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.
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= =
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.
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.
L's 10 days work =
Remaining work =
Now, work is done by K in one day = 1/18
1/3 work is done by K in = 6 days
we know that m1 x d1 = m2 x d2
=> 4 x 6 = 5 x d
where d = no. of days taken by 5 men to paint
d = 24/5 days.
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