Efficiency of P= 100/20= 5% per hour  

Efficiency of Q= 100/25= 4% per hour  

Efficiency of R= 100/40= 2.5% per hour  

Efficiency of S=100/50= 2% per hour 

Cistern filled till 10 am by P, Q and R  

Till 10.00am Pipe P filled 20%Till 10.00am Pipe Q filled 8%Till 10.00am Pipe R filled 2.5%30.5%  

Thus, at 10 am pipe P,Q and R filled 30.5% of the cistern. 

Rest of cistern to be filled = 100 - 30.5 = 69.5%  

Now, the time taken by P,Q,R and S together to fill the remaining capacity of the cistern 

= 69.5 / (5+4+2.5+2) = 5 Hours and 9 minutes(approx).

Therefore, total time =4 hrs + 5hrs 9 mins = 9 hrs and 9 mins

It means cistern will be filled up at 3:09 pm

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 

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

34×(x-2)x=(x+7)(x-10)

⇒x2-6x-280 =0 

=> x= 20   and   x=-14

 so, the acceptable values is x=20 

Therefore, Total work =(x-2)x = 18 x 20 =360 unit

 Now   360 = 30 x k         

=> k=12 days

A + B= 70%   

B + C =50%   

 ∵ (A+B)+(B+C)-(A+B+C)= B

=> B= 20%     A= 50%        and   C=30%

Hence A is most efficient

                       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

Let the initial amount of honey in the jar was K, then 

 512=K1-154    ∵ 20% =20100=15

        or      

512=K454

Therefore, K = 1250

Hence initially the honey in the jar= 1.25 kg