# Pipes and Cisterns Questions & Answers for Competitive Exams - 1

Question: 1

A pipe can fill a tank in 3 hours. There are two outlet pipes from the tank which can empty it in 7 and 10 hours respectively. If all the three pipes are opened simultaneously, then the tank will be filled in

(A) 9 hours

(B) 10 hours

(C) 11 hours

(D) 13 hours

Ans: C

Net part filled in 1 hour = $${1/3}$$ - $$({1/7} +{1/10})$$ = $${1/3} - {17/70}$$ = $${19}/{210}$$

∴ The tank will be filled in $${210}/{19}$$ hrs i.e. $$11{1}/{19}$$ hrs = 11 hrs.

Question: 2

Tap A fills a tank in 4 hours whereas tap B empties the full tank in 24 hours. A and B are opened alternately for 1 hour each. Every 2 hours the level of water is found to increase by 0.5 m. The depth of the tank is

(A) 2.2 m

(B) 2.4 m

(C) 4.8 m

(D) 6.4 m

Ans: B

Part filled in 2 hours = $$({1/4} - {1/24} )$$ = $${5}/{24}$$.

Let the depth of the tank be h metres.

Then, $${5}/{24}$$ h = 0.5 ⇒ h = $$({0.5 × 24} / {5})$$ = 2.4 m.

Question: 3

A town is supplied with water from a big overhead tank which is fed with a constant volume of water regularly. When the tank is full, if 32000 gallons are used daily, the supply fails in 50 days. However, if 37000 gallons are used daily, the supply lasts for 40 days only. How much water can be used daily without the supply ever failing?

(A) 10000 gallons

(B) 12000 gallons

(C) 15000 gallons

(D) 18000 gallons

Ans: B

Let the volume of the overhead tank be x litres and the constant volume being fed per day to the tank by y litres.

Then, x + 50 y = 32000 × 50 ⇒ x + 50y = 1600000

x + 40y =37000 × 40 ⇒ x + 40y = 1480000

Subtracting (ii) from (1) we get 10 y = 120000 or y = 12000.

Clearly, the supply won’t ever fail if the regular demand is equal to the regular supply which is 12000 gallons.

Question: 4

Two pipes A and B can fill a tank in 18 hrs and 6 hrs respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

(A) 4 hrs

(B) $$4{1}/{2}$$hrs

(C) 7 hrs

(D) 9 hrs

Ans: B

Part of tank filled by A in 1 hour = $${1}/{18}$$ part

Part of tank filled by B in 1 hour = $${1}/{6}$$ part

Part of tank filled by pipes A and B in 1 hour = $${1/18} + {1/6}$$ = $${1 + 3} / {18}$$ = $${4}/{18}$$ = $${2}/{9}$$

∴ Required time taken by pipe A and B = $${9}/{2}$$ hours = $$4{1}/{2}$$ hours.

Question: 5

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 fill in

(A) 6 hrs

(B) 7 hrs

(C) 8 hrs

(D) 9 hrs

Ans: B

(A + B)’s I hour’s work = $$({1/12} + {1/15})$$ = $${9}/{60}$$ = $${3}/{20}$$

(A + C)’s I hour’s work = $$({1/12} + {1/20})$$ = $${8}/{60}$$ = $${2}/{15}$$.

Part filled in 2 hrs = $$({3/20} + {2/15})$$ = $${17}/{60}$$;

Part filled in 6 hrs. = $$(3 × {17}/{60})$$ = $${17}/{20}$$.

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) hrs = 7 hrs.

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