Selina Solutions are considered to be very useful when you are preparing for the ICSE Class 9 maths exams. Here, we bring to you detailed answers and solutions to the exercises of Selina Solutions for Class 9 Maths Chapter 19- Mean and Median. These questions have been devised by the subject matter experts as per the syllabus prescribed by the CISCE for the ICSE.

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Exercise 19A page: 238

**1. Find the mean of 43, 51, 50, 57 and 54.**

**Solution:**

Numbers given are 43, 51, 50, 57 and 54.

Mean of given numbers

= 51

**2. Find the mean of first six natural numbers.**

**Solution:**

First six natural numbers are 1, 2, 3, 4, 5, 6.

Mean of first six natural numbers

= 3.5

**3. Find the mean of first ten odd natural number.**

**Solution:**

First ten odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Mean of first ten odd numbers

**4. Find the mean of all factors of 10.**

**Solution:**

All factors of 10 are 1, 2, 5, 10

Mean of all factors of 10

= 4.5

**5. Find the mean of x + 3, x + 5, x + 7, x + 9 and x + 11.**

**Solution:**

Values given are x + 3, x + 5, x + 7, x + 9 and x + 11

Mean of the values

= x + 7

**6. If the different values of variable x are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5 and 11.1; find**

**Solution:**

(i) Numbers given are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5 and 11.1

= 5.8

(ii)

= (9.8 â€“ 5.8) + (5.4 â€“ 5.8) + (3.7 â€“ 5.8) + (1.7 â€“ 5.8) + (1.8 â€“ 5.8) + (2.6 â€“ 5.8) + (8.6 â€“ 5.8) + (10.5 â€“ 5.8) + (11.1 â€“ 5.8)

= 4 â€“ 4 â€“ 2.1 â€“ 4.1 â€“ 4 â€“ 3.2 â€“ 3 + 2.8 + 4.7 + 5.3

= 0

**7. The mean of 15 observations is 32. Find the resulting mean, if each observation is:**

**(i) Increased by 3**

**(ii) Decreased by 7**

**(iii) Multiplied by 2**

**(iv) Divided by 0.5**

**(v) Increased by 60%**

**(vi) Decreased by 20%**

**Solution:**

It is given that

Mean of 15 observations is 32

(i) Resulting mean if each observation is increased by 3 = 32 + 3 = 35

(ii) Resulting mean if each observation is decreased by 7 = 32 â€“ 7 = 25

(iii) Resulting mean if each observation is multiplied by 2 = 32 Ã— 2 = 64

(iv) Resulting mean if each observation is divided by 0.5 = 32/0.5 = 64

(v) Resulting mean if each observation is increased by 60% = 32 + 60/100 Ã— 32

= 32 + 19.2

= 51.2

(vi) Resulting mean if each observation is decreased by 20% = 32 â€“ 20/100 Ã— 32

= 32 â€“ 6.4

= 25.6

**8. The mean of 5 numbers is 18. If one number is excluded, the mean of remaining number becomes 16. Find the excluded number.**

**Solution:**

It is given that

Mean of 5 numbers is 18

Total sum of 5 numbers = 18 Ã— 5 = 90

Excluding an observation, the mean of the remaining 4 number becomes 16 = 16 Ã— 4 = 64

Sum of remaining 4 observations = Total of 5 observations â€“ Total of 4 observations

= 90 â€“ 64

= 26

**9. If the mean of observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find:**

**(i) The value of x;**

**(ii) The mean of first three observations.**

**Solution:**

(i) It is given that

Mean of observations x, x + 2, x + 4, x + 6 and x + 8 is 11

We know that

By taking out 5 as common

11 = [5 (x + 4)]/5

11 = x + 4

By transposing we get

x = 11 â€“ 4

x = 7

(ii) Mean of first three observations

= 9

**10. The mean of 100 observations is 40. It is found that an observation 53 was misread as 83. Find the correct mean.**

**Solution:**

It is given that

Mean of 100 observations is 40

x = 40 Ã— 100 = 4000

Here the incorrect value of x = 4000

So the correct value of x = Incorrect value of x â€“ Incorrect observation + Correct observation

Substituting the values

= 4000 â€“ 83 + 53

= 3970

We know that

= 3970/100

= 39.7

Exercise 19B page: 241

**1. Find the median of:**

**(i) 25, 16, 26, 16, 32, 31, 19, 28 and 35**

**(ii) 241, 243, 347, 350, 327, 299, 261, 292, 271, 258 and 257**

**(iii) 63, 17, 50, 9, 25, 43, 21, 50, 14 and 34**

**(iv) 233, 173, 189, 208, 194, 194, 185, 200 and 220.**

**Solution:**

(i) Arrange the numbers in ascending order

16, 16, 19, 25, 26, 28, 31, 32, 35

As n = 9 (odd)

= 5^{th} term

= 26

Hence, the median is 26.

(ii) Arrange the numbers in ascending order

241, 243, 257, 258, 261, 271, 292, 299, 327, 347, 350

As n = 11 (odd)

= 6^{th} term

= 271

Hence, the median is 271.

(iii) Arrange the numbers in ascending order

9, 14, 17, 21, 25, 34, 43, 50, 50, 63

As n = 10 (even)

Median = Â½ [value of (n/2)^{th} term + value of (n/2 + 1)^{th} term]

= Â½ [value of (10/2)^{th} term + value of (10/2 + 1)^{th} term]

= Â½ [25 + 34]

= Â½ [59]

= 29.5

Hence, the median is 29.5.

(iv) Arrange the numbers in ascending order

173, 185, 189, 194, 194, 200, 204, 208, 220, 223

As n = 10 (even)

Median = Â½ [value of (n/2)^{th} term + value of (n/2 + 1)^{th} term]

= Â½ [value of (10/2)^{th} term + value of (10/2 + 1)^{th} term]

= Â½ [200 + 194]

= Â½ [394]

= 197

Hence, the median is 197.

**2. The following data have been arranged in ascending order. If their median is 63, find the value of x.**

**34, 37, 53, 55, x, x + 2, 77, 83, 89 and 100.**

**Solution:**

Numbers given are 34, 37, 53, 55, x, x + 2, 77, 83, 89 and 100

As n = 10 (even)

Median = Â½ [value of (n/2)^{th} term + value of (n/2 + 1)^{th} term]

= Â½ [value of (10/2)^{th} term + value of (10/2 + 1)^{th} term]

= Â½ [value of 5^{th} term + value of (5 + 1)^{th} term]

= Â½ [value of 5^{th} term + value of 6^{th} term]

Substituting the values

63 = Â½ [x + x + 2]

[2 + 2x]/2 = 63

Taking 2 as common

x + 1 = 63

x = 62

**3. In 10 numbers, arranged in increasing order, the 7 ^{th} number is increased by 8, how much will the median be changed?**

**Solution:**

We know that for any given set of data, the median is the value of its middle term

Total observations n = 10 (even)

Median = Â½ [value of (n/2)^{th} term + value of (n/2 + 1)^{th} term]

= Â½ [value of (10/2)^{th} term + value of (10/2 + 1)^{th} term]

= Â½ [value of 5^{th} term + value of 6^{th} term]

Therefore, if the 7^{th} number is diminished by 8, there will be no change in the median value.

**4. Out of 10 students, who appeared in a test, three secured less than 30 marks and 3 secured more than 75 marks. The marks secured by the remaining 4 students are 35, 48, 66 and 40. Find the median score of the whole group.**

**Solution:**

Total observations n = 10 (even)

Median = = Â½ [value of (10/2)^{th} term + value of (10/2 + 1)^{th} term]

= Â½ [value of 5^{th} term + value of 6^{th} term]

Substituting the values

Median = Â½ [40 + 48]

= 88/2

= 44

Hence, the median score of the whole group is 44.

**5. The median of observations 10, 11, 13, 17, x + 5, 20, 22, 24 and 53 (arranged in ascending order) is 18; find the value of x.**

**Solution:**

Total observations n = 9 (odd)

As n is odd

= 5^{th} term

= x + 5

It is given that, Median = 18

x + 5 = 18

x = 13

Exercise 19C

**1. Find the mean of 8, 12, 16, 22, 10 and 4. Find the resulting mean, if each of the observations, given above be:**

**(i) Multiplied by 3.**

**(ii) Divided by 2.**

**(iii) Multiplied by 3 and then divided by 2.**

**(iv) Increased by 25%.**

**(v) Decreased by 40%.**

**Solution:**

Mean of the given data

= 12

(i) Multiplied by 3

If xÌ„ is the mean of n number of observations x_{1}, x_{2}, x_{3}, â€¦â€¦. x_{n},

Mean of ax_{1}, ax_{2}, ax_{3}, â€¦â€¦. ,ax_{n} is axÌ„

When each of the given data is multiplied by 3, then mean is also multiplied by 3

Mean of the original data = 12

Therefore, the new mean = 12 Ã— 3 = 36

(ii) Divided by 2

If xÌ„ is the mean of n number of observations x_{1}, x_{2}, x_{3}, â€¦â€¦. x_{n},

Mean of x_{1}/a, x_{2}/a, x_{3}/a, â€¦â€¦ x_{n}/a is xÌ„/a

When each of the given data is divided by 2, the mean is also divided by 2.

Mean of the original data = 12

Therefore, the new mean = 12/2 = 6

(iii) Multiplied by 3 and then divided by 2

If xÌ„ is the mean of n number of observations x_{1}, x_{2}, x_{3}, â€¦â€¦. x_{n},

Mean of a/b x_{1}, a/b x_{2}, a/b x_{3}, â€¦.. a/b x_{n} is a/b xÌ„

When each of the given data is multiplied by 3/2, the mean is also multiplied by 3/2.

Mean of original data = 12

Therefore, the new mean = 3/2 Ã— 12 = 36/2 = 18

(iv) Increased by 25%

We know that

New mean = Original mean + 25% of original mean

New mean = 12 + 25% of 12

New mean = 12 + 25/100 Ã— 12

So we get

New mean = 12 + Â¼ Ã— 12

New mean = 12 + 3

New mean = 15

(v) Decreased by 40%

We know that

New mean = 12 â€“ 40% of 12

New mean = 12 â€“ 40/100 Ã— 12

So we get

New mean = 12 â€“ 2/5 Ã— 12

New mean = 12 â€“ 0.4 Ã— 12

New mean = 12 â€“ 4.8

New mean = 7.2

**2. The mean of 18, 24, 15, 2x + 1 and 12 is 21. Find the value of x.**

**Solution:**

We know that

Mean of given data

By cross multiplication

5 Ã— 21 = 70 + 2x

105 = 70 + 2x

On further calculation

2x = 105 â€“ 70

2x = 35

x = 35/2 = 17.5

**3. The mean of 6 numbers is 42. If one number is excluded, the mean of remaining number is 45. Find the excluded number.**

**Solution:**

If xÌ„ is the mean of n number of observations x_{1}, x_{2}, x_{3}, â€¦â€¦. x_{n}

It is given that mean of 6 numbers is 42

x_{1} + x_{2} + x_{3} + â€¦â€¦ +x_{6} = 42 Ã— 6

x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 252 â€“ x_{6} â€¦â€¦ (1)

It is also given that the mean of 5 numbers is 45

x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 45 Ã— 5

x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 225 â€¦â€¦ (2)

From equating both the equations

252 â€“ x_{6} = 225

x_{6} = 252 â€“ 225 = 27

**4. The mean of 10 numbers is 24. If one more number is included, the new mean is 25. Find the included number.**

**Solution:**

If xÌ„ is the mean of n number of observations x_{1}, x_{2}, x_{3}, â€¦â€¦. x_{n}

It is given that the mean of 10 numbers is 24

x_{1} + x_{2} + x_{3} + â€¦â€¦ +x_{10} = 24 Ã— 10

x_{1} + x_{2} + x_{3} + x_{4} + â€¦â€¦ +x_{10} = 240

x_{1} + x_{2} + x_{3} + x_{4} + â€¦â€¦ +x_{10} = 240 + x_{11} â€¦â€¦ (1)

It is also given that the mean of 11 numbers is 25.

x_{1} + x_{2} + x_{3} + x_{4} + â€¦â€¦ +x_{10} +x_{11 }= 11 Ã— 25

x_{1} + x_{2} + x_{3} + x_{4} + â€¦â€¦ +x_{10} +x_{11 }= 275 â€¦.. (2)

From equating both the equations

240 + x_{11 }= 275

x_{11 }= 275 â€“ 240 = 35

**5. The following observations have been arranged in ascending order. If the median of the data is 78, find the value of x.**

**44, 47, 63, 65, x + 13, 87, 93, 99, 110.**

**Solution:**

The data given is

44, 47, 63, 65, x + 13, 87, 93, 99, 110

Total number of observations (n) = 9 which is odd

= 5^{th} term

= x + 13

It is given that the median is 78

x + 13 = 78

x = 78 â€“ 13 = 65

## Selina Solutions for Class 9 Maths Chapter 19- Mean and Median

The Chapter 19, Mean and Median, contains 3 exercises and the Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

19.1 Mean of Ungrouped data

19.2 Properties of mean

19.3 Median

## Selina Solutions for Class 9 Maths Chapter 19- Mean and Median

In chapter 19 of Class 9, the students are taught about a process in which the mean and median of a given data is calculated. Mean is the average of a given data. Study the Chapter 19 of Selina textbook to understand more about Mean and Median. Learn the Selina Solutions for Class 9 effectively to come out with flying colours in the examinations.