Erudite Mumbai: Theory of Averages, Mixtures and Allegations #1 Coaching for CAT, Mh-CET, Bank PO, XAT, JMET, SNAP, CAT2011, MBA Entrance

Theory of “AVERAGES & ALLEGATIONS”

As we all know average can be defined as the sum of all the elements divided by the numbers of elements present. Thus,

So, if I have some elements as 63, 57, 79, 83, 68 what will be the average?

Simple enough,

... RIGHT??? Absolutely!!!!

Now, let me say, the above mentioned is just the way to find the average, Its not the definition.

“Average can be defined as the representation of a group, such that if all the elements are represented and replaced by the average, the sum/totality of the group will not change.” One must understand the definition of the averages to apply it conveniently.

Lets observe,

63	57	79	83	68	Sum = 350
70	70	70	70	70	Sum = 70×5

Lets try to reason it out. Observe carefully.

Elements	63	57	79	83	68	Sum = 350
Average	70	70	70	70	70	Sum= 70×5
Deviations	-7	-13	+9	+13	-2	Sum= 0

Here the last row gives the deviation of the elements from the average, as 63 is 7 less than 70, so -7, 57 is 13 less than 70 and so on. Now here we can observe that the various deviations of the elements gets canceled among themselves (-7 -13 +9 +13 -2 = 0), and hence the overall deviation is “zero”. And hence the sum of the quantities remains same.

We can also observe

So, if we have “n” elements as e₁, e₂, e₃, …......, e_n-1, e_n; and there average is “a”, then

Now, how is this definition helpful?

This definition gives us the idea that average is a pivotal quantity about which the elements balance themselves. This will also lead to the conclusion that average should lie in between the greatest and the lowest value i.e. the average can't be more than the greatest value ( 83 in the above example) and it can't be less than the lowest value (57 in the above example). Lets use this definition to find the averages.

Solved Example: Find the average of 145, 179, 163, 182, 169, 173.

By just observing the numbers, we can understand that average should lie in between 145 and 182, somewhere close to 170. Lets assume the average as 169 (one of the element itself) , and lets try to find the deviation of the sum from the assumed sum.

145	179	163	182	169	173
169	169	169	169	169	169
-24	10	-6	13	0	4	Sum= -3

So total deviation is -3, if equally distributed among all those six 169's, each will get -0.5 and hence the actual average will become 168.5.

Isn't canceling the deviations easier then adding up the big numbers up there? And as you will practice this method more, you will realize that the more accurate is your approximation the lesser calculation will be involved in.

Solved Example: Average percentage scored by Nandini in her seven semesters is 78. How much she must score in her eighth semester so as to have her overall average as 80?

As per the definition discussed above, we can assume that she would have scored 78 in all her seven semesters. Now if she wants to have 80 as the overall average ( that means 80 in all eight semesters), she should compensate for those two marks, apart from scoring 80 in her eighth semester.

So she must score 80 + 2 × 7 = 94.

Solved Example: Average percentage scored by Nandini in her seven semesters is 78. How much she must score in her eighth semester so as to have her overall average as 75?

75 - 3 × 7 = 51

Properties of the Averages:

The average should lie between the largest and the smallest element of the group.
Middle term OR Average of the middle terms is the average of an Arithmetic Progression.
E.g. the average of 121, 129, 127, 125, 123, 131 is 126 i.e. average of 125 and 127 (though the elements are not in the order but they are still the elements of an AP).
If every elements of a group is increased (or decreased) by the a constant, then the average also gets increased (or decreased) by the same constant. E.g. if the average age of a family presently is 35 years, then the average age of the family 5 years hence will be 40 years irrespective of the number of members of the family (Of course subject to condition that no one joined or left the family).
If every element of a group is multiplied (or divided) by a constant, then the average will also get multiplied by the same constant.

Solved Example: 6 years ago the average age of a family of six members was 47 years. 3 years ago one person of the family got married and hence a 22 years old beautiful bride joined the family. This year the grandpa aged 82 passed away. Find the new average age of the family?

The average age 3 years ago, before the marriage, should be 50 (using the concept discussed above).

The age of the new bride is 28 less than the average, equally distributed among the seven members (bride included) of the family, the new average after the marriage will be 50 - 4 = 46.

Three years later i.e. this year, before the grandpa passed away, their average age will be 49.

The age of grandpa is 33 years more than the average, equally distributed among the six member (excluding the grandpa) of the family, the new average will be 49 - 5.5 = 43.5

Weighted Averages:

Now, let me have 2 students in section A, and 3 in section B of a class. Let the average marks scored by each of the students of the section A is 60 and that of the section B is 80. Then what is the over all average of the class?

Conveniently applying the above discussed definition of averages, we can represent the score of students of section A as 60, 60 and that of section B as 80, 80, 80. Total number of elements is 5. Now taking the average of these marks:

Now, if the number of students in section A and B is 20 and 30 respectively, then what will be the overall average of the class?

Almost all of you must have guessed it correct. 60 will repeat 20 times and 80 will repeat 30 times, and the total number of elements in this group would have been 50.

So, what if section A has 40 students and B has 60?

Some of you might have observed that the average in all the three cases is the same. Simple reason being the ratio of the number of students in each of these sections in 2:3. These numbers, 2 and 3, here represents the “weights or strengths” of the section A and B, and if one uses the weights instead of actual number of elements in the group, the concept is called as “Weighted Average”. So instead of considering the values as 40 and 60, one can conveniently take the values as 2 and 3, and get the same results.

We can compare this scenario with a “tug of war game” where both the quantities (60 and 80 as in our previous example) intend to pull the average towards its side. And the position of the average will be decided by strengths of these elements. If the strengths would have been equal i.e. the average would have been exactly at the center (e.g. if the number of students in each of these sections would have been 23 and 23, the ratio of the weights would have been in the ratio 1:1, and the average would have been 70, exactly at the center). Now, obviously if weight or strength of one element is more than that of the other, it will pull average towards its side. In our example, as the strength of 80 is 3 as compared to 2 of 60, hence the average is 72, nearer to 80 than that of 60. So, strength is inversely proportional to the difference (or distance) of the element from the average. Greater the strength, lesser will be the distance.

So, We can find the weighted average as:

Lets assume that the average lies in between 60 and 80, at a distance of d₁ from 60 and d₂ from 80, also we are given with the ratio of the strengths(or weights) as 2:3.

as total distance is 20, d₁ + d₂ = 20; so d₁ = 12; d₂ = 8; average = 72.

So there are two ways of finding the weighted average.

Solved Example: Varun mixed 50 Kgs of sugar costing 20 Rs/kg with 70 Kgs of sugar costing 32 Rs/kg. Find the average cost of the mixture.

As the weights are in the ratio 5:7, the ratio will be considered to calculate the weighted average instead of their actual values.

The difference between 20 and 32 is 12 which has to be divided in the ratio 7:5.

So final average can be 20+7=27 or 32-5=27.

Solved Example: Bapu has 60 ml of 55% concentrated H2SO4. Thaneer has 90 ml of 30% concentrated H₂SO₄. Find the overall concentration of the mixture, if both mix their solutions.

Ratio of the strengths is 2:3 (60 ml : 90 ml).

Solved Example: A container has 20 liters of 78% concentrated milk. If 6 liters of water is added to it. What will be the final concentration?

Nothing great about this problem. Only thing we have to understand is 6, liters of water means, 6 liters of 0% concentrated milk.

OK, now how many kgs of sugar costing 25 Rs/kg should be mixed with 30 Kgs of sugar costing 35 Rs/kg if the overall cost of the mixture should be 31 Rs/kg?

Now in this case, we are given the average of 25 and 35 as 31. Proceeding in the reverse order, as shown in the fig. above , the distances of the average from the element is 6 and 4, so the ratio is 3:2. As the strength is inversely proportional to the distance, ratio of strengths should be 2:3.

Pictorially it can also be expressed in a simpler way as: