7878\frac{7}{8} of full flow in pipe A and 56" role="presentation">5656\frac{5}{6} of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow." /> 7878\frac{7}{8} of full flow in pipe A and 56" role="presentation">5656\frac{5}{6} of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow." /> 7878\frac{7}{8} of full flow in pipe A and 56" role="presentation">5656\frac{5}{6} of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow." />

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Two pipes A and B can fill a cistern in 12 min and 16 min respectively. Both the pipes are opened together for a certain time but due to some obstruction the flow of water was restricted to $\frac{7}{8}$ $\frac{7}{8}$ of full flow in pipe A and $\frac{5}{6}$ $\frac{5}{6}$ of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow.

18 min
23 min
35.6 min
44.5 min

Answer:- 1

Explanation:-

Solution:

Let the obstruction remain for x min.
Hence,
Part of cistern filled in X min + part of cistern filled in 3 min = full cistern

[\frac{7 x}{8 \times 12} + \frac{5 x}{6 \times 16}]

$\left[ {\frac{{7{\text{x}}}}{{8 \times 12}} + \frac{{5{\text{x}}}}{{6 \times 16}}} \right]$

+

$+$

[\frac{3}{12} + \frac{3}{16}]

$\left[ {\frac{3}{{12}} + \frac{3}{{16}}} \right]$ = 1

\frac{12 x}{96} + \frac{7}{16} = 1

$\frac{{12{\text{x}}}}{{96}} + \frac{7}{{16}} = 1$
Thus,
X = 4.5 min.

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of full flow in pipe A and

\frac{5}{6}

$\frac{5}{6}$ of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow.", "text": "Two pipes A and B can fill a cistern in 12 min and 16 min respectively. Both the pipes are opened together for a certain time but due to some obstruction the flow of water was restricted to

\frac{7}{8}

$\frac{7}{8}$ of full flow in pipe A and

\frac{5}{6}

$\frac{5}{6}$ of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow.", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

Let the obstruction remain for x min.
Hence,
Part of cistern filled in X min + part of cistern filled in 3 min = full cistern

[\frac{7 x}{8 \times 12} + \frac{5 x}{6 \times 16}]

$\left[ {\frac{{7{\text{x}}}}{{8 \times 12}} + \frac{{5{\text{x}}}}{{6 \times 16}}} \right]$

+

$+$

[\frac{3}{12} + \frac{3}{16}]

$\left[ {\frac{3}{{12}} + \frac{3}{{16}}} \right]$ = 1

\frac{12 x}{96} + \frac{7}{16} = 1

$\frac{{12{\text{x}}}}{{96}} + \frac{7}{{16}} = 1$
Thus,
X = 4.5 min. ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

Let the obstruction remain for x min.
Hence,
Part of cistern filled in X min + part of cistern filled in 3 min = full cistern

[\frac{7 x}{8 \times 12} + \frac{5 x}{6 \times 16}]

$\left[ {\frac{{7{\text{x}}}}{{8 \times 12}} + \frac{{5{\text{x}}}}{{6 \times 16}}} \right]$

+

$+$

[\frac{3}{12} + \frac{3}{16}]

$\left[ {\frac{3}{{12}} + \frac{3}{{16}}} \right]$ = 1

\frac{12 x}{96} + \frac{7}{16} = 1

$\frac{{12{\text{x}}}}{{96}} + \frac{7}{{16}} = 1$
Thus,
X = 4.5 min. ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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