Rohit crosses a 700 metres wide road in 35 seconds. What is his speed (in km/hr)?

 

Let M, N and O worked together for x days.

From the given data,
M alone worked for 8 days
M,N,O worked for x days
N, O worked for 1 day

But given that
M alone can complete the work in 18 days
N alone can complete the work in 36 days
O alone can complete the work in 54 days

The total work can be the LCM of 18, 6, 54 = 108 units

M's 1 day work = 108/18 = 6 units
N's 1 day work = 108/36 = 3 units
O's 1 day work = 108/54 = 2 units

Now, the equation is
8 x 6 + 11x + 5 x 1 = 108
48 + 11x + 5 = 108
11x = 103 - 48
11x = 55
x = 5 days.

Hence, all M,N and O together worked for 5 days.

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 = 162x8x = 24days.

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.

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.

Work done by Shyam and Rahim in 8 days = 8/32 = 1/4

Remaining work to be done by Shyam and Ram = 1 - 1/4 = 3/4

Given efficieny of Ram is half of Rahim i.e, as Rahim can do the work in 48 days, Ram can do the work in 24 days.

One day work of Ram and Shyam = (1/32 - 1/48) + 1/24 = 5/96

Hence, the total work can be done by Shyam and Ram together in 96/5 days.

Therefore, remaining work 3/4 can be done by them in 3/4 x 96/5 = 72/5 = 14.4 days.

Given 4men, 12 women and 20 children work for  2 days.

Workdone for 2 days by 4men, 12 women and 20 children = 4 x 26 x 12 + 12 x 28 x 18 + 20 x 218 x 10 = 12

Therefore, remaining work = 1 - 12 = 12

To complete the same work by only men in 1 day,

We know that M1 x D1 = M2 x D2

Here M1 = 6 , D1 = 12 and M2 = M , D2 = 1

12 x 6 = M x 1

=> M = 12 x 6 = 72

=> But the remaining work = 1/2

Men required => 1/2 x 72 = 36

Only men required to Complete the remaining work in 1 day = 36.

Given (3 Men + 4 Women + 6 Children) -----> 9 days

But W = 2M and C = M/2

Now, convert Men and Children into Women by

$$\begin{gathered} 3W2 + 4W + 6W4 \\ => 32 + 4 + 32 = 7 Women \end{gathered}$$ 

Therefore, 7 women alone can complete this work in 9 days.

Let the number of days be 'p' 

As the work is same, we know that

M1 x D1 x H1 = M2 x D2 x H2

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.