Mixture and Alligation
Mixture and Alligation: Quantitative Aptitude is a very important section for government recruitment exams like SSC, railway, and banks. In this article, we are going to discuss the rule of mixtures and allegations. Generally, there are questions asked related to basic concepts and formulas of the Mixture and Alligation. The rule of alligation enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of the desired price. We have covered important notes and questions focusing govt. exams that aspirants can practice and prepare for the mixture and alligation.
Mixture: An aggregate of two or more two types of quantities gives us a mixture.
Alligation: It is a method of solving arithmetic problems related to mixtures of ingredients. This rule enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of the desired price.
Mixture and Alligation Formula
- It is a modified form of finding the weighted average. If 2 ingredients are mixed in a ratio and the cost price of the unit quantity of the mixture, called the Mean Price is given then,
The above formula can be represented with the help of a diagram which is easier to understand. Here ‘d’ is the cost of a dearer ingredient, ‘m’ is mean price and ‘c’ is the cost of cheaper ingredient.
Thus, (Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c).
Mixture and Alligation Formula: Repeated Dilution
- This is used to calculate the pure quantity left after ‘n’ number of processes of repeated replacement is done on the pure quantity. Suppose, a container contains ‘x’ units of a liquid from which ‘y’ units are taken out and replaced by water. After ‘n’ operations quantity of pure
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Mixture and Alligation Tricks
The competitive exam is totally based on time management to solve the questions with accuracy. Time management is very important to score more and attempt more questions in less time. Some mixture and allegation tips and tricks are very helpful for solving the word problems in a few minutes. Candidates must try to solve the problems using the tricks. Candidates must note that any tip or trick will be useful only if the candidate practices it a lot.
Mixture and Alligation Questions
1. A container contains 40 liters of milk. From this container, 4 liters of milk were taken out and replaced by water. This process was repeated further two times. How much milk is now contained in the container?
Mixture and Alligation Answers & Solutions
Assume that a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid
Hence milk now contained by the container = 40(1-4/40)^3
=40×9/10×9/10×9/10 =(4×9×9×9)/100 =29.16
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = (3 – 3x/8 + x) litre
Quantity of syrup in new mixture = (5 – 5x/8) litres
So (3 – 3x/8 + x) = (5 – 5x/8) litres
=> 5x + 24 = 40 – 5x
=>10x = 16
=> x = 8/5 .
So, part of the mixture replaced = (8/5 x 1/8) = 1/5
Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1
So their average price = (126+135)/2=130.5
Hence let’s consider that the mixture is formed by mixing two varieties of tea.
one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. Now let’s find out x.
By the rule of alligation, we can write asCost of 1 kg of 1st kind of tea Cost of 1 kg of 2nd kind of tea
(130.50) (x)Mean Price
(153)(x – 153) (22.50)=>(x – 153) : 22.5 = 1 : 1
=>x – 153 = 22.50
=> x = 153 + 22.5 = 175.54(C)Explanation:
Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
Quantity of A in mixture left = (7x – 7/12 x 9)litres
= (7x – 21/4) litres.
Quantity of B in mixture left = (5x – 5/12 x 9) litres
= (5x – 15/4) litres.
So (7x – 21/4)/((5x – 15/4) +9) = 7/9
=> (28x – 21)/(20x + 21) = 7/9
=> 252x – 189 = 140x + 147
=> 112x = 336
=> x = 3.
So, they can contain 21 liters of A.5(D)
Let the Cost Price(CP) of a 1-liter spirit be Rs.1
Quantity of spirit in 1-liter mixture from vessel A = 5/7
Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7
Quantity of spirit in 1-liter mixture from vessel B = 7/13
Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13
Quantity of spirit to be obtained in 1 liter mixture from vessel C = 8/13
Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price
By the rule of alligation, we can write asCP of 1 liter mixture CP of 1 liter mixture
from vessel A (5/7) from vessel B (7/13)
(8/13)8/13 – 7/13 = 1/13 5/7 – 8/13 = 9/91=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio6(C)
By the rule of alligation:
Cost of 1 kg pulses of Cost of 1 kg pulses of
1st kind Rs. (15) 2nd kind Rs.(20)Mean Price
Rs. (16.50)(3.50) (1.50)Required rate = 3.50 : 1.50 = 7 : 3.7(C)
Let initial quantity of wine = x litre
After a total of 4 operations, quantity of wine = x(1-y/x)^n=x(1-8/x)^4
Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65
Hence we can write as (x(1-8/x)^4)/x =16/81
(1-8/x)^4 = (2/3)^4
(1-8/x) = 2/3
(x-8/x) = 2/3
Let C.P. of 1 liter milk be Re. 1
Then, S.P. of 1 liter of mixture = Re. 1, Gain = 25%.
C.P. of 1 litre mixture = Re.(100/125 x 1) = 4/5
By the rule of alligation, we have:
C.P. of 1 liter of milk C.P. of 1 liter of water
Re. (1) (0)
Ratio of milk to water = 4/5 : 1/5 = 4 : 1.
Hence, percentage of water in the mixture = (1/5 x 100)% = 20%
The concentration of alcohol in 1st Jar = 40%
The concentration of alcohol in the 2nd Jar = 19%
After the mixing, the concentration of alcohol in the mixture = 26%
By the rule of alligation,
The concentration of alcohol in
1st Jar(40%) 2nd Jar (19%)
Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2