Time taken by both Meghana and Ganesh to work together is given by =
Where
Therefore, time took by both to work together =
Let workdone 1 boy in 1 day be b
and that of 1 girl be g
From the given data,
4(5b + 3g) = 23
20b + 12g = 23 .......(a)
2(3b + 2g) = 7
6b + 4g = 7 ........(b)
Solving (a) & (b), we get
b = 1, g = 1/4
Let number og girls required be 'p'
6(7 x 1 + p x 1/4) = 45
=> p = 2.
Hence, number of girls required = 2
Given the ratio of efficiencies of P, Q & R are 3 : 8 : 5
Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively
They can do work for 12 days.
=> Total work = 12 x 16x = 192x
Now, the required time taken by Q to complete the job alone = days.
Amount of work K can do in 1 day = 1/16
Amount of work L can do in 1 day = 1/12
Amount of work K, L and M can together do in 1 day = 1/4
Amount of work M can do in 1 day = 1/4 - (1/16 + 1/12) = 3/16 ? 1/12 = 5/48
=> Hence M can do the job on 48/5 days = 9 (3/5) days
Work done by P and Q in the first two hours, working alternately
= First hour P + Second hour Q
work is completed in 2 hours
Then, the total time required to complete the work by P and Q working alternately=2 x 3= 6hours
Thus, work will be completed at 3pm.
Given,
P can fill in 12 hrs
Q can fill in 15 hrs
R can fill in 20 hrs
=> Volume of tank = LCM of 12, 15, 20 = 60 lit
=> P alone can fill the tank in 60/12 = 5 hrs
=> Q alone can fill the tank in 60/15 = 4 hrs
=> R alone can fill the tank in 60/20 = 3 hrs
Tank can be filled in the way that
(P+Q) + (P+R) + (P+Q) + (P+R) + ....
=> Tank filled in 2 hrs = (5+4) + (5+3) = 9 + 8 = 17 lit
=> In 6 hrs = 17 x 6/2 = 51 lit
=> In 7th hr = 51 + (5+4) = 51 + 9 = 60 lit
=> So, total tank will be filled in 7 hrs.
Let x liter be the per day filling and v litr be the capacity of the reservoir, then
90x + v = 40000 * 90 -----(1)
60x + v= 32000 * 60 ------(2)
solving eq.(1) and (2) , we get
x = 56000
Hence , 56000 liters per day can be used without the failure of supply.
Let the number of days be 'p'
As the work is same, we know that
Where M = Men, D = Days, H = Hours per day
Here M1 = 9, D1 = 15, H1 = 7
M2 = 6, D2 = p, H2 = 9
=> 9 x 15 x 7 = 6 x p x 9
=> p = 35/2 = 17.5 days.
We know that,
Here M1 = 1, D1 = 6 min, W1 = 1 and M2 = M, D2 = 90 min, W2 = 1845
=> M = 123
From the given data,
12 children 16 days work,
One child?s one day work = 1/192.
8 adults 12 days work,
One adult?s one day?s work = 1/96.
Work done in 3 days = ((1/96) x 16 x 3) = 1/2
Remaining work = 1 ? 1/2 = 1/2
(6 adults+ 4 children)?s 1 day?s work = 6/96 + 4/192 = 1/12
1/12 work is done by them in 1 day.
1/2 work is done by them in 12 x (1/2) = 6 days.
work done=total number of person x number of days;
half of work done = 140 x 66;
For half of remaining work 25 extra men are added.
Therefore, total men for half work remaining = 140 + 25 = 165;
Let 2nd half work will be completed in K days;
140 x 66 = 165 x K
K = 122;
Hence, extra days => 122-120 = 2days.
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