# average

of 8 quant

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Averages
Definition
Simple Average (or Mean) is defined as the ratio of sum of the quantities to the number of quantities. By Definition, Average = Putting in symbols, X =
Sum of all quantities no. of quantities

x1 + x 2 + x 3 + - - - - - - - - - - - + x n N Here x1, x2, x3, ----------- xn represent the n values of quantity under consideration & x is the mean. Average or mean is said to be a measure of central tendency.

Let us take a very simple example of the first five natural numbers 1, 2, 3, 4 & 5. Their Average =
Sum 15 = = 3. 5 5 21 15 + 6 = = 3. 5+2 7

Now let’s add 2 more 3’s to these 5 numbers. Now, Average =

Ex.1 If a person with age 45 joins a group of 5 persons with an average age of 39 years. What will be the new average age of the group? Sol. Total age will be 45 + 5 × 39 = 240. And there will be 6 persons now. So the average will be 240/6 = 40. (or) Since 45 is 6 more than 39, by joining the new person, the total will increase by 6 and so the average will increase by 1. So, the average is 39 + 1 = 40. Ex.2 Two students with marks 50 and 54 leave class VIII A and move to class VIII B. As a result the average marks of the class VIII A fall from 48 to 46. How many students were there initially in the class VIII A? Sol. The average of all the students of class VIII A is 46, excluding these two students. They have 4 and 8 marks more than 46. So with the addition of these two students, 12 marks are adding more, and hence the average is increasing 2. There should be 6 students in that class including these two. This is the initial number of students, Ex.3 The average of x successive natural numbers is N. If the next natural number is included in the group, the average increases by:(1) Depends on x (2) Depends on the starting number of the series (3) Both (1) and (2) (4)
1 2

Sol.

(5) None of these The average of consecutive numbers is the middle number. If one more number is added to the list, the middle number moves 0.5 towards right. So the answer is (4).

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Weighted Mean
If some body asks you to calculate the combined average marks of both the sections of class X A and X B, when both sections have 60% and 70% average marks respectively? Then your answer will be 65% but this is wrong as you do not know the total number of students in each sections. So to calculate weighted average we have to know the number of students in both the sections. Let N1, N2, N3, …. Nn be the weights attached to variable values X1, X2, X3, …….. Xn respectively. Then the weighted arithmetic mean, usually denoted by
N X + N2 X 2 + N3 X3 + ..... + Nn Xn X = 1 1 N1 + N2 + N3 + ..... + Nn

Do You know ?
Mean has a tendency to tilt towards the side with the greatest number of entries.

X can’t be greater
than the greatest number or less than the least number.

for any two different quantities taken in different ratios. The weighted average is just like a see-saw. More the ratio of a quantity more will be the inclination of the average from mid value towards the value with more ratios.

Ex.4 The average marks of 30 students in a section of class X are 20 while that of 20 students of second section is 30. Find the average marks for the entire class X? Sol. We can do the question by using both the Simple average & weighted average method.
Simple average = = Sum of marks of all students Total number of students

20 × 30 + 30 × 20 = 24. 30 + 20 3 2 × 20 + × 30 5 5

By the weighted mean method, Average =

= 12 + 12 = 24.

1. 2. 3. If each number is increased / decreased by a certain quantity n, then the mean also increases or decreases by the same quantity. If each number is multiplied/ divided by a certain quantity n, then the mean also gets multiplied or divided by the same quantity. If the same value is added to half of the quantities and same value is subtracted from other half quantities then there will not be any change in the final value of the average.

Average Speed
Average Speed =
Total dis tance cov ered Total time taken
Average never

TIP

If d1 & d2 are the distances covered at speeds v1 & v2 respectively and the time taken are t1 & t2 respectively, then the average speed over the entire distance (x1 + x2) is given by d + d2 d1 + d2 Total dis tan ce cov ered = 1 = d1 d2 Total time taken t1 + t 2
v1 + v2

speed double

can or

be

more than double of any of the two speeds.

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If both the distances are equal i.e. d1 = d2 = d then, Average speed = 2 v 1v 2 {i.e. Harmonic mean of two velocities} v1 + v 2
v1 + v 2 {i.e. Algebraic mean of two velocities} 2

But if both the time taken are equal i.e. t1 = t2 = t the Average speed =

Ex.5 The average of 10 consecutive numbers starting from 21 is : Sol. The average is simply the middle number, which is the average of 5th & 6th no. i.e, 25 & 26 i.e. 25.5. Ex.6 There are two classes A and B., each has 20 students. The average weight of class A is 38 and that of class B is 40. X and Y are two students of classes A and B respectively. If they interchange their classes, then the average weight of both the classes will be equal. If weight of x is 30 kg, what is the weight of Y? Sol.

Total weight of class A = 38 × 20, and class B = 40 × 20, if X & Y are interchanged, then the total ages of both the classes are equal.
⇒ 38 × 20 – x + y = 40 × 20 – y + x . 2(y – x) = 2 × 20, ⇒ y = x + 20 = 50 (OR)

Since both the classes have same number of students, after interchange, the average of each class will be 39. Since the average of class ‘A’ is increasing by 1, the total should increase by 20. So, x must be replaced by ‘y’, who must be 20 years elder to ‘x’. So, y must be 50 years old..
Ex.7 The average weight of 10 apples is 0.4 kg. If the heaviest and lightest apples are taken out, the average is 0.41 kg. If the lightest apple weights 0.2 kg, what is the weight of heaviest apple? Sol.

Total weight of the apples is 0.4 × 10 = 4 kg. Weight of apples except heaviest & lightest = 0.41 × 8 = 3.28 kg
∴ Heaviest + lightest = 4 – 3.28 = 0.72 kg. It is given lightest = 0.2 kg. ∴ Heaviest is 0.72 – 0.2 = 0.52 kg.

Ex.8

While finding the average of ‘9’ consecutive numbers starting from X; a student interchanged the digits of second number by mistake and got the average which is 8 more than the actual. What is X?

Sol.

Since the average is 8 more than the actual, the second no will increase by 72 (9 × 8) by interchanging the digits. If ab is the second no, then 10a + b + 72 = 10b + a, ⇒ 9(b – a) = 72. ∴ b – a = 8. The possible number ab is 19. Since the second no is 19. The first no is 18. ∴ X = 18

Ex.9 Sol.

There are 30 consecutive numbers. What is the difference between the averages of first and last 10 numbers? The average of first 10 numbers is the average of 5th & 6th no. Where as the average of last 10

numbers is the average of 25th & 26th no. Since all are consecutive numbers, 25th number is 20 more than the 5th number. We can say that the average of last 10 nos is 20 more than the average of first 10 nos. So, the required answer is 20.

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Instructions for next 3 examples: There are 60 students in a class. These students are divided into three groups A, B, C of 15, 20 & 25 students each. The groups A & C are combined to form group D. Ex.10 What is the average weight of the students in group D? 1) More than the average weight of A. 2) More than the average weight of C. 3) Less than the average weight of C. 4) Less than the average weight of B. 5) Cannot be determined. We know only the no of students in each class, but we don’t know the average weight of any class, Sol. we can’t find the answer. Ex.11 If one student from group A is shifted to group B, which of the following is necessarily true? 1) The average weight of both groups increases 2) The average weight of both groups decreases. 3) The average weight of class remains the same. 4) The average weight of group A decreases and that of group B increases. 5) None of these Sol.

Options (1) & (2) are not possible. Average of both cannot increase or decrease. Option (4) can be eliminated because we are not sure, whether the average of A increases & B decreases or A decreases & B increases or both remains unchanged. It depends on the weight of the student, who shifted from A to B. Option (3) is always true because even the student shifts from one group to other. The average weight of the whole class does not change. Answer: (3)

Ex.12 If all the students of the class have the same weight, which of the following is false? 1) The average weight of all the four groups is same. 2) Total weight of A & C is twice that of B. 3) The average weight of D is greater than that of A. 4) The average weight of class remains same even the students shifts from one group to other. 5) None of these. Sol. Since each student has same weight (1) & (4) are right. Since the number of students in group A & C

together is 15 + 25 = 40. Where as in B, there are only 20 students. So option (2) is also correct. But (3) is false, because, the average weight of each group is same, since all the students have same weight. Answer: (3)

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Mixtures and Alligation
Basics of Fractions OR Chain Rule
(1) When a fraction has its numerator greater than the denominator, its value is greater than one. Let us call it greater fraction. Whenever a number (say x) is multiplied by a greater fraction, it gives a value greater than itself. When a fraction has its numerator less than the denominator, its value is less than one. Let us call it less fraction. Whenever a number (say x) is multiplied by a less fraction, it gives a value less than itself.

(2)

Rule of Fractions OR Chain Rule
Let us try to understand the concept with the help of an example. Ex.1 If 6 men can do a piece of work in 30 days of 9 hours each, how many men will it take to do 10 times the amount of work if they work for 25 days of 8 hours? Sol. Three points arise: (1) (2) (3) Less days, so more men required. Less working hours, so more men required. More work, more men.

By rule of fraction or Chain rule: Step I: We look for our required unit. It is the number of men. So, we write down the number of men given in the question. It is 6. Step II: The number of days gets reduced from 30 to 25, so it will need more men (Reasoning: Less days, more men). It simply means that 6 should be multiplied by a greater fraction because we need a value greater than 6. So, we have: 6 ×

30 25

Step III: Following in the same way, we see that the above figure should be multiplied by a ‘greater fraction’,

i.e., by

9 9 30 × . So, we have: 6 × 25 8 8
30 9 10 10 × × . So, we have: 6 × = 81 men. 25 8 1 1

Step IV: Following in the same way, we see that the above figure should be multiplied by a ‘greater fraction’
i.e. by

Mixtures & Alligation
As the dictionary meaning of Alligation (mixing), we will deal with problems related to mixing of different compounds or quantities. The concept of alligation and weighted average are the same. When two or more quantities are mixed together in different ratios to form a mixture, then ratio of the quantities of the two constituents is given by the following formulae:

Qc d−m = Qd m − c

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Qc Qd c d m

→ → → → →

Cheaper quantity Dearer quantity C.P. of unit qty of 1st constituent.
m c d

C.P. of unit qty of 2 constituent.
Qc d − m = Qd m − c

nd

Mean Cost price of unit qty of mixture

d–m

m–c

Gives us the ratio of quantities in which the two ingredients should be mixed to get the mixture.

Ex.2 A sum of Rs 39 was divided among 45 boys and girls. Each girl gets 50 paise, whereas a boy gets one rupee. Find the number of boys and girls. Sol. Average amount of money received by each =

39 13 = Rs 45 15
1 2

Amount received by each girl = 50 paise = Rs Amount received by each boy = Re. 1 By alligation rule:
13 1 − Number of boys 15 2 11 = = 13 4 Number of girls 1− 15
11 × 45 = 33 11 + 4 Number of girls = 45 – 33 = 12.
∴ Number of boys =

Boys
1

Girls 1 2
13 15

⎡ 13 1 ⎤ ⎢ 15 − 2 ⎥ ⎦ ⎣

⎡ 13 ⎤ ⎢1 − 15 ⎥ ⎦ ⎣

Important Funda
Always identify the ingredients as cheaper & dearer to apply the alligation rule. In the alligation rule, the variables c, d & m may be expressed in terms of percentages (e.g. A 20% mixture of salt in water), fractions (e.g. two-fifth of the solution contains salt) or proportions (e.g. A solution of milk and water is such that milk : Water = 2 : 3) . The important point is to remember is that c & d may represent pure ingredients or mixtures.

(Mixing a Pure component to a solution)
Ex.3 A jar contains a mixture of two liquids A and B in the ratio 4 : 1. When 10 litres of the liquid B is poured into the jar, the ratio becomes 2: 3. How many litres of liquid A were contained in the jar? Sol. Method 1: (weighted average or equation method) Let the quantities of A & B in the original mixture be 4x and x litres.

According to the question

4x 2 = . x + 10 3

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12x = 2x + 20

⇒ 10x = 20

⇒x=2

The quantity of A in the original mixture = 4x = 4 × 2 = 8 litres.
Method 2: (Alligation with composition of B) The average composition of B in the first mixture is 1/5. The average composition of B in the second mixture = 1 The average composition of B in the resultant mixture = 3/5 Hence applying the rule of Alligation we have [1 – (3/5)]/[(3/5) – (1/5)] = (2/5)/(2/5) = 1 So, initial quantity of mixture in the jar = 10 litres.

B in 1st
1 5

B in 2nd
1
3 5

And, quantity of A in the jar = (10 × 4)/5 = 8 litres.
Method 3: (Alligation with percentage of B)

1−

3 5

3 1 − 5 5

The percentage of B in 1st mixture = 20% The percentage of B in 2nd mixture = 100% The percentage of B in Final Mixture = 60% By rule of allegation we have Volume 1st : Volume 2nd = (100% - 60%) : (60% - 20%) V1 : V2 = 1 : 1 Volume of mixture 1st = 10 litres Volume of A in mixture 1st = 80% of 10 litres = 8 litres. Answer

B in 1st 20%

B in 2nd 100% 60%

(100–60)%

(60–20)%

Removal and Replacement
If a vessel contains “x” litres of milk and if “y” litres be withdrawn and replaced by water, then if “y” litres of the mixture be withdrawn and replaced by water, and the operation repeated ‘n’ times in all, then :
Milk left in vessel after nth operation ⎡ x − y ⎤ y⎤ ⎡ =⎢ ⎥ = ⎢1 − x ⎥ Initial quantity of Milk in vessel x ⎣ ⎦ ⎣ ⎦
n n

Ex.4 Nine litres of solution are drawn from a cask containing water. It is replaced with a similar quantity of pure milk. This operation is done twice. The ratio of water to milk in the cask now is 16 : 9. How much does the cask hold? Sol. Let there be x litres in the cask After n operations
Water left in vessel after n operations = (1 – 9/x)n Original quantity of water in vessel

(1 – 9/x)2 = 16/25
∴ x = 45 litres.

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Ex.5 There are two containers A and B of milk solution. The ratio of milk and water in container A is 3 : 1 and in container B, it is 4 : 1. How many liters of container B solution has to be added to 20 lts of container ‘A’ solution such that in the resulting solution; the ratio of milk to water should be 19 : 6? Sol.

In container A, the part of milk = In container B, it is Required = 4 4 = . 4 +1 5

3 3 = . 3 +1 4

19 19 . = 19 + 6 25 Use allegation:
Container A 3 4 Container B 4 5

19 25 1 25
⇒4:1

1 100

It is given 20 lts of container A is added. So, the quantity of container B should be 5 lts.
Ex.6 There are two alloys A and B. Alloy A contains zinc, copper and silver, as 80% 15% and 5% respectively, whereas alloy B also contains the same metals with percentages 70%, 20%, 10% respectively. If these two alloys are mixed such that the resultant will contain 8% silver, what is the ratio of these three metals in the resultant alloy? Sol. Since, the resultant alloy contains 8% silver, first we will find, in what ratio these two alloys A and B

were mixed to form the resultant. Then the resultant zinc percentage is
2 × 80 + 3 × 70 = 74%. 2+3

Alloy A

Alloy B

5 8 2

10

So, copper percentage = 100 – (74 + 8) = 18
∴ The ratio of these metals = 74 : 18 : 8 = 37 : 9 : 4.

3

Ex.7 The cost of an apple is directly proportional to square of its weight in a fruit bazaar. Two friends A and B went there to purchase apples. A got exactly 5 apples per kg and each apple is of same weight. Where as B got exactly 4 apples per kg each weight is exactly same. If B paid Rs. 10 more than A per kg apples, what is the cost of an apple which weighs 1 kg? Sol.

It is given cost ∝ (weight)2
⇒ c = k w2.

A got 5 apples per kg and each apple is of same weight.
∴ His cost per apple = k( /5) .
1 2

⇒ Each apple is 200 gm. = 1/5 kg.

Since, he will get 5 apples per kg, so his cost per kg = 5k(1/5)2 = k/5 Similarly, the cost per B is 4k(1/4)2. It is given 4k (k/4)2 – 5k (k/5)2 = Rs. 10. ⇒ k = 200 Our required answer is k (1)2 = Rs. 200.

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