Model 3 Opening both taps and leaks Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 3 EXERCISES

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

Questions : A tank is fitted with two taps. The first tap can fill the tank completely in 45 minutes and the second tap can empty the full tank in one hour. If both the taps are opened alternately for one minute, then in how many hours the empty tank will be filled completely ?

(a) 5 hours 53 minutes

(b) 4 hours 48 minutes

(c) 2 hours 55 minutes

(d) 3 hours 40 minutes

The correct answers to the above question in:

Answer: (a)

Using Rule 7,

Part of the tank filled in one minute

= $1/45 - 1/60 = {4 - 3}/180 = 1/180$

Since, $1/180$ part is filled in 1 minute

$1 - 1/45 = 44/45$ part is filled in

${2 × 180 × 44}/45$ = 352 minutes

i.e. 5 hours 52 minutes

Remaining $1/45$ part will be filled in 1 minute.

Total time taken = 5 hours 53 minutes

Practice pipes & cisterns (Model 3 Opening both taps and leaks) Online Quiz

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Read more opening both taps and leaks Based Quantitative Aptitude Questions and Answers

Question : 1

A pipe can fill a cistern in 9 hours. Due to a leak in its bottom, the cistern fills up in 10 hours. If the cistern is full, in how much time will it be emptied by the leak ?

a) 100 hours

b) 90 hours

c) 70 hours

d) 80 hours

Answer: (b)

Using Rule 7,

Part of the tank emptied by the leak in 1 hour

= $1/9 - 1/10 = {10 - 9}/90 = 1/90$

Required time = 90 hours

Question : 2

A cistern is normally filled in 8 hours but takes another 2 hours longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in :

a) 40 hours

b) 25 hours

c) 16 hours

d) 20 hours

Answer: (a)

Using Rule 7,

Part emptied by the leak in 1 hour

= $1/8 - 1/10 = {5 - 4}/40 = 1/40$

The leak will empty the cistern in 40 hours.

Question : 3

A pipe can fill a tank in 24 hrs. Due to a leakage in the bottom, it is filled in 36 hrs. If the tank is half full, how much time will the leak take to empty the tank?

a) 24 hrs

b) 36 hrs

c) 48 hrs

d) 72 hrs

Answer: (b)

Using Rule 7,

Part of tank emptied by leak in an hour

= $1/36 - 1/24 = {2 - 3}/72 = {–1}/72$

Time taken in emptying the full tank

= 72 hours

Required time = 36 hours

Question : 4

Pipe A can fill a tank in 4 hours and pipe B can fill it in 6 hours. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full ?

a) 4$2/3$

b) 3$1/4$

c) 4$1/2$

d) 3$1/2$

Answer: (a)

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

Part of tank filled in first two hours

= $1/4 + 1/6 = {3 + 2}/12 = 5/12$

Part of tank filled in first 4 hours

= $10/12 = 5/6$

Remaining part = $1 - 5/6 = 1/6$

This remaining part willl be filled by pipe A.

Time taken by pipe A

= $1/6 × 4 = 2/3$ hour

Total time = 4 + $2/3 = 4{2}/3$ hours

Question : 5

A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When the tank is full, the inlet is opened and due to the leak the tank is empty in 8 hours. Find the capacity of the tank.

a) 24 litres

b) 10 litres

c) 5760 litres

d) 96 litres

Answer: (c)

Using Rule 7,

Part of tank filled by inlet pipe in 1 hour

= $1/6 - 1/8 = {4 - 3}/24 = 1/24$

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

Question : 6

Pipe A can fill a cistern in 6 hours and pipe B can fill it in 8 hours. Both the pipes are opened simultaneously, but after two hours, pipe A is closed. How many hours will B take to fill the remaining part of the cistern ?

a) 4 hrs

b) 2$2/3$ hrs

c) 2 hrs

d) 3$1/3$ hrs

Answer: (d)

Using Rule 1,

Part of the cistern filled in 2 hours by pipe A and B

= $2(1/6 + 1/8) =2({4 + 3}/24) = 7/12$

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

Time taken by pipe B in filling $5/12$ part

= $5/12 × 8 = 10/3 = 3{1}/3$ hours

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