Aptitude - Ratio and Proportion - Discussion

So the ratio using;
n = a + b/c + d.
2 + 1/1 + 2 = 1.
Q = 1
Therefore,
1*60 = 60litres.

The first ratio is 1 : 3.
Quantity of water, 60 * 1/3 = 20,
Another ratio is 2 : 3.
Quantity of water, 60 * 2/3 = 40.
Quantity of water to be added = (20 + 40) = 60 litres.

Initial ratio of milk : water = 2:1.
So (2+1=3) these 3 unit belongs to 60 litres.
Therefore, 2 units of milk belong to 40 litres and 1 unit of water belongs to 20 litres.

Further water is added so the ratio of milk: water = 1:2 => given.
As water is added but the previous quantity of milk is the same which is 40 litres(only ratio change).
So here we consider the 1 unit of milk as 40 litres and therefore 2 units of water is 80 litres.
The final quantity of water -Initial quantity of water = 80 - 20 = 60 litres.
Further water added is 60 litres.

The ratio of milk to water is given as: 2:1.
Let us assume a quantity of milk is 2x ans water is x.
Therefore ,
2x + x = 60 liters (Total mixture).
3x = 60
x = 20 --------(1)

Now milk is 2x therefore 2*20 = 40 liters is milk.
Now new water is added therefore x becomes (20 + x),
now new ration is given 40 /(20+x) = 1/2,
80 = 20 + x.
x = 60 Liters.

60 lit, 2:1.

2x +1x = 60 lit,
3x = 60,
x = 60/3,
x = 20 lit.

The ratio of milk and water in 60L is;
m=2.
w=1 (2+1)=3,
milk = 2/3, water=1/3.

milk in 60L = 2/3 * 60= 40L.
water in 60L = 1/3 * 60= 20L,
40:(20+60)=4:8= 1:2.

@All.

Simply the solution is;

Mixture = 60 litre.
The ratio is 2: 1 , so milk:water=40:20.
New ratio= 1: 2 , means milk : water should be 40:80.
Water in the new mixture should be 80.
Less water already in the mix 20.
Water to be added now is 60.

m:w=2:1
m+w=60
2x+x-=60
x=20.

m=40 w=20.
Added water be y.
40:(20+y) = 1:2.
On cross multiplication
y=60.

The initial quantity of milk=(2/3)*90 = 60,
Initial quantity of water=(1/3)*90 = 30,
If the ratio is to be 2:3 the water to be added;
(60÷(30+x)) = 2/3,
X = 60.

40/20+x = 1/2.

Cross multiple;
80 = 20 + x.
x = 60.