Let the quantities of A and B mixed be 3x kg and 7x kg.Cost of 3x kg of A = 9(3x) = Rs. 27xCost of 7x kg of B = 15(7x) = Rs. 105xCost of 10x kg of the mixture = 27x + 105x = Rs. 132xCost of 5 kg of the mixture = 132x/10x (5) = Rs. 66Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50

Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.

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

Therefore x = 21%

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.

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.

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.

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.