Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 14)
14.
Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
Answer: Option
Explanation:
| (A + B)'s 1 hour's work = |  |
1 | + | 1 |  |
= | 9 | = | 3 | . |
| 12 | 15 | 60 | 20 |
| (A + C)'s hour's work = | |
1 | + | 1 | |
= | 8 | = | 2 | . |
| 12 | 20 | 60 | 15 |
| Part filled in 2 hrs = | |
3 | + | 2 | |
= | 17 | . |
| 20 | 15 | 60 |
| Part filled in 6 hrs = | |
3 x | 17 | |
= | 17 | . |
| 60 | 20 |
| Remaining part = | |
1 - | 17 | |
= | 3 | . |
| 20 | 20 |
| Now, it is the turn of A and B and | 3 | part is filled by A and B in 1 hour. |
| 20 |
Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.
Discussion:
52 comments Page 2 of 6.
Ekta Kaushik said:
5 years ago
Thanks @Roshni.
(2)
Manisha said:
7 years ago
Thanks @Roshni.
(1)
Ekta Kaushik said:
5 years ago
Thanks @Saisam.
(1)
Hema said:
8 years ago
Taking the lcm(12,15,20) = 60 liters (capacity of the tank).
In 1 hr : A can fill 5 liters(60/12), B = 4 liters C = 3 liters.
Its given that A is open all time and B & C alternately.
So, in 1st hr : A+B = 9 liters.
2nd hr : A+C = 8 liters.
3rd hr : A+B = 9 liters.
4th hr : A+C = 8 liters.
5th hr : A+B = 9 liters.
6th hr : A+C = 8 liters.
7th hr : A+B = 9 liters.
Total = 60 liters (Capacity of the tank).
So Ans = C. 7 hours.
In 1 hr : A can fill 5 liters(60/12), B = 4 liters C = 3 liters.
Its given that A is open all time and B & C alternately.
So, in 1st hr : A+B = 9 liters.
2nd hr : A+C = 8 liters.
3rd hr : A+B = 9 liters.
4th hr : A+C = 8 liters.
5th hr : A+B = 9 liters.
6th hr : A+C = 8 liters.
7th hr : A+B = 9 liters.
Total = 60 liters (Capacity of the tank).
So Ans = C. 7 hours.
(1)
Maximus said:
8 years ago
Consider the total capacity = X Litre
Amount filled in 1h by tank "A" = (X/12) Litre
Amount filled in 1h by tank "B" = (X/15) Litre
Amount filled in 1h by tank "C" = (X/20) Litre.
Now the problem says that the tank "A" is kept open and tanks "B" and "C" are opened alternatively for 1h until the tank gets full.
For the first 1 hour, the amount that both "A" and "B" can fill collectively = (X/12)+(X/15)=(27X/180)=(3X/20).
For the next 1hour the amount that both "A" and "C" can fill collectively = (X/12)+(X/20)= (32X/240)=(2X/15).
Total amount of water after 2 hours = {(3X/20)+(2X/15)} = 17X/60.
Now, this amount (17X/60), corresponding to 2hrs, repeats as a unit.
Since the max.amount (Capacity) = X Litre
(17X/60) * n = X
n = 3.5.
17X/60 Litres of 2hrs repeats for 3.5 times which means 3.5*2 = 7 hrs.
Amount filled in 1h by tank "A" = (X/12) Litre
Amount filled in 1h by tank "B" = (X/15) Litre
Amount filled in 1h by tank "C" = (X/20) Litre.
Now the problem says that the tank "A" is kept open and tanks "B" and "C" are opened alternatively for 1h until the tank gets full.
For the first 1 hour, the amount that both "A" and "B" can fill collectively = (X/12)+(X/15)=(27X/180)=(3X/20).
For the next 1hour the amount that both "A" and "C" can fill collectively = (X/12)+(X/20)= (32X/240)=(2X/15).
Total amount of water after 2 hours = {(3X/20)+(2X/15)} = 17X/60.
Now, this amount (17X/60), corresponding to 2hrs, repeats as a unit.
Since the max.amount (Capacity) = X Litre
(17X/60) * n = X
n = 3.5.
17X/60 Litres of 2hrs repeats for 3.5 times which means 3.5*2 = 7 hrs.
(1)
INNA REDDY CHILAKALA said:
9 years ago
@Anil.
Your method is simple and easy to solve the solution.
Thank you.
Your method is simple and easy to solve the solution.
Thank you.
Sadam said:
9 years ago
x/12 + (1/15 + 1/20) x/2 = 1.
x = 7 hours.
x = 7 hours.
Piyush said:
9 years ago
Can I solve it in this way?
Let total time taken to fill the tank be x.
Now ,
(1/12)x + (1/15)(x/2) + (1/20)(x/2) = 1,
(x/12)+(x/30)+(x/40) = 1.
(10x+4x+3x)/120 = 1.
17x = 120.
x = 120/17.
x = 7(approx of 7.0588).
Let total time taken to fill the tank be x.
Now ,
(1/12)x + (1/15)(x/2) + (1/20)(x/2) = 1,
(x/12)+(x/30)+(x/40) = 1.
(10x+4x+3x)/120 = 1.
17x = 120.
x = 120/17.
x = 7(approx of 7.0588).
Siva said:
9 years ago
Thank you, @Alvin.
Your method is easy and really simple.
Your method is easy and really simple.
Anu kanna said:
8 years ago
Taking the lcm(12,15,20) = 60 liters (capacity of the tank).
In 1 hr : A can fill 5 liters(60/12), B = 4 liters C = 3 liters.
It's given that A is open all time and B & C alternately.
So in 1st hr : A+B = 9 liters.
2nd hr : A+C = 8 liters.
3rd hr : A+B = 9 liters.
4th hr : A+C = 8 liters.
5th hr : A+B = 9 liters.
6th hr : A+C = 8 liters.
7th hr : A+B = 9 liters.
Total = 60 liters (Capacity of the tank).
So, Ans = C. 7 hours.
In 1 hr : A can fill 5 liters(60/12), B = 4 liters C = 3 liters.
It's given that A is open all time and B & C alternately.
So in 1st hr : A+B = 9 liters.
2nd hr : A+C = 8 liters.
3rd hr : A+B = 9 liters.
4th hr : A+C = 8 liters.
5th hr : A+B = 9 liters.
6th hr : A+C = 8 liters.
7th hr : A+B = 9 liters.
Total = 60 liters (Capacity of the tank).
So, Ans = C. 7 hours.
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