Pipes and Cisterns Aptitude Solved Examples - Set 01

Question No. 01
Two pipes A and B can fill a cistern in 37½ minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
(A) 5 min
(B) 9 min
(C) 10 min
(D) 15 min
Answer: Option B
Explanation:
Let B be turned off after x minutes. Then,
Part filled by (A + B) in x min. + Part filled by A in (30 - x) min. = 1.
x (2/75 + 1/45) + [(30 - x). 2/75 = 1
=> (11x/225) + [(60 - 2x)/75] = 1
=> 11x + 180 - 6x = 225
=> x = 9

Question No. 02
Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:
(A) 60 gallons
(B) 100 gallons
(C) 120 gallons
(D) 180 gallons
Answer: Option C
Explanation:
Work done by the waste pipe in 1 minute = 1/15 - (1/20 + 1/24)
= (1/15 - 11/120)
= - (1/40) [-ve sign means emptying]
Volume of 1/40 part = 3 gallons.
Volume of whole = (3 × 40) gallons = 120 gallons.

Question No. 03
Two pipes ‘A’ and ‘B’ can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe ‘A’ is turned off. What is the total time required to fill the tank?
(A) 10 min. 20 sec.
(B) 11 min. 45 sec.
(C) 12 min. 30 sec.
(D) 14 min. 40 sec.
Answer: Option D
Explanation:
Part filled in 4 minutes = 4 {(1/15) + (1/20)} = (7/15)
Remaining part = {1 - (7/15)} = (8/15)
Part filled by B in 1 minute = (1/20)
(1/20) : (8/15) :: 1 : x
x = {(8/15) × 1 × 20} = 10⅔ min = 10 min. 40 sec.
 The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

Question No. 04
A large tanker can be filled by two pipes ‘A’ and ‘B’ in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if ‘B’ is used for half the time and ‘A’ and ‘B’ fill it together for the other half?
(A) 15 min
(B) 20 min
(C) 27.5 min
(D) 30 min
Answer: Option 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.

Question No. 05
A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
(A) 3 hrs 15 min
(B) 3 hrs 45 min
(C) 4 hrs
(D) 4 hrs 15 min
Answer: Option 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 - (1/2)} = 1/2
(2/3) : (1/2) :: 1 : x
=> x = [(1/2) × 1 × (3/2)] = (3/4) hours i.e., 45 mins.
So, total time taken = 3 hrs. 45 mins.

Question No. 06
Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, ‘C’ is closed and ‘A’ and ‘B’ can fill the remaining part in 7 hours. The number of hours taken by ‘C’ alone to fill the tank is:
(A) 10
(B) 12
(C) 14
(D) 16
Answer: Option C
Explanation:
Part filled in 2 hours = 2/6 = 1/3
Remaining part = [1 - (1/3)] = 2/3
 (A + B)'s 7 hour's work = 2/3
 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.

Question No. 07
Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P, Q and R respectively. What is the proportion of the solution ‘R’ in the liquid in the tank after 3 minutes?
(A) 5/11
(B) 6/11
(C) 7/11
(D) 8/11
Answer: Option B
Explanation:
Part filled by (A + B + C) in 3 minutes = 3 [(1/30) + (1/20) + (1/10)] = 3 × (11/60) = 11/20
Part filled by C in 3 minutes = 3/10
Required ratio = (3/10) × (20/11) = 6/11

Question No. 08
A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
(A) 6 hours
(B) 10 hours
(C) 15 hours
(D) 30 hours
Answer: Option C
Explanation:
Suppose, first pipe alone takes x hours to fill the tank
Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.
1/x + 1/(x - 5) = 1/(x - 9)
=> (x - 5 + x)/x (x - 5) = 1/(x - 9)
=> (2x - 5)(x - 9) = x(x - 5)
=> x² - 18x + 45 = 0
=> (x - 15)(x - 3) = 0
=> x = 15.    [Neglecting x = 3]

Question No. 09
Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then ‘B’ would have taken 6 hours more than ‘A’ to fill the cistern. How much time will be taken by ‘A’ to fill the cistern separately?
(A) 1 hour
(B) 2 hours
(C) 6 hours
(D) 8 hours
Answer: Option C
Explanation:
Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours.
(1/x) + 1/(x+6) = 1/4
=> [(x + 6 + x)/ x(x + 6)] = 1/4
=> x2 - 2x - 24 = 0
=> (x -6)(x + 4) = 0
=> x = 6.     [Neglecting the negative value of x]

Question No. 10
A tank is filled in 5 hours by three pipes A, B and C. The pipe ‘C’ is twice as fast as ‘B’ and ‘B’ is twice as fast as ‘A’. How much time will pipe ‘A’ alone take to fill the tank?
(A) 20 hours
(B) 25 hours
(C) 35 hours
(D) None of these
Answer: Option C
Explanation:
Suppose pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take x/2 and x/4 hours respectively to fill the tank
(1/x) + (2/x) + (4/x) = (1/5)
=> (7/x) = (1/5)
=> x = 35 hrs.

Pipes and Cisterns:
Formula:                Pipes and Cisterns Formulas
Solved Examples:  Solved Examples: Set 01
Practice Test:         Practice Test: 01
Share on Google Plus

    FB Comments
    Comments

0 comments:

Post a Comment