A) 12 days | B) 14 days |

C) 13 days | D) 16 days |

Explanation:

Now, Total work = LCM(16, 8) = 48

A's one day work = + 48/16 = + 3

B's one day work = - 48/8 = -6

Given A worked for 5 days to build the wall => 5 days work = 5 x 3 = + 15

2days B joined with A in working = 2(3 - 6) = - 6

Remaining Work of building wall = 48 - (15 - 6) = 39

Now this remaining work will be done by A in = 39/3 = 13 days.

A) 3 hrs | B) 3.5 hrs |

C) 2.5 hrs | D) 2 hrs |

Explanation:

Given Prabhas is twice as good a workman as Rana.

Prabhas finishes the work in 3 hrs

=> Rana finishes the work in 6 hrs.

Number of hours required together they could finish the work

**= 3 x 6 / 3 + 6 **

**= 18/9 **

**= 2 hrs.**

A) 2 hrs 24 min | B) 2 hrs 44 min |

C) 1 hrs 24 min | D) 1 hrs 24 min |

Explanation:

Given that three athletes can complete one round around a circular field in 16, 24 and 36 min respectively.

Now, required time after which they met for the first time = LCM of (16, 24 & 36) min

Now, LCM of 16, 24, 36 = 144 minutes = 2 hrs 24 min.

A) 10 | B) 12 |

C) 13 | D) 15 |

Explanation:

One day work of Raghu and Sam together = 1/12 + 1/15 = 9/60 = 3/20

Aru efficiency = 2/3 of (Raghu + Sam)

Number of days required for Aru to do thw work alone = 3/2 x 20/3 = 10 days.

A) 5 | B) 4 |

C) 3 | D) 2 |

Explanation:

Let p be the required number of days.

From the given data,

**12 x 14 = 12 x 5 + (12+3) x p**

12 x 14 = 12[5 + 3p]

14 = 5 + 3p

3p = 9

p = 3 days.

Hence, more **3 days** all of them take to complete the remaining work.

A) 19 : 7 | B) 30 : 19 |

C) 8 : 15 | D) 31 : 17 |

Explanation:

Given Lasya can do a work in 16 days.

Time taken by Rashmi = n days

=> (12 x 20)/(20 - 12) = (12 x 20)/n

=> n= 30 days.

Required ratio of efficiencies of Rashmi and Lasya = 1/30 :: 1/16 = **8 : 15.**

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.**