Brainilist? A tank has four emptying taps, all of equal efficiencies, but each fixed at 1/5th, 2/5th, 3/5th, 4/5th of the height of the tank respectively. Two filling taps, each of which individually can fill the tank in 200 minute, are also connected to the tank. If the ratio of the efficiency of each emptying tap and the efficiency of each filling tap is 1:3, and all the six taps are opened simultaneously, Then in how much time will the empty tank be filled? 1)166 minutes 2)174 minutes 3)182 minutes 4)170 minutes
Answers
Answer:
174 Minutes
Step-by-step explanation:
Tap can fill in Tank in 200 minutes
Tap can fill Tank in 1 Minutes = 1/200
2 Tap Can fill Tank in 1 Minutes = 1/100
Tap can empty tank in 200*3 = 600 min ( 1 : 3 efficiency)
Tap Can empty tank in 1 Min = 1/600
Until tank is filled 1/5 (no emptying tap working)
Tank filed in 1 min = 1/100
1/5 Tank will be filled in 100/5 = 20 min
now from 1/5 to 2/5
Tank filed in 1 min = 1/100 - 1/600 = 5/600 = 1/120
1/5 Tank will be filled in 120/5 = 24 min
now from 2/5 to 3/5 ( two emptying tap will work)
Tank filed in 1 min = 1/100 - 2/600 = 4/600 = 1/150
2/5 to 3/5 tank will be filled in 150/5 = 30 min
Simillarly 3/5 to 4/5 (three Emptying Tap)
1/100 - 3/600 = 3/600 = 1/200
3/5 to 4/5 tank will be filled in 200/5 = 40 min
Similarly 4/5 to 5/5
1/100 - 4/600 = 2/600 = 1/300
3/5 to 4/5 tank will be filled in 300/5 = 60 min
Total Time taken = 60 + 40 + 30 + 24 + 20
= 174 Minutes