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Central Tendency of data is called an Average. Thus an average is a single value that indicates a group of values. Average can be calculated with the help of mean, median and mode. The most commonly used average is the Arithmetic Mean (AM). While some other average geometric mean and harmonic mean are also quite popular.
Arithmetic Mean is calculated by dividing the sum of all the number/quantities by the number of numbers/quantities. The mean of n numbers
x1, x2, x3, … , xn, is denoted by X and Calculated as
= ∑xi / n
where ∑ (sigma) denotes the sum of the term of type I from 1 to n.
Average is a very simple but effective way of representing an entire group by a single value.
(Standard Types In Problem Execution And Solving)
STIPE 01: To find the average when the number of quantities and their sum is given. We have the following formula, Average = Sum of the quantities / Number of quantities.
STIPE 02: To find the sum, when the number of quantities and their average is given. We have the following, Sum of quantities = Average × Number of quantities.
STIPE 03: To find the number of quantities, when the sum of quantities and average are given. We have the following formula, Number of quantities = Sum of quantities / Average.
AVERAGE SHORTCUT METHODS
1. Average = [Total of observations / No. of observations]
2. (i) When a person joins a group in case of increasing average Age
weight of new comer = [ (Previous Age + No. of persons) * Increase in Average ](ii) In case of decreasing Average, Age (or) weight of new comer =
[ (Previous Age - No. of persons) * Decrease in Average ]
3. When a person leaves a group and another person joins the group in the place of person left, then
(i) In case of increasing average, Age (or) weight of new comer =
[ (Age of person left + No. of persons) * Increase in Average ]
(ii) In case of decreasing Average, Age (or) weight of new comer =
[ (Age of person left - No. of persons) * Decrease in Average ]
4.When a person leaves the group but nobody joins this group, then
(i) In the case of increasing Average, Age (or) weight of man left =
[ (Previous Age - No. of present persons) * Increase in Average ]
(ii) In case of decreasing Average, Age (or) weight of new comer =
[ (Previous Age + No. of present persons) * Decrease in Average ]5. If a person travels a distance at a speed of x Km/hr returns to the
original place of y Km/hr then average speed is [ 2.x.y / ( x + y ) ]
6.If half of the journey is travelled at speed of x km/hr and the next half at a speed of x km/hr. Then average speed during the whole journey is [ 2.x.y / (x + y) ]
7. If a person travels 3 equal distances at a speed of x Km/hr, y Km/hr,z km/hr. Then average speed during whole journey is [ 3.x.y / (x.y +y.x +z.x) ]
1. Find the average of all prime numbers between 30 and 50.
Sol. There are five prime numbers between 30 and 50.
They are 31, 37, 41, 43 and 47.
Required average = [(31 + 37 + 41 + 43 + 47) / 5] = 199/5 = 39.8.
2. Find the average of first 40 natural numbers.
Sol. Sum of first n natural numbers = 40 * 41 / 2 = 820.
Required average = 820 / 40 = 20.5
3. Find the average of first 20 multiples of 7.
Sol. Required average = 7(1+2+3 ... + 20) / 20 = [7 * 20 * 21 / 20 * 2] = [147 / 2] = 73.5.
4. The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53kg is replaced by a new man. find the weight of the new man
Sol. Total weight increased = (1.8 * 10)kg = 18kg.
Weight of the new man = (53 + 18)kg = 71kg.
5. A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. find his average after 17th inning.
Sol. Let the average after 17th inning = x.
Then, average after 16th inning = (x - 3)
16(x - 3) + 87 = 17x or x = (87 - 48) = 39.
Sol: Therefore age of the mother
= ( 12 × 7 – 7 × 6)
= 42 years
7. The age of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is
Sol: Therefore excluded number
= (27 × 5) - ( 25 × 4)
= 135 – 100
8. Three years ago, the average age of A and B was 18 years. With C joining them, the average age becomes 22 years. How old is C now?
Sol: Present age of (A + B) = (18 × 2 + 3 × 2) years = 42 years.
Present age of (A + B + C) = (22 × 3) years = 66 years.
Therefore C’s age = (66 – 42) years
= 24 years.
9. The average age of 36 students in a group is 14 years. When teacher’s age is included to it, the average increases by one. What is the teacher’s age in years?
SoL : Age of the teacher = ( 37 × 15 – 36 × 14 ) years = 51 years
10. The average runs of a cricket player of 10 innings was 32. How many runs must he makes in his next innings so as to increase his average of runs by 4?
Sol: Average after 11 innings = 36.
Therefore required number of runs
= (36 × 11) – (32 × 10)
= 396 - 320
11. The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is
Sol: Let the total number of workers be x. Then,
8000x = (12000 × 7) + 6000 ( x – 7)
‹=› 2000x = 42000
‹=› x = 21.
= 76 + 65 + 82 + 67 + 85 / 5)
= (375 / 5) = 75.
13. In the first 10 over’s of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 0vers to reach the target of 282 runs?
Sol: Required run rate = 282 – (3.2 × 10 / 40)
= 240 / 40
14. The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is
Sol: Weight of the teacher
= (35.4 × 25 – 35 × 24)kg
= 45 kg.
15. After replacing an old member by a new member, it was found that the average age of five numbers of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member?
Sol: Age decrease = (5 × 3) years = 15 years.
So, the required difference = 15 years
16. The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
Sol: Total weight increased
= (8 × 2.5) kg
= 20 kg.
Weight of new person
= (65 + 20) kg
= 85 kg.
Sol: Runs scored in the first 10 overs = 10×3.2=32
Total runs = 282, remaining runs to be scored = 282 - 32 = 250
remaining over’s = 40, Run rate needed = 250 40 =6.25
Sol: Let the sale in the sixth month = x
Then 6435+6927+6855+7230+6562+x 6 =6500
=> 6435 + 6927 + 6855+ 7230 + 6562 + x = 6 \times 6500 = 39000
=> 34009 + x = 39000, x = 39000 - 34009 = 4991
Sol: Average of 20 numbers = 0
=>Sum of 20 numbers 20 =0
=> Sum of 20 numbers = 0
Hence at the most, there can be 19 positive numbers.
(Such that if the sum of these 19 positive numbers is x, 20th number will be -x)
20. The average age of a class is 15.8 years. The average age of the boys in the class is 16.4 yrs while that of the girls is 15.4 years. What is the ratio of boys to girls in the class?
21. If a, b, c, d, e are five consecutive odd numbers, their average Is:
22. The average of first five multiples of 3 is
Since the month begins with a Sunday, to there will be five Sundays in the month.
Weight of new person = (65 + 20) kg = 85 kg.
26. The average (arithmetic mean) of 3 numbers is 60. If two of the numbers are 50 and 60, what is the third number?