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Three taps P, Q and R can fill a tank in 12, 15 and 20 hours respectively. If P is open all the time and Q, R are open for one hour each alternatively, the tank will be full in
- a)3 hours
- b)2 hours
- c)7 hours
- d)4 hours
- e)None of the Above
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Three taps P, Q and R can fill a tank in 12, 15 and 20 hours respectiv...
(P + Q)’s 1 hour work = 1/12 + 1/15 = 3/20
(P + R)’s 1 hour work = 1/12 + 1/20 = 2/15
For 2 hrs = (3/20 + 2/15) = 17/60
For 6 hrs = (3*17/60) = 17/20
Remaining Part = 1 – 17/20 = 3/20 filled by P and Q in 1 hour
(P + R)’s 1 hour work = 1/12 + 1/20 = 2/15
For 2 hrs = (3/20 + 2/15) = 17/60
For 6 hrs = (3*17/60) = 17/20
Remaining Part = 1 – 17/20 = 3/20 filled by P and Q in 1 hour
Most Upvoted Answer
Three taps P, Q and R can fill a tank in 12, 15 and 20 hours respectiv...
Solution:
Let's assume that the capacity of the tank is 'x'.
Given, P can fill the tank in 12 hours. So, in 1 hour it can fill 1/12th of the tank.
Also, Q can fill the tank in 15 hours, so in 1 hour it can fill 1/15th of the tank.
Similarly, R can fill the tank in 20 hours, so in 1 hour it can fill 1/20th of the tank.
Now, let's consider the given condition that P is open all the time and Q, R are open for one hour each alternatively.
In the first hour, Q will fill 1/15th of the tank and in the second hour, R will fill 1/20th of the tank.
So, after 2 hours, the total amount of water filled in the tank will be:
(1/12 + 1/15 + 1/20) * 2x = (1/4) * 2x = x/2
Now, in the third hour, Q will fill 1/15th of the tank and in the fourth hour, R will fill 1/20th of the tank.
So, after 4 hours, the total amount of water filled in the tank will be:
(1/12 + 1/15 + 1/20 + 1/15 + 1/20) * 4x = (7/12) * 4x = 7x/3
We can see that after 4 hours, the tank is more than half full.
Therefore, the tank will be full in less than 7 hours.
Option C (7 hours) is the correct answer.
Let's assume that the capacity of the tank is 'x'.
Given, P can fill the tank in 12 hours. So, in 1 hour it can fill 1/12th of the tank.
Also, Q can fill the tank in 15 hours, so in 1 hour it can fill 1/15th of the tank.
Similarly, R can fill the tank in 20 hours, so in 1 hour it can fill 1/20th of the tank.
Now, let's consider the given condition that P is open all the time and Q, R are open for one hour each alternatively.
In the first hour, Q will fill 1/15th of the tank and in the second hour, R will fill 1/20th of the tank.
So, after 2 hours, the total amount of water filled in the tank will be:
(1/12 + 1/15 + 1/20) * 2x = (1/4) * 2x = x/2
Now, in the third hour, Q will fill 1/15th of the tank and in the fourth hour, R will fill 1/20th of the tank.
So, after 4 hours, the total amount of water filled in the tank will be:
(1/12 + 1/15 + 1/20 + 1/15 + 1/20) * 4x = (7/12) * 4x = 7x/3
We can see that after 4 hours, the tank is more than half full.
Therefore, the tank will be full in less than 7 hours.
Option C (7 hours) is the correct answer.
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Community Answer
Three taps P, Q and R can fill a tank in 12, 15 and 20 hours respectiv...
(P + Q)’s 1 hour work = 1/12 + 1/15 = 3/20
(P + R)’s 1 hour work = 1/12 + 1/20 = 2/15
For 2 hrs = (3/20 + 2/15) = 17/60
For 6 hrs = (3*17/60) = 17/20
Remaining Part = 1 – 17/20 = 3/20 filled by P and Q in 1 hour
(P + R)’s 1 hour work = 1/12 + 1/20 = 2/15
For 2 hrs = (3/20 + 2/15) = 17/60
For 6 hrs = (3*17/60) = 17/20
Remaining Part = 1 – 17/20 = 3/20 filled by P and Q in 1 hour
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