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Two pipes P and Q can fill a cistern in 12 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours will the tank be full?
- a)4 hours
- b)5 hours
- c)2 hours
- d)6 hours
- e)None of the Above
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Two pipes P and Q can fill a cistern in 12 hours and 4 hours respectiv...
Pipe P can fill = 1/12
Pipe Q can fill = 1/4
For every two hour, 1/12 + 1/4 = 1/3 Part filled
Total = 6 hours
Pipe Q can fill = 1/4
For every two hour, 1/12 + 1/4 = 1/3 Part filled
Total = 6 hours
Most Upvoted Answer
Two pipes P and Q can fill a cistern in 12 hours and 4 hours respectiv...
Pipe P can fill = 1/12
Pipe Q can fill = 1/4
For every two hour, 1/12 + 1/4 = 1/3 Part filled
Total = 6 hours
Pipe Q can fill = 1/4
For every two hour, 1/12 + 1/4 = 1/3 Part filled
Total = 6 hours
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Two pipes P and Q can fill a cistern in 12 hours and 4 hours respectiv...
Problem:
Two pipes P and Q can fill a cistern in 12 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours will the tank be full?
Solution:
Let us assume the capacity of the tank as 'C' and the rate of filling of pipe P and Q as 'x' and 'y' respectively.
Given,
Pipe P can fill the tank in 12 hours.
Therefore, the rate of filling of pipe P = C/12
Pipe Q can fill the tank in 4 hours.
Therefore, the rate of filling of pipe Q = C/4
Let the time taken by both the pipes working alternatively to fill the tank be 't' hours.
In the first hour, pipe P will fill the tank with a rate of C/12.
In the second hour, pipe Q will fill the tank with a rate of C/4.
In the third hour, pipe P will fill the tank with a rate of C/12.
In the fourth hour, pipe Q will fill the tank with a rate of C/4.
This process will continue until the tank gets filled.
As per the given condition, pipe P is opened first.
Therefore, the tank will be filled by pipe P in the first hour.
So, the total time taken to fill the tank will be:
C/12 + C/4 + C/12 + C/4 + C/12 + C/4 +... (t terms)
= [t/2][(2C/12) + (2C/4)]
= [t/2][(C/6) + (C/2)]
= t(C/3)
As per the question, the tank should be filled in 't' hours.
Therefore,
t(C/3) = C
t = 3 hours
Hence, the tank will be full in 6 hours.
Thus, option (D) is the correct answer.
Two pipes P and Q can fill a cistern in 12 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours will the tank be full?
Solution:
Let us assume the capacity of the tank as 'C' and the rate of filling of pipe P and Q as 'x' and 'y' respectively.
Given,
Pipe P can fill the tank in 12 hours.
Therefore, the rate of filling of pipe P = C/12
Pipe Q can fill the tank in 4 hours.
Therefore, the rate of filling of pipe Q = C/4
Let the time taken by both the pipes working alternatively to fill the tank be 't' hours.
In the first hour, pipe P will fill the tank with a rate of C/12.
In the second hour, pipe Q will fill the tank with a rate of C/4.
In the third hour, pipe P will fill the tank with a rate of C/12.
In the fourth hour, pipe Q will fill the tank with a rate of C/4.
This process will continue until the tank gets filled.
As per the given condition, pipe P is opened first.
Therefore, the tank will be filled by pipe P in the first hour.
So, the total time taken to fill the tank will be:
C/12 + C/4 + C/12 + C/4 + C/12 + C/4 +... (t terms)
= [t/2][(2C/12) + (2C/4)]
= [t/2][(C/6) + (C/2)]
= t(C/3)
As per the question, the tank should be filled in 't' hours.
Therefore,
t(C/3) = C
t = 3 hours
Hence, the tank will be full in 6 hours.
Thus, option (D) is the correct answer.
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Two pipes P and Q can fill a cistern in 12 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours will the tank be full?a)4 hoursb)5 hoursc)2 hoursd)6 hourse)None of the AboveCorrect answer is option 'D'. Can you explain this answer?
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