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) 20 hrs | B) 18 hrs |

C) 22 hrs | D) 20.5 hrs |

Explanation:

Time taken by both Meghana and Ganesh to work together is given by =

$\sqrt{{t}_{1}x{t}_{2}}$

Where

${t}_{1}=32hrs\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{t}_{2}=12\frac{1}{2}hrs=\frac{25}{2}hrs$

Therefore, time took by both to work together =

$\sqrt{32x\frac{25}{2}}=\sqrt{16x25}=4x5=20hrs$

A) 96 days | B) 48 days |

C) 24 days | D) 12 days |

Explanation:

Let number of days Charan can do the same work alone is 'd' days.

According to the given data,

$\frac{\mathbf{5}}{\mathbf{12}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{7}}{\mathbf{16}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{14}}{\mathbf{d}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{14}{\mathrm{d}}=1-\frac{\left(20+21\right)}{48}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{14}{\mathrm{d}}=\frac{48-41}{48}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{d}=\frac{14\mathrm{x}48}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{96}\mathbf{}\mathbf{days}$

Therefore, Charan alone can complete the work in **96 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**