Crack IBPS PO/Clerk 2017 – Quantitative Aptitude (Pipes and Cisterns):
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Dear Readers, IBPS is conducting Online Examination for the recruitment of Probationary Officers (PO) and Clerical Cadre. Preliminary Examination of IBPS PO is scheduled from 7^{th} Oct 2017 and for IBPS Clerk from 2nd Dec 2017. To enrich your preparation here we have providing new series of Practice Questions on Quantitative Aptitude. Candidates those who are appearing in IBPS PO/Clerk Exam can practice these questions daily and make your preparation effective.
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Crack IBPS PO/Clerk 2017 – Quantitative Aptitude (Pipes and Cisterns)
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Question 1 of 10
1. Question
A container of capacity 25 L has an inlet and an outlet tap. If both are opened simultaneously, the container is filled in 5 minutes. But if the outlet flow rate is doubled and taps opened the container never gets filled up. Which of the following can be outlet flow rate?
Correct
Correct Answer is: d)
Let, the inlet pipe fills x l/min
And outlet pipe empties y l/min
Then x –y = 25/5 = 5 … (1)
When outflow rate is doubled, the tank never gets filled up, which means
x – 2y ≤ 0
x ≤ 2y
putting x =2y in Eq. (1)
y = 5
Any value of y greater or equal to 5 will satisfy the given condition.
Incorrect
Correct Answer is: d)
Let, the inlet pipe fills x l/min
And outlet pipe empties y l/min
Then x –y = 25/5 = 5 … (1)
When outflow rate is doubled, the tank never gets filled up, which means
x – 2y ≤ 0
x ≤ 2y
putting x =2y in Eq. (1)
y = 5
Any value of y greater or equal to 5 will satisfy the given condition.

Question 2 of 10
2. Question
Three pipes P1, P2 and P3 can fill a glass tub from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the glass tub is empty, all the three pipes are opened. P1,P2 and P3 discharge chemical solutions C, H and T respectively. What is the proportion of the solution T in the liquid in the pot after 3 minutes?
Correct
Correct Answer is: c)
Part of the glass tub filled by P1 in 1 minute = 1 ⁄30
Part of the glass tub filled by P2 in 1 minute = 1 ⁄20
Part of the glass tub filled by P3 in 1 minute = 1 ⁄10
Here we have to find the proportion of the solution T.
Pipe P3 discharges chemical solution T
Part of the glass tub filled by P3 in 3 minutes = 3 *1/10 = 3/10
Part of the glass tub filled by pipe P1, P2 and P3 together in 1 minute = 1 /30 + 1/20 + 1/10 = 11/60
Part of the glass tub filled by pipe P1, P2 and P3 together in 3 minute = 3 * 11/60 = 11/20
Required proportion = (3/10)/(11/12)= 6/11
Incorrect
Correct Answer is: c)
Part of the glass tub filled by P1 in 1 minute = 1 ⁄30
Part of the glass tub filled by P2 in 1 minute = 1 ⁄20
Part of the glass tub filled by P3 in 1 minute = 1 ⁄10
Here we have to find the proportion of the solution T.
Pipe P3 discharges chemical solution T
Part of the glass tub filled by P3 in 3 minutes = 3 *1/10 = 3/10
Part of the glass tub filled by pipe P1, P2 and P3 together in 1 minute = 1 /30 + 1/20 + 1/10 = 11/60
Part of the glass tub filled by pipe P1, P2 and P3 together in 3 minute = 3 * 11/60 = 11/20
Required proportion = (3/10)/(11/12)= 6/11

Question 3 of 10
3. Question
A tank is filled by three pipes P1, P2, P3 with uniform flow. The P1 and P2 operating simultaneously fill the tank in the same time during which the tank is filled by the P3 alone. The P2 fills the tank 5 hrs faster than the P1 and 4 hrs slower than the P3. The time required by the first pipe, P1 is:
Correct
Correct Answer is: c)
P1 = (x + 5) hr
P2 = x hr
P3 = (x – 4) hr
According to question
1/(x + 5) + (1/x) + 1/(x – 4)
x^2 8x – 20 = 0
x = 10 hr
Time required by first pipe = 15 hr
Incorrect
Correct Answer is: c)
P1 = (x + 5) hr
P2 = x hr
P3 = (x – 4) hr
According to question
1/(x + 5) + (1/x) + 1/(x – 4)
x^2 8x – 20 = 0
x = 10 hr
Time required by first pipe = 15 hr

Question 4 of 10
4. Question
Two water taps T1 and T2 can fill a tank in 900 seconds and 2400 seconds respectively. Both the taps are opened together but after 240 seconds, tap T1 is turned off. What is the total time required to fill the tank?
Correct
Correct Answer is: b)
900 sec = 15 min , 2400 sec = 40 min and 240 sec = 4 min
Part filled by tap T1 in 1 minute = 1/15
Part filled by tap T2 in 1 minute = 1/40
Part filled by tap T1 and tap T2 in 1 minute = 1/15 + 1/40 = 11/120
Tap T1 and tap T2 were open for 4 minutes
Part filled by tap T1 and tap T2 in these 4 minutes = 4 * 11/120 = 11/30
Remaining part to be filled = 1 – 11/30 = 19/30
Time taken by tap T2 to fill this remaining part = (19/30)/(1/40) = 76/3 = 25(1/3)
i.e. 25 min 20 sec
Incorrect
Correct Answer is: b)
900 sec = 15 min , 2400 sec = 40 min and 240 sec = 4 min
Part filled by tap T1 in 1 minute = 1/15
Part filled by tap T2 in 1 minute = 1/40
Part filled by tap T1 and tap T2 in 1 minute = 1/15 + 1/40 = 11/120
Tap T1 and tap T2 were open for 4 minutes
Part filled by tap T1 and tap T2 in these 4 minutes = 4 * 11/120 = 11/30
Remaining part to be filled = 1 – 11/30 = 19/30
Time taken by tap T2 to fill this remaining part = (19/30)/(1/40) = 76/3 = 25(1/3)
i.e. 25 min 20 sec

Question 5 of 10
5. Question
Two pipes P and Q can fill a container in 25 and 30 minutes respectively and a waste pipe R can empty 3 gallons per minute. All the three pipes P,Q and R working together can fill the container in 15 minutes. The capacity of the container is:
Correct
Correct Answer is: b)
Part filled by pipe P in 1 minute=1/25
Part filled by pipe Q in 1 minute=1/30
Let the waste pipe R can empty the full container in x minutes
Then, part emptied by waste pipe R in 1 minute=1/x
All the three pipes P,Q and R can fill the container in 15 minutes
i.e. part filled by all the three pipes in 1 minute = 1/15
1/25 + 1/30 – 1/x = 1/15
x = 150
i.e, the waste pipe R can empty the full container in 150 minutes
Given that waste pipe R can empty 3 gallons per minute
ie, in 150 minutes, it can empty 150 * 3 = 450 gallons
Hence, the volume of the container = 450 gallons
Incorrect
Correct Answer is: b)
Part filled by pipe P in 1 minute=1/25
Part filled by pipe Q in 1 minute=1/30
Let the waste pipe R can empty the full container in x minutes
Then, part emptied by waste pipe R in 1 minute=1/x
All the three pipes P,Q and R can fill the container in 15 minutes
i.e. part filled by all the three pipes in 1 minute = 1/15
1/25 + 1/30 – 1/x = 1/15
x = 150
i.e, the waste pipe R can empty the full container in 150 minutes
Given that waste pipe R can empty 3 gallons per minute
ie, in 150 minutes, it can empty 150 * 3 = 450 gallons
Hence, the volume of the container = 450 gallons

Question 6 of 10
6. Question
Two taps T1 and T2 together can fill the tank in 2160 seconds. If the tap T1 can fill a tank four times as fast as another tap T2, then what is the time taken by the slower tap to fill the tank alone?
Correct
Correct Answer is: d)
2160 seconds = 36 min
Let the slower tap T2 alone can fill the tank in x minutes
Then the faster tap T2 can fill the tank in minutes = x/4
Part filled by the slower tap in 1 minute =1/x
Part filled by the faster tap in 1 minute =4/x
Part filled by both the taps in 1 minute = 1/x + 4/x
It is given that both the taps together can fill the tank in 2160 seconds i.e. 36 minutes
Part filled by both the taps in 1 minute =1/36
1/x +4/x =1/36
x = 5 × 36 = 180
i.e. the slower tap T2 alone fill the tank in 180 minutes
Incorrect
Correct Answer is: d)
2160 seconds = 36 min
Let the slower tap T2 alone can fill the tank in x minutes
Then the faster tap T2 can fill the tank in minutes = x/4
Part filled by the slower tap in 1 minute =1/x
Part filled by the faster tap in 1 minute =4/x
Part filled by both the taps in 1 minute = 1/x + 4/x
It is given that both the taps together can fill the tank in 2160 seconds i.e. 36 minutes
Part filled by both the taps in 1 minute =1/36
1/x +4/x =1/36
x = 5 × 36 = 180
i.e. the slower tap T2 alone fill the tank in 180 minutes

Question 7 of 10
7. Question
A large tanker can be filled by two pipes P1 and P2 in 3600 seconds and 40 minutes respectively. How many seconds will it take to fill the tanker from empty state if P2 is used for half the time and P1 and P2 fill it together for the other half?
Correct
Correct Answer is: d)
Part filled by pipe P1 in 1 minute =1/60
Part filled by pipe P2 in 1 minute =1/40
Part filled by pipes P1 and pipe P2 in 1 minute = 1/60 + 1/40 = 1/24
Suppose the tank is filled in x minutes
Then, to fill the tanker from empty state, P2 is used for x/2 minutes
And P1 and P2 is used for the rest x/2 minutes
x/2 * 1/40 + x/2 * 1/24 = 1
x/2(1/40 + 1/24) =1
x = 30 min = 1800 sec.
Incorrect
Correct Answer is: d)
Part filled by pipe P1 in 1 minute =1/60
Part filled by pipe P2 in 1 minute =1/40
Part filled by pipes P1 and pipe P2 in 1 minute = 1/60 + 1/40 = 1/24
Suppose the tank is filled in x minutes
Then, to fill the tanker from empty state, P2 is used for x/2 minutes
And P1 and P2 is used for the rest x/2 minutes
x/2 * 1/40 + x/2 * 1/24 = 1
x/2(1/40 + 1/24) =1
x = 30 min = 1800 sec.

Question 8 of 10
8. Question
A small hole in the bottom of a cistern can empty the full container in 360 minutes. An inlet pipe fills water at the rate of 4 liters a minute. When the container is full, the inlet is opened and due to the leak, the cistern is empty in 24 hours. How many liters does the cistern hold?
Correct
Correct Answer is: c)
360 min = 6 hr
Part emptied by the leak in 1 hour =1/6
Net part emptied by the leak and the inlet pipe in 1 hour =1/24
Part filled by the inlet pipe in 1 hour =1/6 – 1/24 =1/8
i.e., inlet pipe fills the cistern in 8 hours = (8 × 60) minutes = 480 minutes
Given that the inlet pipe fills water at the rate of 4 liters a minute
Hence, water filled in 480 minutes = 480 × 4 = 1920 litre
i.e. The cistern can hold 1920 litre
Incorrect
Correct Answer is: c)
360 min = 6 hr
Part emptied by the leak in 1 hour =1/6
Net part emptied by the leak and the inlet pipe in 1 hour =1/24
Part filled by the inlet pipe in 1 hour =1/6 – 1/24 =1/8
i.e., inlet pipe fills the cistern in 8 hours = (8 × 60) minutes = 480 minutes
Given that the inlet pipe fills water at the rate of 4 liters a minute
Hence, water filled in 480 minutes = 480 × 4 = 1920 litre
i.e. The cistern can hold 1920 litre

Question 9 of 10
9. Question
Three taps T1, T2 and T3 can fill a tank in 720, 900 and 1200 minutes respectively. If T1 is open all the time and T2 and T3 are open for one hour each alternately, the tank will be full in:
Correct
Correct Answer is: d)
720 min =12 hr, 900 min = 15 hr and 1200 min = 20 hr
Part filled by pipe T1 in 1 hour =1/12
Part filled by pipe T2 in 1 hour =1/15
Part filled by pipe T3 in 1 hour =1/20
In first hour, T1 and T2 is open
In second hour, T1 and T3 is open
then this pattern goes on till the tank fills
Part filled by pipe T1 and pipe T2 in 1 hour =1/12 +1/15 = 3/20
Part filled by pipe T1 and pipe T3 in 1 hour =1/12 + 1/20 = 2/15
Part filled in 2 hour = 3/20 +2/15 =17/60
Part filled in 6 hour =17/60 * 3 = 17/20
Remaining part = (1 – 17/20) = 3/20
Now, 6 hours are over and only 3/20 part needed to be filled.
At this 7th hour, T1 and T2 is open
Time taken by pipe T1 and T2 to fill this 3/20 part = (3/20)/(3/20) = 1 hr
Total time taken = 6 hr + 1 hr = 7 hr
Incorrect
Correct Answer is: d)
720 min =12 hr, 900 min = 15 hr and 1200 min = 20 hr
Part filled by pipe T1 in 1 hour =1/12
Part filled by pipe T2 in 1 hour =1/15
Part filled by pipe T3 in 1 hour =1/20
In first hour, T1 and T2 is open
In second hour, T1 and T3 is open
then this pattern goes on till the tank fills
Part filled by pipe T1 and pipe T2 in 1 hour =1/12 +1/15 = 3/20
Part filled by pipe T1 and pipe T3 in 1 hour =1/12 + 1/20 = 2/15
Part filled in 2 hour = 3/20 +2/15 =17/60
Part filled in 6 hour =17/60 * 3 = 17/20
Remaining part = (1 – 17/20) = 3/20
Now, 6 hours are over and only 3/20 part needed to be filled.
At this 7th hour, T1 and T2 is open
Time taken by pipe T1 and T2 to fill this 3/20 part = (3/20)/(3/20) = 1 hr
Total time taken = 6 hr + 1 hr = 7 hr

Question 10 of 10
10. Question
A booster pump can be used for filling as well as for emptying a container. The capacity of the container is 2400. The emptying of the container is 10 per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the container than it needs to fill it. What is the filling capacity of the pump?
Correct
Correct Answer is: b)
Let the filling capacity of the pump = x m^3/ min.
Then the emptying capacity of the pump = (x + 10) m^3 / min.
Time required for filling the container = 2400/x min.
Time required for emptying the container = 2400/(x+10) min.
Pump needs 8 minutes lesser to empty the container than it needs to fill it
2400/x – 2400/(x+10)=8
+ 10x – 3000 = 0
x = 50 or 60
i.e. filling capacity of the pump = 50 m^3 / min.
Incorrect
Correct Answer is: b)
Let the filling capacity of the pump = x m^3/ min.
Then the emptying capacity of the pump = (x + 10) m^3 / min.
Time required for filling the container = 2400/x min.
Time required for emptying the container = 2400/(x+10) min.
Pump needs 8 minutes lesser to empty the container than it needs to fill it
2400/x – 2400/(x+10)=8
+ 10x – 3000 = 0
x = 50 or 60
i.e. filling capacity of the pump = 50 m^3 / min.