Tips And Tricks For Alligation And Mixtures
Tips and Tricks for Alligation and Mixtures will be discussed on this page for solution of problems regarding the mixing of ingredients in a shortcut way.
Mixture It is the process of mixing two or more elements/ingredients together.
AlligationIt is a process or a rule that tells us how to mix two or more ingredients/mixtures to get the desired mixture of a certain price & concentration.
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Tips and Tricks and Shortcuts Alligations and Mixtures:-
Type 1. Tips and Tricks for Alligation and Mixtures
In most of these questions , you are supposed to find the ratio , or any one of the values either M(mean price) , or C(C.P of cheaper) or D(C.P of dearer) where the other two values and the ratio is given.
To find that , we have the best trick for you.
When two commodities are mixed then,
\left ( \frac{ \text{ Quantity of Cheaper }}{\text{ Quantity of dearer }} \right ) = \left ( \frac{ \text{ C.P of Dearer (d) – Mean price (m)}}{ \text{ Mean price(m) – C.P. of cheaper(c) }} \right)
This equation can also be developed using the pictorial diagram.
Question 1:
Two varieties of wheat are mixed in the ratio of 4:5. The price of the mixture is ₹15 per Kg. The price of the variety having lesser weight is ₹12 per Kg. Calculate the price of the other variety.
Solution:
First of all we will identify and substitute the values in the diagram
Now substituting this in the formula
\left ( \frac{ \text{ Quantity of Cheaper }}{\text{ Quantity of dearer }} \right ) = \left ( \frac{ \text{ C.P of Dearer (d) – Mean price (m)}}{ \text{ Mean price(m) – C.P. of cheaper(c) }} \right)
\frac{4}{5} = \frac{(N – 15)}{(15 – 12)}
\frac{4}{5} = \frac{N – 15}{3}
So , 5N – 75 = 12
N = Rs. 17.4 per Kg
Type 2: Shortcuts, Tips and Tricks for Alligation and Mixtures
Calculate quantity of pure Liquid after ‘n’ successive operations:
If a Container contains ‘x’ units of pure liquid , and we replace the liquid with ‘y’ units of water :
Then after ‘n’ successive operations, the units of pure liquid left is
\left ( x \left( 1 – \frac{y}{x} \right)n \right)
Below is an example to explain this concept.
Question 2:
A vessel contains 60L of milk, out of which 15L litres of milk is taken out and replaced by water. This process is repeated two times. Find the amount of milk left in the container.
Solution:
From the question we have ,
Total Milk (x) = 60L
Milk taken out in one round (y) = 15L
No. of rounds (N) = 2
So , Using the above formula ,
Amount of milk left in the container
=\left ( x \left( 1 – \frac{y}{x} \right)n \right)
=\left ( 60 \left( 1 – \frac{15}{60} \right)2 \right)
=\left ( 60 \left( \frac{3}{4} \right)2 \right)
Solving this we get the answer as 33.75L
Question 3:
A dealer has 1000 kg sugar and he sells a part of it at 8% profit and the rest of it at 18% profit. The overall profit he earns is 14%. What is the quantity which is sold at 18% profit?
Solution :
As per the rule of alligation,
Quantity of Dearer: Quantity of Cheaper = (18-14) : (14-8) = 4:6 = 2:3
Quantity of sugar sold at 18% profit = 3/5 × 1000 = 600kg
Question 4 :
How much coffee of variety A, costing Rs. 5 a kg should be added to 20 kg of Type B coffee at Rs. 12 a kg so that the cost of the two coffee variety mixture be worth Rs. 7 a kg?
Solution :
As per the rule of alligation,
Quantity of Dearer: Quantity of Cheaper = (12-7) : (7-5) = 5:2
Quantity of Variety A coffee that needs to be mixed ⇒ 5:2 = x:20
⇒ x =50 kg
Question 5 :
A mixture of 20 kg of water and spirit contains 10% water. How much water must be added to this mixture to raise the percentage of water to 25%?
Solution :
Water = \left ( \frac{10}{100} \right )\times100=2 and spirit = 18kg.
In the second mixture:
75 kg spirit is contained in a mixture of 100 kg
18 kg spirit is contained in a mixture of\left ( \frac{100}{75} \right )\times18=24kg
So, water to be added=24-20=4 kg
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How to Solve Quickly | - Compound Interest – Questions |
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How to Solve Quickly |
Others
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