# Average

**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 x_{1},x_{2},……x_{n} is

A = (x_{1}+ x_{2}........x_{n})/ n

**Important points:**

1. Sum of first n natural number =n(n + 1)/2Average of first n natural number =(n + 1)/22. Sum of square of first n natural number =n(n+1)(2n+1)/6Avg. of square of first n natural number =(n+1)(2n+1)/63. Sum of cube of first n natural number =[n(n+1)/2]Avg. of cube of first n natural number =^{2}n(n+1)4. Sum of first n natural odd number =^{2/4}nAvg. of first n natural odd number =^{2}n5. Sum of first n natural even number =n(n+1)Avg. of first n natural even number =n+1

**Sample problems**

**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

= 187/2

**Question 2:** The average of 9 observations is 87. If the average of the first five observations is 79 and the average of the next three is 92. Find the 9th observation. **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 the Husband and wife was 25 years, today the average age of the 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 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

The child was born after 1 year of marriage so present age of the 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 are 31^{o}, and the average temp. of Tuesday, Wednesday, Thursday, and Friday are 29.5^{o}. 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 = 30^{o}

**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 calculated as 61 and 34. Find his correct average it was also found that the total number is 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 in school is 13 years and of girls is 12 years. If the total number of boys is 240, then find the number of girls if the average of school is 12 years 8 months. **Solution:** Average age of 240 boys = 13 years

Sum of the age of 240 boys = 240 x 13

Let a be the number of girls in school.

than Sum of the 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**. **Alternate solution:**

**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 numbers 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.

Given

(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 a speed of 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

= 240/5

= 48

Hence, the average speed is **48 km/hr**.