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
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
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.
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.
Amount of milk left after 3 operations
litres
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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