LRDI Questions [Difficult] Set 9

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 $Q = \frac { 800 } { 3 } - \frac { 800 } { 3 } \times \frac { 5 a } { 100 } = 150$

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

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

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

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