank by 1st tap and 2nd tap in 1 minute=(2+1)

                                           =3 unit

[Concept: From fig.

pipe and cistern problem shortcut tricks

According to question 1st tap and second tap को  5 minutes तक एक साथ खुला रखा गया है तो दोनों 5 minutes में कितना water fill करेगा वो निकाल लेंगे| 5 minutes के बाद तो 1st tap बंद कर दिया जा रहा है तो सिर्फ अब tank के remaining unit  को fill करने के लिए 2nd tap ही बचा है तो वो कितना time लेगा वो निकाल लेंगे. ]

Rate of Filling the tank by 1st tap and 2nd tap in 5 minute=5×3=15 unit

Remaining=60-15=45 unit that is filled by 2nd tap

So, total time taken by 2nd tap=45/1

                                      =45 minutes Ans.

(Q2)Three pipes A, B and C can fill a tank in 6 hours, 9 hours and 12 hours respectively. B and C are opened for half an hour, then A is also opened. The time taken by the three pipes together to fill the remaining part of the tank is.

Solution: Fig.

pipe and cistern problem shortcut tricks

Rate of Filling the tank by A+B+C in 1 hour=(6+4+3)unit

                                =13 unit

[ Concept: From fig.

pipe and cistern problem shortcut tricks

half an hour तक B and C tank fill करेगा |half an hour के बाद A को open किया जाता है तो अब A, B and C मिलकर करेगा| ]

Rate of Filling the tank by B+C in 1 hour=(4+3)unit

                                             =7 unit

Rate of Filling the tank by B+C in half an hour=7/2 unit

Remaining unit=36-(7/2)=65/2 unit

So, total time taken by (A+B+C)=(65/2)/13

=2 ½ hours Ans.

(Q3)Three pipes A, B and C can fill a tank in 6 hours. After working together for 2 hours, C is closed and A and B fill the tank in 8 hours. The time(in hours) in which the tank can be filled by pipe C alone is.

Solution:

Let tank fill by A+B+C in 1 hour= 1 unit

Tank fill by A+B+C in 6 hour= 6 unit

After working together for 2 hours then

Tank fill by A+B+C in 2 hour= 2 unit

Remaining = 6 – 2=4 unit that is filled by A and B because C is closed

Rate of filling of A+B in 1 hour=4/8=0.5 unit

Rate of filling of C in 1 hour=1-0.5=0.5 unit

Time taken by C = 6/0.5= 12 hours Ans.

Common Type Of Pipe And Cistern Problem Shortcut Tricks That Are Frequently Asked In Various Competitive Exam like SSC CGL, SSC CHSL, FCI etc. Type-1.2  

(Q1)A tank has two pipes. The first pipe can fill it in 4 hours and the second can empty it in 16 hours. If two pipes are opened together at a time, then the tank will be filled in.

Solution: Fig.

pipe and cistern problem shortcut tricks

Rate of filling the tank by 1st pipe= 4 unit per hour

Rate of emptying the tank by 2nd pipe= 1 unit per hour

Net rate of filling the tank=4-1=3unit

so required time=16/3=5 ¹/3 hours Ans.

[concept:जब एक pipe tank को fill(भर)कर रहा हो or दूसरा empty(खाली) कर रहा हो तो जो pipe tank को fill कर रहा होगा, तो उसका Rate of filling the tank (+ve) में लेना है or जो pipe tank को empty कर रहा होगा तो उसका Rate of emptying the tank   ( -ve) में लेना है| इसलिए figure में +ve or -ve लिया गया है.]

(Q2)Two pipes A and B can fill a cistern in 3 hours and 5 hours respectively. Pipe C can empty in 2 hours. If all the three pipes are open, in how many hours the cistern will be full?

Solution: Fig.

pipe and cistern problem shortcut tricks

Rate of filling a cistern by A= 10 unit per hour

Rate of filling a cistern by B= 6 unit per hour

Rate of emptying a cistern by C= -15 unit per hour

Net rate of filling the tank by (A+B-C)=10+6-15

                           =1 unit per hour

so total time taken by (A+B-C)=30/1

                          =30 hours Ans.

Common Type Of Pipe And Cistern Problem Shortcut Tricks That Are Frequently Asked In Various Competitive Exam like SSC CGL, SSC CHSL, FCI etc. Type-1.3

(Q1)Two pipes A and B can fill a tank with water in 30 minutes and 45 minutes respectively. Another pipe C can empty the tank in 36 minutes. First A and B are opened. After 12 minutes, C is also opened. The tank is filled up in.

Solution: Fig.

pipe and cistern problem shortcut tricks

[ Concept: From fig.

pipe and cistern problem shortcut tricks ]

Rate of filling the tank by (A+B) in 1 minute=(6+4)=10 unit

Rate of filling the tank by (A+B) in 12 minute=12×10=120 unit

Remaining=180-120=60 unit that is filled by A, B and C

Rate of emptying the tank by C = -5 unit per minute

Net rate of filling the tank by (A+B-C)=(6+4-5)

                            =5unit per minute

so required time taken by (A+B-C)=60/5

                           =12 minutes

Total time=12+12=24 minutes Ans.

I have already shared a Time and Work Problem Shortcut Tricks, Basic concept and frequently asked question if you have not read that yet, you should read it the right way here is the link.

Time And Work Problem Shortcut Tricks Type-1

If you have any queries regarding this topic fell free to ask me in the comments below.

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