Question :1. A tap fills tank in 3 hours and another tap can fI'll it

1.	Question :1. A tap fills tank in 3 hours and another tap can fI'll it in 6 hours. If both taps are opened together, how long it take for the tank to be filled.2. Tap 'A' fills the tank in 8 hours and another tap 'B' empties it in 12 hours. If the tank is empty and both the tanks are opened together, how long will it take to fill the tank?3. A pipe can fill a cistern in 3 hours. Due to a leak at the bottom it is filled in 4 hours. When the Cistern is full. in how much time will it be emptied by the leak.
Answer» SOLUTION : 1 $\sf \: Time \: taken \: by \: a \: tap \: to \: fill \: a \: tank \: = \: <klux>3</klux> \: hours.$ $\sf \: So, \: in \: 1 \: hours \: it \: fills \: \: \dfrac{1}{3} \: \: of \: the \: tank.$ $\sf \: Time \: taken \: by \: the \: second \: tap \: to \: fill \: the \: tank \: = \: 6 \: hours.$ $\sf \: So, \: in \: 1 \: hours \: it \: fills \: \: \dfrac{1}{6} \: \: of \: the \: tank.$ When both are OPENED TOGETHER $\sf \: Work \: done \: in \: 1 \: hours \: = \: \dfrac{1}{3} \: + \: \dfrac{1}{6} \: = \: \dfrac{<klux>2</klux> \: + 1}{6} \: = \: \dfrac{1}{2}$ $\sf \: Thus, \: Time \: taken \: by \: both \: the \: taps \: to \: fill \: the \: tank \: = \: 1 \: \div \: \dfrac{1}{2} \: = \: 2 \: hours.$ _____________________________________ Solution : 2 $\sf \: Time \: taken \: by \: tap \: A \: to \: fill \: the \: tank \: = 8 \: hours.$ $\sf \: Work \: done \: by \: tap \: A \: in \: 1 \: hours \: = \: \dfrac{1}{8}$ $\sf \: Time \: taken \: by \: tap \: B \: to \: empty \: the \: tank \: = \: 12 \: hours.$ $\sf \: Work \: done \: by \: tap \: B \: in \: one \: hours \: = \: - \: \dfrac{1}{12}$ $\sf \: Work \: done \: by \: tap \: (A + B) \: :$ $\sf \longrightarrow \: \dfrac{1}{8} \: - \: \dfrac{1}{12} \: = \: \dfrac{3 \: - \: 2}{24} \: = \: \dfrac{1}{24}$ $\sf \: Thus, \: Time \: taken \: by \: tap \: A \: and \: B \: to \: fill \: the \: tank \: = \:1 \: + \: \dfrac{1}{24} \: = \: 24 \: hours.$ _____________________________________ Solution : 3 $\sf \: When \: there \: is \: no \: leakage$ $\sf \: Time \: taken \: by \: pipe \: to \: fill \: the \: cistern \: = \: 3 \: hours.$ $\sf \: Work \: done \: by \: pipe \: in \: 1 \: hours \: = \: \dfrac{1}{3}$ $\sf \: When \: there \: is \: leakage$ $\sf \: Time \: taken \: by \: pipe \: to \: fill \: the \: tank \: = \: 4 \: hours.$ $\sf \: Work \: done \: by \: pipe \: in \: 1 \: hours \: = \: - \: \dfrac{1}{4}$ $\sf \: Work \: done \: by \: the \: leak \: in \: 1 \: hour \: :$ $\sf \: \dfrac{1}{3} \: - \: \dfrac{1}{4} \: = \: \dfrac{4 \: - \: 3}{12} \: = \: \dfrac{1}{12}$ $\sf \: Hence \: the \: Cistern \: will \: be \: emptied \: by \: the \: leakage \: in \: 12 \: hours.$

Question :1. A tap fills tank in 3 hours and another tap can fI'll it in 6 hours. If both taps are opened together, how long it take for the tank to be filled.2. Tap 'A' fills the tank in 8 hours and another tap 'B' empties it in 12 hours. If the tank is empty and both the tanks are opened together, how long will it take to fill the tank?3. A pipe can fill a cistern in 3 hours. Due to a leak at the bottom it is filled in 4 hours. When the Cistern is full. in how much time will it be emptied by the leak.

Answer»

$\sf \: Time \: taken \: by \: a \: tap \: to \: fill \: a \: tank \: = \: <klux>3</klux> \: hours.$

$\sf \: So, \: in \: 1 \: hours \: it \: fills \: \: \dfrac{1}{3} \: \: of \: the \: tank.$

$\sf \: Time \: taken \: by \: the \: second \: tap \: to \: fill \: the \: tank \: = \: 6 \: hours.$

$\sf \: So, \: in \: 1 \: hours \: it \: fills \: \: \dfrac{1}{6} \: \: of \: the \: tank.$

$\sf \: Work \: done \: in \: 1 \: hours \: = \: \dfrac{1}{3} \: + \: \dfrac{1}{6} \: = \: \dfrac{<klux>2</klux> \: + 1}{6} \: = \: \dfrac{1}{2}$

$\sf \: Thus, \: Time \: taken \: by \: both \: the \: taps \: to \: fill \: the \: tank \: = \: 1 \: \div \: \dfrac{1}{2} \: = \: 2 \: hours.$

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Solution : 2

$\sf \: Time \: taken \: by \: tap \: A \: to \: fill \: the \: tank \: = 8 \: hours.$

$\sf \: Work \: done \: by \: tap \: A \: in \: 1 \: hours \: = \: \dfrac{1}{8}$

$\sf \: Time \: taken \: by \: tap \: B \: to \: empty \: the \: tank \: = \: 12 \: hours.$

$\sf \: Work \: done \: by \: tap \: B \: in \: one \: hours \: = \: - \: \dfrac{1}{12}$

$\sf \: Work \: done \: by \: tap \: (A + B) \: :$

$\sf \longrightarrow \: \dfrac{1}{8} \: - \: \dfrac{1}{12} \: = \: \dfrac{3 \: - \: 2}{24} \: = \: \dfrac{1}{24}$

$\sf \: Thus, \: Time \: taken \: by \: tap \: A \: and \: B \: to \: fill \: the \: tank \: = \:1 \: + \: \dfrac{1}{24} \: = \: 24 \: hours.$

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Solution : 3

$\sf \: When \: there \: is \: no \: leakage$

$\sf \: Time \: taken \: by \: pipe \: to \: fill \: the \: cistern \: = \: 3 \: hours.$

$\sf \: Work \: done \: by \: pipe \: in \: 1 \: hours \: = \: \dfrac{1}{3}$

$\sf \: When \: there \: is \: leakage$

$\sf \: Time \: taken \: by \: pipe \: to \: fill \: the \: tank \: = \: 4 \: hours.$

$\sf \: Work \: done \: by \: pipe \: in \: 1 \: hours \: = \: - \: \dfrac{1}{4}$

$\sf \: Work \: done \: by \: the \: leak \: in \: 1 \: hour \: :$

$\sf \: \dfrac{1}{3} \: - \: \dfrac{1}{4} \: = \: \dfrac{4 \: - \: 3}{12} \: = \: \dfrac{1}{12}$

$\sf \: Hence \: the \: Cistern \: will \: be \: emptied \: by \: the \: leakage \: in \: 12 \: hours.$

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