100+ Pipes and Cisterns Questions for Bank Exams - 1
Question: 1
Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened A, B and C discharge chemical solutions P, Q and R respectively. What is the proportion of solution R in the liquid in the tank after 3 minutes?
(A) $${5}/{11}$$.
(B) $${6}/{11}$$.
(C) $${7}/{11}$$.
(D) $${8}/{11}$$.
Ans: B
Part filled by (A + B + C) in 3 minutes = 3$$({1}/{30} + {1}/{20} + {1}/{10})$$
= $$(3 × {11}/{60})$$ = $${11}/{20}$$.
Part filled by C in 3 minutes = $${3}/{10}$$
∴ Required ratio = $$({3}/{10} × {20}/ {11})$$ = $${6}/{11}$$.
Question: 2
Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 30 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?
(A) 9 hrs
(B) 18 hrs
(C) 36 hrs
(D) 48 hrs
Ans: C
Part filled by (A + B) in 1 hour = $$({1}/{5} + {1}/{20})$$ = $${1}/{4}$$
So, A and B together can fill the tank in 4 hours.Work done by the leak in 1 hour = $$({1}/{4} - {2}/{9})$$ = $${1}/{36}$$
∴ Leak will empty the tank in 36 hrs.
Question: 3
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
(B) 11
(C) 12
(D) 14
Ans: D
Part filled in 2 hours = $${2}/{6}$$ = $${1}/{3}$$.
Remaining part = $$(1 – {1}/{3})$$ = $${2}/{3}$$.
∴ (A + B)’s 7 hour’s work = $${2}/{3}$$
(A + B)’s 1 hour’s work =$${2}/{21}$$.
∴ C’s 1 hour’s work= (A + B + C)’s 1 hour’s work – (A + B)’s 1 hour’s work) = $$({1}/{6} - {2}/{21})$$ = $${1}/{14}$$
∴ C alone can fill the tank in 14 hours.
Question: 4
A tap can fill a tank in 48 minutes whereas another tap can empty it in 2 hours. If both the taps are opened at 11 : 40 A.M., then the tank will be filled at
(A) 12 : 40 P.M.
(B) 1 : 00 P.M.
(C) 1 : 20 P.M.
(D) 1 : 30 P.M.
Ans: B
Net part filled in 1 hour = $$({1}/{48} - {1}/{120})$$ = $${3}/{240}$$ = $${1}/{80}$$.
∴ The tank will be filled 80 mins i.e. 1 hour 20 min. After 11 : 40 A.M. i.e. at 1 P.M.
Question: 5
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 = $${Quantity removed} / {Total capacity}$$ = $${a} / {g}$$.
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