From figures (i) and (ii) we conclude that the number 1, 2, 3 and 4 appear adjacent to 6. Thus, the number 5 will appear opposite 6. Therefore, when six is at the bottom, then 5 will be at the top.

We shall assume the dice in fig. (ii) to be rotated so that the 5 dots appear at the same position as in fig. (i) i.e. on RHS face (i.e. on face II as per activity 1) and 1 dot appears at the same position as in fig; (i) i.e. on Front face (i.e. on face I). Then, from the, two figures, 2 dots appear on the top face (i.e. on face V) and 4 dots appear on the Bottom face (i.e. on face VI).

Since, these two faces are opposite to each other, therefore, two dots are contained on the face opposite to that containing four dots.

Let C.P. be Rs. x.

Then,= >1920-xx*100=x-1280x*100

=>  1920 - x = x - 1280

=>  2x = 3200

=>  x = 1600

 Required S.P. = 125% of Rs. 1600 =Rs(125/100*1600) = Rs2000

If the total area of pump=1 part

The pumop take 2 hrs to fill 1 part 

The pumop take1 hour to fill 1/2 portion 

Due to lickage  

The pumop take 7/3 hrs to fill 1 part 

The pumop take1 hour to fill 3/7 portion 

Now the difference of area = (1/2-3/7)=1/14 

This 1/14 part of water drains in 1 hour 

Total area=1 part of water drains in (1x14/1)hours= 14 hours 

So the leak can drain all the water of the tank in 14 hours. 

Leak will empty the tank in 14 hrs

Time taken by one tap to fill half of the tank = 3 hrs.

  Part filled by the four taps in 1 hour =4*1/6 =2/3

Remaining part =1-12=12

23:12::1:x

=> x = 12*1*32=34

So, total time taken = 3 hrs. 45 mins.

Part filled in 2 hours = 2/6=1/3

Remaining part =1-13=23

 (A + B)'s 7 hour's work = 2/3

(A + B)'s 1 hour's work = 2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=1/6-2/21 = 1/14

C alone can fill the tank in 14 hours.

Let the required number of days be x.

Less cows, More days (Indirect Proportion)

Less bags, Less days (Direct Proportion)

$$\begin{gathered} cows 1:40 :40:x \\ bags 40:1 :40:x \\ \end{gathered}$$

1 x 40 x x = 40 x 1 x 40

=>  x = 40.

There is a meal for 200 children. 150 children have taken the meal.

Remaining meal is to be catered to 50 children.

Now, 200 children = 120 men.

50 children = 12020050 =30men