Mean and Median Class-6 RS Aggarwal ICSE Maths Goyal

Mean and Median Class-6 RS Aggarwal ICSE Maths Goyal Brothers Prakashan Chapter-26 Solutions. We provide step by step Solutions of Exercise / lesson-26 Mean and Median of Plane Figure for ICSE Class-6 RS Aggarwal Maths.

Our Solutions contain all type Questions with Exe-26 A and Exe-26 B with Notes on Mean and Median to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6 Mathematics.

Board ICSE
Publications Goyal brothers Prakshan
Subject Maths
Class 6th
Chapter-26 Mean and Median 
Writer RS Aggrawal
Book Name Foundation
Topics Solution of Exe-26 A and Exe-26 B
Academic Session 2021-2022

Mean and Median Class-6 RS Aggarwal ICSE Maths Goyal Brothers Prakashan Chapter-26 Solutions


– : Select Topics :-

Exe-26 A

Exe-26 B

Notes


What is the need of Mean Median Mode Formula for Student?

No concept in mathematics is useful if it is not applied to the real-world problem and the same is true for mean median mode too. They are suitable for a plenty of real-life problems in different situations. You have to analyze the situation deeply to check where to use Mean, median, and mode and find the appropriate one for that particular situation

Mean

Mean always lies between the greatest and smallest observation of the data.

Range is the difference between the highest and the lowest observation of the data. i.e. Range = Highest observation – Lowest observation

Median

Median refers to the value which lies in the middle of the data with half of the observations above it and the other half below it.

e.g. 24, 36,46,17,18, 25, 35 is given data.
Firstly, data is to arranged in ascending order i.e. 17,18, 24,25, 35,36, 46.
Since the median is the middle observation, therefore 25 is the median.

If the data has an odd number of items, then the median is the middle number.

If the data has an even number of items, then the median is mean of two middle numbers.

What is Mean Median ?

In statistics, mean, median, are different types of averages and used commonly to solve complex problems in real world. Mean is the average where you add all the number and divide the sum of numbers by count of numbers.

Median is taken as the middle value of the list. To find the median, you need to write numbers from smaller to largest and then find the median.

List of Basic Median Formula

Median is the middle value in a data set when numbers are arranged in increasing order i.e. from smaller to larger. There are certain basic median formulas and steps to find the median of the list.

  • First of all, arrange the numbers in perfect order i.e. smaller to largest.
  • If the total number of the count in the list is odd then finding median is easy.
  • In case, the total number of the count in the list is even then median would be average of two middle points on the list.

The median is good to use when data include exceptionally higher or lower values that are difficult to calculate without basic median formulas.

If the total number of numbers(n) is an odd number, then the formula is

Median = (n+1)th /2 term value

If the total number of the numbers(n) is an even number, then the formula is

median


Exe-26 A

Mean and Median Class-6 RS Aggarwal ICSE Maths Goyal Brothers Prakashan Solutions

Page 283-284

Question 1:

Find the mean of :

(i) 11, 13, 17, 19, 23

(ii) 22, 24, 26, 28, 30, 32, 34, 36

(iii) 1/4, 3(1/4), 4(3/4), 5(1/4), 7(1/4)

(iv) 6.5, 8.2, 9.4, 4.6, 7.8, 4.9

Answer :

(i) 11, 13, 17, 19, 23

Sum of observations 11 + 13 + 17 + 19 + 23 = 83

Number of observations = 5

So. Mean = Sum of observations/Number of observations = 83/5

= 16.6

(ii) 22, 24, 26, 28, 30, 32, 34, 36

Sum of observations 22 + 24 + 26 + 28 + 30 + 32 + 34 + 36 = 232

Number of observations = 8

So. Mean = Sum of observations/Number of observations = 232/8

= 29

(iii) 1/4, 3(1/4), 4(3/4), 5(1/4), 7(1/4)

Sum of observations 1/4 + 3(1/4) + 4(3/4) + 5(1/4) + 7(1/4)

1/4 + 13/4 + 19/4 + 21/4 + 15/2 = (1 + 13 + 19 + 21 + 30)/4

= 84/4

= 21

Number of observations = 5

So. Mean = Sum of observations/Number of observations = 21/5

= 4(1/5)

(iv) 6.5, 8.2, 9.4, 4.6, 7.8, 4.9

Sum of observations 6.5 + 8.2 + 9.4 + 4.6 + 7.8 + 4.9 = 41.4

Number of observations = 6

So. Mean = Sum of observations/Number of observations = 41.4/6

= 6.9

Question 2:

Find the mean of :

(i) first eight natural numbers.

(ii) first five multiples of 6.

(iii) all prime numbers between 20 and 40.

Answer :

(i) the first eight natural numbers are: 1, 2, 3, 4, 5, 6, 7, 8

Sum of observations = 1 + 2 +3 + 4 +5 + 6 + 7 + 8 = 36

So, Number of observations = 8

Mean = 36/8

= 4.5

(ii) The first five multiples of 6 are : 6, 12, 18, 24, 30

Sum of observations = 6 + 12 + 18 + 24 + 30 = 90

Number of observations = 5

So, Mean = 90/5

= 18

(iii) Prime numbers between 20 and 40 are 23, 29, 31, 37

Sum of observations = 23 + 29 + 31 + 37 = 120

Number of observations = 4

So,  Mean = Sum of observations/Number of observations = 120/4

= 30

Question 3:

The number of children in 10 families of a locality are :

2, 4, 3, 4, 2, 0, 3, 5, 1, 6

Find the mean number of children per family

Answer :

Sum of observations = 2 + 4 + 3 + 4+ 2+ 0 + 3 + 5 + 1 + 6 = 30

Number of observations = 10

So,  Mean = Sum of observations/Number of observations = 30/10

= 3

Question 4:

The following are the number of books issued in a school library during a week :

105, 216, 322, 167, 273, 405, 346

Find the mean number of books issued per day

Answer :

Sum of observations 105 + 216 + 322 + 167 + 273 + 405 + 346 = 1834

Number of observations = 7

So,  Mean = Sum of observations/Number of observations = 1834/7

= 262

Question 5:

The daily temperature recorded (in degree F) at a place during a week was as under :

Monday Tuesday Wednesday Thursday Friday Saturday
35.5 30.8 27.3 32.1 23.8 29.9

Find the mean temperature :

Answer :

Sum of observations = 35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9 = 179.4

Number of observations = 6

So,  Mean = Sum of observations/Number of observations = 179.4/6

= 29.9°F

Question 6:

The mean of 9, 14, x, 16, 7 and 18 is 11.5. Find the value of x.

Answer :

The given mean = 11.5

The given observations is 9, 14, x, 16, 7, 18

Sum of observations = 9 + 14 + x + 16 + 7+ 18 = 64 + x

Number of observations = 6

So,  Mean = Sum of observations/Number of observations

11.5 = 64 + x

= 69 = 64 + x

x = 69 – 64

x = 5

Question 7:

The mean of 7, 9, x + 3, 12, 2x- 1 and 3 is 9. Find the value of x.

Answer :

The given mean = 9

The given observations is 7, 9, x + 3, 12, 2x – 1 and 3

Sum of observation = 7 + 9 + x + 3 + 12 + 2x – 1 + 3 =  3x + 33

Number of observations = 6

So,  Mean = Sum of observations/Number of observations

=> 9 = (3x + 33)/6

=> 54 =3x + 33

=> 3x = 21:

=> x = 7


Exe-26 B

Mean and Median Class-6 RS Aggarwal ICSE Maths Goyal Brothers Prakashan Solutions

Page 285

Question 1:

Find the median of each of the following data :

(i) 7, 11, 20, 6, 3, 16, 15, 21, 12

(ii) 9, 25, 32, 51, 7, 16, 37, 50, 0, 13, 19

(iii) 5.6, 7.2, 1.8, 4.3, 9.1, 2.6, 3.4

(iv) 122, 127, 109, 118, 125, 108

Answer :

(i) 7, 11, 20, 6, 3, 16, 15, 23, 12

Arranging the given series in ascending order ex. 3, 6,7, 11, 12, 15, 16, 20, 23

Number of terms (n) = 9

Middle term = 1/2(n + 1)th term

= 1/2(9 + 1)th term

5th term = 12

So, Median 12

(ii) 9, 25, 32, 51, 7, 16, 37, 50, 0, 13, 19

Arranging the given series in ascending order ex, 0, 7, 9, 13, 16, 19, 25, 32, 37, 50, 51

Number of terms (n) = 11

Middle term = 1/2(n + 1)th term

= 1/2(11 + 1)th term

6th term  = 19

So, Median = 19

(iii) 5.6, 7.2, 1.8, 4.3, 9.1, 2.6, 3.4

Arranging the given series in ascending order ex, 1.8, 2.6, 3.4, 4.3, 5.6, 7.2, 9.1
Number of terms (n) = 7

Middle term = 1/2(n + 1)th term

= 1/2(7 + 1)th term

4th term = 4.3

So, Median = 4.3

(iv) 122, 127, 109, 118, 125, 108

Arranging the given series in ascending order ex, 108, 118, 122, 125, 127
Number of terms (n) = 6
so, 3rd and 4th terms are the two middle terms
So, Median = Mean of 3rd and 4th terms
= 1/2(118 + 122)
= 1/2 x 240 = 120
Hence, Median = 120
Question 2:
The marks of 7 students in an examination are :
25, 19, 17, 24, 31, 26, 40

Find the median score.

Answer :

Arranging in ascending order we get :

17, 19, 24, 25, 26, 31, 40

Number of terms = 7 which is odd

So, Middle term = 1/2(7+ 1 )th term = 4th term = 25

Median  = 25

Question 3:
The runs scored by 11 members of a cricket team are :
15, 29, 43, 13, 31, 50, 20, 0, 27, 56, 34
Find the median score.
Answer :
Arranging in ascending order we get :
0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56Number of terms = 11 which odd

Middle term = 1/2(11 + 1)th term 4th term = 29

Median = 29

Question 4:

The weights (in kg) of 8 children are :
13.4, 10.6, 12.7, 17.2, 14.3, 15, 16.5, 9.8
Find the median weight.

Answer :Arranging in ascending order we get :
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2

Number of terms = 8
So, 4th and 5th terms are the two middle terms

So, Median = Mean of  4th and  5th terms = 1/2 (13.4) + 14.3)
= 27.7/2
= 13.85 kg
Question 5:
The heights (in cm) of 9 girls are :
144.2, 148.5, 143.7, 149.6, 150, 146.5, 145, 147.3, 152. 1
Find the median height.Answer :

Arranging in ascending order we get :143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1

Number of terms = 9Middle term = 1/2(9+ 1)th term

= 5th term = 147.3 cm

So, Median 147.3 cm

-: End of Mean and Median Class-6 RS Aggarwal Solutions :–

Return to-  RS Aggarwal Solutions for ICSE Class-6 Goyal Brothers Prakashan


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