A) 7 hours | B) 8 hours |

C) 12 hours | D) 14 hours |

Explanation:

If the total area of pump=1 part

The pumop take 2 hrs to fill 1 part

The pumop take1 hour to fill 1/2 portion

Due to lickage

The pumop take 7/3 hrs to fill 1 part

The pumop take1 hour to fill 3/7 portion

Now the difference of area = (1/2-3/7)=1/14

This 1/14 part of water drains in 1 hour

Total area=1 part of water drains in (1x14/1)hours= 14 hours

So the leak can drain all the water of the tank in 14 hours.

Leak will empty the tank in 14 hrs

A) 6 min | B) 8 min |

C) 12 min | D) 10 min |

Explanation:

Let pipe A takes p min to fill

Then,

pipe B takes 3p min to fill

=> 3p - p = 32

=> p = 16 min => 3p = 48 min

Required, both pipes to fill = (48 x 16)/(48 + 16) min = 12 min.

A) 16 min | B) 12 min |

C) 14 min | D) 20 min |

Explanation:

Let the efficiencies of filling pipes is 4p and 5p respectively.

Efficiency of pipe which empty the tank = 2/3 x 9p/2 = 3p

Total work = 3p x 36 = 108p

Time to fill the tank by both the pipes = 108p/9p = 12 min.

A) 26 min | B) 32 min |

C) 36 min | D) 42 min |

Explanation:

Given that the diameters of the three pipes are 2 cm, 3 cm and 4 cm

From the given data,

Amount of water from three pipes is 4 units, 9 units and 16 units.

Let the capacity of cistern be 'p' units.

∴ p/58 = 16

⇒ p = 928 units.

In 1 minute, quantity to be filled by 3 pipes = 29 units

∴ Total time required = 928/29 = **32 minutes.**

A) 34 lit/min | B) 25 lit/min |

C) 22 lit/min | D) 18 lit/min |

Explanation:

Given that the waste tap can empty the filled tank in **32 min.**

Now, the rate at which the waste tap can empty the tank = (40 + 60)8/32 = 100/4 = **25 lit/min**.

A) 46 cub.m/min | B) 44 cub.m/min |

C) 48 cub.m/min | D) 50 cub.m/min |

Explanation:

Let filling capacity of the pump be **'F'** ${\mathbf{m}}^{\mathbf{3}}\mathbf{/}\mathbf{min}$

Then, the filling capacity of the pump will be **(F + 10)** ${\mathbf{m}}^{\mathbf{3}}\mathbf{/}\mathbf{min}$

From the given data,

$\frac{\mathbf{2400}}{\mathbf{F}}\mathbf{}\mathbf{-}\mathbf{}\frac{\mathbf{2400}}{\mathbf{F}\mathbf{+}\mathbf{10}}\mathbf{=}\mathbf{}\mathbf{8}\phantom{\rule{0ex}{0ex}}\Rightarrow 2400\left[\frac{\mathrm{F}+10-\mathrm{F}}{\mathrm{F}(\mathrm{F}+10)}\right]=8\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{F}}^{2}+10\mathrm{F}-3000=0\phantom{\rule{0ex}{0ex}}\mathbf{\Rightarrow}\mathbf{}\mathbf{F}\mathbf{}\mathbf{=}\mathbf{}\mathbf{50}\mathbf{}{\mathbf{m}}^{\mathbf{3}}\mathbf{/}\mathbf{min}$

A) 34 hrs | B) 36 hrs |

C) 38 hrs | D) 40 hrs |

Explanation:

The time taken by the leak to empty the tank = $\frac{\mathbf{1}}{\mathbf{8}}\mathbf{}\mathbf{-}\mathbf{}\frac{\mathbf{1}}{\mathbf{10}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{5}\mathbf{}\mathbf{-}\mathbf{}\mathbf{4}}{\mathbf{40}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{40}}$

Therefore, the leak empties the tank in **40 hours.**

A) 70 lit | B) 170 lit |

C) 90 lit | D) 190 lit |

Explanation:

Given A alone can fill the tank of capacity 240 lit in 16 hrs.

=> A can fill in 1 hr = 240/16 = 15 lit

=> B alone can fill the tank of capacity 240 lit in 12 hrs.

=> B can fill in 1 hr = 240/12 = 20 lit

Now, (A + B) in 1 hr = 15 + 20 = 35 lit

But they are opened for 2 hrs

=> 2 x 35 = 70 lit rae filled

Remaining water to be filled in tank of 240 lit = 240 - 70 = 170 lit.