Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
6 2/3hours
7 hours
6 hours
7 1/2hours
Explanation:
LCM(12,15,20)=60
Take capacity of the tank as 60 litre. Then,
Quantity filled by pipe A in 1 hour =60/12=5 litre.
Quantity filled by pipe B in 1 hour =60/15=4 litre.
Quantity filled by pipe C in 1 hour =60/20=3 litre.
In first hour, A and B are open and fill 5+4=9 litre. In second hour, A and C are open and fill 5+3=8 litre.
Therefore, in 2 hours, 9+8=17 litre is filled.
This pattern goes on.
In 2×3=6 hours, 17×3=51 litre is filled.
In 7th hour, A and B are open and fill 9 more litre. Thus, tank is totally filled.
(∵51+9=60)
Therefore, required time =7 hour.
LCM(12,15,20)=60
Take capacity of the tank as 60 litre. Then,
Quantity filled by pipe A in 1 hour =60/12=5 litre.
Quantity filled by pipe B in 1 hour =60/15=4 litre.
Quantity filled by pipe C in 1 hour =60/20=3 litre.
In first hour, A and B are open and fill 5+4=9 litre. In second hour, A and C are open and fill 5+3=8 litre.
Therefore, in 2 hours, 9+8=17 litre is filled.
This pattern goes on.
In 2×3=6 hours, 17×3=51 litre is filled.
In 7th hour, A and B are open and fill 9 more litre. Thus, tank is totally filled.
(∵51+9=60)
Therefore, required time =7 hour.