- Staff Selection Commission - Pipes and Cisterns (1999-2017)

# Staff Selection Commission Mathematics - PIPE AND CISTERN

TYPE-I

Ans .

2

1. Explanation :

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

2. $$\frac{1}{20} + \frac{1}{30}$$

Required Time = $$\frac{1}{\frac{1}{20}+\frac{1}{30}}$$

$$\frac{20 * 30}{50}$$ = 12 minutes

Ans .

1

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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. Explanation :

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

2. $$\frac{1}{10} - \frac{1}{15}$$ = $$\frac{3+2}{30}$$ = $$\frac{1}{6}$$

Therefore, The cistern will be filled in 6 hours.

Ans .

2

1. 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

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

1. 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

1. 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

1. Explanation :

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

2. $$\frac{1}{3} + \frac{1}{5} - \frac{1}{2}$$ = $$\frac{10+6-15}{30}$$ = $$\frac{1}{30}$$

Hence, the cistern will be filled in 30 hours.

Ans .

1

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

1. Explanation :

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

$$\frac{1}{x} + \frac{1}{x-10}$$ = $$\frac{1}{12}$$

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

=> x2 – 10x = 24x – 120

=> x2 – 34x + 120 = 0

=> x2 – 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

1. 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

1. 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

1. 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

1. 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

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

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

1. 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

1. 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

1. 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

1. 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

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

1. 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

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

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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

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

1. 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

1. 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

1. 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

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

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

1. 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

1. Explanation :

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

Time α $$\frac{1}{ \frac{π}{4} }$$ d2

[Being inversely related]

t1 = 40 minutes, d1 = d, d2 = 2d

Ans .

3

1. 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

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

1. Explanation :

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

V = πr2h [ 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 ∝ r2

Ans .

4

1. 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

1. Explanation :

(2) M1 D1 = M2 D2

=> 9 × 20 = M2 × 15

M2 = $$\frac{9 × 20}{15}$$ = 12 pipe

Note : Same relation as men and days is applicable

Ans .

2

1. 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

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

TEST YOURSELF

Ans .

3

1. 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

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

1. 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

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

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

1. 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

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

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

1. 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

1. 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

1. 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

1. 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