Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 1)
1.
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?
Answer: Option
Explanation:
| Part filled by (A + B + C) in 3 minutes = 3 |  |
1 | + | 1 | + | 1 |  |
= | |
3 x | 11 | |
= | 11 | . |
| 30 | 20 | 10 | 60 | 20 |
| Part filled by C in 3 minutes = | 3 | . |
| 10 |
 Required ratio = |
|
3 | x | 20 | |
= | 6 | . |
| 10 | 11 | 11 |
Discussion:
75 comments Page 3 of 8.
Archisha said:
9 years ago
@Drishti Anand.
In the end, we are doing (3/10) / (11/20).
Which is equal to (3/10) * (20/11)
That's why we take it in reciprocal.
In the end, we are doing (3/10) / (11/20).
Which is equal to (3/10) * (20/11)
That's why we take it in reciprocal.
(1)
Zara said:
8 years ago
The section is the great help for me. Thanks.
Doug said:
9 years ago
Just to make it easier for you guys let me show you my trick to solve this particular question.
So, first we have to find out the total capacity of the tank so for that we have to find out the lcm of the above 3 no's.
i.e 30/20/10 so lcm would be 60 and this 60 is the capacity of the tank.
Now let us find out the speed at which the tank will be filled so just divide the total capacity with the given time.
i.e : 60/30, 60/20, 60/10
Now we have the the speed at which this tank will be filled i.e 2lt/min, 3lt/min and 6lt/min.
If a pipe can fill 2 lt of chemical in 1 min so how much will be the amount of chemical after 3 min. i.e 6lt similarly for b it would be 9lt and for c it will be 18lt and we have the calculate the ratio of chemical l after 3 min.
So the ratio can be calculated as shown below:
18(amount of chemical after 3min)/18 + 9 ++6 = 18/33ie 6/11 (solution).
So, first we have to find out the total capacity of the tank so for that we have to find out the lcm of the above 3 no's.
i.e 30/20/10 so lcm would be 60 and this 60 is the capacity of the tank.
Now let us find out the speed at which the tank will be filled so just divide the total capacity with the given time.
i.e : 60/30, 60/20, 60/10
Now we have the the speed at which this tank will be filled i.e 2lt/min, 3lt/min and 6lt/min.
If a pipe can fill 2 lt of chemical in 1 min so how much will be the amount of chemical after 3 min. i.e 6lt similarly for b it would be 9lt and for c it will be 18lt and we have the calculate the ratio of chemical l after 3 min.
So the ratio can be calculated as shown below:
18(amount of chemical after 3min)/18 + 9 ++6 = 18/33ie 6/11 (solution).
Hrishi said:
9 years ago
Great & better solution @Niranjankumar.
JISHNU said:
9 years ago
Regardless of the minutes the ratio will always the same,
i.e, the ratio is [1/10] / [(1/30) + (1/20) + (1/10)] = (1/10)/(11/60) = 6/11.
i.e, the ratio is [1/10] / [(1/30) + (1/20) + (1/10)] = (1/10)/(11/60) = 6/11.
Raji said:
9 years ago
You are right @Shakthi.
Sravanthi said:
9 years ago
Is it liters or minutes for a, b, c can you explain the question?
Jothi said:
8 years ago
1 =? =>7*3/2=21/2hrs.
Please explain in detail.
Please explain in detail.
Moon said:
8 years ago
Part filled by (A + B + C) in 3 minutes = 3 = 1 + 1 + 1.
3 x 11 = 11.
30 20 10 60 20
Part filled by C in 3 minutes = 3.
10
Required ratio = 3x20= 6.
10 11 11
3 x 11 = 11.
30 20 10 60 20
Part filled by C in 3 minutes = 3.
10
Required ratio = 3x20= 6.
10 11 11
Mohsin said:
8 years ago
Why made reciprocal of 11/20? please explain it.
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