Aptitude - Alligation or Mixture - Discussion

Let's say after removing some part of the mixture.

The ratio of the remaining mixture is the same i.e. 3:5 for water to syrup.
Now consider this mixture is 8L in total (3+5).
Now once we add 2L of water in the mixture the ratio becomes 5:5 which was desired.
5+5 =10.

So we can say that even in the original mixture the quantity was 10L as 2L was removed from that.
(2L mixture removed and 2L water added).
This satisfies the condition.
Now,
The question is how much of the mixture is to be removed and replaced.
2L/10L i.e.1/5.
1/5 will be the answer.

Initially we have 3 :5 ratio of w:s.

We are replacing only water so syrup quantity will remain same.
we want mixture in half that is mixture ratio
of w: s will be 1:1
but we know syrup quantity won't change so multiply mixture ratio in order to get same quantity of syrup as before i.e. 5
so multiply mixture ratio by 5
we get final ratio of water to syrup (w:s)=5:5.

So,
initially 3:5
finally 5:5
from above we can say that 2 part of water changing
and no of a part in the final mixture is 5+5=10 part.
So,
2/10 =1/5.

I am getting 8/5 as the amount to be replaced. If there is 8litres of the total liquid then 3 litres is water and 5 litres is syrup. Now suppose if you remove x litres of the liquid the remaining liquid is 8-x.

In this 8-x, 3/8(8-x) is water and 5/8(8-x) is syrup.
Now if we add x litres of water to it the equation we get is
3/8(8-x)+x=5/8(8-x) since both the syrup and the liquid equals 4/8

Solving this equation we get x=8/5

Where did 1/5 come from? Please somebody explain. I am highly confused.

Attention to those who want to understand the solution provided with the question.

Suppose the vessel initially contains 8 litres of liquid. ------> vessel contains => 8l mixture.

Let x litres of this liquid be replaced with water.

Quantity of water in new mixture = 3 - (3x/8) + x litres.

------ >[3lit water it originally contained] - [ (3x/8) l water which was removed when you removed 'x' l mixture] + ['x' litre water that you added].

Hope that'll justify the remaining solution too.

A 10 litre container holds a mixture of water and sugar, the volume of sugar being 15% of total volume. A few litres of the mixture is released and an equal amount of water is added. Then the same amount of the mixture as before is released and replaced with water for a second time. As a result, the sugar content becomes 10% of total volume. What is the approximate quantity of mixture released each time? please solve this sum?

Can anyone solve this?

Given : 3 part is water.
5 part is syrup.

Therefore the total mixture content becomes 8.

We want the half of this complete mixture means half water and half syrup.
So if we add 1 in 3 parts we get 4 i.e. 4/8 = 1/2.
And if we subtract 1 from 5 again we get 4 i.e 4/8= 1/2.
Therefore we get the half-half content of the water and syrup.
In percentage, we added 10% and reduced 10% therefore 10% + 10% = 20%.
And 20% = 20/100.
So, we get 1/5.

Let be x liters of mixture.

From equation quantity of pure liquid = x*(1-y/x)^n.

Where why is quantity of water used.

N is no. of times water replaces the liquid.

x liters of mixture and why liters of water final quantity of pure liquid = 1/2*x.

So from equation 1/2*x = 5/8*x(1-y/x).

Where 5/8*x is the initial quantity of syrup.

We need to find out y/x.

So, 1/2 = 5/8*(1- y/x).

1-(y/x) = 4/5.

y/x = 1-(4/5).

y/x = 1/5.

We know that water is 3 part and syrup is 5 part.
So total amount of mixture must be multiple of 8,
If we take 40 lit mixture.
It must contain 15 lit water and 25 lit syrup.
Now we have to make it 1:1.

i.e-20 lit water and 20 lit syrup.

Now we have to draw 5 lit of syrup from the mix.
When we drew out 8 lit of mix then only it contains 5 lit syrup.
8 lit of water added to the mix makes water and syrup 1:1.
8 out of 40 is 1/5.

Let the volume of the mixture be 8 litres, in which 3 litre is water and 5 litres is syrup.

Now add 2 litre of water in the mixture, the volume becomes 10 litre and the ratio becomes 1:1.

Since the earlier mixture was 8 litre in volume.

Therefore we have to subtract 2 litres from the new mixture, so as it becomes 8 litres.

Now calculate how much part of the mixture was subtracted from the new mixture i.e. (2/10 or 1/5).

For 62.5% water, to get to 50%, 12.5 must go which is exactly 20% of 62.5.

Syrup is to water = 3 : 5.

% water in mixture = 5/8*100 = 62.5%.

Desired % of water = 1/2*100 = 50%.

Difference = 62.5-50 = 12.5%.

To get 50% of water the 12.5% must go.

Hence Formula = Difference/Current Water in Mixture.

= 12.5/62.5 = 1/5 = 20 %.

5 syrup parts of 8 = 62.5% syrup, to get to 50%, 12.5 must go which is exactly 20% of 62.5.