100+ Pipes and Cisterns Problems with Solutions Pdf - 1

Question: 1

Two pipes A and B can fill a tank is 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

(A) 10 min 20 sec

(B) 11 min 45 sec

(C) 12 min 30 sec

(D) 14 min 40 sec

Ans: D

Part filled in 4 minutes = 4$$({1/15} + {1/20})$$ = $${7}/{15}$$.

Remaining part = $$(1 – {7}/{15})$$ = $${8}/{15}$$

Part filled by B in 1 minute = $${1}/{20} : {1/20} : {8/15} : : 1 : x$$

or x = $$({8}/{15} × 1 × 20)$$ = $$10{2}/{3}$$ min = 10 min 40 sec.

∴ The tank will be fill in (4 min + 10 min. 40 sec) = 14 min. 40 sec.

Question: 2

A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m3. The emptying capacity of the tank is 10 m3 per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?

(A) 50 m3 / min

(B) 60 m3 / min

(C) 72 m3 / min

(D) 82 m3 / min

Ans: A

Let the filling capacity of the pump be x m3/min.

Then, emptying capacity of the pump = (x + 10) m3/min

So, $${2400}/{x}$$ – $${2400}/{(x + 10)}$$ = 8

⇒ x2 + 10x – 3000 = 0 ⇒ (x - 50)

⇒ (x + 60) = 0 ⇒ x = 50.

Hence filling capacity of the pump = 50m3/min.

Question: 3

The petrol tank of an automobile can hold g litres. If a litres was removed when the tank was full, what part of the full tank was removed?

(A) g - a

(B) $${g}/{a}$$

(C) $${a}/{g}$$

(D) $${g - a}/{a}$$

Ans: C

Required part = $${Quantity removed}/{Total capacity}$$ = $${a}/{g}$$

Question: 4

Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?

(A) 1 hr

(B) 2 hrs

(C) 6 hrs

(D) 8 hrs

Ans: C

Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

∴ $${1}/{x} + {1}/{(x + 6)}$$ = $${1}/{4}$$

⇒ $${ x + 6 + x}/{x (x + 6)}$$ = $${1}/{4}$$

⇒x2 - 2x - 24 = 0

⇒ (x – 6) (x + 4) = 0 ⇒ x = 6.

So, A alone will fill the cistern in 6 hrs.

Question: 5

Pipe A can fill a tank in 30 hours and pipe B in 45 hours. If both the pipes are opened in an empty tank, how long much time will they take to fill it?

(A) 16 hours

(B) 18 hours

(C) 19 hours

(D) 20 hours

Ans: B

Part filled by (A + B) in 1 hour = $$({1/30} + {1/45})$$ = $$({10}/{180})$$ = $${1}/{18}$$.

Hence, pipes A and B together will fill the tank in 18 hours.

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