Aptitude - Alligation or Mixture - Discussion

w:s
3:5

Consider in tens

30 : 50
3 : 5

1:1
40:40

To get 40 syrup you need to remove 10.
as the ratio is 3:5.

So, to remove 10 you need to remove 8x2=16 volume of liquid = 16/80 = 1/5.

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?

Thanks for your active cooperation. It's very helpful.

Let us consider the mixture to be of 8 litres (3 Litres of water + 5 litres of Syrup).
Amount of Water in the 8 litre Mixture : 3/8.
Amount of Syrup in the 8 litre Mixture : 5/8.
For easy calculation, I am considering the mixture to be of 10 litres.
Hence Amount of Water in the 10 litre Mixture : 3/8 * 10 = 3.75.
Amount of Syrup in the 10 litre Mixture : 5/8 * 10 = 6.25.


We need to replace 1.25 litres of syrup from the mixture and replace with water 50 make it 5 / 5. Hence we need to replace 2 litres of mixture (.625+.375) * 2.

2 Litres out of 10 litres is 20% or 1/5.

Very good explanation, Thanks @Surendrareddy.

Formula.

Final volume or the ratio of solute A=initial volume or ratio solution A(1-solute to be replaced/total volume)^n.
let A=syrup, B=water.

Putting in formula:
assuming volume is 1litre.
A=5/8, B=3/8.
applying formula on solute A because it will be reduced after an operation.
so,
4/8=5/8(1-x/1),
x=1/5.

How did you get that equation 5x+24=40-5x?

1/2(y-x)/y.
=1/2(5-3)/5,
= 1/2 * 2/5,
= 1/5.

Thanks @Surendrareddy.

3 5
\ /
8
/ \
3 5
ratio 3:5.
we need to calculate for water only,
3/5*1/3 = 1/5,

1/3 because one portion of water is 1/3.