How To Solve Allegation And Mixtures Problems Quickly
How to solve Allegation and Mixtures Ques Quickly
Definition of Allegation?
The rule of allegation help to find the ratio in which two or more variety of ingredients of a given price must be mixed to produce a mixture of desired price.It is a rule for the solution of problems concerning the compounding or mixing of ingredients
Definition of Mixture?
A mixture contains two or more commodities of certain quantity mixed together to get the desired quantity.
What is called Mean Price?
The cost of a unit quantity of the mixture is called the mean price.
Type 1 : How to Solve Quickly Allegation and Mixtures (When two quantities are mixed then)
|Quantity of cheaper||=||C.P. of dearer(d) – Mean Price(m)|
|Quantity of dearer||Mean price(m) – C.P. of cheaper(c)|
Therefore, (Cheaper Quantity) : (Dearer Quantity) = (d – m) : (m-c)
Ques 1. Calculate the average price of the resulting mixture when two variety of sugar at ₹12 per Kg and ₹15 per Kg are mixed in the ratio of 2:3.
Solution: Let average price be Aw
2/3 = (15-Aw) / (Aw-12)
2Aw-24 = 45-3Aw
5Aw = 69
Aw = 69/5 = ₹13.8 per Kg
Ques 2.Two varieties of rice are in the ratio of 3:2 such that the average price of the resulting mixture is ₹15 per Kg. The price of one of the varieties is ₹10 per Kg. Find the price of the other variety of rice.
Solution: Let Price of another variety is A2
3/2= (A2-15)/ (15-10)
15 = 2A2-30
45/2 = A2
A2 = ₹22.5 per Kg
Ques 3 : Two varieties of tea are mixed in some ratio. The cost of the first variety is ₹20 per Kg and that of second variety is ₹30 per Kg. If the average cost of the resulting mixture is ₹25 per Kg, find the ratio.
Solution: x/y= (30-25)/ (25-20)
x/y= 5/5 = 1:1
Type 2 : Solve Allegations and Mixtures Ques Quickly
If a container has x unit of liquid A from which y units are taken out and replaced by water. This process is repeated n number of time, then the quantity of pure liquid will be given by:
|After n operations, the quantity of pure liquid =||x||1 –||y||n||units.|
Ques.1 A 420L of the mixture contains pure milk and water in the ratio of 2:5. Now 80 L of water is added to the mixture. Calculate the ratio of milk and water in the resulting mixture.
The quantity of milk in initial mixture =2/7* 420 = 120 L
Quantity of water = 420-120 = 300L
Water added = 80L
Concentration of water in the resulting mixture = 300 + 80 = 380L
The ratio of the resulting mixture 120/380 = 6:19.
Ques 2 . A milkman mixes 30L of water in 90L of milk. He then sells 1/4th of this mixture. Now he adds water to replenish the quantity of milk sold. Find the current proportion of milk and water.
The initial ratio of milk and water = 90:30 = 3:1
Now, 1/4th of the mixture is sold, that is the total volume of the mixture is reduced by 25%.
In other words, both water and milk are reduced by 25%.
So, the volume of milk and water is 67.5L and 22.5 respectively.
Now, 30L (25% of the total mixture volume) of water is added to the mixture.
The volume of milk = 67.5L
Volume of water = 22.5+30= 52.5L
Current ratio = 67.5/52.5 = 9:7
Ques 3. A 20 litres mixture of milk and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removal and replacement, what is the ratio of milk and water in the resultant mixture?
In 20 liters of mixture
Milk = 3/5 * 20 = 12 l
Water = 8 litres
In 10 liters of mixture.
Milk = 6 liters
Water = 4 litres
On adding 10 liters of milk,
Milk = 12 – 6 + 10 = 16 l
Water = 8 – 4 = 4 l
Again. in 10 liters of mixture,
Milk = 4/5 * 10 = 8 l
Water = 2 litres
On adding 10 litres of milk,
Milk =16 – 8 + 10 = 18 litres
Water = 2 litres
Therefore, Required ratio = 18 : 2 = 9 : 1
Read Also: Formula for Allegation & mixture question