Two bottles A and B contain diluted acid. In bottle A, the amount of water is double the amount of acid, while in bottle B, the amount of acid is 3 times that of water. How much mixture (in liters) should be taken from each bottle A and B respectively in order to prepare 5 liters of diluted acid containing an equal amount of acid and water?

Let's use the method of allegations focusing on the concentration of acid.
Acid concentration in Bottle A:
Water : Acid = 2:1⟹Acid fraction=
1+2
1
​
=
3
1
​


Acid concentration in Bottle B:
Acid : Water = 3:1⟹Acid fraction=
3+1
3
​
=
4
3
​


Target Acid concentration in the final mixture:
Equal amount of acid and water ⟹ Acid : Water = 1:1⟹Target acid fraction=
1+1
1
​
=
2
1
​


Apply Alligation Rule:

Ratio of Mixture A to Mixture B=

​

2
1
​
−
3
1
​


​


​

4
3
​
−
2
1
​


​

​


​

4
3
​
−
2
1
​


​
=
4
1
​


​

2
1
​
−
3
1
​


​
=
6
1
​

Ratio=
1/6
1/4
​
=
4
6
​
=
2
3
​

Divide the total quantity (5 liters) in the ratio 3:2:

Quantity from Bottle A=
3+2
3
​
×5=3 liters
Quantity from Bottle B=
3+2
2
​
×5=2 liters
Hence, 3 liters from bottle A and 2 liters from bottle B should be taken.