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31. Two pipes A and B can fill a tank in 20 and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is one - third full, a leak develops in the tank through which one - third water supplied by both the pipes gose out. The total time taken to fill the tank is?

A. 12 hours

B. 14 hours

C. 16 hours

D. 18 hours

Discuss

Solution:

Part filled by (A + B) in 1 hour

\begin{aligned} = (\frac{1}{20} + \frac{1}{30}) \\ = \frac{1}{12} \end{aligned}

So, A and B together can fill the tank in 12 hrs,

\frac{1}{3}

part is filled by (A + B) in

(\frac{1}{3} \times 12) =  4 hrs

Since the leak empties one - third water, so time taken to fill the tank
= Time taken by (A + B) to fill the whole tank + Time taken by (A + B) to fill one - third tank
= (12 + 4)
= 16 hours

32. Two pipes can fill a tank in 40 and 48 minutes respectively and a waste pipe can empty 3 gallons per minutes. All the three pipes working together can fill the tank in 30 minutes, The capacity of the tank is-

A. 60 gallons

B. 120 gallons

C. 180 gallons

D. 240 gallons

Discuss

Solution:

Work done by the waste pipe in 1 minute

\begin{aligned} = \frac{1}{30} - (\frac{1}{40} + \frac{1}{48}) \\ = (\frac{1}{30} - \frac{11}{220}) \\ = - \frac{1}{80} [- Nagetive sign means emptying] \\ ∴ Volume of \frac{1}{80} part  =  3 galons \\ Volume of whole tank \\ = (3 \times 80) gallons \\ =  240 gallons \end{aligned}

33. An outlet pipe can empty a cistern in 3 hours. In what time will empty $\frac{2}{3}$ of the cistern?

A. 3 hours

B. 5 hours

C. 2 hours

D. 4 hours

Discuss

Solution:

The outlet pipe empties the one complete cistern in 3 hours
Time taken to empty

\frac{2}{3}

Part of the cistern

\begin{aligned} = \frac{2}{3} \times 3 \\ = 2 hours \end{aligned}

34. A tank is 7 metre long and 4 meter wide wide. At what speed should water run through a pipe 5 cm broad and 4 cm deep so that in 6 hours and 18 minutes water level in the tank rises by 4.5 meter?

A. 10 km/hours

B. 12 km/hours

C. 8 km/hours

D. None of these

Discuss

Solution:

Rate of flow of water = x cm/minute
∴ Volume of water that flowed in the in 1 minutes
= (5 × 4 × x) = 20 x cu.cm.
∴ Volume of water that flowed in the tank in 6 hours 18 minutes.
i.e. (6 × 60 + 18) = 378 minutes
= 2x × 378 cu. cm.
According to question,

\begin{aligned} 20 x \times 378 = 700 \times 400 \times 450 \\ \Rightarrow x = (\frac{700 \times 400 \times 450}{20 \times 378}) cm /minutes \\ \Rightarrow x = (\frac{700 \times 400 \times 450 \times 60}{100000 \times 20 \times 378}) km/hours \\ \Rightarrow x =  10 km/hours \end{aligned}

35. Two pipes can fill a tank in 12 hours and 16 hours respectively. A third pipe can empty the tank in 30 hours. If all three pipes are opened and functions simultaneously, how much time will the tank take to be full?( in hours )

A. $10 \frac{4}{9}$

B. $9 \frac{1}{2}$

C. $8 \frac{8}{9}$

D. $7 \frac{2}{9}$

Discuss

Solution:

First pipe fill the tank in 1 hour =

\frac{1}{12}

part of tank
Second pipe fill the tank in 1 hour =

\frac{1}{16}

part of tank
Third pipe empty the tank in 1 hour =

\frac{1}{30}

part of tank
When all three pipes are opened simultaneously, part of the tank filled in 1 hour

= \frac{1}{12} + \frac{1}{16} - \frac{1}{30}

LCM of 12, 16 and 30 = 240

\begin{aligned} = \frac{20 + 15 - 8}{240} \\ = \frac{27}{240} \end{aligned}

∴ Required time taken by all the three pipes

= \frac{240}{27} = \frac{80}{9} = 8 \frac{8}{9} Hours

36. A tap can completely fill a water tank in 8 hours. The water tank has a hole in it through which the water leaks out. The leakage will cause the full water tank to get empty in 12 hours. How much time will it take for the tap the the tank completely with the hole?

A. 16 hours

B. 18 hours

C. 24 hours

D. None of these

Discuss

Solution:

Net part filled in 1 hour

\begin{aligned} = (\frac{1}{8} - \frac{1}{12}) = \frac{1}{24} \end{aligned}

∴ The tank will be filled in 24 hours

37. A tank can be filled by pipe A in 2 hours and pipe B in 6 hours. At 10 A.M. pipe A was opened. At what time will the tank be filled if pipe B is opened at 11 A.M. ?

A. 12.45 A.M.

B. 5 P.M.

C. 11.45 A.M.

D. 12 P.M.

Discuss

Solution:

Pipe A will fill 3 units till 11 A.M. Remaining capacity
= 6 - 3
= 3 units
Now both pipes will fill the tank in

\frac{Total Capacity}{Efficiency} = \frac{3}{(3 + 1)} = \frac{3}{4} hours

So,

(11 + \frac{3}{4})

A.M., tank will be filled = 11.45 A.M.

38. A water tank can be filled by a tap in 30 minutes and another tap can fill it in 60 minutes. If both taps are kept open for 5 minutes and then the first tap is closed, how long will it take foe the tank to be filled ?

A. 20 minutes

B. 25 minutes

C. 30 minutes

D. 45 minutes

Discuss

Solution:

(A + B)'s filling (2 + 1) = 3 units/min
In 5 minutes they will fill 3 × 5 = 15 units
Remaining capacity = 60 - 15 = 45 units
Second pipe (B) fills it in

\begin{aligned} = \frac{Remaining capacity}{efficiency of B} \\ = \frac{45}{1} \\ = 45 minutes \end{aligned}

39. A pipe can fill a tank in x hours and another can empty it in y hours. In how many hours they together fill it in ( y > x) ?

A. (x - y) hours

B. (y - x) hours

C. $(\frac{x y}{x - y}) hours$

D. $(\frac{x y}{y - x}) hours$

Discuss

Solution:

Time will be taken by both of them to fill the tank

= \frac{x y}{y - x}

40. A water tap fills a tub in 'p' hours and a sink at the bottom empties it in 'q' hours. If p < q and both tap and sink are opened the tank is filled in 'r' hours, then the relation between p, q, r :

A. $\frac{1}{r} = \frac{1}{p} + \frac{1}{q}$

B. $\frac{1}{r} = \frac{1}{p} - \frac{1}{q}$

C. $r = p + q$

D. $r = p - q$

Discuss

Solution:

Net efficiency = q - p (∵ q > p)
Time required

\begin{aligned} r = \frac{p q}{q - p} \\ o r \frac{1}{r} = \frac{1}{p} - \frac{1}{q} \end{aligned}

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