A) 10 min. 20 sec. | B) 11 min. 45 sec. |

C) 12 min. 30 sec. | D) 14 min. 40 sec. |

Explanation:

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 : 8/15 :: 1:x

x = (8/15*1*20) = $10\frac{2}{3}min=10min40sec$

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

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.

A) 200 hrs | B) 240 hrs |

C) 300 hrs | D) 270 hrs |

Explanation:

Volume of water collected in the tank in 1 hour

⇒ (0.3 × 0.2 × 20km × 1000mts) = 1200 m cubic

If after t hours, the water is at height of 12m,

1200t=200×150×12

⇒ t = 300 Hours.

A) 32.5 hrs | B) 29.25 hrs |

C) 30.30 hrs | D) 31 hrs |

Explanation:

Let the leak will empty the tank in x hrs

Then, 1/9 - 1/x = 1/13

x = 29.25 hrs

A) 4 hrs 15 min | B) 3 hrs 24 min |

C) 4 hrs 51 min | D) 3 hrs 45 min |

Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the taps in 1 hour = 4 x 1/6 = 2/3

Remaining part = 1 - 1/2 = 1/2

2/3 : 1/2 :: 1 : p

p = 1/2 x 1 x 3/2 = 3/4 hrs. i.e., 45 min

So, total time taken = 3 hrs 45 min.

A) 31 min | B) 29 min |

C) 28 min | D) 30 min |

Explanation:

Part filled by (A + B) in 1 minute = (1/60 + 1/40) = 1/24

Suppose the tank is filled in x minutes.

Then, x/2(1/24 + 1/40) = 1

(x/2) * (1/15) = 1 => x = 30 min.