Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135)/2= Rs.130.50

So, Now the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say Rs. 'x' per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find 'x'.

Cost of 1 kg tea of 1st kind         Cost of 1 kg tea of 2nd kind 

x-153/22.50 = 1  => x - 153 = 22.50  => x=175.50.

 Hence, price of the third variety = Rs.175.50 per kg.

Suppose the vessel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.
Quantity of water in new mixture =   3 - 3 x 8 + xlitres.

Quantity of syrup in new mixture =   5 - 5 x 8 litres.

   3 - 3 x 8 + x = 5 - 5 x 8

=>   5x + 24 = 40 - 5x 

=>   10x = 16    => x = 8/5

So, part of the mixture replaced =  8 5 × 1 8 = 1/5.

Therefore x = 21%

Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters

Let us Assume that another 'P' liters of water are added to the mixture to make water 25% of the new mixture. So, the total amount of water becomes (30 + P) and the total volume of the mixture becomes (150 + P)

Thus, (30 + P) = 25% of (150 + P)

Solving, we get P = 10 liters

By the rule of alligation:

C.P. of 1 kg sugar of 1st kind      C.P. of 1 kg sugar of 2nd kind 

Therefore, Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3. 

Let x kg of sugar of 1st kind be mixed with 27 kg of 2nd kind. 

Then, 7 : 3 = x : 27 or x = (7 x 27 / 3) = 63 kg.

Let the price of the mixed variety be Rs. x per kg. By the rule of alligation, we have :

Cost of 1 kg of type 1 rice           Cost of 1 kg of type 2 rice  

?(20-x)/(x-15) = 2/3 

? 60 - 3x = 2x - 30

  ? x = 18.

           Wine                         Water

            8L                              32L

            1                :               4

           20 %                           80% (original ratio)

           30 %                           70% (required ratio)

In  ths case, the percentage of water being reduced when the mixture is being replaced with wine.

so the ratio of left quantity to the initial quantity is 7:8

Therefore ,  7 8 = 1 - K 40

=> K = 5 Lit

pool : kerosene 

   3  :  2(initially) 

   2  :  3(after replacement)

  R e m a i n i n g Q u a n t i t y I n i t i a l Q u a n t i t y = 1 - R e p l a c e d Q u a n t i t y T o t a l Q u a n t i t y

(for petrol)    2 3 = 1 - 10 k

 => K = 30

 Therefore the total quantity of the mixture in the container is 30 liters.

Profit (%) = 9.09 % = 1/11

Since the ratio of water and milk is  1 : 11,

Therefore the ratio of water is to mixture = 1:12

Thus the quantity of water in mixture of 1 liter = 1000 x (1/12) = 83.33 ml