Milk = 3/5 x 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 x 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
Now 10 lts milk is added => total milk = 18 lts
The required ratio of milk and water in the final mixture obtained
= (8 + 10):2 = 18:2 = 9:1.
It means (since 343 + 169 = 512)
Thus,
Thus the initial amount of wine was 120 liters.
Let the tin contain 5x litres of liquids
=> 5(4x - 36) = 2(x + 36)
=> 20x - 180 = 2x + 72
=> x = 14 litres
Hence, the initial quantity of mixture = 70l
Quantity of liquid B
=
= 50 litres.
Water in 60 gm mixture=60 x 75/100 = 45 gm. and Milk = 15 gm.
After adding 15 gm. of water in mixture, total water = 45 + 15 = 60 gm and weight of a mixture = 60 + 15 = 75 gm.
So % of water = 100 x 60/75 = 80%.
Given mixture = 48 lit
Milk in it = 48 x 5/8 = 30 lit
=> Water in it = 48 - 30 = 18 lit
Let 'L' lit of water is added to make the ratio as 3:5
=> 30/(18+L) = 3/5
=> 150 = 54 + 3L
=> L = 32 lit.
Let the initial amount of honey in the jar was K, then
or
Therefore, K = 1250
Hence initially the honey in the jar= 1.25 kg
Customer ratio of Milk and Water is given by
Milk :: Water
6.4 0
=> Milk : Water = 110 : 11 = 10 : 1
Therefore, the proportionate of Water to Milk for Customer is 1 : 10
Now, take percentage of milk and applying mixture rule
25 100
50
50 25 = 2 : 1
Hence required answer = 1/3 or 33.33%
Given that container has 50 litres of milk.
After replacing 8 litres of milk with water for three times, milk contained in the container is:
= 29.63 litres.
Copper in 4 kg = 4/5 kg and Zinc in 4 kg = 4 x (4/5) kg
Copper in 5 kg = 5/6 kg and Zinc in 5 kg = 5 x (5/6) kg
Therefore, Copper in mixture = kg
and Zinc in the mixture = kg
Therefore the required ratio = 49 : 221
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