Model 3 Opening both taps and leaks Practice Questions Answers Test with Solutions & More Shortcuts
pipes & cisterns PRACTICE TEST [3 - EXERCISES]
Model 1 Basic Pipes & Cisterns problems
Model 2 Filling tank by parts or fractions
Model 3 Opening both taps and leaks
Question : 16
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) 24
b) 36
c) 30
d) 12
Answer »Answer: (a)
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
Question : 17 [SSC MTS 2013]
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) 5$1/2$ hours
b) 4$2/3$ hours
c) 1$3/4$ hours
d) 2$3/5$ hours
Answer »Answer: (b)
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
Question : 18 [SSC CHSL 2013]
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) 9
b) 8
c) 10
d) 12
Answer »Answer: (d)
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
Question : 19 [SSC CGL Prelim 2003]
A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely ?
a) 3 hours 45 minutes
b) 3 hours 15 minutes
c) 4 hours
d) 4 hours 15 minutes
Answer »Answer: (a)
A tap can fill the tank in 6 hours. In filling the tank to its half, time required = 3 hours.
Remaining part = $1/2$
Since, 1 tap takes 6 hours to fill the tank
Time taken by 4 taps take to fill $1/2$ of the tank
= $6/4 × 1/2 = 3/4$ hour
Total time = $3 + 3/4$
= $3{3}/4$ hours = 3 hours 45 minutes
Question : 20
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) 16
b) 14
c) 10
d) 12
Answer »Answer: (b)
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.
IMPORTANT quantitative aptitude EXERCISES
Model 3 Opening both taps and leaks Shortcuts »
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