Question

A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes.both the taps are kept open for 10 minutes and then the first tap is shut off.after this,the tank will be completely in what time

Answer 1

Hii...
Here is your answer..

Solution:-
A can fill tab in 20 min
In one min=1/20
B can fill tap in 60min
In one min=1/60

(A+B)'s one minute work=
$= > \frac{1}{20} + \frac{1}{60} \\ = > \frac{3 + 1}{60} = \frac{4}{60}$
Now,
Amount of work completed in 10 min=
$= > \frac{4}{60} \times 10 \\ = > \frac{4}{6} = \frac{2}{3}$
Remain work =
$= > 1 - \frac{2}{3} \\ = > \frac{3 - 2}{3} \\ = > \frac{1}{3}$
Now,
The left work has to be done by Tap B because A is shut off.
Work will be completed by Tap B in=
$= > \frac{1}{3} \times 60 \\ = > 20min$
Hence,
Tap B will then can fill the tank in 20min.

Hope it helps uh...✌️✌️✌️

Answer 2

Explanation:

Hii...

Here is your answer..

Solution:-

A can fill tab in 20 min

In one min=1/20

B can fill tap in 60min

In one min=1/60

(A+B)'s one minute work=

\begin{gathered} = > \frac{1}{20} + \frac{1}{60} \\ = > \frac{3 + 1}{60} = \frac{4}{60} \end{gathered}

=>

20

1

+

60

1

=>

60

3+1

=

60

4

Now,

Amount of work completed in 10 min=

\begin{gathered} = > \frac{4}{60} \times 10 \\ = > \frac{4}{6} = \frac{2}{3} \end{gathered}

=>

60

4

×10

=>

6

4

=

3

2

Remain work =

\begin{gathered} = > 1 - \frac{2}{3} \\ = > \frac{3 - 2}{3} \\ = > \frac{1}{3} \end{gathered}

=>1−

3

2

=>

3

3−2

=>

3

1

Now,

The left work has to be done by Tap B because A is shut off.

Work will be completed by Tap B in=

\begin{gathered} = > \frac{1}{3} \times 60 \\ = > 20min\end{gathered}

=>

3

1

×60

=>20min

Hence,

Tap B will then can fill the tank in 20min.

Hope it helps uh...✌️✌️✌️