Practice Opening both taps and leaks - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   Three pipes A, B and C can fill a tank in 6 hours. After working together for 2 hours, C is closed and A and B fill the tank in 8 hours. The time (in hours) in which the tank can be filled by pipe C alone is

(a)

(b)

(c)

(d)

Explanation:

Part of the tank filled by

(A + B + C) in 1 hour = $1/6$

Part of tank filled by these in 2 hours

= $2/6 = 1/3$

Remaining part = $1 - 1/3 = 2/3$

Time taken by A and B in filling

$2/3$ rd part = 8 hours

Time taken by A and B in filling the whole tank

= ${8 × 3}/2$ = 12 hours

Part of tank filled by C in an hour

= $1/6 - 1/12 = 1/12$

Hence, required time = 12 hours

Q-2)   Two pipes A and B can fill a tank with water in 30 minutes and 45 minutes respectively. The water pipe C can empty the tank in 36 minutes. First A and B are opened. After 12 minutes C is opened. Total time (in minutes) in which the tank will be filled up is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1 and 2,

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

= $1/30 + 1/45 = {3 + 2}/90 = 1/18$ part

Part of tank filled in 12 minutes

= $12/18 = 2/3$ part

Remaining part

= $1 - 2/3 = 1/3$ part

When pipe C is opened,

Part of tank filled by all three pipes

= $1/30 + 1/45 - 1/36$

= ${6 + 4 - 5}/180 = 5/180 = 1/36$

Time taken in filling $1/3$ part

= $1/3$ × 36 = 12 minutes

Total time = 12 + 12 = 24 miuntes

Q-3)   Three pipes A, B and C can fill a tank in 6 hours. After working together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is

(a)

(b)

(c)

(d)

Explanation:

A, B and C together fill the tank in 6 hours.

Part of the tank filled in 1 hour by (A + B + C) = $1/6$

Part of the tank filled in 2 hours by all three pipes

= $2/6 = 1/3$

Remaining empty part

= $1 - 1/3 = 2/3$

This $2/3$ part is filled by (A + B).

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

= ${7 × 3}/2 = 21/2$ hours

Part of tank filled by C in 1 hour

= $1/6 - 2/21 = {7 - 4}/42 = 3/42 = 1/14$

∴ Required time = 14 hours.

Q-4)   A tap can fill a cistern in 40 minutes and a second tap can empty the filled cistern in 60 minutes. By mistake without closing the second tap, the first tap was opened. In how many minutes will the empty cistern be filled ?

(a)

(b)

(c)

(d)

Explanation:

Using Rule 7,

Tricky Approach

Part of the cistern filled in 1 minute by both the taps

= $1/40 - 1/60 = {3 - 2}/120 = 1/120$

Empty cistern will be filled in 120 minutes.

Q-5)   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)

(b)

(c)

(d)

Explanation:

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

Q-6)   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)

(b)

(c)

(d)

Explanation:

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

Q-7)   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)

(b)

(c)

(d)

Explanation:

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

Q-8)   A tap takes 36 hours extra to fill a tank due to a leakage equivalent to half of its inflow. The inflow can fill the tank in how many hours ?

(a)

(b)

(c)

(d)

Explanation:

Using Rule 7,

Let the inflow fill the tank in x hours.

$1/x - 1/{2x} = 1/36$

[leakage being half of inflow]

= ${2 - 1}/{2x} = 1/36$

2x = 36

$x = 36/2$ = 18 hours

Q-9)   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)

(b)

(c)

(d)

Explanation:

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.

Q-10)   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)

(b)

(c)

(d)

Explanation:

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.