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

Question

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.​

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Athena 4 weeks 2021-09-18T22:12:39+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-09-18T22:13:46+00:00

    Solution : 1

     \sf \: Time \:  taken \:  by \:  a \:  tap \:  to \:  fill \:  a  \: tank  \: =  \: 3  \: 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{2 \:  + 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.

    0
    2021-09-18T22:14:20+00:00

    part filled by tap A in 1 hr=1/4

    part filled by tap B in 1 hr=1/6

    both are opened s0

    =1/4+1/6

    =10/24

    =5/12

    take the reciprocal

    so 12/5

    so 2 hour (2/5) minutes

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