Pipes and Cisterns Questions & Answers for Competitive Exams - 1

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.

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.

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.

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.

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.