Time and Work problems

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.

Ratio of efficiencies of P, Q and R = 2 : 3 : 4

From the given data,

Number of working days of P, Q, R = 5 : 10 : 5

Hence, ratio of amount of p, Q, R = 2x5 : 3x10 : 4x5 = 10 : 30 : 20

Amounts of P, Q, R = 200, 600 and 400.

Total Water reqduired = 5000 × 150 lit = 750,000 litres = 750 cu.m.

Volume of tank = 20 × 15 × 5 = 1500 Cu.m.

Number of days required =1500/750 = 2 days.

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 work done by 1 man in i day be m

and Let work done by 1 boy in 1 day be b

From the given data,

4(5m + 3b) = 23

20m + 12b = 23....(1)

2(3m + 2b) = 7

6m + 4b = 7 ....(2)

By solving (1) & (2), we get

m = 1, b = 1/4

Let the number of required boys = n

6(7 1 + n x 1/4) = 45

=> n = 2.

Raghu can complete the work in (12 x 9)hrs = 108 hrs.

 Arun can complete the work in (8 x 11)hrs = 88 hrs.

Raghu's 1 hrs work = 1/108 and Arun's 1 hrs work = 1/88

(Raghu + Arun)'s 1 hrs work =  1 108 + 1 88 = 49 2376

 
So, both Raghu and Arun will finish the work in  2376 49 h r s

Number of days of 12 hours each= 2376 49 × 1 12 =  198 49 = 4 3 49 d a y s

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.

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

According to the given data,

$$\begin{gathered} 5 12 + 7 16 + 14 d = 1 \\ 14 d = 1 - 20 + 21 48 \\ 14 d = 48 - 41 48 \\ d = 14 x 48 7 \\ d = 96 days \end{gathered}$$

Therefore, Charan alone can complete the work in 96 days.