A) 64000 liters | B) 56000 liters |

C) 78000 liters | D) 60000 liters |

Explanation:

Let x liter be the per day filling and v litr be the capacity of the reservoir, then

90x + v = 40000 * 90 -----(1)

60x + v= 32000 * 60 ------(2)

solving eq.(1) and (2) , we get

x = 56000

Hence , 56000 liters per day can be used without the failure of supply.

A) 4 | B) 3 |

C) 2 | D) 1 |

Explanation:

Let workdone 1 boy in 1 day be b

and that of 1 girl be g

From the given data,

4(5b + 3g) = 23

**20b + 12g = 23 .......(a)**

2(3b + 2g) = 7

**6b + 4g = 7 ........(b)**

Solving (a) & (b), we get

**b = 1, g = 1/4**

Let number og girls required be 'p'

6(7 x 1 + p x 1/4) = 45

=> p = 2.

Hence, **number of girls required = 2**

A) 80 | B) 8 |

C) 0.8 | D) 0.08 |

Explanation:

Given for year = 70000

=> 365 days = 70000

=> 365 x 24 hours = 70000

=> 1 hour = ?

70000/365x24 = 7.990 = 8

A) 1/5 | B) 1/6 |

C) 1/7 | D) 1/8 |

Explanation:

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 =** 1/6.**

A) 215 days | B) 225 days |

C) 235 days | D) 240 days |

Explanation:

Given that

**(10M + 15W) x 6 days = 1M x 100 days**

=> 60M + 90W = 100M

=> 40M = 90W

=> **4M = 9W.**

From the given data,

1M can do the work in 100 days

=> 4M can do the same work in 100/4= 25 days.

=> 9W can do the same work in 25 days.

=> 1W can do the same work in **25 x 9 = 225 days.**

Hence, 1 woman can do the same work in** 225 days.**

Given A,B,C can complete a work in 15,20 and 30 respectively.

The total work is given by the LCM of 15, 20, 30 i.e, 60.

A's 1 day work = 60/15 = 4 units

B's 1 day work = 60/20 = 3 units

C's 1 day work = 60/30 = 2 units

(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units

Let B + C worked for x days = (3 + 2) x = 5x units

C worked for 2 days = 2 x 2 = 4 units

Then, 18 + 5x + 4 = 60

22 + 5x = 60

5x = 38

x = 7.6

Therefore, total number of days taken to complete the work **= 2 + 7.6 + 2 = 11.6 = 11 3/5 days.**

A) 4 | B) 5 |

C) 3 | D) 6 |

Explanation:

Let M, N and O worked together for x days.

From the given data,

M alone worked for 8 days

M,N,O worked for x days

N, O worked for 1 day

But given that

M alone can complete the work in 18 days

N alone can complete the work in 36 days

O alone can complete the work in 54 days

The total work can be the LCM of 18, 6, 54 = 108 units

M's 1 day work = 108/18 = 6 units

N's 1 day work = 108/36 = 3 units

O's 1 day work = 108/54 = 2 units

Now, the equation is

8 x 6 + 11x + 5 x 1 = 108

48 + 11x + 5 = 108

11x = 103 - 48

11x = 55

x = 5 days.

Hence, all M,N and O together worked for 5 days.

A) 36 days | B) 30 days |

C) 24 days | D) 22 days |

Explanation:

Given the ratio of efficiencies of P, Q & R are **3 : 8 : 5**

Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively

They can do work for 12 days.

=> Total work = **12 x 16x = 192x**

Now, the required time taken by Q to complete the job alone = $\frac{\mathbf{162}\mathbf{x}}{\mathbf{8}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{24}\mathbf{}$days.

A) 6 | B) 4 |

C) 2 | D) 3 |

Explanation:

Let work done by 1 man in i day be m

and Let work done by 1 boy in 1 day be b

From the given data,

4(5m + 3b) = 23

20m + 12b = 23....(1)

2(3m + 2b) = 7

6m + 4b = 7 ....(2)

By solving (1) & (2), we get

m = 1, b = 1/4

Let the number of required boys = n

6(7 1 + n x 1/4) = 45

=> n = 2.