Time and Work problems

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

Time taken by both Meghana and Ganesh to work together is given by =

t 1 x t 2

Where

$$\begin{gathered} t 1 = 32 h r s \\ t 2 = 12 1 2 h r s = 25 2 h r s \end{gathered}$$

Therefore, time took by both to work together = 

32 x 25 2 = 16 x 25 = 4 x 5 = 20 h r s

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

M 1 x D 1 x H 1 = M 2 x D 2 x H 2

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,  M 1 × D 1 W 1 = M 2 × D 2 W 2

Here M1 = 1, D1 = 6 min, W1 = 1 and M2 = M, D2 = 90 min, W2 = 1845

1 × 6 1 = M × 90 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.

Given 2000 ---- 54 days

The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days.

i.e, 2000 ----- 39 days == m ---- 20 days

=> m x 20 = 2000 x 39

m = 3900

So total men for 20 days is 3900

=> 2000 old and 1900 new reinforcement.

Hence, reinforcement = 1900.

The mother completes the job in x hours.
So, the daughter will take 2x hours to complete the same job.

In an hour, the mother will complete 1/x of the total job.
In an hour, the daughter will complete 1/2x of the total job.

So, if the mother and daughter work together, in an hour they will complete 1/x + 1/2x of the job.
i.e., in an hour they will complete 3/2x of the job.

The question states that they complete the entire task in 6 hours if they work together.
i.e., they complete 1/6 th of the task in an hour.

Equating the two information, we get 3/2x = 1/6
By solving for x, we get 2x = 18 or x = 9.

The mother takes 9 hours to complete the job.

One day work of Raghu and Sam together = 1/12 + 1/15 = 9/60 = 3/20

Aru efficiency = 2/3 of (Raghu + Sam)

Number of days required for Aru to do thw work alone = 3/2 x 20/3 = 10 days.