GATE Mechanical 2019 Set 2 | GA | Question: 7

Question

Two pipes $P$ and $Q$ can fill a tank in $6$ hours and $9$ hours respectively, while a third pipe $R$ can empty the tank in $12$ hours.  Initially, $P$ and $R$ are open for $4$ hours, Then $P$ is closed and $Q$ is opened. After $6$ more hours $R$ is closed. The total time taken to fill the tank (in hours) is  ____

• A. $13.50$
• B. $14.50$
• C. $15.50$
• D. $16.50$

Comments

Dear admin, Option A is mentioned as answer. Please rectify this.

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We can solve this using \% as well.

Initially \(P\) is open for 4 hours it fills the tank in 6 hours that means
\[
\frac{4}{6} = 66.6\%
\]
of tank will get filled.

\(R\) is open for 4 hours too but it empties the tank in 12 hours so
\[
\frac{4}{12} = -33.3\%.
\]

Now \(P\) filled \(R\) emptied that means \(33.3\%\) remained in the tank
\[
(66.6 - 33.3).
\]

Doing this for other \(Q, R\)

\(Q\) opened for 6 hours
\[
\frac{6}{9} = 66.6\%
\]

\(R\) closed after 6 hours
\[
\frac{6}{12} = 50\%
\]

subtracting because \(R\) emptied
\[
66.6 - 50 = 16.6\%.
\]

Earlier we filled \(33\%\) now \(16.6\%\) so total is \(49.6\%\) that means we've to fill \(50.4\%\) using \(Q\).

\(Q\) takes 9 hours to fill tank completely i.e \(100\%\) so for  \(50.4\%\) will take roughly \(4.5\) hours.

After all this math

4 hours for \(P, R\)

6 hours for \(Q, R\)

and then \(4.5\) hours to fill it further using \(Q\)

\[
\boxed{\text{Total time required } = 14.5 \text{ hours}}
\]

Answer 1

P can fill the tank in $6$ hours

$\Rightarrow$ In $1$ hr P can fill $\frac{1}{6}$ of the tank.

Q can fill the tank in $9$ hours

$\Rightarrow$ In $1$ hr Q can fill $\frac{1}{9}$ of the tank.

R can empty the tank in $12$ hours

$\Rightarrow$ In $1$ hr R can empty $\frac{1}{12}$ of the tank.

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P and R are opened for $4$ hours

$\Rightarrow$ They fill $4*\left ( \frac{1}{6}-\frac{1}{12} \right ) = 4*\frac{1}{12}=\frac{1}{3}$ of the tank.

$\Rightarrow$ $1-\frac{1}{3}=\frac{2}{3}$ of the tank is still empty.

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Then P is closed and Q is opened. After $6$ more hours R is closed.

$\Rightarrow$ Q and R are opened together for $6$ hours.

$\Rightarrow$ They fill $6*\left ( \frac{1}{9}-\frac{1}{12} \right ) = 6*\frac{1}{36}=\frac{1}{6}$ of the tank.

$\Rightarrow$ $\frac{2}{3}-\frac{1}{6}=\frac{4-1}{6}=\frac{1}{2}$ of the tank is still empty.

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Now only Q is opened

$\because$ Q can fill a tank in $9$ hr

$\Rightarrow$ Q can fill $\frac{1}{2}$ of the tank in $9*\frac{1}{2}$ hours $=4.5$ hours.

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$\therefore$ Total Time to fill the tank $=4$ hours $+6$ hours $+ 4.5$ hours $=14.5$ hours

So, Option B. $14.50$  is the correct answer.

Comments

There is a mistake in the last step. Q can fill tank in 9 hours. So, answer would be 4+6+4.5=14.5 hours. Hence, B is the answer.

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Corrected now :)

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this question comes under pipes and cistern or time and work?

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pipes and cisterns is the subtopic under the time and work.

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why subtraction done when two pipes are opened?

Answer 2

This method can be helpful..

Comments

this question comes under pipes and cistern or time and work?

Answer 3

$Let\ Capacity=36L$

$P\rightarrow 6H-\ 6L/H$

$Q\rightarrow 9H-\ 4L/H$

$R\rightarrow 12H-3L/H$

$|\underbrace{4H:P+R=3L/H}||\underbrace{6H:Q+R=1L/H}||\underbrace{?H:Q=4L/H}|$

               $12L$                              $6L$                        $18L$

$1H\leftarrow4L$

$?\leftarrow18L$

$?=4.5H$

$ans:4+6+4.5=14.5H$

Answer 4

P can fill the tank in $6$ hours

$\Rightarrow$ In $1$ hr P can fill $\frac{1}{6}$ of the tank.

Q can fill the tank in $9$ hours

$\Rightarrow$ In $1$ hr Q can fill $\frac{1}{9}$ of the tank.

R can empty the tank in $12$ hours

$\Rightarrow$ In $1$ hr R can empty $\frac{1}{12}$ of the tank.

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P and R are opened for $4$ hours

$\Rightarrow$ They fill $4*\left ( \frac{1}{6}-\frac{1}{12} \right ) = 4*\frac{1}{12}=\frac{1}{3}$ of the tank.

$\Rightarrow$ $1-\frac{1}{3}=\frac{2}{3}$ of the tank is still empty.

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Then P is closed and Q is opened. After $6$ more hours R is closed.

$\Rightarrow$ Q and R are opened together for $6$ hours.

$\Rightarrow$ They fill $6*\left ( \frac{1}{9}-\frac{1}{12} \right ) = 6*\frac{1}{36}=\frac{1}{6}$ of the tank.

$\Rightarrow$ $\frac{2}{3}-\frac{1}{6}=\frac{4-1}{6}=\frac{1}{2}$ of the tank is still empty.

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Now only Q is opened

$\because$ Q can fill a tank in $6$ hr

$\Rightarrow$ Q can fill $\frac{1}{2}$ of the tank in $6*\frac{1}{2}$ hours = $3$ hours = $3$ hours.

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$\therefore$ Total Time to fill the tank = $4$ hours+$6$ hours+$3$ hours= $13$ hours

So Option A. $13.50$  is the correct answer.

Comments

Answer is B 14.5 Hrs

Last remaining half tank takes 4.5 hrs to fill.

Answer 5

Total work 100%

P can fill 100/6 = 16.67% in 1 hr

similarly , Q fills 11.11%

and R fills   -8.33%

First 4 hrs tanks is filled P+R =>(16.67-8.33)*4 =33.36%

next 6 hr anks is filled Q+R => (11.11-8.33)*6 =16.68%

at 10 hr  total filled 50 %

time required to fill the remaining = 50% / 11.11% = 4.5 hrs

Total Time = 10+4.5 =14.5 hrs

Answer 6

By LCM method

P--->6hrs , Q--->9hrs and R--->12hrs

LCM(6,9,12)= 36 =Total work
So, one hour work of: P is 6 units , Q is 4 units & R is -3 units.

let x be the time when only Q is opened.

equation will be :  4[P+R] + 6[Q+R] + x[Q] = LCM(6,9,12)= 36 =Total work

4[6-3] + 6[4-3] + x[4] = 36

12+6+4x=36

then x = 4.50

 therefore total time to fill the tank is 4+6+4.50 = 14.50 hrs

 

By Fraction method:

one hour work of P=1/6 ,Q=1/9,R=1/12

total work in fraction method = 1.

let x be the time when only Q is opened.

So, 4[1/6-1/12]+ 6[1/9-1/12] + x[1/9]=1

1/3+1/6+x/9= 1

x/9= 1-[1/2]

x=9/2=4.50 hrs

therefore, total time to fill the tank is 4+6+4.50 = 14.50 hrs

So, Option B is the correct answer.