**Sol : **In this kind of questions we find the work force required to complete the work in 1 day (or given unit of time) then we equate the work force to find the relationship between the efficiencies (or work rate) between the different workers.

Therefore, 6B+8G = 6 days

=> 6(6B+8G)= 1 day (inversely proportional)

=> 36B+48G =1 ( by unitary method)

Again 14B+10G = 4days

=> 56B+40G =1

so, here it is clear that either we employ 36B and 48G to finish the work in 1 day or 56B and 40G to finish the same job in 1 day. thus , we can say

=> 36B+48G = 56B+40G

=> G= 2.5B

Thus a Girl is 2.5 times as efficient as a boy.

Now, since 36B+48G = 1

=> 36B+48(2.5 B)=1

=>156B=1

i.e., to finish the job in 1 day 156 boys are required or the amount of work is 156 boys-days

Again 1G+1B=2.5B+1B=3.5B

Now, since 156 boys can finish the job in 1 day

so 1 boy can finish the job in 156 days

Therefore, 3.5 boys can finish the job in $\frac{1\times 156}{3.5}=\frac{4}{7}$ days.

A) 6 days | B) 8 days |

C) 11 days | D) 13 days |

Explanation:

M = 10 days

The ratio of efficiency of M & N are 3 : 2

Hence, the time rquired for N alone = 15 days

=> Required time taken t=by both to complete the work = M x N / M + N

= 10 x 15/ 10 + 15

= 150/25

= 6 days.

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.**