There are three taps A,B and C.they can fill the tank in 5,30,and 105 hrs respectively.At first all of them are opened simultaneously.Then after 1 hr tap C is closed and A and B are kept running.After another one hour tap B is closed.The remaining work is done by tap A alone.Find the percentage of the work done by Tap A by itself?
Answers
Total % of work done by A is 92.38%
Explanation :
A takes 5 hrs to fill the tank
=> work done by A in one hour = 1/5
B takes 30 hrs to fill the tank
=> work done by B in one hour = 1/30
C takes 105 hrs to fill the tank
=> work done by C in one hour = 1/105
Work done by all 3 taps in 1st hour = 1/5 + 1/30 + 1/105
= 51/210
work done by A and B in 2nd hour = 1/5 + 1/30 = 7/30
Total work done after 2 hours = 51/210 + 7/30 = 100/210
Hence Rest of the work done by A = 1 - 100/210
= 110/210
Total work done by A = 1/5 + 1/5 + 110/210 = 194/210
Hence % of work done by A = 194 x 100/210 = 92.38 %
HEY Buddy.....!! here is ur answer
Answer : 92.38%
Step by step solution :
Given that : Taps A,B and C can fill the tank in 5,30 and 105 hrs respectively.
LCM of 5,30 and 105 = 210
Then, Let the total work = 210
A can fill the tank in 1 hour = 210/5 = 42
B can fill the tank in 1 hour = 210/30 = 7
and C can fill the tank in 1 hour = 210/105 = 2
Given that : Till 1 hour they work simultaneously it means they complete the work in 1 hour = 42+7+2 = 51
After 1 hour tank C is left then till next one hour A and B fill the tank then they both complete the work in next 1 hour = 42+7 = 49
Remaining work = 210–(51+49) = 210–100 = 110
Remaining work is done by A.
Total work is done by A = 42+42+110 = 194
Percentage of work which is done by A = (194×100)210 = 92.38%
Thus, A done 92.38% of total work.
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