Total work is given by L.C.M of 72, 48, 36
Total work = 144 units
Efficieny of A = 144/72 = 2 units/day
Efficieny of B = 144/48 = 3 units/day
Efficieny of C = 144/36 = 4 units/day
According to the given data,
2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100
3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54
p = 198/16.5
p = 12 days.
Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day
(C+D) in p days = (4 + 6) x 12 = 120 unit
Remained part of work = (144-120)/144 = 1/6.
4/10 + 9/x = 1 => x = 15
Then both can do in
1/10 + 1/15 = 1/6 => 6 days
(5m + 9w)10 = (6m + 12w)8
=> 50m + 90w = 48w + 96 w => 1m = 3w
5m + 9w = 5m + 3m = 8m
8 men can do the work in 10 days.
3m +3w = 3m + 1w = 4m
So, 4 men can do the work in (10 x 8)/4 = 20 days.
K's work in a day(1st day) = 1/30
L's work in a day(2nd day)= -1/60(demolishing)
hence in 2 days, combined work= 1/30 - 1/60 =1/60
since both works alternatively, K will work in odd days and L will work in even days.
1/60 unit work is done in 2 days
58/60 unit work will be done in 2 x 58 days = 116 days
Remaining work = 1-58/60 = 1/30
Next day, it will be K's turn and K will finish the remaining 1/30 work.
hence total days = 116 + 1 = 117.
Girl eats 112 chocolates in 30 sec
so she can eat in 12 sec is 12 x 112/30 = 44.8 chocolates.
Her boy friend can eat one-half of 112 in twice of 30 sec
so he can eat 56 in 60 sec
Then he can eat in 12 sec is 56 x 12/60 = 11.2 chocolates.
Hence, together they can eat = 44.8 + 11.2 =56 chocolates in 12 seconds.
Let the total women in the group be 'W'
Then according to the given data,
W x 20 = (W-12) x 32
=> W = 32
Therefore, the total number of women in the group = 32
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.
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