Milk Water
74% 26% (initially)
76% 24% ( after replacement)
Left amount = Initial amount
24 = 26
=> k = 91
Let the capacity of the pot be 'P' litres.
Quantity of milk in the mixture before adding milk = 4/9 (P - 8)
After adding milk, quantity of milk in the mixture = 6/11 P.
6P/11 - 8 = 4/9(P - 8)
10P = 792 - 352 => P = 44.
The capacity of the pot is 44 liters.
Let quantity of mixture be x liters.
Suppose a container contains x units of liquid from which y units are taken out and replaced by Water. After operations , the quantity of pure liquid = units, Where n = no of operations .
So, Quantity of Milk =
Given that, Milk : Water = 9 : 16
=> Milk : (Milk + Water) = 9 : (9+16)
=> Milk : Mixture = 9 : 25
Therefore,
=> x = 15 liters
From the given data,
let the initial quantity of the mixture = 5x
Then,
Then the initial quantity of the mixture = 5x = 5 x 16 = 80 lit.
It means
Thus ,
=> K = 120
Thus the initial amount of wine was 120 liters.
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.
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