Model 1 Basic Pipes & Cisterns problems Section-Wise Topic Notes With Detailed Explanation And Example Questions

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The following question based on pipes & cisterns topic of quantitative aptitude

Questions : Three taps A, B, C can fill an overhead tank in 4, 6 and 12 hours respectively. How long would the three taps take to fill the tank if all of them are opened together ?

(a) 5 hrs.

(b) 2 hrs.

(c) 4 hrs.

(d) 3 hrs.

The correct answers to the above question in:

Answer: (b)

Using Rule 2,

Part of the tank filled by all three taps in an hour

= $1/4 + 1/6 + 1/12 = {6 + 4 + 2}/24 = 1 2$

Hence, the tank will be filled in 2 hours.

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Read more basic pipes and cisterns problems Based Quantitative Aptitude Questions and Answers

Question : 1

Two pipes A and B can separately fill a tank in 2 hours and 3 hours respectively. If both the pipes are opened simultaneously in the empty tank, then the tank will be filled in

a) 1 hour 20 minutes

b) 1 hour 12 minutes

c) 2 hours 30 minutes

d) 1 hour 15 minutes

Answer: (b)

Using Rule 1,

Part of tank filled by pipes A and B in 1 hour

= $1/2 + 1/3 = {3 + 2}/6 = 5/6$ parts

Required time = $6/5$ hours

= 1 hour $1/5$ × 60

= 1 hour 12 minutes

Question : 2

Pipe A can fill an empty tank in 6 hours and pipe B in 8 hours. If both the pipes are opened and after 2 hours pipe. A is closed, how much time B will take to fill the remaining tank?

a) 3$1/3$ hours

b) 7$1/2$ hours

c) 2$2/5$ hours

d) 2$1/3$ hours

Answer: (a)

Using Rule 1,

Part of tank filled by pipes A and B in 2 hours

= $2(1/6 + 1/8)$

=$2({4 + 3}/24) = 7/12$

Remaining part = $1 - 7/12 = 5/12$

This part is filled by pipe B.

Required time = $5/12$ × 8

= $10/3$ hours = 3$1/3$ hours

Question : 3

Two pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. If both pipes are opened together, the time taken to fill the tank is :

a) 15 minutes

b) 50 minutes

c) 12 minutes

d) 25 minutes

Answer: (c)

Part of the tank filled by both pipes in one minute

= $1/20 + 1/30$

Required time = $1/{1/20 + 1/30}$

= ${20 × 30}/50$ = 12 minutes

Using Rule 1,
Two taps 'A' and 'B' can fill a tank in 'x' hours and 'y' hours respectively. If both the taps are opened together, then how much time it will take to fill the tank?
Required time = $({xy}/{x + y})$ hrs

Here, x = 20, y = 30

Required time = $({xy}/{x + y})$ minutes

= $({20 × 30}/{20 + 30})$ minutes = 12 minutes.

Question : 4

If two pipes function simultaneously, a tank is filled in 12 hours. One pipe fills the tank 10 hours faster than the other. How many hours does the faster pipe alone take to fill the tank?

a) 12 hrs

b) 20 hrs

c) 18 hrs

d) 15 hrs

Answer: (b)

Using Rule 1,

If the slower pipe fills the tank in x hours, then

$1/x + 1/{x - 10} = 1/12$

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

$x^2 - 10x = 24x - 120$

$x^2 - 34x + 120$ = 0

$x^2 - 30x - 4x + 120$ = 0

$x (x - 30) - 4 (x - 30)$ = 0

$(x - 4) (x - 30)$ = 0

$x$ = 30 because $x ≠ 4$

Required time = 30 - 10 = 20 hours

Question : 5

Two pipes can fill a cistern in 3 hours and 4 hours respectively and a waste pipe can empty it in 2 hours. If all the three pipes are kept open, then the cistern will be filled in :

a) 12 hours

b) 5 hours

c) 8 hours

d) 10 hours

Answer: (a)

Using Rule 2,

Part of the cistern filled in 1 hour

= $1/3 + 1/4 - 1/2$

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

= ${4 + 3 - 6}/12 = 1/12$

Hence, the cistern will be filled in 12 hours.

Question : 6

Two pipes X and Y can fill a cistern in 24 minutes and 32 minutes respectively. If both the pipes are opened together, then after how much time (in minutes) should Y be closed so that the tank is full in 18 minutes ?

a) 5

b) 10

c) 8

d) 6

Answer: (c)

If pipe y be closed after x minutes, then

$18/24 + x/32$ = 1

$x/32 = 1 - 18/24 = 1- 3/4 = 1/4$

$x = 32/4$ = 8 minutes

Using Rule 8,
Two taps A and B can fill a tank in x hours and y hours respectively. If both the pipes are opened together, then the time after which pipe B should be closed so that the tank is full in t hours
Required time = $[y(1 –{t/x})]$ hours

x = 24, y = 32, t = 18

Required time = $[y(1 –{t/x})]$ minutes

= $[32(1 - {18/24})]$ minutes

= $[32(1 - {3/4})] = 32 × 1/4$ = 8 minutes

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