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.
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.
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.
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.
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.
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 =
So, both Raghu and Arun will finish the work in
Number of days of 12 hours each= =
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,
Therefore, Charan alone can complete the work in 96 days.
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.
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