Question

Two taps can fill a tank in 10 min and 15 Mim separately. A pipe can empty the full tank in 9 mins. If they are all opened together at the same time,in how much time will the tank be full​

Answer 1

Given:

• Two taps can fill a tank in 10 min and 15 Min
• A pipe can empty the full tank in 9 mins

To Find:

• If they are all opened together at the same time,in how much time will the tank be full

Solution:

$\sf: \implies { \orange{rate \: of \: work \: done \: by \: tap \: 1}} = \frac{1}{10} \\ \\ \sf: \implies { \orange{rate \: of \: work \: done \: by \: tap \: 2}} = \frac{1}{15}$

we consider the values positive here as they are filling the tank

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$\sf: \implies { \orange{rate \: of \: work \: done \: by \: pipe}} = \frac{ - 1}{9}$

we consider the value negetive here as it's emptying the tank

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Time taken when they are together operated:

$\sf: \implies { \blue{time: }} \: \frac{1}{10} + \frac{1}{15} - \frac{1}{9} \: \: \: \\ \\\sf: \implies { \blue{time: }} \: \frac{9}{90} + \frac{6}{90} - \frac{10}{90} \\ \\ \sf: \implies { \blue{time: }} \: \frac{9 + 6 - 10}{90} \: \: \: \: \: \: \: \\ \\ \sf: \implies { \blue{time: }} \: \cancel\frac{5}{90} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf: \implies { \blue{time: }} \frac{1}{18} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\blue{ \underline{ \boxed{ \green{ \mathfrak{ \therefore \: time \: taken = 18mins}}}}}$

$\sf{ \red{ - @mrcostheta}}$

Answer 2

Given:

↬Two taps can fill a tank in 10 min and 15 Min separately.

↬A pipe can empty the full tank in 9 mins.

To Find:

↬they are all opened together at the same time,in how much time will the tank be full.?

Solution:

let's calculate the rate of work done by them respectively per min:

$\sf \: rate \: of \: work \: done \: by \: tap \: a = \frac{1}{10} \\ \\ \sf \: rate \: of \: work \: done \: by \: tap \: b = \frac{1}{15} \\ \\ \sf \: rate \: of \: work \: done \: by \: pipe = \frac{1}{9}$

To find the rate of work done by them all together we add up the values:

$: \implies \tt \: \frac{1}{10} + \frac{1}{15} + ( - \frac{1}{9} ) \: \: \: \: \: \: \: \: \\ \\ {\blue {\tt{ \: \: \: \: \: \: \: ⇝\: l.c.m \: of \: 10,15,9 = 90}}} \\ \\ : \implies \tt \frac{9}{90} + \frac{6}{90} - \frac{10}{90} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies \tt \: \frac{9 + 6 - 10}{90} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies \tt \cancel\frac{5}{90} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies \frac{1}{18} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\blue{ \underline{ \boxed{ \pink{ \mathfrak{ \therefore \:it \: took \: 18 \: mins \: to \: fill \: the \: tank \: respectivly}}}}}$

hope this helps.!!