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# LRDI Questions [Difficult] Set 9

A water-tank ‘T’ supplies 3000 litres of water. It supplies equal volume of water to the pipelines connecting the sub-stations P, Q & R. Similarly, sub-stations Q & R supply equal volume of water to the pipelines connecting them to their respective mini-stations. Sub-station P supplies water such that equal volume of water is received at mini-stations P1 & P2. The numbers indicate the length of the pipelines in km. It is observed that there is a loss of ‘x’% and ‘2x’% of water in the pipelines joining the tank to the sub-station and the pipelines joining the sub-stations to mini-stations respectively. (where ‘x’ represents the length of the pipeline).

Q1. How much water (in litres) is received at mini-station P1?

Q2. Find the length of pipe $\mathrm { Q } \rightarrow \mathrm { Q } _ { 3 }$ (in km.) if the sum of the volume of water received at $\mathrm { Q } _ { 1 }$ and $\mathrm { Q } _ { 3 }$ is 380 litres.

Q3. $\mathrm { L } _ { 1 }$ and $\mathrm { L } _ { 2 }$ are the lengths of $\mathrm { R } \rightarrow \mathrm { R } _ { 1 }$ and $\mathrm { R } \rightarrow \mathrm { R } _ { 2 }$ pipelines respectively. Also, it is known that $\mathrm { L } _ { 1 } + \mathrm { L } _ { 2 } = 25 \mathrm { km }$ . The sum of the volume of water received at mini-stations $\mathrm { R } _ { 1 }$ and $\mathrm { R } _ { 2 }$ is 600 litres. Find the length of the pipeline $\mathrm { T } \rightarrow \mathrm { R }$ .

1. 10$\mathrm { km }$
2. 30$\mathrm { km }$
3. 25$\mathrm { km }$
4. 20$\mathrm { km }$

Q4. Due to scarcity of water at sub-station Q, a new pipeline is fitted between sub-station R and sub-station Q. A water loss of 5x% is observed in the new pipeline where ‘x’ is the length of the pipeline (in km). Sub-station Q now supplies 300 litres of water in each pipeline to its mini-stations. Find the length of the new pipeline if R supplies equal volume of water to each pipeline. (Use data from the previous question if necessary)

1. 12.5$\mathrm { km }$
2. 11.25$\mathrm { km }$
3. 17.5$\mathrm { km }$
4. 8.75$\mathrm { km }$

Q1. As the water gets equally divided in the pipelines from the tank, water entering pipeline

$\mathrm { T } \rightarrow \mathrm { P } = \frac { 3000 } { 3 } = 1000$ litres

$\rightarrow$ Water received at sub-station $\mathrm { P } = 85 \%$ of 1000 litres $= 850$ litres

Let a and b represent the volume of water entering pipeline $P \rightarrow P _ { 1 }$ and $P \rightarrow P _ { 2 }$

respectively.

$\rightarrow a + b = 850$

As mini-stations $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 }$ receive equal volume of water,

90$\%$ of $\mathrm { a } = 80 \%$ of $\mathrm { b }$

Therefore, $9 \mathrm { a } = 8 \mathrm { b }$

Therefore, $9 \mathrm { a } - 8 \mathrm { b } = 0$

Solving $( 1 )$ and $( 2 )$ simultaneously, we get $a = 400$ and $b = 450$

Hence, volume of water received at mini-station $P _ { 1 } = 400 \times 0.9 = 360$ litres

Therefore, the required answer is 360 .

Q2. Let the length of pipe $\mathrm { Q } \rightarrow \mathrm { Q } _ { 3 }$ be 'a' $\mathrm { km }$ .

Volume of water received by sub-station $\mathrm { Q } = 75 \%$ of 1000 litres $= 750$ litres.

Quantity of water received by mini-station $\mathrm { Q } _ { 1 } = 84 \%$ of $\left( \frac { 750 } { 3 } \right) = 210$ litres

Volume of water received by mini-station $\mathrm { Q } _ { 3 } = \frac { 750 } { 3 } - \frac { 750 } { 3 } \times \frac { 2 \mathrm { a } } { 100 } = 5 ( 50 - \mathrm { a } )$

$\Rightarrow 210 + 5 ( 50 - \mathrm { a } ) = 380$

$\Rightarrow \quad ( 50 - \mathrm { a } ) = 170$

$\Rightarrow ( 50 - \mathrm { a } ) = 34$

$\Rightarrow \mathrm { a } = 16 \mathrm { km }$

Q3. Let 'x' represent the length of pipeline $\mathrm { T } \rightarrow \mathrm { R }$ .

Therefore Volume of water received by sub-station $\mathrm { R } = 1000 \frac { ( 100 - \mathrm { x } ) } { 100 } = 10 ( 100 - \mathrm { x } )$

Therefore Volume of water distributed by $\mathrm { R } = \frac { 10 ( 100 - \mathrm { x } ) } { 2 } = 5 ( 100 - \mathrm { x } )$

Volume of water received by mini-stations:

$R _ { 1 } = 5 ( 100 - x ) \frac { \left( 100 - 2 \mathrm { L } _ { 1 } \right) } { 100 }$ and

$R _ { 2 } = 5 ( 100 - x ) \frac { \left( 100 - 2 \mathrm { L } _ { 2 } \right) } { 100 }$

$\rightarrow \frac { 5 ( 100 - \mathrm { x } ) } { 100 } \left[ 200 - 2 \left( \mathrm { L } _ { 1 } + \mathrm { L } _ { 2 } \right) \right] = 600$

$\Rightarrow ( 100 - \mathrm { x } ) ( 200 - 50 ) = 12000$

$\Rightarrow \quad ( 100 - \mathrm { x } ) = \frac { 1200 } { 15 } = 80$

$\Rightarrow \mathrm { x } = 20 \mathrm { km }$

Q4. Since sub-station Q distributes 300 litres of water to each of its mini-stations, it receives 300 x 3 = 900 litres of water.

As it receives 750 litres from T, it receives the remaining 150 litres from R.

Let the length of the new pipeline be ‘a' km.

At sub-station $\mathrm { R } ,$ there are 3 pipelines. Thus, $\mathrm { R }$ distributes $\frac { 800 } { 3 }$ litres of water in each pipeline.

Volume of water received by $Q = \frac { 800 } { 3 } - \frac { 800 } { 3 } \times \frac { 5 a } { 100 } = 150$

$\Rightarrow \quad \frac { 800 } { 3 } \frac { ( 100 - 5 a ) } { 100 } = 150$

$\Rightarrow ( 100 - 5 a ) = \frac { 150 \times 3 } { 8 }$

$\Rightarrow 5 a = 100 - \frac { 150 \times 3 } { 8 } = \frac { 800 - 450 } { 8 } = \frac { 350 } { 8 }$

$\Rightarrow a = 8.75 \mathrm { km }$

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