Aptitude - Alligation or Mixture - Discussion

From where did you get 1/8 in finding the final result?

How we find the % of thing?

There has to be a direct formula for this.

Anyone has it?

Mean litre, m = (l+s)/2.
m = 4.

Proportion x = (l-m)/l.
x = 5-4/5.
x = 1/5.

Pure and Simple.

I cannot understand. Its so difficult for me. Please explain in another way.

Let there be 8 liters of solution, in which 3/8 liter be water and 5/8 lt be syrup. As we can see syrup is more than water so of course some syrup is to be replaced with water. As it is a solution and we can't remove only syrup or water so we have to remove the whole solution and replace it with water.

Now using ratio and proportion (5-(5x/8))/(3-(3x/8)+x) = 1/1.

As some part of syrup will be removed we also have to remove water as i have told that we are replacing some part of whole solution with water thats why we are subtracting 5x/8, 3x/8 from numerator and denominator respectively, now after removing solution we have add equal quantity of water thats why we will add water in denominator which will be equivalent to the quantity of the solution removed i.e 5x/8+3x/8=x. On the RHS side value will be as we want both the syrup and water in same proportion i.e 1/1.

1. Let volume of vessel be x.

Now the ratio is 3:5 initially.
(3/8)x water and (5/8)x syrup.

2. Now let y liter of mixture drawn off.

Means now {(3/8)x - (3/8)y} water and {(5/8)x - (5/8)y} syrup in remaining mixture.

3. Now y liters of water mix to make the equal quantity of both water and syrup,

Then, 3/8(x-y)+y = 5/8(x-y),
y = 1/5x.

What is the another way to solve this problem even more easier?

Can anyone solve this equation for me?

3-3x/8+x = 5-5x/8

Let there be 1kg of mixture with water concentration 3/8 i.e. 3/8 kg water present.

X kg mixture is removed so water removed = 3X/8.

Now X kg pure water is added.

In the required mixture water = 1/2 kg. So balancing over water.

3/8 " 3X/8 + X = 1/2 solving we get X = 1/5.