Let the number of workers be x.

Now, Using work equivalence method,

X + (X-1) + (X-2)+ . . . . + 1 = X *55% of X

=> [X * (X+1)] / 2 = X * (55X/100)    [because, Series is in AP. Sum of AP = {No. of terms (first term+ last term)/2} ]

Therefore, X = 10 

Since 20% i.e 1/5 typists left the job. So, there can be any value which is multiple of 5 i.e, whose 20% is always an integer. Hence, 5 is the least possible value.

Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days.

D * x +(100- D) * 2x= 175x 

=>  D= 25 days 

Now , the work done in 25 days = 25x 

Total work = 175x

Therefore, workdone before increasing the no of workers = 25x175x×100 % = 1427%

T                 C              B

16              10             15

8                12             12

128            120           180             <------- in one hour

1280          1200         1800            <------- in 10 hours

Since, restriction is imposed by composers i.e,since only 1200 books can be composed i 10 hours so not more than 1200 books can be finally pepared.

Efficiency of kaushalya = 5%

Efficiency of kaikeyi  = 4%

Thus, in 10 days working together they will complete only 90% of the work.

          [(5+4)*10] =90

Hence, the remaining work will surely done by sumitra, which is 10%.

Thus, sumitra will get 10% of Rs. 700, which is Rs.70

                       A   :   C    

Efficiency      5    :   3    

No of days   3x   :  5x     

Given that, 5x-6 =3x  => x = 3  

Number of days taken by A = 9  

Number of days taken by C = 15     

           B  :  C    

Days   2  :  3  

Therefore, Number of days taken by B = 10   

Work done by B and C in initial 2 days = 2110+115= 1/3  

Thus,  Rest work =2/3  

Number of days required by A to finish 2/3 work = (2/3) x 9 = 6 days

                                                A    :    B    :    C   

Ratio of efficiency               3     :    1    :    2   

Ratio of No.of days            1/3  :   1/1  :   1/2    

     or                                       2    :    6    :    3   

Hence A is correct.  

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 1×1563.5= 47 days.