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 : Three pipes A, B and C can fill a cistern in 6 hours. After working at it together for 2 hours, C is closed and A and B fill it in 7 hours more. The time taken by C alone to fill the cistern is

(a) 17 hours

(b) 16 hours

(c) 14 hours

(d) 15 hours

The correct answers to the above question in:

Answer: (c)

Part of the cistern filled by pipes A, B and C in 1 hour = $1/6$

Part of the cistern filled by all three pipes in 2 hours = $1/3$

Remaining part =$1– 1/3 = 2/3$

Now, pipe A and B fill $2/3$ part of the cistern in 7 hours

Pipe A and B will fill the cistern in ${7 × 3}/2 = 21/2$ hours

Part of the cistern filled by Aand B in 1 hour = $2/21$

So Part of the cistern filled by C in 1 hour = $1/6 - 2/21$

= ${7 - 4}/42 = 1/14$

Pipe C will fill the cistern in 14 hours.

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 tank with water in 3 hours. Due to leakage in bottom, it takes 3$1/2$ hours to fill it . In what time the leak will empty the fully filled tank ?

a) 10$1/2$ hours

b) 6$1/2$ hours

c) 12 hours

d) 21 hours

Answer: (d)

Using Rule 7,

Let the leak empty the full tank in x hours.

$1/3 - 1/x = 2/7$

$1/x = 1/3 - 2/7 = {7 - 6}/21$

$1/x = 1/21 ⇒ x = 21$ hours

Question : 2

A pump can fill a tank with water in 2 hours. Because of a leak in the tank it was taking 2$1/3$ hours to fill the tank. The leak can drain all the water off the tank in :

a) 14 hours

b) 4$1/3$ hours

c) 8 hours

d) 7 hours

Answer: (a)

Using Rule 7,
A tap 'A' can fill a tank in 'x' hours and 'B' can empty the tank in 'y' hours. Then (a) time taken to fill the tank
when both are opened = $({xy}/{x - y})$ : x > y
b) time taken to empty the tank
when both are opened = $({xy}/{y- x})$ : y > x

Work done in 1 hour by the filling pump = $1/2$

Work done in 1 hour by the leak and the filling pump = $3/7$

Work done by the leak in 1 hour

= $1/2 - 3/7 = {7 - 6}/14 = 1/14$

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

Question : 3

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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