Two taps a and b can fill a tank in 20 min and 30 min and tap c can empty the tank. A man open tap a and b after coming back he saw tap c was also openedthen he closes tap c.Then tank is filled in 3min.In how much time c can empty the full tank
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Part filled by A in 1 min = 124124
Part filled by B in 1 min = 132132
Let B is closed after x min. Then, [Part filled by (A + B) in x mm] + [Part filled by A in (9-x) min] = 1
Therefore,
x(124+132)+(9−x)×124=1x(124+132)+(9−x)×124=1
=> x(4+396)+(9−x)24=1x(4+396)+(9−x)24=1
=> 7x96+(9−x)24=17x96+(9−x)24=1
=> 7x+4(9−x)96=17x+4(9−x)96=1
=> 7x+4(9−x)=967x+4(9−x)=96
=> 7x+36-4x=96
=> 7x - 4x = 96 - 36
= 3x = 60 => x = 603603 = 20
Part filled by B in 1 min = 132132
Let B is closed after x min. Then, [Part filled by (A + B) in x mm] + [Part filled by A in (9-x) min] = 1
Therefore,
x(124+132)+(9−x)×124=1x(124+132)+(9−x)×124=1
=> x(4+396)+(9−x)24=1x(4+396)+(9−x)24=1
=> 7x96+(9−x)24=17x96+(9−x)24=1
=> 7x+4(9−x)96=17x+4(9−x)96=1
=> 7x+4(9−x)=967x+4(9−x)=96
=> 7x+36-4x=96
=> 7x - 4x = 96 - 36
= 3x = 60 => x = 603603 = 20
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