if a tank was filled by two taps A and B can fill the tank in 20 hours and B can fill the tank in 30 hours and third type C can empty the tank in 1 hour then 10 me how much time will taken by A and B to fill the tank at once
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Answer:
Part of the tank filled when all the three pipes are opened for 10 hours
=10×(
15
1
+
20
1
−
25
1
)=10×(
300
20+15−12
)=
300
230
=
30
23
Remaining empty part = 1−
30
23
=
30
7
Part of the tank filled in 1 hour by pipes (A and B) =
15
1
+
20
1
=
60
4+3
=
60
7
∵
60
7
part is filled in 1 hour
30
7
part is filled in 2 hours
∴ Total time to fill the tank =10+2=12 hours
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