Alligation or Mixture problems
In a 40 litre mixture of alcohol & water, the ratio of alcohol and water is 5 : 3. If 20% of this mixture is taken out and the same amount of water is added then what will be the ratio of alcohol and water in final mixture?
Quantity of alohol in the mixture = 40 x 5/8 = 25 lit Quantity of water = 40 - 25 = 15 lit According to question, Required ratio = 20 - 40 x 20 100 x 5 8 15 - 40 x 20 100 x 3 8 + 40 x 20 100 = 20 15 - 3 + 8 = 1 : 1
From the given data, let the initial quantity of the mixture = 5x Then, 2 x - 16 3 x - 24 + 40 = 1 4 8 x - 64 = 3 x + 16 5 x = 80 x = 16 lit Then the initial quantity of the mixture = 5x = 5 x 16 = 80 lit.
Here withdrawal of liquid A and B result into making the container empty. Hence percentage of two liquids withdrawn are two components of the percentage by which the container becomes empty. Applying the rule of alligation, we getA : B = 10 : 20 or 1 : 2 Quantity of liquid = 1 (1 + 2) × 90 = 30 liters Quantity of liquid B = 90 ? 30 = 60 liters.
w i n e ( l e f t ) w i n e ( a d d e d ) = 343 169 It means w i n e ( l e f t ) w i n e ( i n i t i a l a m o u n t ) = 343 512 (since 343 + 169 = 512) Thus, 343 x = 512 x 1 - 15 k 3 343 512 = 7 8 3 = 1 - 15 k 3 1 - 15 k = 7 8 = 1 - 1 8 Thus the initial amount of wine was 120 liters.
Let the amount of juice and water in original mixture '4x' litre and '3x' litre respectively.
According to given data,
4x / 3x + 6 = 8/7
28x = 24x + 48
28 x ? 24x = 48
4x = 48
x = 12
Amount of juice = 4x = 4×12 = 48 litre.
EXAMIANS