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Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is opened all the time and B and C are opened one hour each alternatively , the tank will be full in:
- a)4
- b)4.5
- c)5
- d)7
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectiv...
Solution:
Let us assume that the tank is initially empty.
Let the total time taken to fill the tank be T.
In the first hour, tap A can fill 1/12th of the tank.
In the second hour, tap A and tap B can fill 1/12 + 1/15 = 3/60 + 4/60 = 7/60th of the tank.
In the third hour, tap A, tap B, and tap C can fill 1/12 + 1/20 = 5/60 + 3/60 = 8/60th of the tank.
In the fourth hour, tap A and tap B can fill 1/12 + 1/15 = 3/60 + 4/60 = 7/60th of the tank.
Similarly, in the fifth hour, tap A, tap B, and tap C can fill 8/60th of the tank.
So, we can observe that after every 3 hours, the amount of water filled is 7/60 + 8/60 = 15/60 = 1/4th of the tank.
Therefore, if A is opened all the time and B and C are opened alternatively for 1 hour each, then the tank will be full in 3n + 2 hours, where n is an integer.
Let us assume that the tank is full after 'T' hours.
So, (T-2) is a multiple of 3.
12a + 15b + 20c = T
a + b + c = T/3
a + b = (T-2)/3
If (T-2) is a multiple of 3, then (T-2)/3 is also an integer.
So, a+b is an integer.
Hence, T/3 is also an integer.
Therefore, T is a multiple of 3.
Let T = 3n.
12a + 15b + 20c = 3n
4a + 5b + 6c = n
Now, we need to find the smallest value of 'n' for which the equation 4a + 5b + 6c = n has a solution.
We can observe that the smallest value of 'n' for which the equation 4a + 5b + 6c = n has a solution is when n = 6.
So, the tank will be full in 3n + 2 = 3(6) + 2 = 20 hours.
Therefore, the correct option is (d) 7.
Let us assume that the tank is initially empty.
Let the total time taken to fill the tank be T.
In the first hour, tap A can fill 1/12th of the tank.
In the second hour, tap A and tap B can fill 1/12 + 1/15 = 3/60 + 4/60 = 7/60th of the tank.
In the third hour, tap A, tap B, and tap C can fill 1/12 + 1/20 = 5/60 + 3/60 = 8/60th of the tank.
In the fourth hour, tap A and tap B can fill 1/12 + 1/15 = 3/60 + 4/60 = 7/60th of the tank.
Similarly, in the fifth hour, tap A, tap B, and tap C can fill 8/60th of the tank.
So, we can observe that after every 3 hours, the amount of water filled is 7/60 + 8/60 = 15/60 = 1/4th of the tank.
Therefore, if A is opened all the time and B and C are opened alternatively for 1 hour each, then the tank will be full in 3n + 2 hours, where n is an integer.
Let us assume that the tank is full after 'T' hours.
So, (T-2) is a multiple of 3.
12a + 15b + 20c = T
a + b + c = T/3
a + b = (T-2)/3
If (T-2) is a multiple of 3, then (T-2)/3 is also an integer.
So, a+b is an integer.
Hence, T/3 is also an integer.
Therefore, T is a multiple of 3.
Let T = 3n.
12a + 15b + 20c = 3n
4a + 5b + 6c = n
Now, we need to find the smallest value of 'n' for which the equation 4a + 5b + 6c = n has a solution.
We can observe that the smallest value of 'n' for which the equation 4a + 5b + 6c = n has a solution is when n = 6.
So, the tank will be full in 3n + 2 = 3(6) + 2 = 20 hours.
Therefore, the correct option is (d) 7.
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Community Answer
Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectiv...
- Tank filled in one hour when the taps A and B are open = (1/12)+(1/15)= 9/60= 3/20.
- Tank filled in one hour when the taps A and C are open = (1/12) + (1/20) = 8/60 = 2/15
- Therefore, part of the tank filled in two hours = (3/20) + (2/15) = 17/60.
- Part filled in 6 hours = ( 17/60) * 3 = 51/60
- Remaining part = 9/60 = 3/20
After 6 hours Now it is (A+B)' s turn
- Time taken by the A+B to fill 3/20 part = 1 hour
- Therefore, total time taken = 6+1 = 7 hours
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Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is opened all the time and B and C are opened one hour each alternatively , the tank will be full in:a)4b)4.5c)5d)7Correct answer is option 'D'. Can you explain this answer?
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