Alligation is nothing but a faster technique of solving problems based on the weighted average situation as applied to the case of two groups being mixed together.
THEORY
To recollect, the weighted average is used when a number of smaller groups are mixed together to form one larger group.
If the average of the measured quantity was
A1 for group 1 containing n1 elements
A2 for group 2 containing n2 elements
A3 for group 3 containing n3 elements
Ak for group k containing nk elements
We say that the weighted average, Aw is given by:
Aw = (n1A1 + n2 A2 + n3 A3 + ……. + nkAk)/ (n1 + n2 + n3 … + nk)
That is, the weighted average
If the average of the measured quantity was
A1 for group 1 containing n1 elements
A2 for group 2 containing n2 elements
A3 for group 3 containing n3 elements
Ak for group k containing nk elements
We say that the weighted average, Aw is given by:
Aw = (n1A1 + n2 A2 + n3 A3 + ……. + nkAk)/ (n1 + n2 + n3 … + nk)
That is, the weighted average
In the case of the situation where just two groups are being mixed, we can write this as:
Aw = (n1A1 + n2 A2 )/(n1 + n2)
Rewriting this equation we get: (n1 + n2) Aw = n1A1 + n2A2
n1(Aw – A1) = n2 (A2 – Aw)
or n1/n2 = (A2 – Aw)/(Aw – A1) Æ The alligation equation.
Aw = (n1A1 + n2 A2 )/(n1 + n2)
Rewriting this equation we get: (n1 + n2) Aw = n1A1 + n2A2
n1(Aw – A1) = n2 (A2 – Aw)
or n1/n2 = (A2 – Aw)/(Aw – A1) Æ The alligation equation.
The Alligation Situation
Two groups of elements are mixed together to form a third group containing the elements of both the groups. If the average of the first group is A1 and the number of elements is n1 and the average of the second group is A2 and the number of elements is n2, then to find the average of the new group formed, we can use either the weighted average equation or the alligation equation.
As a convenient convention, we take A1 < A2. Then, by the principal of averages, we get A1 < Aw < A2.
Two groups of elements are mixed together to form a third group containing the elements of both the groups. If the average of the first group is A1 and the number of elements is n1 and the average of the second group is A2 and the number of elements is n2, then to find the average of the new group formed, we can use either the weighted average equation or the alligation equation.
As a convenient convention, we take A1 < A2. Then, by the principal of averages, we get A1 < Aw < A2.
Illustration 1
Two varieties of rice at ` 10 per kg and ` 12 per kg are mixed together in the ratio 1 : 2. Find the average price of the resulting mixture.
Solution
Solution
1/2 = (12 – Aw)/(Aw – 10) Æ Aw – 10 = 24 – 2Aw
fi 3Aw = 34 fi Aw = 11.33 `/kg.
fi 3Aw = 34 fi Aw = 11.33 `/kg.
Illustration 2
On combining two groups of students having 30 and 40 marks respectively in an exam, the resultant group has an average score of 34. Find the ratio of the number of students in the first group to the number of students in the second group.
Solution
Solution
n1/n2 = (40 – 34)/(34 – 30) = 6/4 = 3/2
Graphical Representation of Alligation
The formula illustrated above can be represented by the following cross diagram:
The formula illustrated above can be represented by the following cross diagram:
[Note that the cross method yields nothing but the alligation equation. Hence, the cross method is nothing but a graphical representation of the alligation equation.]
As we have seen, there are five variables embedded inside the alligation equation. These being: the three averages Æ A1, A2 and Aw and the two weights Æ n1 and n2
Based on the problem situation, one of the following cases may occur with respect to the knowns and the unknown, in the problem.
As we have seen, there are five variables embedded inside the alligation equation. These being: the three averages Æ A1, A2 and Aw and the two weights Æ n1 and n2
Based on the problem situation, one of the following cases may occur with respect to the knowns and the unknown, in the problem.
Now, let us try to evaluate the effectiveness of the cross method for each of the three cases illustrated
above:
Case 1: A1, A2, Aw are known; may be one of n1 or n2 is known.
To find: n1 : n2 and n2 if n1 is known OR n1 if n2 is known.
Let us illustrate through an example:
above:
Case 1: A1, A2, Aw are known; may be one of n1 or n2 is known.
To find: n1 : n2 and n2 if n1 is known OR n1 if n2 is known.
Let us illustrate through an example:
Illustration 3
On mixing two classes of students having average marks 25 and 40 respectively, the overall average
obtained is 30 marks. Find
(a) The ratio of students in the classes
(b) The number of students in the first class if the second class had 30 students.
obtained is 30 marks. Find
(a) The ratio of students in the classes
(b) The number of students in the first class if the second class had 30 students.
Solution
(a) Hence, solution is 2 : 1.
(b) If the ratio is 2 : 1 and the second class has 30 students, then the first class has 60 students.
(b) If the ratio is 2 : 1 and the second class has 30 students, then the first class has 60 students.
Note: The cross method becomes pretty effective in this situation when all the three averages are known and the ratio is to be found out.
Case 2: A1, A2, n1 and n2 are known, Aw is unknown.
Case 2: A1, A2, n1 and n2 are known, Aw is unknown.
Illustration 4
4 kg of rice at ` 5 per kg is mixed with 8 kg of rice at ` 6 per kg. Find the average price of the mixture.
Solution
= (6 – Aw)/(Aw – 5) = 4/8 Æ12 – 2 Aw = Aw – 5
3Aw = 17
Aw = 5.66 `/kg.
Note: The cross method becomes quite cumbersome in this case, as this method results in the formula being written. Hence, there seems to be no logic in using the cross method in this case.
Case 3: A1, Aw, n1 and n2 are known; A2 is unknown.
Illustration 5
5 kg of rice at ` 6 per kg is mixed with 4 kg of rice to get a mixture costing ` 7 per kg. Find the price of the costlier rice.
Solution
Solution
Using the cross method:
(x – 7)/1 = 5/4 Æ 4x – 28 = 5
x = ` 8.25.
Note: The cross method becomes quite cumbersome in this case since this method results in the formula being written. Hence, there seems to be no logic in using the cross method in this case.
The above problems can be dealt quite effectively by using the straight line approach, which is explained below.
The above problems can be dealt quite effectively by using the straight line approach, which is explained below.
The Straight Line ApproachAs we have seen, the cross method becomes quite cumbersome in Case 2 and Case 3. We will now proceed to modify the cross method so that the question can be solved graphically in all the three cases.
Consider the following diagram, which results from closing the cross like a pair of scissors. Then the
positions of A1, A2, Aw, n1 and n2 are as shown.
Consider the following diagram, which results from closing the cross like a pair of scissors. Then the
positions of A1, A2, Aw, n1 and n2 are as shown.
Visualise this as a fragment of the number line with points A1, Aw and A2 in that order from left to right.
Then,
(a) n2 is responsible for the distance between A1 and Aw or n2 corresponds to Aw – A1
(b) n1 is responsible for the distance between Aw and A2. or n1 corresponds to A2 – Aw
(c) (n1 + n2) is responsible for the distance between A1 and A2. or (n1 + n2) corresponds to A2 – A1.
The processes for the 3 cases illustrated above can then be illustrated below:
Then,
(a) n2 is responsible for the distance between A1 and Aw or n2 corresponds to Aw – A1
(b) n1 is responsible for the distance between Aw and A2. or n1 corresponds to A2 – Aw
(c) (n1 + n2) is responsible for the distance between A1 and A2. or (n1 + n2) corresponds to A2 – A1.
The processes for the 3 cases illustrated above can then be illustrated below:
Illustration 6
On mixing two classes of students having average marks 25 and 40 respectively, the overall average
obtained is 30 marks. Find
(a) the ratio in which the classes were mixed.
(b) the number of students in the first class if the second class had 30 students.
obtained is 30 marks. Find
(a) the ratio in which the classes were mixed.
(b) the number of students in the first class if the second class had 30 students.
Solution
Hence, ratio is 2 : 1, and the second class has 60 students.
Case 2 A1, A2, n1 and n2 are known; Aw is unknown.
Illustration 7
4 kg of rice at ` 5 per kg is mixed with 8 kg of rice at ` 6 per kg. Find the average price of the mixture.
Solution
Then, by unitary method:
n1 + n2 corresponds to A2 – A1
Æ 1 + 2 corresponds to 6 – 5
That is, 3 corresponds to 1
\ n2 will correspond to
n1 + n2 corresponds to A2 – A1
Æ 1 + 2 corresponds to 6 – 5
That is, 3 corresponds to 1
\ n2 will correspond to
In this case (1/3) × 2 = 0.66.
Hence, the required answer is 5.66.
Hence, the required answer is 5.66.
Note: In this case, the problem associated with the cross method is overcome and the solution becomes graphical.
Case 3: A1, Aw, n1 and n2 are known; A2 is unknown.
Case 3: A1, Aw, n1 and n2 are known; A2 is unknown.
Illustration 8
5 kg of rice at ` 6 per kg is mixed with 4 kg of rice to get a mixture costing ` 7 per kg. Find the price of the costlier rice.
Using straight line method:
Using straight line method:
4 corresponds to 7 – 6 and 5 corresponds to x – 7.
The thought process should go like:
The thought process should go like:
4 Æ1
\ 5 Æ1.25
Hence, x – 7 = 1.25
and x = 8.25
\ 5 Æ1.25
Hence, x – 7 = 1.25
and x = 8.25
SOME TYPICAL SITUATIONS WHERE ALLIGATIONS CAN BE USED
Given below are typical alligation situations, which we should be able to recognize. This will help us improve upon the time required in solving questions.
The following situations should help us identify alligation problems better as well as spot the way A1, A2, n1 and n2 and Aw are mentioned in a problem.
In each of the following problems the following magnitudes represent these variables:
A1 = 20, A2 = 30, n1 = 40, n2 = 60
Each of these problems will yield an answer of 26 as the value of Aw.
1. A man buys 40 kg of rice at ` 20/kg and 60 kg of rice at ` 30/kg. Find his average price. (26/kg)
2. Pradeep mixes two mixtures of milk and water. He mixes 40 litres of the first containing 20% water and 60 litres of the second containing 30% water. Find the percentage of water in the final mixture. (26%)
3. Two classes are combined to form a larger class. The first class having 40 students scored an average of 20 marks on a test while the second having 60 students scored an average of 30 marks on the same test. What was the average score of the combined class on the test. (26 marks)
4. A trader earns a profit of 20% on 40% of his goods sold, while he earns a profit of 30% on 60% of his goods sold. Find his percentage profit on the whole. (26%)
5. A car travels at 20 km/h for 40 minutes and at 30 km/h for 60 minutes. Find the average speed of the car for the journey. (26 km/hr)
6. 40% of the revenues of a school came from the junior classes while 60% of the revenues of the school came from the senior classes. If the school raises its fees by 20% for the junior classes and by 30% for the senior classes, find the percentage increase in the revenues of the school. (26%)
Given below are typical alligation situations, which we should be able to recognize. This will help us improve upon the time required in solving questions.
The following situations should help us identify alligation problems better as well as spot the way A1, A2, n1 and n2 and Aw are mentioned in a problem.
In each of the following problems the following magnitudes represent these variables:
A1 = 20, A2 = 30, n1 = 40, n2 = 60
Each of these problems will yield an answer of 26 as the value of Aw.
1. A man buys 40 kg of rice at ` 20/kg and 60 kg of rice at ` 30/kg. Find his average price. (26/kg)
2. Pradeep mixes two mixtures of milk and water. He mixes 40 litres of the first containing 20% water and 60 litres of the second containing 30% water. Find the percentage of water in the final mixture. (26%)
3. Two classes are combined to form a larger class. The first class having 40 students scored an average of 20 marks on a test while the second having 60 students scored an average of 30 marks on the same test. What was the average score of the combined class on the test. (26 marks)
4. A trader earns a profit of 20% on 40% of his goods sold, while he earns a profit of 30% on 60% of his goods sold. Find his percentage profit on the whole. (26%)
5. A car travels at 20 km/h for 40 minutes and at 30 km/h for 60 minutes. Find the average speed of the car for the journey. (26 km/hr)
6. 40% of the revenues of a school came from the junior classes while 60% of the revenues of the school came from the senior classes. If the school raises its fees by 20% for the junior classes and by 30% for the senior classes, find the percentage increase in the revenues of the school. (26%)
Some Keys to spot A1, A2 and Aw and differentiate these from n1 and n2
1. Normally, there are 3 averages mentioned in the problem, while there are only 2 quantities. This isn’t foolproof though, since at times the question might confuse the student by giving 3 values for quantities representing n1, n2 and n1 + n2 respectively.
2. A1, A2 and Aw are always rate units, while n1 and n2 are quantity units.
3. The denominator of the average unit corresponds to the quantity unit (i.e. unit for n1 and n2).
4. All percentage values represent the average values.
1. Normally, there are 3 averages mentioned in the problem, while there are only 2 quantities. This isn’t foolproof though, since at times the question might confuse the student by giving 3 values for quantities representing n1, n2 and n1 + n2 respectively.
2. A1, A2 and Aw are always rate units, while n1 and n2 are quantity units.
3. The denominator of the average unit corresponds to the quantity unit (i.e. unit for n1 and n2).
4. All percentage values represent the average values.
A Typical Problem
A typical problem related to the topic of alligation goes as follows:
4 litres of wine are drawn from a cask containing 40 litres of wine. It is replaced by water. The process is repeated 3 times
(a) What is the final quantity of wine left in the cask.
(b) What is the ratio of wine to water finally.
If we try to chart out the process, we get: Out of 40 litres of wine, 4 are drawn out.
This leaves 36 litres wine and 4 litres water. (Ratio of 9 : 1)
Now, when 4 litres are drawn out of this mixture, we will get 3.6 litres of wine and 0.4 litres of water (as the ratio is 9 : 1). Thus at the end of the second step we get: 32.4 litres of wine and 7.6 litres of water. Further, the process is repeated, drawing out 3.24 litres wine and 0.76 litres water leaving 29.16 litres of wine and 10.84 litres of water.
This gives the final values and the ratio required.
A closer look at the process will yield that we can get the amount of wine left by:
40 × 36/40 × 36/40 × 36/40 = 40 × (36/40)3
fi 40 × (1 – 4/40)3
This yields the formula:
Wine left : Capacity × (1 – fraction of wine withdrawn)n for n operations.
Thus, you could have multiplied:
40 × (0.9)3 to get the answer
That is, reduce 40 by 10% successively thrice to get the required answer.
Thus, the thought process could be:
40 – 10% Æ 36 – 10% Æ 32.4 – 10% Æ 29.16
A typical problem related to the topic of alligation goes as follows:
4 litres of wine are drawn from a cask containing 40 litres of wine. It is replaced by water. The process is repeated 3 times
(a) What is the final quantity of wine left in the cask.
(b) What is the ratio of wine to water finally.
If we try to chart out the process, we get: Out of 40 litres of wine, 4 are drawn out.
This leaves 36 litres wine and 4 litres water. (Ratio of 9 : 1)
Now, when 4 litres are drawn out of this mixture, we will get 3.6 litres of wine and 0.4 litres of water (as the ratio is 9 : 1). Thus at the end of the second step we get: 32.4 litres of wine and 7.6 litres of water. Further, the process is repeated, drawing out 3.24 litres wine and 0.76 litres water leaving 29.16 litres of wine and 10.84 litres of water.
This gives the final values and the ratio required.
A closer look at the process will yield that we can get the amount of wine left by:
40 × 36/40 × 36/40 × 36/40 = 40 × (36/40)3
fi 40 × (1 – 4/40)3
This yields the formula:
Wine left : Capacity × (1 – fraction of wine withdrawn)n for n operations.
Thus, you could have multiplied:
40 × (0.9)3 to get the answer
That is, reduce 40 by 10% successively thrice to get the required answer.
Thus, the thought process could be:
40 – 10% Æ 36 – 10% Æ 32.4 – 10% Æ 29.16
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