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 3 of 6.
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.
Ashwin R said:
8 years ago
A in one hour = 1/12 part,
B in one hour = 1/15 part,
C in one hour = 1/20 part,
Now, A is opened continuously and B and C one hour alternately. So, 1st hour A+B, then for the next hour A+C.
A+B in one hour= 1/12 + 1/15= 3/20 parts,
A+C in one hour= 1/12 +1/20= 2/15 parts,
Now, in the first 2 hours ( ie; A+B for first hour and A+C for second hour)= 3/20 + 2/15 parts= 17/60 parts.
For the tank to be filled the part must become 60/60.
We know, 17 * 3 = 51 (which is the multiple less than 60),
So, 2 * 3 hours = 6 hours.
In 6 hours, part filled is 51/60. Now, simplifying it we get 17/20.
17/20 part is filled, so the rest part to be filled is 1-17/20= 3/20.
A+B in first hour.
A+C in second hour.
A+B in third hour.
A+C in fourth hour.
A+B in fifth hour.
A+C in sixth hour.
Now, it is the turn for A+B.
A+B can fill the rest in 1 hour, since part filled by A+B in one hour is 3/20.
So, the total time taken is 6+1 hours= 7 hours.
Hopes all got it.
B in one hour = 1/15 part,
C in one hour = 1/20 part,
Now, A is opened continuously and B and C one hour alternately. So, 1st hour A+B, then for the next hour A+C.
A+B in one hour= 1/12 + 1/15= 3/20 parts,
A+C in one hour= 1/12 +1/20= 2/15 parts,
Now, in the first 2 hours ( ie; A+B for first hour and A+C for second hour)= 3/20 + 2/15 parts= 17/60 parts.
For the tank to be filled the part must become 60/60.
We know, 17 * 3 = 51 (which is the multiple less than 60),
So, 2 * 3 hours = 6 hours.
In 6 hours, part filled is 51/60. Now, simplifying it we get 17/20.
17/20 part is filled, so the rest part to be filled is 1-17/20= 3/20.
A+B in first hour.
A+C in second hour.
A+B in third hour.
A+C in fourth hour.
A+B in fifth hour.
A+C in sixth hour.
Now, it is the turn for A+B.
A+B can fill the rest in 1 hour, since part filled by A+B in one hour is 3/20.
So, the total time taken is 6+1 hours= 7 hours.
Hopes all got it.
(5)
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).
Sadam said:
9 years ago
x/12 + (1/15 + 1/20) x/2 = 1.
x = 7 hours.
x = 7 hours.
Siva said:
9 years ago
Thank you, @Alvin.
Your method is easy and really simple.
Your method is easy and really simple.
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.
Brajesh said:
9 years ago
It was a nice solution @Alvin .
Thank you.
Thank you.
Narendra said:
10 years ago
Let's assume B works for x hours and C for y hours. So A will be working for x+y hours.
= (x+y)/12 + x/15 + y/20 = 1 (tank having 1 part).
= 9x + 8y = 60.
x = 4; y = 3.
= (x+y)/12 + x/15 + y/20 = 1 (tank having 1 part).
= 9x + 8y = 60.
x = 4; y = 3.
Swetha said:
10 years ago
Please say short cut method for problem 14 it is length process.
Arun said:
1 decade ago
1 hr tank can fill 3/20 -> (1/12+1/15).
2nd hr tank can fill 3/20+2/15 = 17/60.
For 2 hr it fills 17/60 then for 1 hr it fills 17/120 -> (1/2*17/60).
Consider 17 table 17*6 = 102, so in six hours it it fills 102/120 and next hr B Pipe will be open and.
Fills as 102/120+3/20 = 1. So 7 hrs it will take.
Another way:
For 2 hrs it fills 17/60.
Consider next 2 hrs i.e. 4 hrs 2*17/60 = 34/60.
Consider next 2 hrs i.e 6 hrs 51/60 only 9 is less for 60 i.e to full.
So next 1 hr b pipe open 51/60+9/60 = 1.
2nd hr tank can fill 3/20+2/15 = 17/60.
For 2 hr it fills 17/60 then for 1 hr it fills 17/120 -> (1/2*17/60).
Consider 17 table 17*6 = 102, so in six hours it it fills 102/120 and next hr B Pipe will be open and.
Fills as 102/120+3/20 = 1. So 7 hrs it will take.
Another way:
For 2 hrs it fills 17/60.
Consider next 2 hrs i.e. 4 hrs 2*17/60 = 34/60.
Consider next 2 hrs i.e 6 hrs 51/60 only 9 is less for 60 i.e to full.
So next 1 hr b pipe open 51/60+9/60 = 1.
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