A) 15 | B) 30 |

C) 25 | D) 10 |

Explanation:

Combined efficiency of all the three boats = 60 passenger/trip

Now, consider option(a)

15 trips and 150 passengers means efficiency of B1 = 10 passenger/trip

which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10-5 = 5) in which B2 and B3 together carried remaining 250 (300 - 50 = 250) Passengers.

Therefore the efficiency of B2 and B3 = 250/5 = 50 passenger/trip

Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct.

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.**

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.**