  #### Answer to Aptitude Question For SSC Bank -Pipe & Cistran

###### 2020-01-08 09:52:26

Rate of filling of Pipe A – 1/8 of tank in 1 hour
Rate of filling of Pipe B – 1/6 of tank in 1 hour
Rate of filling when both are opened together
1/8 + 1/6 = 7/24 of tank in 1 hour
Since both are open for 2 hours, 2 x(7/24) = 7/12 of tank gets filled
Left over portion i.e. 5/12 of tank is to be filled by B, which fills 1/6 ( or 2/12) tank in 1 hour.
So it takes 2 ½ hours to fill 5/12 of tank.

Rate of filling of the pump without the open tap – 1/8 of tank in 1 hour
Rate of filling of the pump with the open tap – 1/10 of tank in 1 hour
Rate of emptying of the tap (1/8 – 1/10) of tank in 1 hour
Or, the tap empties 1/40 of the tank in 1 hour
Thus the tap would empty a full tank in 40 hours.

Rate of filling when both the pipes are open
1/15 + 1/20 or 7/60 of the tank in 1 minute
Rate of emptying the tank by the leak – 1/12 of the tank in 1 minute
So with both the taps open and the active leak, the rate of fill would be
(7/60 – 1/12) = 2/60 or 1/30 of the tank in 1 minute.
Thus, the tank would be full in 30 minutes.

Let Pipe A fill the cistern in H hours
&there4 Pipe B can fill it in H + 6 hours
both the pipes can fill the 1/H + 1/(H + 6) of the cistern in 1 hour
1/H + 1/H + 6 = 1/4
or H + 6 + H/H(H + 6) = 1/4
or H2 – 2H – 24 = 0
(H – 6)(H + 4) = 0 or H = 6
Pipe A can fill the cistern in 6 hours.

If the faster tap can fill the drum in M minutes, the slower tap fills it in 3M minutes
Both the pipes can fill (1/M + 1/3M) of the drum in 1 minute
1/M + 1/3M = 1/36 or 4/3M = 1/36 or 3M = 144 or M = 48
Faster tap can fill the drum in 48 minutes.
Slower tap can fill the drum in 3 x 48 = 144 minutes.

The leak can empty 1/10 of the cistern in 1 hour
With the tap open, the 1/15 cistern is emptied in 1 hour
Thus, the rate of filling of the tap is 1/101/15 = 1/30 of the cistern in 1 hour
In other words the tap can fill the cistern in 30 hours or 1800 minutes
Since the tap supplies 8 litres of water per minute, the capacity of the cistern is 1800 x 8 = 14,400 litres

Taps A and B can fill 1/40 of the tank in 1 minute
Tap A can fill 1/60 of the tank in 1 minute
Therefore Tap B can fill 1/401/60 of the tank in 1 minute
Or Tap B can fill 1/120 of the tank in 1 minute
Therefore Tap B can fill the tank in 120 minutes

8 taps can fill a tank in 30 minutes
6 taps can fill the tank in M minutes
M = (8 x 30)/6 = 40 minutes

When both the pipes are open (1/30 + 1/45) of the tank is filled in 1 minute.
In the first 10 minutes (1/30 + 1/45) x 10 = 10/18 of the tank gets filled ....(1).
In the next 10 minutes, 10/45 of the tank is filled by pipe B ....(2)
Thus, adding (1) and (2) we get 7/9
Thus, 2/9 of the tank is to be filled by A
A takes 2/9 x 30 = 6 2/3 minutes to fill the tank

Pipe A fills 1/12 of the tank in 1 hour
Pipe B can empty 1/36 in 1 hour or 1/144 in 15 minutes
From 8 a.m. to 9 a.m. 1/12 of the tank gets filled
Thus, 11/12 of the tank is empty
From 9 a.m. to 10 a.m. (1/121/144) = 11/144 of the tank gets filled
i.e. in 1 hour 11/144 of the tank gets filled
So to fill, 11/12 of the tank it would take 11/12 x 144/11 = 12 hours
Total time required is 1 + 12 = 13 hours
The tank would be full at 8 a.m. + 13 = 9 p.m.  