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Aptitude Questions: Pipes & Cistern Set 5

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Hello Aspirants. Welcome to Online Quantitative Aptitude Section with explanation in AffairsCloud.com. Here we are creating question sample in Pipes & Cistern which is common for all competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!

  1. Two taps can separately fill the tank in 10m and 15min respectively. They fill the tank in 12 minutes when a third pipe which empties the tank is also opened. What is the time taken by the third pipe to empty the whole tank?
    A) 14 minutes
    B) 15 minutes
    C) 12 minutes
    D) 20 minutes
    E) 16 minutes
    C) 12 minutes
    Explanation:

    1/10 + 1/15 – 1/x = 1/12
    Solve, x = 12

  2. Two pipes A and B can fill a tank in 12 hours and 15 hours respectively. If they are opened on alternate hours with pipe A opened first, then in how many hours the tank will be full?
    A) 13 hrs
    B) 14 1/2 hrs
    C) 12 hrs
    D) 12 1/2 hrs
    E) 10 2/3 hrs
    D) 12 1/2 hrs
    Explanation:

    A = 12 hours, B = 15 hours
    Total work = LCM(12,15) = 60
    So efficiency of A = 60/12 = 5, efficiency of B = 60/15 = 4
    2 hrs work of (A+B) = 5+4 = 9
    2*6(12) hours work of (A+B) = 9*6 = 54
    So remaining work = 60-54 = 6
    Now A’s turn at 13th hour, he will do remaining work(6) in 6/12 hr
    So total 12 1/2 hrs

  3. Pipes P and Q can fill the tank in 24 minutes and 32 minutes respectively. Both piped are opened together. To have the tank full in 18 minutes, after how many minutes the pipe P must be closed?
    A) 22 minutes
    B) 21 minutes
    C) 15 minutes
    D) 12.5 minutes
    E) 10.5 minutes
    E) 10.5 minutes
    Explanation:

    P is to be closed before 18 minutes, let it is closed after x minutes, then Q worked for all 18 minutes. So,
    (1/24)*x + (1/32)*18 = 1
    Solve, x = 10.5

  4. Three pipes, A, B and C are opened to fill a tank such that A and B cam fill the tank alone in 36 min. and 45 min. respectively and C can empty it in 30 min. After 6 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) 24
    E) 30
    D) 24
    Explanation:

    Let the tank full in x minutes, then A and B opened for x minutes and C for 6 minutes.
    (1/36 + 1/45)*x – (1/30)*6 = 1
    (1/20)*x = 6/5
    Solve, x = 24

  5. A and B are pipes such that A can empty the tank in 60 minutes and B can fill in 30 minutes. The tank is full of water and pipe A is opened. If after 18 minutes, pipe B is also opened, then in how much total time the tank will be full again?
    A) 32 minutes
    B) 29 minutes
    C) 36 minutes
    D) 23 minutes
    E) 18 minutes
    B) 36 minutes
    Explanation:

    Emptying pipe A is opened first for 18 minutes, so in 18 minutes the part of tank it has emptied is (1/60)*18 = 9/30
    Now filling pipe is also opened, now since only 9/30 of the tank is empty so 9/30 is only to be filled by both pipes, let it take now x minutes, so
    (1/30 – 1/60)*x = 9/30
    Solve, x= 18
    So total = 18+18 = 36 minutes [total time is asked – 18 minutes when emptyimh pipe was only opened, 18 minutes when both were operating.]

  6. Two pipes A and B can alone fill a tank in 20 minutes and 30 minutes respectively. But due to a leak at the bottom of tank, it took 3 more minutes to fill the tank. In how many hours, the leak can alone empty the full tank?
    A) 60
    B) 30
    C) 48
    D) 56
    E) 72
    A) 60
    Explanation:

    A and B can fill tank in (1/20 + 1/30) = 1/12 so 12 minutes
    But it took 3 more minutes, this means the tank got full in 12+3 = 15 minutes
    So (1/20 + 1/30 – 1/x) = 1/15
    Solve, x = 60

  7. Pipes A and B can fill a cistern in 15 hours together. But if these pipes operate separately A takes 40 hours less than B to fill the tank. In how many hours the pipe A will fill the cistern working alone?
    A) 60
    B) 20
    C) 40
    D) 15
    E) 25
    B) 20
    Explanation:

    Let A takes x hours, then B = (x+40) hours
    1/x + 1/(x+40) = 1/15
    Solve, x = 20

  8. Three pipes A, B and C can fill the cistern in 10, 12, and 15 hours respectively. In how much time the cistern will be full if A is operated for the whole time and B and C are operated alternately which B being first?
    A) 10 hours 32 minutes
    B) 6 hours
    C) 5 1/2 hours
    D) 5 7/10 hours
    E) 9 2/11 hours
    D) 5 7/10 hours
    Explanation:

    In first hour, part of cistern filled is (1/10 + 1/12) = 11/60
    In second hour, part of cistern filled is (1/10 + 1/15) = 1/6
    So in 2 hours, part of cistern filled is 11/60 + 10/60 = 21/60 = 7/20
    now in 2*2 (4) hours, part of cistern filled is (7/20)*2 = 14/20 = 7/10
    now in the 5th hour, A+B’s turn which fill 11/60 in that hour, but the cistern remaining to be filled is (1 – 7/10) = 3/10, since 3/10 is more than 11/60, so after 5th hour remaining part to be filled is 3/10 – 11/60 = 7/60
    now in 6th hour, (A+C)’s turn, it will fill remaining 7/60 in (7/60)*(6/1) = 7/10
    so total 5 7/10 hours

  9. A cistern is 1/4th full. Two pipes which fill the cistern in 15 minutes and 20 minutes respectively are opened simultaneously. After 5 minutes, a third pipe which empties the full cistern in 30 minutes is also opened. In how many minutes the cistern will be full?
    A) 6 1/2
    B) 2
    C) 5
    D) 7
    E) 8
    D) 7
    Explanation:

    Since 1/4th is already filled, 3/4th is to filled now.
    So
    (1/15 + 1/20)*(5+x) – (1/30)*x = 3/4
    (7/60)*5 + (7/60 – 1/30)*x = 3/4
    (5/60)*x = 2/12
    Solve, x = 2 mins
    So total 7 minutes

  10. Pipes A, B and C which fill the tank together in 6 hours are opened for 2 hours after which pipe C was closed. Find the number of hours taken by pipe C to fill the tank if the remaining tank is filled in 7 hours.
    A) 16
    B) 14
    C) 20
    D) 22
    E) Cannot be determined
    B) 14
    Explanation:

    1/A + 1/B + 1/C = 1/6
    Now given that first all open for 2 hours, then C closed and A+B completes in 7 hours, so
    (1/A + 1/B + 1/C) *2 + (1/A + 1/B)*7 = 1
    Put 1/A + 1/B = 1/6 – 1/C
    (1/6 – 1/C + 1/C) *2 + (1/6 – 1/C)*7 = 1
    2/6 + 7/6 – 7/C = 1
    Solve, C = 14