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Pipes and cistern questions are mostly the trickiest questions to solve in an examination. To handle the questions in the best possible manner, you can start practicing them today. Practice makes a man perfect and more accurate in solving problems. These types of questions generally appear in the banking exams, SSC, RRB, insurance, Olympiads, and others. To secure your place in this competitive examination, practice the questions today.
Table of Contents
Formulas To Solve Pies and Cistern Questions
here are some basic formulas that will help you to solve the Pipes and cistern questions with them. Remember these formulas may not be enough for each question, you have to practice more questions to bring flow in solving questions.
- If it will take x hours to fill a tank, then half the tank will be filled in 1 hr =1/x
- In the event that y hours are expected to discharge the tank, partly exhausted in 1 hour = 1/y
- If a line can fill a tank in x hours and can discharge a similar tank in y hours. At the point when both the lines are opened simultaneously, then, at that point, the net piece of the tank filled in 1 hr = {(xy)/(y-x)}, gave y>x
- If a line can fill a tank in x hours and can exhaust a similar tank in y hours. At the point when both the lines are opened simultaneously, then, at that point, the net piece of the tank filled in 1 hr = {(xy)/(x-y)}, gave x>y
- Net work done = (Amount of work done by Gulfs) – (Amount of work done by Outlets)
- One gulf can fill the tank in x hr and the other channel can fill the similar tank in y hrs, assuming both the bays are opened simultaneously, the time taken to fill the entire tank = {(xy)/(y+x)}
- If two lines take x and y hours separately to fill a tank of water and a third line is opened which takes z hours to exhaust the tank, then, at that point, the time taken to fill the tank = {1/(1/x)+(1/y)+(1/z)} and the net piece of the tank filled in 1 hr = (1/x)+(1/y)- (1/z).
Pipes and Cistern Questions and Answers
Below you can find different questions and answers related to Pipes and Cisterns:
85 Min.
95 Min.
105 Min.
105 Min.
Answer: D) 100 min
Explanation:
Work done by the third pipe in 1 min = 1/50 – (1/60 + 1/75) = – 1/100.
[-ve sign means emptying]
The third pipe alone can empty the cistern in 100 min.
6 min.to empty
9 min. to empty
10min. to fill
6 min. to fill
Answer: A
Explanation:
Clearly,pipe B is faster than pipe A and so,the tank will be emptied.
part to be emptied = 2/5
part emptied by (A+B) in 1 minute = (16−110)=11516-110=115
115:25::1:x115:25::1:
25*15=6mins25*15=6
so, the tank will be emptied in 6 minutes`
2hrs 45 min
3hrs 25 min
1hr 20 min
2hr 20min
Answer: B
Explanation:
Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the four taps in 1 hour =4*1/6 =2/3
Remaining part =(1−12)=121-12=12
23:12::1:x23:12::1
=> x = (12*1*32)=3412*1*32=34
So, total time taken = 3 hrs. 45 mins.
11 Hours
10 Hours
9 Hours
14 Hours
Answer: D
Explanation:
If the total area of pump=1 part
The pumop take 2 hrs to fill 1 part
The pumop take1 hour to fill 1/2 portion
Due to lickage
The pumop take 7/3 hrs to fill 1 part
The pumop take1 hour to fill 3/7 portion
Now the difference of area = (1/2-3/7)=1/14
This 1/14 part of water drains in 1 hour
Total area=1 part of water drains in (1×14/1)hours= 14 hours
So the leak can drain all the water of the tank in 14 hours.
Leak will empty the tank in 14 hrs
20 Hours
10 Hours
9 Hours
21 Hours
Answer: A) 20 hrs
Explanation:
1/8 – 1/x = 1/10
=> x = 40 hrs
23
43
14
18
Answer: C
Explanation:
Part filled in 2 hours = 2/6=1/3
Remaining part =(1−13)=231-13=23
(A + B)’s 7 hour’s work = 2/3
(A + B)’s 1 hour’s work = 2/21
C’s 1 hour’s work = { (A + B + C)’s 1 hour’s work } – { (A + B)’s 1 hour’s work }
=1/6-2/21 = 1/14
C alone can fill the tank in 14 hours.
400 lit
480 lit
500 lit
580 lit
Answer: B
Explanation:
1/k – 1/20 = -1/24
k = 120
120 x 4 = 480
Therefore, the capacity of the cistern is 480 liters.
15min.
10min.
24min.
20min.
Answer: C
Explanation:
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.
50 min.
45 min.
60 min.
30 min.
Answer: D
Explanation:
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.
10
12
30
18
Answer: D
Explanation:
Capacity of the tank =(12 x 13.5) liters =162 liters.
Capacity of each bucket =9 liters
Number of buckets needed = 162/9 =18.
200 hours
100 hours
300 hours
400 hours
Answer: C
Explanation:
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.
7(1/2) hrs
4(4/5) hrs
3 (2/7) hrs
1(1/5) hrs
Answer: B
Explanation: Speed of the man (upstream) = (4 / 40 * 60) km/hr
= 6 km/hr
Speed of the man (downstream) = ( 4 / 36 * 60) km/hr
= (60 / 9)km/hr
= 20/3 km/h
Rate of the current = ½[downstream speed – upstream speed]
= (1/2) [ 20 / 3 – 6] km/hr
Rate of the current = 1/3 km/hr
120 gallons
240 gallons
450 gallons
840 gallons
Answer:D
Explanation:
Work done by waste pipe in 1 min = (Part filled by total pipes together) -(part filled by first pipe + part filled by second pipe)
= 1/6 – [(1/10) + (1/14)] = 1/6 – (24/140) = -1/210
Here, negative (-) sign indicates emptying of tank.
To find the capacity, we need to determine the volume of 1/210 part.
Therefore, volume of 1/210 part = 4 gallons ———(given condition)
Hence, the capacity of tank = volume of whole = 4 x 210 = 840 gallons.
25
50
75
100
Answer : B
Explanation:
Assume that the filling capacity of the pump = x m3/min
Given condition : Emptying capacity of the tank is 10 m3 per minute higher than its filling capacity. This means that the emptying capacity of the pump is = x + 10 m3/min
We need to filling the capacity of pump from the given relation of filling & emptying capacities. Hence, we can write that,
1200 / x – 1200/ x+ 10 = 4
[1/x – 1/ x+10] = 1/300
By solving the above equation, we get the simplified form of quadratic equation.
So, x2 + 10x -3000 = 0
(x + 60) (x-50) = 0
Therefore, we get two values of ‘x’; i.e. x = -60 & x =50
Since we are asked to find the filling capacity, which is always positive; we will neglect the negative value of x = –60.
Therefore, the filling capacity of the pump = 50 m3/min.
8 hrs
16 hrs
21 hrs
36 hrs
Answer: D
Explanation:
Consider a pipe fills the tank in ‘x’ hrs. If there is a leakage in the bottom, the tank is filled in ‘y’ hrs.
The time taken by the leak to empty the full tank =
xyhrs / y – x
35 hrs
70 hrs
180 hrs
300 hrs
The time taken by the leak to empty the full tank =
xy hrs / y – x
Time taken to empty the tank by the leak = 4 x (9/2) / (9/2 ) – 4 = (36/2 ) / ½ = 18/ ½ = 18 x 2 = 36 hrs
But, direct values of x & y are not given. So, we need to find the work done by the two pipes in 1 hr = (1/15) + (1/20) = 7/60
Hence, the time taken by these pipes to fill the tank = 60/7 hrs = 8 .57 hrs = 8 hrs 34 min ——–( by multiplying ‘0.57’ hrs x 60 = 34 minutes)
Due to leakage, time taken = 8 hrs 34 min + 26 min = 8 hrs 60 min = 9 hrs ———-( because 60 min = 1 hr)
Thus, work done by (two pipes + leak) in 1 hr = 1/9
Hence, work done by leak in 1 hr = work done by two pipes – 1/9
= 7 /60 – 1/9 = 3/ 540 = 1/180
Therefore, leak will empty the full cistern in 180 hours.
12 hrs
30 hrs
44 hrs
60 hrs
Answer: D
Explanation:
Assume that the reservoir is filled by first pipe in ‘x’ hours.
So, the reservoir is filled by second pipe in ‘x + 20’ hours.
Now, from these above conditions, we can form the equations as,
1/x + 1/ (x + 20) = 1/24
[x + 20 + x] / [x(x + 20)] = 1/24
x2 – 28x – 480 = 0
By solving this quadratic equation , we get the factors (x – 40) (x+12) = 0
Hence, we get two values : x = 40 & x = -6
Since filling of reservoir is positive work , we can neglect the negative value of ‘x’.
Thus, x = 40
This means that the second pipe will take (x+ 20) hrs = 40 + 20 = 60 hrs to fill the reservoir.
6 hrs
7 hrs
8 hrs
14 hrs
Answer: B
Explanation:
Pipe A’s work in 1 hr = 1/8
Pipe B’s work in 1 hr = 1/6
Pipes (A+B)’s work in first 2 hrs when they are opened alternately = 1/8 + 1/6 = 7 /24
Now,
In 4 hrs they fill : 2 X (7/24) = 7/12
In 6 hrs they fill : 3 X (7/24) = 7/8
After 6 hrs, part left empty = ⅛
Now it is A’s turn to open up.
In one hr it fills 1/8 of the tank.
So, the tank will be full in = 6 hrs + 1 hr = 7 hrs
3 hrs
7 hrs
12 hrs
20 hrs
Answer: C
Explanation:
Assume that the tank will be full in ’10 + x’ hrs.
Part filled by pipe A in 1 hr = 1/15
Part filled by pipe B in 1 hr = 1/20
Part emptied by pipe C in 1 hr = 1/ 25
(A+ B)’s part filled in 1 hr = 1/15 + 1/20 = 7 /60
As Pipe C is closed after 10 hours, let us find the part of tank filled in 10 hrs.
Tank filled in 10 hrs = 10 (part filled by A in 1 hr + part filled by B in 1 hr– part emptied by C in 1 hr)
= 10 [1/15 + 1/20 – 1/25] = 23 /30
Remaining part = 1 – part filled in 10 hours
= 1 – 23/30 = 7/30
We know that,
Part filled by (A + B) in 1 hr : Total part filled
Remaining part : Time taken
So, we get the ratio as ,
7/60 : 7/30 :: 1 : x
So, the value of x = 2/30 x 1 x 60/7 = 2 hrs.
Hence, the tank will be full in x + 2 = 10 + 2 hrs = 12 hrs
4 min
6 min
14 min
20 min
Answer: B
Explanation:
Assume the total time required = x + 2 min
Part filled by pipe A in 1min = 1/5 & part filled by B in 1 min = 1/10
After 2 min, part filed by A & B together = 2 [ 1/5 + 1/10] = 3/5
Remaining part = 1 – 3/5 = 2/5
Part filled by pipe B in 1 min = 1/10
As pipe A is turned off after 2 minutes, we get the relation as,
Part filled by (A + B) in 1 hr : Total part filled
Remaining part : Time taken
1/10 : 2/5 :: 1: x
x = (2/5) x 1 x 10 = 4 min
Tank will be full in ( 2 min + 4 min) = 6 min
FAQs
Read the question multiple times and try to answer the query with the application of simplest formulas and using common sense.
There are various exams like bank exams, SSC, RRB, Insurance, Olympiads and others in which these types of questions actually comes in.
No by using common sense and correct formulas you can easily solve these types of questions.
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