Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A water tank has inlets of two types A

Quantitative Aptitude – Arithmetic - Pipes and Cisterns – A water tank has inlets of two types A

Slot -2 – Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A water tank has inlets of two types A

A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?

Answer: 48

Suppose inlets of type A fill A liters per minute and type B fills B liters per minute.

So, capacity of tank = 30(10A + 45B) = 60(8A + 18B)

⇒10A + 45B = 16A + 36B

⇒6A = 9B

⇒A = 1.5B

⇒Capacity of Tank = 30(15B + 45B) = 30*60B = 1800 B

Time taken to fill the tank with 7A and 27B, which is 10.5B and 27B, which is 37.5B = 1800 B / 37.5 B = 48 minutes

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Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A tank is emptied every day

Quantitative Aptitude – Arithmetic - Pipes and Cisterns – A tank is emptied every day

Slot -2 – Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A tank is emptied every day at a fixed time

A tank is emptied every day at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed filling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?

a) 4:24 pm

b) 4:12 pm

c) 4:48 pm

d) 4:36 pm

Answer: a) 4 : 24 pm

Solution:

Let pump A alone can fill the tank in t hours so time taken by pump B alone = t-2 hours

As per question , (t-3)/t+2/(t-2)=1

t^2-3t+6=t^2-2t
Or t=6

So the time at which pumps tank is emptied = 8 pm – 6 hours = 2 pm

Time taken by both pumps together to fill the tank =(6×4)/(6+4)=2.4 hours=2 hour 24 minutes

Thus tank will be filled by 2 pm + 2 hour 24 minutes=4:24 pm

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Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A tank is fitted with pipes

Quantitative Aptitude – Arithmetic - Pipes and Cisterns – A tank is fitted with pipes

Slot -1 – Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A tank is fitted with pipes

A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on? TITA

Answer: 10

Solution:
Let each filling pipe alone can fill the tank in F hours and each draining tank alone can empty the tank in D hours. So

6/F-5/D=1/6———-1)
5/F-6/D=1/60———-2)

by solving the above equations, F=12 and D=15
So 2/F-1/D=2/12-1/15=1/10

Thus required time = 10/1 = 10 hours

Answer: 10

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