** TYPE-I**

**Ans . **

2

**Explanation :**(2) Part of the tank filled by both pipes in one minute

\( \frac{1}{20} + \frac{1}{30} \)

Required Time = \( \frac{1}{\frac{1}{20}+\frac{1}{30}} \)

\( \frac{20 * 30}{50} \) = 12 minutes

**Ans . **

1

**Explanation :**(2) 1 hour = 60 minutes. Rate of emptying the tank by the two taps are \( \frac{1}{60} \) and \( \frac{1}{30} \) of the tank per minute respectively. Rate of emptying the tank when both operate simultaneously \( \frac{1}{60} + \frac{1}{30} \) = \( \frac{1+2}{60} \) = \( \frac{30}{60} \) = \( \frac{1}{20} \)

**Ans . **

3

**Explanation :**(3) According to the question Cistern filled in 1 hrs \( \frac{1}{5} \) part Cistern emptied in 1 hour \( \frac{1}{4} \) part When the both pipes are opened, simultaneously ; Cistern emptied in 1 hour \( \frac{1}{4} - \frac{1}{5} \) = \( \frac{5-4}{20} \) = \( \frac{1}{20} \)

therefore, The time in which it will be emptied = 20 hours.

**Ans . **

2

**Explanation :**(2) Using Rule 2, Let the third pipe empty the cistern in x minutes. Part of cistern filled in 1 minute when all three pipes are opened simultaneously = \( \frac{1}{60} + \frac{1}{75} - \frac{1}{x} \)

According to the question,

\( \frac{1}{60} + \frac{1}{75} - \frac{1}{x} \) = \( \frac{1}{50} \)

\( \frac{1}{x} \) = \( \frac{1}{60} + \frac{1}{75} - \frac{1}{50} \)

\( \frac{5+4-6}{300} \) = \( \frac{3}{300} \) => \( \frac{1}{x} \) = \( \frac{3}{300} \)

Therefore x= \( \frac{300}{3} \) = 100 minutes

**Ans . **

4

**Explanation :**(4) Using Rule 2, Part of the cistern filled in 1 hour

\( \frac{1}{3} + \frac{1}{4} - \frac{1}{2} \)

[Cistern filled by 1st pipe + Cistern filled by 2nd pipe – Cistern emptied by 3rd pipe]

\( \frac{4+3-6}{12} \) = \( \frac{1}{12} \)

Hence, the cistern will be filled in 12 hours.

**Ans . **

2

**Explanation :**(2) Using Rule 2, Part of tank filled in 1 hour when all three pipes are opened simultaneously

\( \frac{1}{15} + \frac{1}{20} - \frac{1}{30} \)

\( \frac{4+3-2}{60} \) = \( \frac{5}{60} \) = \( \frac{1}{12} \)

Hence, the tank will be filled in 12 hours.

**Ans . **

3

**Explanation :**(3) Part of the cistern filled in 1 hour = \( \frac{1}{8} \)

Part of the cistern emptied in 1 hour = \( \frac{1}{16} \)

When both the taps are opened simultaneously, part of cistern filled in 1 hour

\( \frac{1}{8} - \frac{1}{16} \) = \( \frac{2-1}{16} \) = \( \frac{1}{16} \)

Hence, the cistern will be filled in 16 hours.

**Ans . **

4

**Explanation :**(4) Part of the tank filled in 1 hour= \( \frac{1}{x} \)

Part of the tank emptied in 1 hour = \( \frac{1}{y} \)

Part of the tank filled in 1 hour when both are opened

\( \frac{1}{x} - \frac{1}{y} \) = \( \frac{y-x}{xy} \)

therefore Tank will be filled in = \( \frac{xy}{y-x} \) hrs

**Ans . **

3

**Explanation :**Let x be number of pumps

Therefore 9 : 6 :: 12 : x = 12 : 15 :: 12 : x

9 × 12 × x = 6 × 12 × 15

x = \( \frac{6 × 12 × 15}{9 × 12} \) = 10

**Ans . **

4

**Explanation :**(4) Using Rule 2 and 7, Part of the cistern filled in 1 hour when pipes P and S are open

\( \frac{1}{4} - \frac{1}{10} \) = \( \frac{5-2}{20} \) = \( \frac{3}{20} \)

Hence, the cistern will be filled in \( \frac{20}{3} \) » 6.6 hours

Part of the cistern filled in 1 hour when pipes P, R and S are open

\( \frac{1}{4} + \frac{1}{12} \ - \frac{1}{10} \) = \( \frac{15+5-6}{60} \) = \( \frac{14}{60} \) = \( \frac{7}{30} \)

Hence, the cistern will be filled in \( \frac{30}{7} \) hrs

Part of the cistern filled in I hour when pipes P, Q and S are open

\( \frac{1}{4} + \frac{1}{8} \ - \frac{1}{10} \)

\( \frac{10+5-4}{40} \) = \( \frac{11}{40} \)

Hence, the cistern will be filled in \( \frac{40}{11} \) hrs

therefore, Cistern can be filled faster when P, Q & S are open

**Ans . **

1

**Explanation :**(1) Using Rule 1, Part of the cistern filled by both pipes in 1 hour

\( \frac{1}{10} - \frac{1}{15} \) = \( \frac{3+2}{30} \) = \( \frac{1}{6} \)

Therefore, The cistern will be filled in 6 hours.

**Ans . **

2

**Explanation :**(2) Using Rule 1 and 7, Part of the cistern filled by taps A, B and C in 1 minute = \( \frac{1}{10} \)

Part of the cistern filled by taps A and B in 1 minute

\( \frac{1}{10} - \frac{7}{120} \) = \( \frac{12-7}{120} \) = \( \frac{5}{120} \) = \( \frac{1}{24} \)

Therefore, Tap C will fill the cistern in 24 minutes.

**Ans . **

1

**Explanation :**(1) Using Rule 7, Part of the tank filled when both taps are opened together

\( \frac{1}{40} - \frac{1}{60} \) = \( \frac{3-2}{120} \) = \( \frac{1}{120} \)

Hence, the tank will be filled in 120 minutes 2 hours.

**Ans . **

3

**Explanation :**(3) Part of the tank filled in 1 hour by pipe A =

\( \frac{1}{2} \)

Part of the tank filled by both pipes in 1 hour

\( \frac{1}{2} + \frac{1}{6} \) = \( \frac{3+1}{6} \) = \( \frac{2}{3} \)

So, Time taken to fill = \( \frac{2}{3} \) parts = 60 minutes

Time taken to fill \( \frac{1}{2} \) parts

\( \frac{60 × 3}{2} × \frac{1}{2} \) = 45 minutes

**Ans . **

2

**Explanation :**(2) Using Rule 7, Part of the cistern filled by pipe Q in 1 minute

\( \frac{1}{20} - \frac{1}{30} \) = \( \frac{3-2}{60} \) = \( \frac{1}{60} \)

Therefore, Required time = 60 minutes

**Ans . **

4

**Explanation :**(4) Using Rule 2, Part of cistern filled by three pipes in an hour

\( \frac{1}{3} + \frac{1}{5} - \frac{1}{2} \) = \( \frac{10+6-15}{30} \) = \( \frac{1}{30} \)

Hence, the cistern will be filled in 30 hours.

**Ans . **

1

**Explanation :**(1) Using Rule 2, Part of the tank filled by all three taps in an hour

\( \frac{1}{4} + \frac{1}{6} + \frac{1}{12} \) = \( \frac{6+4+2}{24} \) = \( \frac{1}{2} \)

Hence, the tank will be filled in 2 hours.

**Ans . **

1

**Explanation :**(1) ) Using Rule 1, If the slower pipe fills the tank in x hours, then

\( \frac{1}{x} + \frac{1}{x-10} \) = \( \frac{1}{12} \)

\( \frac{x-10+x}{x(x-10)} \) = \( \frac{1}{12} \)

=> x

^{2}– 10x = 24x – 120=> x

^{2}– 34x + 120 = 0=> x

^{2}– 30x – 4x + 120 ==> x (x – 30) – 4 (x – 30) = 0

=> (x – 4) (x – 30) = 0

Therefore, x = 30 because x ¹ 4

therefore, Required time = 30 – 10 = 20 hours

**Ans . **

2

**Explanation :**(2) If pipe y be closed afterx minutes, then

\( \frac{18}{24} - \frac{x}{32} \) = 1

\( \frac{x}{32} \) = 1 - \( \frac{18}{24} \) = 1 - \( \frac{3}{4} \) = \( \frac{1}{4} \)

x = \( \frac{32}{4} \) = 8 minutes

**Ans . **

3

**Explanation :**(3) Using Rule 7, Part of the tank filled in first two minutes =

\( \frac{1}{20} - \frac{1}{30} \) = \( \frac{3-2}{60} \) = \( \frac{1}{60} \)

Therefore, Part of tank fillled in 114 minutes \( \frac{57}{60} \) = \( \frac{19}{20} \)

therefore, Remaining part of cistern will be filled in 115th minute

**Ans . **

4

**Explanation :**(4) Using Rule 2, Part of the tank filled by both taps in 5 minutes

5 \( (\frac{1}{30} + \frac{1}{60} ) \)

5 \( \frac{2+1}{60} \) = 5 × \( \frac{3}{60} \) = \( \frac{1}{4} \)

Remaining part = 1 - \( \frac{1}{4} \) = \( \frac{3}{4} \) that is filled by second tap

Time taken =\( \frac{3}{4} \) × 60 = 45 minutes

**Ans . **

2

**Explanation :**(2) Part of the tank filled by pipes A and B in 1 minute

\( \frac{1}{36} - \frac{1}{45} \) = \( \frac{5+4}{180} \) = \( \frac{9}{180} \) = \( \frac{1}{20} \)

Part of the tank filled by these pipes in 7 minutes = \( \frac{7}{20} \)

Remaining unfilled part = 1 - \( \frac{7}{20} \) = \( \frac{20-7}{20} \) = \( \frac{13}{20} \)

When all three pipes are opened

\( \frac{1}{20} - \frac{1}{30} \) = \( \frac{3-2}{60} \) = \( \frac{1}{60} \)

Time taken in filling \( \frac{13}{20} \) parts

= \( \frac{13}{20} \) × 60 = 39 minutes

Required time = 39 + 7 = 46 minutes

**Ans . **

1

**Explanation :**(1) Using Rule 1, Part of tank filled by pipes A and B in 1 hour

\( \frac{1}{2} - \frac{1}{3} \) = \( \frac{3+2}{6} \) = \( \frac{5}{6} \)

Required time = \( \frac{6}{5} \) hrs

= 1 hr \( \frac{1}{5} \) × 60

= 1 hour 12 minutes

**Ans . **

1

**Explanation :**(1) When both pipes are opened simultaneously, part of the tank filled in 1 hour

\( \frac{1}{x} - \frac{1}{y} \) = \( \frac{y-x}{xy} \)

Therefore, Required time = \( \frac{xy}{y-x} \) hrs

**Ans . **

2

**Explanation :**(2) Using Rule 1, Part of tank filled by pipes A and B in 2 hours

2 \( (\frac{1}{6} + \frac{1}{8}) \)

2 \( (\frac{4+3}{24}) \) = \( \frac{7}{12} \)

Remaining part = 1 - \( \frac{7}{12} \) = \( \frac{5}{12} \)

This part is filled by pipe B.

Required time = \( \frac{5}{12} \) × 8

\( \frac{10}{3} \) hrs

3 \( \frac{1}{3} \) hrs

** TYPE-II**

**Ans . **

2

**Explanation :**(2) Let the capacity of the tank be x litres then

\( \frac{x}{3} \) = 80

therefore, x= 240

Therefore \( \frac{x}{2} \) = \( \frac{240}{2} \)= 120 litres

**Ans . **

2

**Explanation :**(2) Part of cistern emptied in 1 hour

\( \frac{1}{5} - \frac{1}{8} \) = \( \frac{8-5}{40} \) = \( \frac{3}{40} \)

Since, \( \frac{3}{40} \) part is emptied in 1 hour..

Therefore, \( \frac{3}{4} \) part is emptied in \( \frac{40}{3} \) × \( \frac{3}{4} \) = 10 hrs

**Ans . **

3

**Explanation :**(3) Let the capacity of the tank be x litres. According to the question,

\( \frac{3x}{4} \) = 30

=> 3x = 30 × 4

\( \frac{30 × 4}{3} \) = 40 litres

**Ans . **

1

**Explanation :**(1) Using Rule 7, Part of the tank filled in 1 hour

\( \frac{1}{12} - \frac{1}{20} \) = \( \frac{5-3}{60} \) = \( \frac{1}{30} \)

therefore, Tank will be filled in 30 hours

**Ans . **

2

**Explanation :**(2) Using Rule 2, Part of tank filled in 1 hour when all three pipes are opened

\( \frac{1}{10} + \frac{1}{12} - \frac{1}{6} \) = \( \frac{6+5-10}{60} \) = \( \frac{1}{60} \)

Therefore, The tank will be filled in 60 hours.

Therefore, One fourth of the tank will be

filled in 15 hours \( \frac{1}{4} \) × 60 i.e the tank will be filled at 10 p.m.

**Ans . **

1

**Explanation :**(1) Time taken to fill the \( \frac{3}{5} \) of the cistern = 60 seconds

Therefore, Time taken in fill \( \frac{2}{5} \) parts \( \frac{60 × 5}{3} \) × \( \frac{2}{5} \) = 40 seconds

**Ans . **

1

**Explanation :**(1) Using Rule 1, Part of the tank filled in an hour by both pumps

\( \frac{1}{8} - \frac{1}{5} \) = \( \frac{5+4}{40} \) = \( \frac{9}{40} \)

Therefore, Part of the tank filled in 4 hours \( \frac{4 × 9}{40} \) = \( \frac{9}{10} \)

**Ans . **

3

**Explanation :**(3) Using Rule 1 and 2, Part of the tank filled by B and C in half an hour

\( \frac{1}{2} ( \frac{1}{9} + \frac{1}{12} ) \)

\( \frac{1}{2} ( \frac{4+3}{36} ) \) = \( \frac{7}{32} \)

Remaining part

\( 1 - \frac{7}{72} \) = \( \frac{72-7}{72} \) = \( \frac{65}{72} \)

Part of tank filled by three pipes in an hour

\( \frac{1}{6} + \frac{1}{9} + \frac{1}{12} \) = \( \frac{6+4+3}{36} \) = \( \frac{13}{36} \)

Therefore, Time to fill remaining part

\( \frac{65}{72} \) × \( \frac{36}{13} \) = \( \frac{5}{2} \) = 2\( \frac{1}{2} \) hrs

** TYPE-III**

**Ans . **

3

**Explanation :**(3) Using Rule 1, Part filled by A and B in 1 hour

= \( \frac{1}{12} + \frac{1}{15} \) = \( \frac{5+4}{60} \) = \( \frac{3}{20} + ..... \) (i)

Part filled by A and C in the next 1 hour = \( \frac{1}{12} + \frac{1}{20} \) = \( \frac{5+3}{60} \) = \( \frac{2}{15} \)

Part filled in 2 hours

\( \frac{3}{20} + \frac{2}{15} \) = \( \frac{9+8}{60} \) = \( \frac{17}{60} \)

=> Part filled in 6 hours = \( \frac{51}{60} \)

Remaining part

1 - \( \frac{51}{60} \) = \( \frac{9}{60} \) = \( \frac{3}{20} \)

This part will be filled by (A+B) in 1 hour. [By (i)]

Therefore, Total time taken = 7 hours

**Ans . **

4

**Explanation :**(4) ) Using Rule 7, Work done in 1 hour by the filling pump = \( \frac{1}{2} \)

Work done in 1 hour by the leak and the filling pump = \( \frac{3}{7} \)

Therefore, Work done by the leak in 1 hour

\( \frac{1}{2} - \frac{3}{7} \) = \( \frac{7-6}{14} \) = \( \frac{1}{14} \)

Hence, the leak can empty the tank in 14 hours.

**Ans . **

2

**Explanation :**(2) Using Rule 7, Let the leak empty the full tank inx hours.

\( \frac{1}{3} - \frac{1}{x} \) = \( \frac{2}{7} \)

\( \frac{1}{x} \) = \( \frac{1}{3} - \frac{2}{7} \)= \( \frac{7-6}{21} \)

\( \frac{1}{x} \) = \( \frac{1}{21} \) => 21 hrs

**Ans . **

4

**Explanation :**(4) A tap can fill the tank in 6 hours. In filling the tank to its half, time required = 3 hours.

Remaining part = \( \frac{}{} \)

Since, 1 tap takes 6 hours to fill the tank

\ Time taken by 4 taps take to fill \( \frac{1}{2} \) of the tank

\( \frac{6}{4} \) × \( \frac{1}{2} \) = \( \frac{3}{4} \) hrs

Total time = 3 + \( \frac{3}{4} \)

3\( \frac{3}{4} \)

= 3 hours 45 minutes

**Ans . **

4

**Explanation :**(4) Pipe A fills the tank in \( \frac{75}{2} \) minutes.

Therefore, Part of the tank filled by A in 30 minutes

\( \frac{2}{75} \) × 30 = \( \frac{4}{5} \)

Remaining part = 1 - \( \frac{4}{5} \) = \( \frac{1}{5} \)

Now, 1 part is filled by pipe B in 45 minutes

Therefore, \( \frac{1}{5} \) part is filled in

45 × \( \frac{1}{5} \) = 9 minutes

Hence, the pipe B should be turned off after 9 minutes.

**Ans . **

4

**Explanation :**(4) Using Rule 7, Part of the tank filled in one minute

\( \frac{1}{45} - \frac{1}{60} \) = \( \frac{4-3}{180} \) = \( \frac{1}{180} \)

Therefore, \( \frac{1}{180} \) part is filled in 1 minute

1 - \( \frac{1}{45} \) = \( \frac{44}{45} \) part is filled in

\( \frac{2 × 180 × 44}{45} \) = 352 minutes

i.e. 5 hours 52 minutes

Remaining \( \frac{1}{45} \) part will be filled in 1 minute.

Therefore, Total time taken = 5 hours 53 minutes

**Ans . **

2

**Explanation :**(2) Let the first pipe be closed after x minutes

\( \frac{x}{20} - \frac{18}{30} \) = 1

\( \frac{x}{20} \) = 1 - \( \frac{18}{30} \) = 1 - \( \frac{3}{5} \) = \( \frac{2}{5} \)

x = = \( \frac{2}{5} \) × 20 = 8 minutes

**Ans . **

4

**Explanation :**(4) Using Rule 7, Let the inflow fill the tank in x hours.

\( \frac{1}{x} - \frac{1}{2x} \) = \( \frac{1}{36} \)

[leakage being half of inflow]

\( \frac{2-1}{2x} \) = \( \frac{1}{36} \)

2x = 36

x = \( \frac{36}{2} \) = 18 hrs

**Ans . **

4

**Explanation :**(4) Let the pipe B fill the tank in x minutes. Part of the tank filled by pipes A and B in 1 minute = \( \frac{1}{36} \)

Therefore, Part of the tank filled by pipe A in 1 minute = \( \frac{1}{36} \) - \( \frac{1}{x} \)

According to the question,

30 × \( \frac{1}{x} \) + 40 (\( \frac{1}{36} \) - \( \frac{1}{x} \)) = 1

\( \frac{30}{x} + \frac{10}{9} - \frac{40}{x} \) = 1

\( \frac{30}{x} - \frac{40}{x} \) = \( \frac{10}{9} \) - 1

\( \frac{10}{x} \) = \( \frac{1}{9} \) => x = 90 minutes

**Ans . **

4

**Explanation :**(4) Let the capacity of the tank = x litres

According to the question, Quantity of water emptied by the leak in 1 hour = \( \frac{x}{10} \) litres

Qunatity of water filled by the tap in 1 hour = 240 litres

According to the question,

\( \frac{x}{10} - \frac{x}{15} \)= 240

\( \frac{3x-2x}{30} \) = 240

=> \( \frac{x}{30} \) = 240

=> x = 240 × 30 = 7200 litres

**Ans . **

3

**Explanation :**(3) Part of the tank filled in first 2 hours

\( \frac{1}{4} + \frac{1}{6} \) = \( \frac{3+2}{12} \) = \( \frac{5}{12} \)

Therefore, Part of the tank filled in first 4 hours

\( \frac{2 × 5}{12} \) parts = \( \frac{5}{6} \)

Remaining part = 1 - \( \frac{5}{6} \) = \( \frac{1}{6} \)

Now it is the turn of pipe A Time taken to fill \( \frac{1}{4} \) part = 1 hr

Time taken to fill \( \frac{1}{6} \) part

\( \frac{1}{6} \) × 4 = \( \frac{2}{3} \) hrs

Total time = 4 + \( \frac{2}{3} \)

4\( \frac{2}{3} \) hrs

**Ans . **

1

**Explanation :**(1) ) Part of the cistern filled by pipes A, B and C in 1 hour \( \frac{1}{6} \)

Therefore, \ Part of the cistern filled by all three pipes in 2 hours = \( \frac{1}{3} \)

Therefore, Remaining part = \( \frac{1}{3} \) = \( \frac{2}{3} \)

Now, pipe A and B fill \( \frac{2}{3} \) part of the cistern in 7 hours

Therefore, Pipe A and B will fill the cis tern in \( \frac{7 × 3}{2} \) = \( \frac{21}{2} \) hrs

Therfore, Part of the cistern filled by A and B in 1 hour = \( \frac{2}{21} \)

So Part of the cistern filled by C in 1 hour =\( \frac{1}{6} \) - \( \frac{2}{21} \)

= \( \frac{7-4}{42} \) = \( \frac{1}{14} \)

Therefore, Pipe C will fill the cistern in 14 hours.

**Ans . **

4

**Explanation :**(4) Using Rule 7, Part of the cistern filled in 1 minute by both the taps

\( \frac{1}{40} - \frac{1}{60} \) = \( \frac{3-2}{120} \) = \( \frac{1}{120} \)

Therefore, Empty cistern will be filled in 120 minutes.

**Ans . **

4

**Explanation :**(4) Using Rule 1, Part of the tank filled in 3 minutes by pipes P and Q

= 3 (\( \frac{1}{12} + \frac{1}{15} \))

3 \( \frac{5+4}{60} \) = \( \frac{3 × 9}{60} \) = \( \frac{9}{20} \)

So, Remaining part

1 - \( \frac{9}{20} \) = \( \frac{11}{20} \)

Therefore, Time taken by Q

\( \frac{11}{20} \) × 15 = \( \frac{33}{4} \) = 8\( \frac{1}{4} \)

**Ans . **

4

**Explanation :**(4) Using Rule 7, Part emptied by the leak in 1 hour

\( \frac{1}{8} - \frac{1}{10} \) = \( \frac{5-4}{40} \) = \( \frac{1}{40} \)

Therefore, The leak will empty the cistern in 40 hours.

**Ans . **

3

**Explanation :**(3) Using Rule 1, Part of the cistern filled in 2 hours by pipe A and B

2( \( \frac{1}{6} - \frac{1}{8} \)) = 2( \( \frac{4+3}{24} \)) = \( \frac{7}{12} \)

Remaining part = 1 - \( \frac{7}{12} \) = \( \frac{5}{12} \)

Time taken by pipe B in filling \( \frac{5}{12} \) part

\( \frac{5}{12} \) 8 = \( \frac{10}{3} \) = 3\( \frac{1}{3} \) hrs

**Ans . **

4

**Explanation :**(4) Part of the tank filled in 4 hours by pipe A /p>

\( \frac{4}{6} \) = \( \frac{2}{3} \)

Time taken by pipe B in filling

\( \frac{1}{3} \) parts = \( \frac{8}{3} \) hrs

**Ans . **

2

**Explanation :**(2) Part of the tank filled by (A + B + C) in 1 hour = \( \frac{1}{6} \)

Part of tank filled by these in 2 hours = \( \frac{2}{6} \) = \( \frac{1}{3} \)

Remaining part = 1 - \( \frac{1}{3} \) = \( \frac{2}{3} \)

Time taken by A and B in filling \( \frac{2}{3} \)rd part

= 8 hours

Therefore, Time taken by A and B in filling the whole tank

\( \frac{8 × 3}{2} \) = 12 hrs

Therfore, Part of tank filled by C in an hour

\( \frac{1}{6} - \frac{1}{12} \) = \( \frac{1}{12} \)

Hence, required time = 12 hours

**Ans . **

3

**Explanation :**(3) Using Rule 7, Part of the tank emptied by the leak in 1 hour =

\( \frac{1}{9} - \frac{1}{10} \) = \( \frac{10-9}{90} \) = \( \frac{1}{90} \)

Therefore, Required time = 90 hours

**Ans . **

3

**Explanation :**(3) A, B and C together fill the tank in 6 hours.

Therefore, Part of the tank filled in 1 hour by (A + B + C) = \( \frac{1}{6} \)

Part of the tank filled in 2 hours by all three pipes \( \frac{2}{6} \) = \( \frac{1}{3} \)Remaining empty part = 1 - \( \frac{1}{3} \) = \( \frac{2}{3} \)

This \( \frac{2}{3} \) part is filled by (A + B).

Therefore, Time taken by (A + B) to fill the fully empty tank

\( \frac{7 × 3}{2} \) = \( \frac{21}{2} \) hours

Part of tank filled by C in 1 hour

\( \frac{1}{6} - \frac{2}{21} \) = \( \frac{7-4}{42} \) = \( \frac{3}{42} \) = \( \frac{1}{14} \)

Therefore, Required time = 14 hours.

**Ans . **

4

**Explanation :**(4) Using Rule 7, Part of tank filled by both the pipes in 1 hour =

\( \frac{1}{4} - \frac{1}{16} \) = \( \frac{4-1}{16} \) = \( \frac{3}{16} \)

Required time = \( \frac{16}{3} \)

= 5 \( \frac{1}{3} \) hrs

**Ans . **

4

**Explanation :**(4) Using Rule 1, Part of tank filled in first two hours

\( \frac{1}{4} - \frac{1}{6} \) = \( \frac{3+2}{12} \) = \( \frac{5}{12} \)

Part of tank filled in first 4 hours

\( \frac{10}{12} \) = \( \frac{5}{6} \)

Remaining part

1- \( \frac{5}{6} \) = \( \frac{1}{6} \)

This remaining part willl be filled by pipe A.

Time taken by pipe A \( \frac{1}{6} \) × 4 = \( \frac{2}{3} \) hrs

Total time

4 + \( \frac{2}{3} \) = 4 \( \frac{2}{3} \) hrs

**Ans . **

4

**Explanation :**(4)) Using Rule 1 and 2, Part of tank filled by pipes A and B in 1 minute

\( \frac{1}{30} - \frac{1}{45} \) = \( \frac{3+2}{90} \) = \( \frac{1}{18} \)

Part of tank filled in 12 minutes

= \( \frac{12}{18} \) = \( \frac{2}{3} \) part

Remaining part 1 - \( \frac{2}{3} \) = \( \frac{1}{3} \) part

When pipe C is opened, Part of tank filled by all three pipes =

\( \frac{1}{30} +\frac{1}{45} - \frac{1}{36} \) = \( \frac{6+4-5}{180} \) = \( \frac{5}{180} \) = \frac{1}{36} \)

Time taken in filling \( \frac{1}{3} \) part

\( \frac{1}{3} \) × 36 = 12 minutes

Total time = 12 + 12 = 24 miuntes

**Ans . **

1

**Explanation :**(1) Using Rule 7, Part of tank filled by inlet pipe in 1 hour

\( \frac{1}{6} - \frac{1}{8} \) = \( \frac{4-3}{24} \) = \( \frac{1}{24} \)

Hence, if there is no leak, the inlet pipe will fill the tank in 24 hours.

Capacity of the tank

= 24 × 60 × 4

= 5760 litres

**Ans . **

3

**Explanation :**(3)) Using Rule 7, Part of tank emptied by leak in an hour =

\( \frac{1}{36} - \frac{1}{24} \) = \( \frac{2-3}{72} \) = \( \frac{-1}{72} \)

Therefore, Time taken in emptying the full tank = 72 hours

Therefore, Required time = 36 hours

**Ans . **

2

**Explanation :**(2) Since, P < q

Therefore, On opening pipe and sink together, Part of the tub filled in 1 hour =

\( \frac{1}{p} - \frac{1}{q} \)

Clearly, \( \frac{1}{p} - \frac{1}{q} \) = \( \frac{1}{r} \)

** TYPE-IV**

**Ans . **

3

**Explanation :**(3) Using Rule 1, Let time taken by faster pipe be x minutes.

\( \frac{1}{x} + \frac{1}{3x} \) = \( \frac{1}{36} \)

\( \frac{3+13x}{} \) = \( \frac{1}{36} \)

=> 3x = 36 × 4

=> x = 48

Therefore, Time taken by slower pipe to fill the tank = 3x = 3 × 48 = 144 minutes

**Ans . **

1

**Explanation :**(1) Using Rule 1, Here, the diameter of the second pipe is twice that of first pipe.

\ Volume of water emptied by the second pipe will be 4 times to that of first pipe.

Hence, time taken will b \( \frac{1}{4} \) of the first pipe.

Therefore, Second pipe will empty the tank in \( \frac{1}{4} \) × 40 = 10 minutes

When both the pipes are open, the part of the tank emptied in 1 minute = \( \frac{1}{40} + \frac{1}{10} \) = \( \frac{1+4}{40} \) = \( \frac{1}{8} \)

Hence, the tank will be emptied in 8 minutes.

**Ans . **

1

**Explanation :**(1) Using Rule 2, Part filled by A from 8 a.m to 11 a.m. = \( \frac{3}{15} \) = \( \frac{1}{5} \)

Part filled by B from 9 a.m. to 11 a.m. = \( \frac{2}{12} \) = \( \frac{1}{6} \)

Total Part filled till 11 a.m.

\( \frac{1}{5} + \frac{1}{6} \) = \( \frac{6+5}{30} \) = \( \frac{11}{30} \)

At 11 a.m. pipe C is opened to empty it.

\ Part of tank emptied in 1 hour

\( \frac{1}{4} - \frac{1}{15} - \frac{1}{12} \) = = \( \frac{15-4-5}{60} \) = \( \frac{1}{10} \)

Therefore, \( \frac{11}{30} \) part will be emptied in

\( \frac{11}{30} \) 10 = \( \frac{11}{3} \) hrs or 3 \( \frac{2}{3} \)

i.e. 3 hours 40 minutes i.e. at 11.40 a.m.

**Ans . **

3

**Explanation :**(3) Water filled by the boy and girl in 1 minute

\( \frac{4}{3} - \frac{3}{4} \) = \( \frac{16+9}{12} \) = \( \frac{25}{12} \) litres

Therefore, Time taken to fill 100 litres

\( \frac{100}{25} \) × 12 = 48 minutes

**Ans . **

2

**Explanation :**(2) Time α \( \frac{1}{cross sectional area of the pipe } \)

Time α \( \frac{1}{ \frac{π}{4} } \) d

^{2}[Being inversely related]

t

_{1}= 40 minutes, d_{1}= d, d_{2}= 2d

**Ans . **

3

**Explanation :**(3) Let the capacity of the tank be x gallons. Quantity of water filled in the tank in 1 minute when all the pipes A, B and C are opened simultaneously = \( \frac{x}{20} - \frac{x}{24} \) -3

According to the question,

\( \frac{x}{20} - \frac{x}{24} \) -3 = \( \frac{x}{15} \)

\( \frac{x}{20} - \frac{x}{24} - \frac{x}{15} \) = 3

\( \frac{6x+5x-8x}{120} \) = 3

=> 3 x = 3 × 120

=> x = \( \frac{3 × 120}{3} \) = 120 gallons

**Ans . **

1

**Explanation :**(1) Time taken by pipe B = 2x hours

Time taken by pipe A = x hours

Therefore, Time taken by pipe C

\( \frac{2}{ \frac{1}{2x} + \frac{1}{x} } \) = \( \frac{2}{ \frac{1+2}{2x} } \)

= \( \frac{4x}{3} \) hrs

\( \frac{1}{x} + \frac{1}{2x} + \frac{3}{4x} \)

\( \frac{1}{ 6 + \frac{40}{60} } \) = \( \frac{1}{ 6 + \frac{2}{3} } \)

\( \frac{4+2+3}{4x} \) = \( \frac{3}{20} \)

=> 9 × 20 = 4x × 3

x = \( \frac{9 × 20}{4 × 3} \) = 15 hrs

**Ans . **

1

**Explanation :**(1) If the flow of water per unit time be x km, then

V = πr

^{2}h [ Since, Pipe is in cylinderical shape]Greater the radius, larger the capacity of pipe.

Radius is greatest in (i) i.e. 30cm. Hence, pipe with 60 cm diameter will empty the pool fastest.

Therefore, V ∝ r

^{2}

**Ans . **

4

**Explanation :**(4) 300 days = (300 × 24) hours

= (300 × 24 × 60 × 60) seconds

Therefore, Number of drops

= 300 × 24 × 60 × 60

Since, 600 drops = 100 ml.

Therefore, 300 × 24 × 60 × 60 drops

\( (\frac{300 × 24 × 60 × 60}{6}) \) ml

= (1200 × 60 × 60) ml.

\( (\frac{1200 × 60 × 60}{1000}) \) litre

= 4320 litre

**Ans . **

2

**Explanation :**(2) M

_{1}D_{1}= M_{2}D_{2}=> 9 × 20 = M

_{2}× 15M

_{2}= \( \frac{9 × 20}{15} \) = 12 pipeNote : Same relation as men and days is applicable

**Ans . **

2

**Explanation :**(2) Using Rule 6, Part of tank emptied by both pipes in 1 minute

\( \frac{1}{30} + \frac{1}{45} \) = \( \frac{3+2}{90} \) = \( \frac{}{} \)

= \( \frac{5}{90} \) = \( \frac{1}{18} \)

Therefore, Required time = 18 minutes

**Ans . **

1

**Explanation :**(1) Part of bucket filled by both pipes in 5 minutes

5 ( \( \frac{1}{20} + \frac{1}{25} \) )

5 ( \( \frac{5+4}{100} \) ) = \( \frac{9}{20} \)

Remaining part = 1 - \( \frac{9}{20} \)

= \( \frac{11}{20} \)

This remaining part will be filled by first pipe.

Required time = \( \frac{11}{20} \) × 20

= 11 minutes

**Ans . **

3

**Explanation :**(3) Work done in 1 hour by all 3 pipes =

\( \frac{1}{6} + \frac{1}{9} - \frac{1}{18} \) = ( \frac{4}{18} \)

Time required to fill the tank completely = ( \frac{18}{4} \) = 4.5 hrs

**Ans . **

1

**Explanation :**(1)) Tank filled by A & B in 5 hours

(\( \frac{1}{12} - \frac{1}{15} \))5 = \( \frac{9 × 5}{60} \) = \( \frac{3}{4} \)

Work done in 1 hour when all 3 pipes are opened

\( \frac{1}{12} + \frac{1}{15} - \frac{1}{6} \) = \( \frac{9-10}{60} \) = \( \frac{-1}{60} \)

Since the result (or net effect) is negative, hence tank would be emptied

So, \( \frac{1}{60} \) is emptied in 1 hour \( \frac{3}{4} \) would be emptied in

\( \frac{1}{\frac{1}{60}} \) × \( \frac{3}{4} \) = 45 hrs

**Ans . **

4

**Explanation :**(4) Tank filled by all 3 in 6 hours

\( \frac{1}{24} \)6 = \( \frac{1}{4} \)

Remaining = ( 1 - \( \frac{1}{4} \) ) = \( \frac{3}{4} \) is filled by, Pipe 1 and 2 in 30 hours So, the entire tank would be filled by 1 and 2 in 40 hours

1 hour work of all 3 = \( \frac{1}{24} \)

1 hour work of Pipe 1 & 2 = \( \frac{1}{40} \)

Hence, 1 hour work of Pipe

3 = \( \frac{1}{24} \) - \( \frac{1}{40} \) = \( \frac{2}{120} \) = \( \frac{1}{60} \)

Therefore, Pipe 3 alone would fill the tank in 60 hours.

**Ans . **

1

**Explanation :**(1)) When Pipe x is turned off (after 6 minutes),

Work done by x and y in 6 minutes

\( \frac{1}{24} +\frac{1}{30} \)6 = \( \frac{9}{20} \)

Remaining work

1 - \( \frac{9}{20} \) = \( \frac{11}{20} \) which would be done by Pipe y alone. 1 work is done by Pipe y (alone) in 30 minutes

work is done by Pipey (alone) in 30 × \( \frac{11}{20} \) = \( \frac{33}{2} \) = minutes = 16.5 minutes.

**Ans . **

1

**Explanation :**(1) Consider the case when there is no leak – Then in one hour, work done = \( \frac{1}{18} \) and in 6 hrs \( \frac{6}{18} \) = \( \frac{1}{3} \)

This means \( \frac{1}{3} \)rd of the tank is emptied because of the leakage in 18 + 6 = 24 hours.

So,\( \frac{1}{3} \)rd is emptied in 24 hours, full tank would be emptied in 24 × 3 = 72 hours.

**Ans . **

3

**Explanation :**(3) In 1 hour, water filled = \( \frac{1}{4} \)th of the tank. \( \frac{1}{4} \)th is emptied by leakage in 5 hours.

Full tank would be emptied in 20 hours (i.e. 5 × 4).

**Ans . **

1

**Explanation :**(1) Let the time for which Pipe 1 is turned on be ‘x’ hours, hence Pipe 1 has worked for ‘x’ hours and Pipe 2 has worked for 12 hours.

\( \frac{1}{18} \)(x) + \( \frac{1}{24} \)(12) = 1

\( \frac{x}{18} - \frac{1}{2} \) = 1 or \( \frac{x}{18} \) = \( \frac{1}{2} \) => x = 9

Therefore, Pipe 1 was turned on for 9 hours.

**Ans . **

1

**Explanation :**(1) We are given that V = k (r)2 where V is volume of water and ‘r’ is radius of pipe and K is a constant. The smallest pipe takes 30 hours to fill the tank alone, hence work done in 1 hour = \( \frac{1}{30} \)

radius = \( \frac{diameter}{2} \) = \( \frac{2}{2} \) = 1

\( \frac{1}{30} \) = k(1)

^{2}so, k = \( \frac{1}{30} \)Work done in 1 hour by Pipe 2

= \( \frac{1}{30} \) \( \frac{4}{2} \)

^{2}= \( \frac{4}{30} \)Work done in 1 hour by Pipe 3

= \( \frac{1}{30} \) \( \frac{6}{2} \)

^{2}= \( \frac{9}{30} \)Work done in 1 hour by Pipe 4

= \( \frac{1}{30} \) \( \frac{8}{2} \)

^{2}= \( \frac{16}{30} \)In 1 hour, work done by all 4 pipes

\( \frac{1}{30} + \frac{4}{30} + + \frac{9}{30} + + \frac{16}{30} \) = \( \frac{30}{30} \) = 1

Hence, the whole tank gets filled in 1 hour.

**Ans . **

4

**Explanation :**(4) Part of the tank filled in 1 minute when all the three pipes are opened simultaneously

\( \frac{1}{10} + \frac{1}{30} - \frac{1}{20} \) = \( \frac{6+2-3}{60} \) = \( \frac{5}{60} \) = \( \frac{1}{12} \)

Hence, the tank will be filled in 12 minutes.

**Ans . **

4

**Explanation :**(4) In 1 minute both pipes can fill

= \( \frac{1}{20} + \frac{1}{30} \) part of the cistern In 10 minutes, second pipe can fill

\( \frac{1}{30} \) × 10 = \( \frac{1}{3} \)part

Cistern filled by both pipes

1 - \( \frac{1}{3} \) = \( \frac{2}{3} \)

Therefore, Time taken by both the pipes to fill \( \frac{2}{3} \) part of cistern \( \frac{12}{2} \) × 3 = 8 minutes

Therefore, the first pipe can be turned off after 8 minutes.

**Ans . **

4

**Explanation :**(4) Net amount of water filled in the tank in 1 hour when all three taps are opened simultaneously,

= 42 + 56 – 48 litres = 50 litres

The tank gets completely filled in 16 hours.

Therefore, Capacity of the tank

= 16 × 50 = 800 litres

**Ans . **

4

**Explanation :**(4)) Let conical tank contain ‘x’ litres of fuel, then cylindrical tank would hold (x + 500) litres.

So,

(x – 200)2 = x + 500 – 200

2x – 400 = x + 300 Þ x = 700

Hence, cylindrical tank would hold

700 + 500 = 1200 L