Practice Pipes and cisterns - quantitative aptitude Online Quiz (set-2) 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 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)
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-2) 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)
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-3) 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)
(b)
(c)
(d)
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
Q-4) 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)
(b)
(c)
(d)
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.
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)
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) A cistern has two pipes. One can fill it with water in 8 hours and other can empty it in 5 hours. In how many hours will the cistern be emptied if both the pipes are opened together when $3/4$ of the cistern is already full of water ?
(a)
(b)
(c)
(d)
Part of cistern emptied in 1 hour
= $1/5 - 1/8 = {8 - 5}/40 = 3/40$
Since, $3/40$ part is emptied in 1 hour.
$3/4$ part is emptied in
$40/3 × 3/4$ = 10 hours.
Q-7) One tap can fill a water tank in 40 minutes and another tap can make the filled tank empty in 60 minutes. If both the taps are open, in how many hours will the empty tank be filled ?
(a)
(b)
(c)
(d)
Using Rule 7,
Part of the tank filled when both taps are opened together
= $1/40 - 1/60 = {3 - 2}/120 = 1/120$
Hence, the tank will be filled in 120 minutes 2 hours.
Q-8) A tap can fill an empty tank in 12 hours and another tap can empty half the tank in 10 hours. It both the taps are opened simultaneously, how long would it take for the empty tank to be filled to half its capacity ?
(a)
(b)
(c)
(d)
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 tankwhen both are opened = $({xy}/{x - y})$ : x > yb) time taken to empty the tankwhen both are opened = $({xy}/{y- x})$ : y > x
Part of the tank filled in 1 hour
= $1/12 - 1/20 = {5 - 3}/60 = 1/30$
Tank will be filled in 30 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)
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)
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 tankwhen both are opened = $({xy}/{x - y})$ : x > yb) time taken to empty the tankwhen 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.
Q-11) 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)
(b)
(c)
(d)
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
Q-12) 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)
(b)
(c)
(d)
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
Q-13) Two pipes A and B can fill a cistern in 37$1/2$ minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled just in half an hour, if the pipe B is turned off after :
(a)
(b)
(c)
(d)
Pipe A fills the tank in $75/2$ minutes.
Part of the tank filled by A in 30 minutes
= $2/75 × 30 = 4/5$
Remaining part
= $1- 4/5 = 1/5$
Now, 1 part is filled by pipe B in 45 minutes
$1/5$ part is filled in
= 45 × $1/5$ = 9 minutes
Hence, the pipe B should be turned off after 9 minutes.
Q-14) Two pipes A and B can fill a tank in 6 hours and 8 hours respectively. If both the pipes are opened together, then after how many hours should B be closed so that the tank is full in 4 hours?
(a)
(b)
(c)
(d)
Part of the tank filled in 4 hours by pipe A = $4/6 = 2/3$
Remaining part = ${1 - 2}/3 = 1/3$
Time taken by pipe B in filling $1/3$ part = $8/3$ hours
Using Rule 8,
Here, x = 6, y = 8, t = 4
Required time = $[y(1 –{t/x})]$ hours
= $[8(1 - 4/6)]$ hours = $8/3$ hours
Q-15) A water tap fills a tub in ‘p’ hours and a sink at the bottom empties it in ‘q’ hours. If p < q and both tap and sink are open, the tank is filled in ‘r’ hours; then
(a)
(b)
(c)
(d)
Since, P < q,
On opening pipe and sink together,
Part of the tub filled in 1 hour
= $1/P - 1/q$
Clearly, $1/P - 1/q = 1/r$
Q-16) A tank has two pipes. The first pipe can fill it in 4 hours and the second can empty it in 16 hours. If two pipes be opened together at a time, then the tank will be filled in
(a)
(b)
(c)
(d)
Using Rule 7,
Part of tank filled by both the pipes in 1 hour
= $1/4 - 1/16 = {4 - 1}/16 = 3/16$
Required time = $16/3 = 5{1}/3$ hours
Q-17) An empty tank can be filled by pipe A in 4 hours and by pipe B in 6 hours. If the two pipes are opened for 1 hour each alternately with first opening pipe A, then the tank will be filled in
(a)
(b)
(c)
(d)
Part of the tank filled in first 2 hours
= $1/4 + 1/6 = {3 + 2}/12 = 5/12$ Part
Part of the tank filled in first 4 hours
= ${2 × 5}/12$ parts= $5/6$ parts
Remaining part = $1 - 5/6 = 1/6$
Now it is the turn of pipe A
Time taken to fill $1/4$ part = 1 hour
Time taken to fill $1/6$ part
= $1/6 × 4 = 2/3$ hour
Total time = 4 + $2/3 = 4{2}/3$ hours
Q-18) If $1/3$ of a tank holds 80 litres of water, then the quantity of water that $1/2$ tank holds is :
(a)
(b)
(c)
(d)
Let the capacity of the tank be x litres then
$x/3$ = 80 ⇒ x = 240
$x/2 = 240/2$ = 120 litres
Q-19) 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)
(b)
(c)
(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
Q-20) Two pipes, P and Q can fill a cistern in 12 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes, P is turned off. In how many more minutes will Q fill the cistern ?
(a)
(b)
(c)
(d)
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 the tank filled in 3 minutes by pipes P and Q
= $3(1/12 + 1/15)$
= $3({5 + 4}/60) = {3 × 9}/60 = 9/20$
So, Remaining part
= 1– $9/20 = 11/20$
Time taken by Q
= $11/20 × 15 = 33/4 = 8{1}/4$ minutes