Three taps A, B, C fill up a tank independently in 10 hr, 20 hr, 30 hr respectively. Initially, the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for an hour. What is the minimum number of hours required to fill the tank?
(a) 8
(b) 9
(c) 10
(d) 11
Answer
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Hint: Find the capacity of the tank filled by each pair of taps AB, BC and CA in 1 hour and then use the pair of taps which fills the highest capacity of tank in 1 hour to fill the remaining tank.
Given that three taps A, B, C fill up the tank independently in 10 hr, 20 hr, 30 hr respectively. We have to find the minimum number of hours required to fill the tank if each pair of taps is open for at least an hour.
Since 60 is LCM of 10, 20 and 30, therefore let us take the total capacity of the tank as L = 60 liters.
So, we are given that A can fill the full tank in 10 hours.
That is, in 10 hours, A can fill 60 liters.
So, in 1 hour, A can fill
\[\dfrac{60}{10}=6\text{ liters}\]
Similarly, in 20 hours B can fill 60 liters.
So, in 1 hour, B can fill
\[\dfrac{60}{20}=3\text{ liters}\]
Similarly, in 30 hours C can fill 60 liters.
So, in 1 hour, C can fill
\[\dfrac{60}{30}=2\text{ liters}\]
Therefore, in 1 hour, a pair of A and B can fill = 6 + 3 = 9 liters.
Similarly, in 1 hour, pair of B and C can fill = 3 + 2 = 5 liters
Similarly, in 1 hour, pair of C and A can fill = 6 + 2 = 8 liters
It is given in the question that each pair is open for at least an hour. Therefore, if we open AB for first hour, BC for second hour and CA for third hour,
Then in total 3 hours, they can fill = 9 + 5 + 8 = 22 liters.
Since we know that total capacity is 60 liters, we still have 60 – 22 = 38 liters to be filled.
It is given in question that the tank must be filled in minimum hours. We have already found that the pair AB fills the highest capacity of the tank in 1 hour. This capacity is 9 liters. Therefore, we will use the pair AB to fill the remaining part of the tank. This remaining part of the tank, as we have found already is 38 liters.
Let us consider that AB takes x hours to fill the remaining tank,
Then, we get 9x = 38 liters. Or,
\[x=\dfrac{38}{9}\text{ hours}\]
\[x=4.22\text{ hours}\]
Therefore, total time taken to fill the tank
\[\begin{align}
& =3+4.22 \\
& =7.22\text{ hours} \\
\end{align}\]
As the tank gets filled in the 8th hour, therefore option (a) is correct.
Note: In these types of questions, students take the total capacity of the tank to be x and get confused with fractions. So it is advisable to take the total capacity of the tank to be a number preferably the one which is the LCM of the given hours so that, easily it gets cancelled. Also, students must remember that each pair of taps must get opened for at least 1 hour.
Given that three taps A, B, C fill up the tank independently in 10 hr, 20 hr, 30 hr respectively. We have to find the minimum number of hours required to fill the tank if each pair of taps is open for at least an hour.
Since 60 is LCM of 10, 20 and 30, therefore let us take the total capacity of the tank as L = 60 liters.
So, we are given that A can fill the full tank in 10 hours.
That is, in 10 hours, A can fill 60 liters.
So, in 1 hour, A can fill
\[\dfrac{60}{10}=6\text{ liters}\]
Similarly, in 20 hours B can fill 60 liters.
So, in 1 hour, B can fill
\[\dfrac{60}{20}=3\text{ liters}\]
Similarly, in 30 hours C can fill 60 liters.
So, in 1 hour, C can fill
\[\dfrac{60}{30}=2\text{ liters}\]
Therefore, in 1 hour, a pair of A and B can fill = 6 + 3 = 9 liters.
Similarly, in 1 hour, pair of B and C can fill = 3 + 2 = 5 liters
Similarly, in 1 hour, pair of C and A can fill = 6 + 2 = 8 liters
It is given in the question that each pair is open for at least an hour. Therefore, if we open AB for first hour, BC for second hour and CA for third hour,
Then in total 3 hours, they can fill = 9 + 5 + 8 = 22 liters.
Since we know that total capacity is 60 liters, we still have 60 – 22 = 38 liters to be filled.
It is given in question that the tank must be filled in minimum hours. We have already found that the pair AB fills the highest capacity of the tank in 1 hour. This capacity is 9 liters. Therefore, we will use the pair AB to fill the remaining part of the tank. This remaining part of the tank, as we have found already is 38 liters.
Let us consider that AB takes x hours to fill the remaining tank,
Then, we get 9x = 38 liters. Or,
\[x=\dfrac{38}{9}\text{ hours}\]
\[x=4.22\text{ hours}\]
Therefore, total time taken to fill the tank
\[\begin{align}
& =3+4.22 \\
& =7.22\text{ hours} \\
\end{align}\]
As the tank gets filled in the 8th hour, therefore option (a) is correct.
Note: In these types of questions, students take the total capacity of the tank to be x and get confused with fractions. So it is advisable to take the total capacity of the tank to be a number preferably the one which is the LCM of the given hours so that, easily it gets cancelled. Also, students must remember that each pair of taps must get opened for at least 1 hour.
Last updated date: 14th Sep 2023
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