RD Sharma Solutions for Class 7 Maths Chapter 23–Data Handling – II Central Values

Class 7: Maths Chapter 23 solutions. Complete Class 7 Maths Chapter 23 Notes.

RD Sharma Solutions for Class 7 Maths Chapter 23–Data Handling – II Central Values

RD Sharma 7th Maths Chapter 23, Class 7 Maths Chapter 23 solutions

Exercise 23.1 Page No: 23.6

1. Ashish studies for 4 hours, 5 hours and 3 hours on three consecutive days. How many hours does he study daily on an average?

Solution:

Given Ashish studies for 4 hours, 5 hours and 3 hours on three consecutive days

Average number of study hours = sum of hours/ number of days

Average number of study hours = (4 + 5 + 3) ÷ 3

= 12 ÷ 3

= 4 hours

Thus, Ashish studies for 4 hours on an average.

2. A cricketer scores the following runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100.

Find the mean score.

Solution:

Given runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100

Mean score = total sum of runs/number of innings

The mean score = (58 + 76 + 40 + 35 + 48 + 45 + 0 + 100) ÷ 8

= 402 ÷ 8

= 50.25 runs.

3. The marks (out of 100) obtained by a group of students in science test are 85, 76, 90, 84, 39, 48, 56, 95, 81 and 75. Find the

(i) Highest and the lowest marks obtained by the students.

(ii) Range of marks obtained.

(iii) Mean marks obtained by the group.

Solution:

In order to find the highest and lowest marks, we have to arrange the marks in ascending order as follows:

39, 48, 56, 75, 76, 81, 84, 85, 90, 95

(i) Clearly, the highest mark is 95 and the lowest is 39.

(ii) The range of the marks obtained is: (95 – 39) = 56.

(iii) From the following data, we have

Mean marks = Sum of the marks/ Total number of students

Mean marks = (39 + 48 + 56 + 75 + 76 + 81 + 84 + 85 + 90 + 95) ÷ 10

= 729 ÷ 10

= 72.9.

Hence, the mean mark of the students is 72.9.

4. The enrolment of a school during six consecutive years was as follows:

1555, 1670, 1750, 2019, 2540, 2820

Find the mean enrolment of the school for this period.

Solution:

Given enrolment of a school during six consecutive years as follows

1555, 1670, 1750, 2019, 2540, 2820

The mean enrolment = Sum of the enrolments in each year/ Total number of years

The mean enrolment = (1555 + 1670 + 1750 + 2019 + 2540 + 2820) ÷ 6

= 12354 ÷ 6

= 2059.

Thus, the mean enrolment of the school for the given period is 2059.

5. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

(i) Find the range of the rainfall from the above data.

(ii) Find the mean rainfall for the week.

(iii) On how many days was the rainfall less than the mean rainfall.

Solution:

(i) The range of the rainfall = Maximum rainfall – Minimum rainfall

= 20.5 – 0.0

= 20.5 mm.

(ii) The mean rainfall = (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.3 + 1.0) ÷ 7

= 41.1 ÷ 7

= 5.87 mm.

(iii) Clearly, there are 5 days (Mon, Wed, Thu, Sat and Sun), when the rainfall was less than the mean, i.e., 5.87 mm.

6. If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.

Solution:

The mean height = Sum of the heights /Total number of persons

= (140 + 150 + 152 + 158 + 161) ÷ 5

= 761 ÷ 5

= 152.2 cm.

7. Find the mean of 994, 996, 998, 1002 and 1000.

Solution:

Mean = Sum of the given numbers/Total number of given numbers

Mean = (994 + 996 + 998 + 1002 + 1000) ÷ 5

= 4990 ÷ 5

= 998.

8. Find the mean of first five natural numbers.

Solution:

We know that first five natural numbers = 1, 2, 3, 4 and 5

Mean of first five natural numbers = (1 + 2 + 3 + 4 + 5) ÷ 5

= 15 ÷ 5

= 3

9. Find the mean of all factors of 10.

Solution:

We know that factors of 10 are 1, 2, 5 and 10

Arithmetic mean of all factors of 10 = (1 + 2 + 5 + 10) ÷ 4

= 18 ÷ 4

= 4.5

10. Find the mean of first 10 even natural numbers.

Solution:

The first 10 even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20.

Mean of first 10 even natural numbers = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20) ÷ 10

= 110 ÷ 10

= 11

11. Find the mean of x, x + 2, x + 4, x + 6, x + 8

Solution:

Mean = Sum of observations ÷ Number of observations

Mean = (x + x + 2 + x + 4 + x + 6 + x + 8) ÷ 5

Mean = (5x + 20) ÷ 5

Mean = 5 (x + 4) ÷ 5

Mean = x + 4

12. Find the mean of first five multiples of 3.

Solution:

The first five multiples of 3 are 3, 6, 9, 12 and 15.

Mean of first five multiples of 3 are = (3 + 6 + 9 + 12 + 15) ÷ 5

= 45 ÷ 5

= 9

13. Following are the weights (in kg) of 10 new born babies in a hospital on a particular day: 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6 Find the mean ¯¯¯¯¯XX¯

Solution:

We know that¯¯¯¯¯XX¯ = sum of observations/ number of observations¯¯¯¯¯XX¯ = sum of weights of babies/ number of babies¯¯¯¯¯XX¯ = (3.4 + 3.6 + 4.2 + 4.5 + 3.9 + 4.1 + 3.8 + 4.5 + 4.4 + 3.6) ÷ 10¯¯¯¯¯XX¯ = (40) ÷ 10¯¯¯¯¯XX¯ = 4 kg

14. The percentage of marks obtained by students of a class in mathematics are:

64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1 Find their mean.

Solution:

Mean = sum of the marks obtained/ total number of students

= (64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1) ÷ 12

= 474 ÷ 12

= 39.5%

15. The numbers of children in 10 families of a locality are:

2, 4, 3, 4, 2, 3, 5, 1, 1, 5 Find the mean number of children per family.

Solution:

Mean number of children per family = sum of total number of children / total number of families

= (2 + 4 + 3 + 4 + 2 + 3 + 5 + 1 + 1 + 5) ÷ 10

= 30 ÷ 10

= 3

Thus, on an average there are 3 children per family in the locality.

16. The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.

Solution:

Given n = the number of observations = 100, Mean = 40

Mean = sum of observations/total number of observations

40 = sum of the observations/ 100

Sum of the observations = 40 x 100

Thus, the incorrect sum of the observations = 40 x 100 = 4000.

Now,

The correct sum of the observations = Incorrect sum of the observations – Incorrect observation + Correct observation

The correct sum of the observations = 4000 – 83 + 53

The correct sum of the observations = 4000 – 30 = 3970

Correct mean = correct sum of the observations/ number of observations

= 3970/100

= 39.7

17. The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

Solution:

We know that

Mean = sum of five numbers/5 = 27

So, sum of the five numbers = 5 x 27 = 135.

Now,

The mean of four numbers = sum of the four numbers/4 = 25

So, sum of the four numbers = 4 x 25 = 100.

Therefore, the excluded number = Sum of the five number – Sum of the four numbers

The excluded number = 135 – 100

= 35.

18. The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.

Solution:

We know that

Mean = sum of weights of students/ number of students

Let the weight of the seventh student be x kg.

Mean = (52 + 54 + 55 + 53 + 56 + 54 + x)/ 7

55 = (52 + 54 + 55 + 53 + 56 + 54 + x)/ 7

55 x 7 = 324 + x

385 = 324 + x

x = 385 – 324

x = 61 kg.

Therefore weight of seventh student is 61kg.

19. The mean weight of 8 numbers is 15 kg. If each number is multiplied by 2, what will be the new mean?

Solution:

Let x₁, x₂, x₃…x₈ be the eight numbers whose mean is 15 kg. Then,

15 = x₁ + x₂ + x₃+……+ x₈ /8

x₁ + x₂ + x₃ + …+ x₈ = 15 × 8

x₁ + x₂ + x₃ +…+ x₈ = 120.

Let the new numbers be 2x₁, 2x₂, 2x₃ …2x₈.

Let M be the arithmetic mean of the new numbers.

Then,

M = 2x₁ + 2x₂ + 2x₃ +…+ 2x₈/8

M = 2 (x₁ + x₂ + x₃ + …+ x₈)/8

M = (2 × 120)/8

= 30

20. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

Solution:

Let x₁, x₂, x₃, x₄ and x₅ be five numbers whose mean is 18. Then,

18 = Sum of five numbers ÷ 5

Hence, sum of five numbers = 18 × 5 = 90

Now, if one number is excluded, then their mean is 16.

So,

16 = Sum of four numbers ÷ 4

Therefore sum of four numbers = 16 × 4 = 64.

The excluded number = Sum of five observations – Sum of four observations

The excluded number = 90 – 64

Therefore The excluded number = 26.

21. The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

Solution:

Given n = Number of observations = 200

Mean = sum of observations/ number of observations

50 = sum of observations/ 200

Sum of the observations = 50 x 200 = 10,000.

Thus, the incorrect sum of the observations = 50 x 200

Now,

The correct sum of the observations = Incorrect sum of the observations – Incorrect observations + Correct observations

Correct sum of the observations = 10,000 – (92 + 8) + (192 + 88)

Correct sum of the observations = 10,000 – 100 + 280

Correct sum of the observations = 9900 + 280

Correct sum of the observations = 10,180.

Therefore correct mean = correct sum of the observations/ number of observations

= 10180/200

= 50.9

22. The mean of 5 numbers is 27. If one more number is included, then the mean is 25. Find the included number.

Solution:

Given Mean = Sum of five numbers ÷ 5

Sum of the five numbers = 27 × 5 = 135.

Now, New mean = 25

25 = Sum of six numbers ÷ 6

Sum of the six numbers = 25 × 6 = 150.

The included number = Sum of the six numbers – Sum of the five numbers

The included number = 150 – 135

Therefore the included number = 15.

23. The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean.

Solution:

Let x₁, x₂, x₃…x₇₅ be 75 numbers with their mean equal to 35. Then,

35 = x₁ + x₂ + x₃ + …..+ x₇₅ /75

x₁ + x₂ + x₃ + …..+ x₇₅ = 35 × 75

x₁ + x₂ + x₃ +…+ x₇₅ = 2625

The new numbers are 4 x 1, 4 x 2, 4 x 3…4 x 75

Let M be the arithmetic mean of the new numbers. Then,

M = 4x₁ + 4x₂ + 4x₃ +…+ 4x₇₅/75

M = 4 (x₁ + x₂ + x₃ + …+ x₇₅)/75

M = (4 × 2625)/75

= 140

Exercise 23.2 Page No: 23.12

1. A die was thrown 20 times and the following scores were recorded:

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

Prepare the frequency table of the scores on the upper face of the die and find the mean score.

Solution:

The frequency table for the given data is as follows:

To compute arithmetic mean we have to prepare the following table:

Mean score = Σ f_i x_i/ Σ f_i

= 73/20

= 3.65

2. The daily wages (in Rs) of 15 workers in a factory are given below:

200, 180, 150, 150, 130, 180, 180, 200, 150, 130, 180, 180, 200, 150, 180

Prepare the frequency table and find the mean wage.

Solution:

To compute arithmetic mean we have to prepare the following table:

Mean score = Σ f_i x_i/ Σ f_i

= 2540/15

= 169.33

3. The following table shows the weights (in kg) of 15 workers in a factory:

Calculate the mean weight.

Solution:

Calculation of mean:

Mean score = Σ f_i x_i/ Σ f_i

= 975/15

= 65 kg

4. The ages (in years) of 50 students of a class in a school are given below:

Find the mean age.

Solution:

Calculation of mean:

Mean score = Σ f_i x_i/ Σ f_i

= 770/50

= 15.4 years

5. Calculate the mean for the following distribution:

Solution:

Mean score = Σ f_i x_i/ Σ f_i

= 281/40

= 7.025

6. Find the mean of the following data:

Solution:

Mean score = Σ f_i x_i/ Σ f_i

= 2650/106

= 25

7. The mean of the following data is 20.6. Find the value of p.

Solution:

Mean score = Σ f_i x_i/ Σ f_i

20.6 = 530 + 25p/50

530 + 25 p = 20.6 x 50

25 p = 1030 – 530

p = 500/25

p = 20

8. If the mean of the following data is 15, find p.

Solution:

Mean score = Σ f_i x_i/ Σ f_i

15 = 445 + 10p/27 + p

445 + 10 p = 405 + 15p

5 p = 445 – 405

p = 40/5

p = 8

9. Find the value of p for the following distribution whose mean is 16.6

Solution:

Mean score = Σ f_i x_i/ Σ f_i

16.6 = 1228 + 24p/100

1228 + 24 p = 16.6 x 100

24 p = 1660 – 1228

p = 432/24

p = 18

10. Find the missing value of p for the following distribution whose mean is 12.58

Solution:

Mean score = Σ f_i x_i/ Σ f_i

12.58 = 524 + 7p/50

524 + 7 p = 12.58 x 50

7 p = 629 – 524

p = 105/7

p = 15

11. Find the missing frequency (p) for the following distribution whose mean is 7.68

Solution:

Mean score = Σ f_i x_i/ Σ f_i

7.68 = 303 + 9p/41 + p

303 + 9 p = 314.88 + 7.68p

1.32 p = 314.88 – 303

p = 11.88/1.32

p = 9

12. Find the value of p, if the mean of the following distribution is 20

Solution:

Mean score = Σ f_i x_i/ Σ f_i

20 = [(295 + (20 + p) 5p)]/ 15 + 5p

295 + 100 p + 5p² = 300 + 100p

5p² = 300 – 295

5p²= 5

p² = 1

p = 1

Exercise 23.3 Page No: 23.16

Find the median of the following data (1 – 8)

1. 83, 37, 70, 29, 45, 63, 41, 70, 34, 54

Solution:

First we have to arrange given data into ascending order,

29, 34, 37, 41, 45, 54, 63, 70, 70, 83

Given number of observations, n = 10 (even)

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 5^th term + value of 6^th term)/2

= (45 + 54)/2

= 49.5

Hence median for given data = 49.5

2. 133, 73, 89, 108, 94,104, 94, 85, 100, 120

Solution:

First we have to arrange given data into ascending order,

73, 85, 89, 94, 100, 104, 108, 120, 133

Given number of observations, n = 10 (even)

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 5^th term + value of 6^th term)/2

= (94 + 100)/2

= 97

Hence median for given data = 97

3. 31, 38, 27, 28, 36, 25, 35, 40

Solution:

First we have to arrange given data into ascending order

25, 27, 28, 31, 35, 36, 38, 40

Given number of observations, n = 8 (even)

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 4^th term + value of 5^th term)/2

= (31 + 35)/2

= 33

Hence median for given data = 33

4. 15, 6, 16, 8, 22, 21, 9, 18, 25

Solution:

First we have to arrange given data into ascending order

6, 8, 9, 15, 16, 18, 21, 22, 25

Given number of observations, n = 9 (odd)

Therefore median = ((n+1)/2)^th term

Median = value of 5^th term

= 16

5. 41, 43,127, 99, 71, 92, 71, 58, 57

Solution:

First we have to arrange given data into ascending order

41, 43, 57, 58, 71, 71, 92, 99, 127

Given number of observations, n = 9 (odd)

Therefore median = ((n+1)/2)^th term

Median = value of 5^th term

= 71

6. 25, 34, 31, 23, 22, 26, 35, 29, 20, 32

Solution:

First we have to arrange given data into ascending order,

20, 22, 23, 25, 26, 29, 31, 32, 34, 35

Given number of observations, n = 10 (even)

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 5^th term + value of 6^th term)/2

= (26 + 29)/2

= 27.5

Hence median for given data = 27.5

7. 12, 17, 3, 14, 5, 8, 7, 15

Solution:

First we have to arrange given data into ascending order,

3, 5, 7, 8, 12, 14, 15, 17

Given number of observations, n = 8 (even)

Therefore median = (n/2)^th term + ((n +1)/2)^th term

Median = (value of 4^th term + value of 5^th term)/2

= (8 + 12)/2

= 10

Hence median for given data = 10

8. 92, 35, 67, 85, 72, 81, 56, 51, 42, 69

Solution:

First we have to arrange given data into ascending order,

35, 42, 51, 56, 67, 69, 72, 81, 85, 92

Given number of observations, n = 10 (even)

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 5^th term + value of 6^th term)/2

= (67 + 69)/2

= 68

Hence median for given data = 68

9. Numbers 50, 42, 35, 2x +10, 2x – 8, 12, 11, 8, 6 are written in descending order and their median is 25, find x.

Solution:

Here, the number of observations n is 9.

Since n is odd, the median is the n+12th observation, i.e., the 5^th observation.

As the numbers are arranged in the descending order, we therefore observe from the last.

Median = 5^th observation.

=> 25 = 2x – 8

=> 2x = 25 + 8

=> 2x = 33

=> x = (33/2)

x = 16.5

10. Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?

Solution:

Arranging the given data in ascending order, we have:

33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

Here, the number of observations n is 11 (odd).

Since the number of observations is odd, therefore,

Therefore median = ((n+1)/2)^th term

Median = value of 5^th term

= 58.

Hence, median = 58.

If 92 is replaced by 99 and 41 by 43, then the new observations arranged in ascending order are:

33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99

New median = Value of the 6^th observation = 58.

11. Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is replaced by 85, what will be the new median?

Solution:

Arranging the given data in ascending order, we have:

41, 43, 57, 58, 61, 71, 92, 99,127

Here, the number of observations, n, is 9(odd).

Therefore median = ((n+1)/2)^th term

Median = value of 5^th term

Hence, the median = 61.

If 58 is replaced by 85, then the new observations arranged in ascending order are:

41, 43, 57, 61, 71, 85, 92, 99, 12

New median = Value of the 5^th observation = 71.

12. The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.

Solution:

Arranging the given data in ascending order, we have:

27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45

Here, the number of observations n is 15(odd).

Since the number of observations is odd, therefore,

Therefore median = ((n+1)/2)^th term

Median = value of 8^th term

Hence, median = 35 kg.

If 44 kg is replaced by 46 kg and 27 kg by 25 kg, then the new observations arranged in ascending order are:

25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46

∴ New median = Value of the 8^th observation = 35 kg.

13. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

Solution:

Here, the number of observations n is 10. Since n is even,

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 5^th term + value of 6^th term)/2

63 = x + (x + 2)/2

63 = (2x + 2)/2

63 = 2 (x + 1)/2

63 = x + 1

x = 63 – 1

x = 62

Exercise 23.4 Page No: 23.20

1. Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14

By using the empirical relation also find the mean.

Solution:

Arranging the data in ascending order such that same numbers are put together, we get:

12, 12, 13, 13, 14, 14, 14, 16, 19

Here, n = 9.

Therefore median = ((n+1)/2)^th term

Median = value of 5^th term

Median = 14

Here, 14 occurs the maximum number of times, i.e., three times. Therefore, 14 is the mode of the data.

Now,

Mode = 3 Median – 2 Mean

14 = 3 x 14 – 2 Mean

2 Mean = 42 – 14 = 28

Mean = 28 ÷ 2

= 14.

2. Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34

Solution:

Arranging the data in ascending order such that same numbers are put together, we get:

32, 32, 34, 35, 35, 38, 42

Here, n = 7

Therefore median = ((n+1)/2)^th term

Median = value of 4^th term

Median = 35

Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.

3.  Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4

Solution:

Arranging the data in ascending order such that same values are put together, we get:

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

Here, 2, 3 and 4 occur three times each. Therefore, 2, 3 and 4 are the three modes.

4. The runs scored in a cricket match by 11 players are as follows:

6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10

Find the mean, mode and median of this data.

Solution:

Arranging the data in ascending order such that same values are put together, we get:

6, 8, 10, 10, 15, 15, 50, 80, 100, 120

Here, n = 11

Therefore median = ((n+1)/2)^th term

Median = value of 6^th term

Median = 15

Here, 10 occur three times. Therefore, 10 is the mode of the given data.

Now,

Mode = 3 Median – 2 Mean

10 = 3 x 15 – 2 Mean

2 Mean = 45 – 10 = 35

Mean = 35 ÷ 2

= 17.5

5. Find the mode of the following data:

12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14

Solution:

Arranging the data in ascending order such that same values are put together, we get:

10, 12, 12, 14, 14, 14, 14, 14, 14, 16, 18

Here, clearly, 14 occurs the most number of times.

Therefore, 14 is the mode of the given data.

6. Heights of 25 children (in cm) in a school are as given below:

168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 163, 164, 163, 160, 165, 163, 162

What is the mode of heights?

Also, find the mean and median.

Solution:

Arranging the data in tabular form, we get:

Therefore median = ((n+1)/2)^th term

Median = value of 13^th term

Median = 163 cm

Here, clearly, 163 cm occurs the most number of times. Therefore, the mode of the given data is 163 cm.

Mode = 3 Median – 2 Mean

163 = 3 x 163 – 2 Mean

2 Mean = 326

Mean = 163 cm.

7. The scores in mathematics test (out of 25) of 15 students are as follows:

19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20

Find the mode and median of this data. Are they same?

Solution:

Arranging the data in ascending order such that same values are put together, we get:

5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25

Here, n = 15

Therefore median = ((n+1)/2)^th term

Median = value of 8^th term

Median = 20

Here, clearly, 20 occurs most number of times, i.e., 4 times. Therefore, the mode of the given data is 20.

Yes, the median and mode of the given data are the same.

8. Calculate the mean and median for the following data:

Using empirical formula, find its mode.

Solution:

Calculation of mean

Mean = Σ f_i x_i/ Σ f_i

= 332/25

= 13.28

Here, n = 25, which is an odd number. Therefore,

Therefore median = ((n+1)/2)^th term

Median = value of 13^th term

Median = 13

Now, by using empirical formula we have,

Mode = 3Median – 2 Mean

Mode = 3 (13) – 2 (13.28)

Mode = 39 – 26.56

Mode = 12.44.

9. The following table shows the weights of 12 persons.

Find the median and mean weights. Using empirical relation, calculate its mode.

Solution:

Calculation of mean

Mean = Σ f_i x_i/ Σ f_i

= 612/12

= 51 kg

Here n = 12

Therefore median = (n/2)^th term + ((n + 1)/2)^th term

Median = (value of 6^th term + value of 7^th term)/2

= (50 + 50)/2

= 50

Now by empirical formula we have,

Now,

Mode = 3 Median – 2 Mean

Mode = 3 x 50 – 2 x 51

Mode = 150 – 102

Mode = 48 kg.

Thus, Mean = 51 kg, Median = 50 kg and Mode = 48 kg.

RD Sharma Solutions for Class 7 Maths Chapter 23: Download PDF

RD Sharma Solutions for Class 7 Maths Chapter 23–Data Handling – II Central Values

Download PDF: RD Sharma Solutions for Class 7 Maths Chapter 23–Data Handling – II Central Values PDF

Chapterwise RD Sharma Solutions for Class 7 Maths :

• Chapter 1–Integers
• Chapter 2–Fractions
• Chapter 3–Decimals
• Chapter 4–Rational Numbers
• Chapter 5–Operations On Rational Numbers
• Chapter 6–Exponents
• Chapter 7–Algebraic Expressions
• Chapter 8–Linear Equations in One Variable
• Chapter 9–Ratio And Proportion
• Chapter 10–Unitary Method
• Chapter 11–Percentage
• Chapter 12–Profit And Loss
• Chapter 13–Simple Interest
• Chapter 14–Lines And Angles
• Chapter 15–Properties of Triangles
• Chapter 16–Congruence
• Chapter 17–Constructions
• Chapter 18–Symmetry
• Chapter 19–Visualising Solid Shapes
• Chapter 20–Mensuration – I (Perimeter and area of rectilinear figures)
• Chapter 21–Mensuration – II (Area of Circle)
• Chapter 22–Data Handling – I (Collection and Organisation of Data)
• Chapter 23–Data Handling – II Central Values
• Chapter 24–Data Handling – III (Constructions of Bar Graphs)
• Chapter 25–Data Handling – IV (Probability)

About RD Sharma

RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.

Read More

• RD Sharma Solutions for Class 10 Maths Chapter 7–Statistics
• NCERT Solutions for 11th Class Economics: Chapter 5-Measures of Central Tendency
• RD Sharma Solutions for Class 9 Maths Chapter 24–Measure of Central Tendency
• NCERT Solutions for 9th class Maths : Chapter 14 Statistics
• RD Sharma Solutions for Class 11 Maths Chapter 32–Statistics