Students who are getting confused by understanding the AM Properties with proof. If that’s the concern then have a look at this **Properties Questions on Arithmetic Mean** article. In this article, you will see how to solve the different example questions on average/mean/AM by using properties of arithmetic mean.

Simply follow the steps covered in the solution of Problems Based on Average & understand the explanations for solving various complex questions on the arithmetic mean properties. Also, Improve your math proficiency by answering the various example word problems on Mean/Average/Arithmetic Mean in Statistics.

Do check:

### Example Questions on Arithmetic Mean Properties with Solutions PDF

**Example 1: **

The mean of nine numbers is 40. If four is subtracted from each number, what will be the new average.

**Solution: **

Let the given numbers be x_{1}, x_{2}, x_{3},…., x_{9}

Then, the mean of these numbers = (x_{1}+x_{2}+x_{3} + …. +x_{9})/9

Hence, (x_{1}+x_{2}+x_{3} + …. +x_{9})/9 = 40

⇒ (x_{1}+x_{2}+x_{3} + …. +x_{9}) = 40 x 9

⇒ (x_{1}+x_{2}+x_{3} + …. +x_{9}) = 360 ……(i)

The new numbers are (x_{1} – 4), (x_{2} – 4),…., (x_{9} – 4)

Mean of the new numbers = {(x_{1} – 4)+ (x_{2} – 4)+….+ (x_{9} – 4)} / 9

= [(x_{1}+x_{2}+x_{3} + …. +x_{9}) – 36] / 9

= [360-36]/9, [using (i)]

= 324/9

= 36

Hence, the new mean is 36.

**Example 2:
**Find the average of the first five prime numbers.

**Solution:**

Given quantities are First five prime numbers

The first five prime numbers are 2, 3, 5, 7 and 11

The formula of Arithmetic mean or the Average = Sum of the quantities/Number of quantities

= 2+3+5+7+11 / 5

= 28/5

= 5.6

**Example 3: **

Find the second term of the given A.M terms \(\frac{1}{log_{3}3} \), X , \(\frac{1}{log_{18}3}\)

**Solution:
**Given terms are \(\frac{1}{log_{3}3} \), X , \(\frac{1}{log_{18}3} and this can be rewritten as log

_{3}3, X, log

_{3}18

⇒ 2X = log

_{3}3 + log

_{3}21 = log

_{3}3 + log

_{3}(6×3)

⇒ 2X = 2 ( log

_{3}3) + 2 log

_{3}3

⇒ 2X = 2 ( log

_{3}3) + (2 x 1 )

⇒ 2X = 2(1) + 2

⇒ X = 4/2

⇒ X = 2

**Example 4:
**The average of 10 numbers is 5. If 2 is added to every number, what will be the new average?

**Solution:**

Let the given numbers be x

_{1}, x

_{2}, x

_{3},…., x

_{10}

Then, the average of given numbers = x

_{1}, x

_{2}, x

_{3},…., x

_{10}/ 10

Hence, (x

_{1}+x

_{2}+x

_{3}+…+x

_{10})/10 = 5

⇒ (x

_{1}+x

_{2}+x

_{3}+…+x

_{10}) = 50 …….(i)

The new numbers are (x

_{1}– 4), (x

_{2}– 4),…., (x

_{10}+ 2)

Mean of the new numbers = (x

_{1}+ 2), (x

_{2}+ 2),…., (x

_{10}+ 2) / 10

= (x

_{1}+x

_{2}+x

_{3}+…+x

_{10}) + 20 / 10

= (50 + 20)/10 [Using (i)]

= 70/10

= 7.

Hence, the new mean is 9.

**Example 5: **

In the exams, the mean of marks scored by 30 students was calculated as 60. Next, it was identified that the marks of one student were wrongly copied as 57 instead of 75. Find the correct mean?

**Solution:**

Given that Mean of marks = [latex]\frac { Incorrect sum of marks of 30 students }{ 30 } \)

\(\frac { Incorrect sum of marks of 30 students }{ 30 } \) = 60

An incorrect sum of marks of 40 students = 60 x 30 = 1800

Considering that the marks of one student were wrongly copied as 57 instead of 75,

Then, the correct sum of marks of 30 students = 1800 – 57 + 75 = 1818

Finally, correct mean = 1818 / 30 = 60.6

**Example 6: **

The mean marks of two batches of students having 60 and 40 students respectively are 35 and 65. Find the average marks of all the 100 students, taken together.

**Solution:
**Let x be the average marks of all 100 students taken together.

Batch – I (\(\overline{x}\))

Marks (x

_{1}) = 35

No. of students n

_{1}= 60

Batch – II (\(\overline{x}\))

Marks (x

_{2})= 65

No. of students n

_{2}= 40

\(\overline{x}\) = \(\frac {n

_{1}[latex]\overline{x

_{1}}\) + n

_{2}\(\overline{x

_{2}}\)}{ n

_{1}+n

_{2}} [/latex]

= \(\frac { 60×35 + 40×65 }{ 60+40 } \)

= \(\frac { 4700 }{ 100 } \)

= 47.

Therefore, \(\overline{x}\) = 47 marks.