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## Monday, 29 April 2019

24.  Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
A. 623 hours            B. 6 hours
C. 712 hours           D.  7 hours
Explanation :
Part filled by pipe A in 1 hour = 1/12
Part filled by pipe B in 1 hour = 1/15
Part filled by pipe C in 1 hour = 1/20
In first hour, A and B is open In second hour, A and C is open then this pattern goes on till the tank fills
Part filled by pipe A and pipe B in 1 hour
= 1/12+1/15=9/60=3/20
Part filled by pipe A and pipe C in 1 hour
= 1/12+1/20=8/60=2/15
Part filled in 2 hour = 3/20+2/15=17/60
Part filled in 6 hour = (17/60) x3=17/20
Remaining part = (1-(17/20) =3/20
Now, 6 hours are over and only 3/20 part needed to be filled. At this 7th hour, A and B is open Time taken by pipe A and B to fill this 3/20 part
= (3/20) + (3/20)
= 1 hour
Total time taken = 6 hour + 1 hour = 7 hour

25.  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 liters a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 24 hours. How many liters does the cistern hold?
A. 4010 litre B. 2220 litre C. 1920 litre D. 2020 litre
Explanation :
Part emptied by the leak in 1 hour = 1/6
Net part emptied by the leak and the inlet pipe in 1 hour = 1/24
Part filled by the inlet pipe in 1 hour = 1/6"1/24=1/8
i.e., inlet pipe fills the tank in 8 hours = (8 x 60) minutes = 480 minutes
Given that the inlet pipe fills water at the rate of 4 liters a minute
Hence, water filled in 480 minutes = 480 x 4 = 1920 litre
i.e., The cistern can hold 1920 litre

26. A cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 8 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled?
A. 4.8 hr                     B. 2.4 hr
C. 3.6 hr                     D. 1.8 hr
Explanation:
Part filled by the first tap in 1 hour = 1/3
Part emptied by the second tap 1 hour = 1/8
Net part filled by both these taps in 1 hour = 1/3-1/8=5/24
i.e, the cistern gets filled in 24/5 hours = 4.8 hours

27. Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 40 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?
A. 28 hr                      B. 16 hr
C. 22 hr                      D. 32 hr
Explanation :
Part filled by pipe A in 1 hour = 1/5
Part filled by pipe B in 1 hour = 1/20
Part filled by pipe A and B in 1 hour = 1/5+1/20=1/4
i.e., A and B together can fill the tank in 4 hours
Given that due to the leakage, it took 40 minutes more to fill the tank.
i.e., due to the leakage, the tank got filled in 4(40/
60) hour =4(2/3) hour=14/3 hour
==> Net part filled by pipe A and B and the leak in 1 hour = 3/14
==>Part emptied by the leak in 1 hour =1/4"3/14=1/28
i.e., the leak can empty the tank in 28 hours

28. Bucket P has thrice the capacity as bucket Q. It takes 80 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P and Q, having each turn together to fill the empty drum?
A. 30                            B. 45
C. 60                            D. 80
Explanation :
Let capacity of bucket P = x
Then capacity of bucket Q = x/3
Given that it takes 80 turns for bucket P to fill the empty drum
=> capacity of the drum = 80x
Number of turns required if both P and Q are used
= 80x+(x+(x/3))
=240x+(3x+x)
=240+4
=60

29. Two pipes A and B can separately fill a cistern in 40 minutes and 30 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 20 minutes. In how much time, the third pipe alone can empty the cistern?
A. 120 min                B. 100 min
C. 140 min                D. 80 min
Explanation :
Part filled by pipe A in 1 minute = 1/40 Part filled by pipe B in 1 minute = 1/30
Net part filled by pipe A, pipe B and the third pipe in 1 hour = 1/20
===>Part emptied by the third pipe in 1 hour = (1/40)+(1/30)"(1/20) =(3+4"6)/120=1/120
i.e., third pipe alone can empty the cistern in 120 minutes

30. Two pipes A and B can fill a tank in 9 hours and 3 hours respectively. If they are opened on alternate hours and if pipe A is opened first, how many hours, the tank shall be full?
A. 4 hr                     B. 5 hr
C. 2 hr                     D. 6 hr
Explanation :
Part filled by pipe A in 1hour =1/9
Part filled by pipe B in     1hour =1/3
Pipe A and B are opened alternatively.
Part filled in every 2 hour = 1/9+1/3=(1+3)/9 =4/9
Part filled in 4 hour = 2x(4/9)=8/9 remaining part =1-(8/9)
=1/9
Now it is pipe A's turn.
Time taken by pipe A to fill the remaining 19 part =(1/9)+ (1/9)=1 hour
Total time taken = 4 hour + 1hour = 5 hour

* MATHS (SOLVED IN STEPS ==> Ratio and Proportion
* MATHS (SOLVED IN STEPS ==> Progressions
* MATHS (SOLVED IN STEPS ==> Percentage
* MATHS (SOLVED IN STEPS ==> Surds and Indices
* MATHS (SOLVED IN STEPS ==> Calendar
* MATHS (SOLVED IN STEPS ==> Square and cube
* MATHS (SOLVED IN STEPS ==> Simplification
* MATHS (SOLVED IN STEPS ==> Problems on Average
* MATHS (SOLVED IN STEPS ==> HCF & LCM
* MATHS (SOLVED IN STEPS ==> Profit and Loss
* MATHS (SOLVED IN STEPS ==> Problems on Trains
* MATHS (SOLVED IN STEPS ==>
* MATHS (SOLVED IN STEPS ==>
* MATHS (SOLVED IN STEPS ==>

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