Aptitude :: Pipe And CisternPage 3

17.

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
(a)6 hours
(b)10 hours
(c)15 hours
(d)30 hours
Answer is: CSuppose, first pipe alone takes X hours to fill the tank.
Then, second and third pipes will take (X – 5) and (X – 9) hours respectively to fill the tank
∴ 1/X + 1/(X-5) = 1/(X-9)
⇒ (X – 5 + X)/X (X – 5) = 1/(X – 9)
⇒ (2X – 5) (X – 9) = X (X – 5)
⇒ X² - 18X + 45 = 0
⇒ (X – 15) (X – 3) = 0
⇒ X = 15. [neglecting X = 3]

18.

A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time, and A and B fill it together for the other half?
(a)15 min.
(b)20 min.
(c)27.5 min.
(d)30 min.
Answer is: DPart filled by (A + B) in 1 minute = (1/60 + 1/40) = 1/24
Suppose the tank is filled in X minutes
Then, X/2 (1/24 + 1/40) = 1
⇒ X/2 x 1/15 = 1
⇒ X = 30 min.

19.

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)6 hours
(b)6 hours 40 minutes
(c)7 hours
(d)7 hours 30 minutes
Answer is: C(A + B)'s 1 hour's work = (1/12 + 1/15) = 9/60 = 3/20
(A + C)'s 1 hour's work = (1/12 + 1/20) = 8/60 = 2/15
Part filled in 2 hours = (3/20 + 2/15) = 17/60
Part filled in 6 hours = 3 x (17/60) = 17/20
Remaining part = (1 – 17/20) = 3/20
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
∴ Total time taken to fill the tank = (6 + 1) = 7 hours.

20.

Three pipes A, B and C can fill a tank in 6 hours. After working at it 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)10 hours
(b)12 hours
(c)14 hours
(d)16 hours
Answer is: CPart filled in 2 hours = 2/6 = 1/3
Remaining part = (1 – 1/3) = 2/3
∴ (A + B)'s 7 hours work = 2/3
∴ (A + B)'s 1 hours work = 2/21
∴ C's 1 hours work = [(A + B + C)'s 1 hour's work – (A + B)'s 1 hour's work]
∴ C's 1 hours work = (1/6 – 2/21) = 1/14
∴ C alone can fill tank in 14 hours.

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