Average a number expressing the central or typical value in a set of data, in particular the mode, median, or (most commonly) the mean, which is calculated by dividing the sum of the values in the set by their number.
The basic formula for the average of n numbers x1,x2,……xn is
A = (x1 + x2 ........xn)/ n
1. Sum of first n natural number = n(n + 1)/2 Average of first n natural number = (n + 1)/2 2. Sum of square of first n natural number = n(n+1)(2n+1)/6 Avg. of square of first n natural number = (n+1)(2n+1)/6 3. Sum of cube of first n natural number = [n(n+1)/2]2 Avg. of cube of first n natural number = n(n+1)2/4 4. Sum of first n natural odd number = n2 Avg. of first n natural odd number = n 5. Sum of first n natural even number = n(n+1) Avg. of first n natural even number = n+1
Question :1 Find the average of the square of first 16 natural number is:
Solution : We know that
Sum of square of first n natural number = n(n+1)(2n+1)/6
Avg. of square of first n natural number = (n+1)(2n+1)/6
So, Avg. of square of first 16 natural number = (16+1)(2×16+1)/6
= 17 x 33 /6
Question 2: The average of 9 observations is 87. If the average of first five observations is 79 and the average of next three is 92. Find the 9th observations.
Solution : Average of 9 observations = 87
So, Sum of 9 observations = 87 x 9 = 783
Average of first 5 observations = 79
The sum of first 5 observations = 79 x 5 = 395
The sum of 6th,7th and 8th = 92 x 3 = 276
9th number = 783 – 395 – 276 = 112
Question 3: Five years ago the average of Husband and wife was 25 years, today the average age of Husband, wife and child is 21 years. How old is the child.
Solution : H + W = 25
Sum of ages of both 5 years before = 25×2 = 50
Today, sum of there ages is = 50 + 5 + 5 = 60
Today avg. of H + W + C = 21
Sum of ages of H , W and C = 21×3 = 63
Age of child = 63 – 60 = 3 years
Question 4: The average of mother, father and son was 44 years at the time of marriage of the son. After 1 year an infant was born and after 5 years of the marriage the average age of the family becomes 37 years. Find the age of the bride at the time of the marriage.
Solution : Sum of ages of F + M + S = 44 x 3 = 132 years
Sum of ages of M + F + S + D + C = 37 x 5 = 185 years
Sum of ages of (M + F + S) = 132 + 3×5 = 147 years
Sum of ages of D + C after 5 years = 185 – 147 = 38 years
D + C = 38
Child was born after 1 year of marriage so present age of child is 4 year
then D + 4 = 38
D = 34 years at present
So,at the time of marriage bride age was= 34 – 5 = 29 years
Question 5: The average temp. of Monday,Tuesday ,Wednesday and Thursday is 31o and the average temp. of Tuesday,Wednesday,Thursday and Friday is 29.5o. If the temp of Friday is 4/5 times of Monday. Find the temp of Monday.
Solution :Find Sum
Sum of temp. of M + T + W + Th = 31 x 4 = 124……….(1)
Sum of temp. of T + W + Th + F = 29.5 x 4 = 118………(2)
Subtract (2) for (1)
M – F = 6
Given F = (4/5)M
M/F = 5/4
then 5x – 4x = 6
x = 6
Temp. of Monday = 5 x 6 = 30o
Question 6: There are 42 students in a hostel. If the number of students increased by 14. The expense of mess increased by Rs 28 per day. While the average expenditure per head decreased by Rs 2. Find the original expenditure.
Solution : Total students after increment = 42 + 14 = 56
Let the expenditure of students is A Rs/day.
Increase in expenditure Rs 28/day.
Acc. to question
42A + 28 = 56(A – 2)
42A + 28 = 56A – 112
14A = 140
A = 10
Hence, the original expenditure of the student was Rs 10/day.
Question 7: The average of 200 numbers is 96 but it was found that 2 numbers 16 and 43 are mistakenly calculate as 61 and 34. Find his correct average it was also found that total number are only 190.
Solution : Average of 200 numbers = 96
Sum of 200 numbers = 96 x 200 = 19200
Two numbers mistakenly calculated as 61 and 34 instead of 16 and 43.
So, 61 + 34 = 95
16 + 43 = 59
Diff = 95 – 59 = 36
So,Actual sum of 200 numbers = 19200 – 36 = 19164
total numbers are also 190 instead of 200.
So, correct average = 19164/190 = 100.86
Question 8: The average age of boys of school is 13 years and of girls is 12 years. If the total number of boys are 240 , then find the number of girls if the average of school 12 years 8 months.
Solution : Average age of 240 boys = 13 years
Sum of age of 240 boys = 240 x 13
Let a be the number of girls in school.
then Sum of age of girls = 12a
Total number of boys and girls in school = (240 + a)
Total sum of age of boys and girls = (240 + a)12 year 8 month
Acc. to question
240x13y + 12a = (240 + a)12y8m
change years into month
240x13x12 + (12×12)a = (240 + a)(12×12 + 8)
37440 + 144a = 36480 + 152a
8a = 960
a = 120
Hence, the number of girls in school is 120.
Question 9: A batsman scored 120 runs in his 16th innings due to this his average increased by 5 runs. Find his current average.
Solution :Let the average of 15 innings is A.
Acc. to question
15A + 120 = 16(A + 5)
=>15A + 120 = 16A + 80
=>A = 40
Hence, current average of the batsman is (40 + 5) = 45
Question 10: There are three natural numbers if the average of any two number is added with the third number 48,40 and 36 will be obtained. Find all the natural numbers.
Solution : Let a,b and c are the numbers.
(a+b)/2 + c = 48
=> a + b + 2c = 96 ………(1)
(b+c)/2 + a = 40
=> 2a + b + c = 80 ……….(2)
(c+a)/2 + b = 36
=> a + 2b + c = 72 ……….(3)
Add (1)(2)(3), we get
4(a + b + c) = 248
a + b + c = 62
Put value of (a+b+c) in (1)(2) and (3)to get individual values
1)(a+b+c) + c = 96
62 + c = 96
c = 34
2)a + (a+b+c) = 80
a + 62 = 80
a = 18
3) b + (a+b+c) = 72
b + 62 = 72
b = 10
Question 11: A biker travels at speed of 60 km/hr from A to B and returns with speed 40 km/hr. What is the average speed of the total journey?
Solution : Let a is the distance between A and B.
Total distance travel in journey = 2a
Time to travel from A to B = Distance/speed = a/60
Time to travel from B to A = Distance/speed = a/40
Total time of journey = a/60 + a/40
Average speed = Total distance/total time
=2a / (a/60 + a/40)
=240 x 2a /10a
Hence, the average speed is 48 km/hr.
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