Latest NIIT Aptitude Question SOLUTION: Four pipes fill a cistern in 15,20,30 and 60 hours respectively.The first was opened at 6 am,the second at 8 am,the third at 10 am and fourth at noon.when will the reservoir will b

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(#M40032832) NIIT Numerical Ability question PIPES Keep an EYE

Four pipes fill a cistern in 15,20,30 and 60 hours respectively.The first was opened at 6 am,the second at 8 am,the third at 10 am and fourth at noon.when will the reservoir will be filled?

Asked In NIIT Harsha Raghav Mucherla (10 years ago)

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S.SHAFI  (10 years ago)

ans) 2pm
Pipe a will be opened for 6am-12pm -> 6hours ===> 6/15
Pipe b will be opened for 8am-12pm -> 4hours ===> 4/20
Pipe c will be opened for 10am-12pm -> 2hours ==> 2/30
From noon, all pipes are opened...
Up-to 12pm, the amount of the reservoir is 2/3
6/15 + 4/20 + 2/30 = 40/60 = 2/3

In 1 hour, the four pipes are going to fill 1/6th of reservoir
1/15 + 1/20 + 1/30 + 1/60 = 10/60 = 1/6

So, The remaining 1/3 will be filled by 4 pipes in 2 hours...
So, the reservoir will be filled by 2PM...

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Anitesh Kumar  (8 years ago)

(1/15)*2 +(1/15+1/20)*2+(1/15+1/20+1/30)*2+(1/15+1/20+1/30+1/60)*x=1
so that x=2;
so that reservoir will be filled at 2 o'clock

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shwetma  (5 years ago)

Loss matches = X
Win matches = X+3
2 points for + (-1) points for loss
2*(X+3) + (-1)*X = 23
X+6= 23
X= 17.
Therefore loss matches =17
Now as per question
total matches = loss matches + won matches
(X )+ (X+3) = 2x + 3
2(17)+3= 37.
Hence 37 matches they played.

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