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 =
litres.
Quantity of syrup in new mixture = litres.
=> 5x + 24 = 40 - 5x
=> 10x = 16 => x = 8/5
So, part of the mixture replaced = = 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 ,
=> K = 5 Lit
pool : kerosene
3 : 2(initially)
2 : 3(after replacement)
(for petrol)
=> 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
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