ML Aggarwal Solutions For Class 10 Maths Chapter 21 Measures Of Central Tendency

ML Aggarwal Solutions For Class 10 Maths Chapter 21 Measures Of Central Tendency give error free solutions, which are created by our subject experts. This chapter deals with mean, mode and median. The clear diagrams given in our solutions help students better understand the topic. Students can easily download ML Aggarwal Solutions from our website in PDF format for free. These solutions help students to improve their confidence during exams and thus score more marks.

In Statistics, central tendency is the central value for a data set or distribution. In Chapter 21, we come across ogive and histogram. Students are recommended to practise ML Aggarwal Solutions For Class 10 to improve their conceptual understanding. Sign in with BYJU’S to practice these problems.

ML Aggarwal Solutions for Class 10 Maths Chapter 21 :-Click Here to Download PDF

 

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Access answers to ML Aggarwal Solutions for Class 10 Maths Chapter 21- Measures Of Central Tendency

Exercise 21.1

1. (a) Calculate the arithmetic mean of 5.7, 6.6, 7.2, 9.3, 6.2.
(b) The weights (in kg) of 8 new born babies are 3, 3.2, 3.4, 3.5, 4, 3.6, 4.1, 3.2. Find the mean weight of the babies.

Solution:

(a) Given observations are 5.7, 6.6, 7.2, 9.3, 6.2.

Number of observations = 5

Mean = sum of observations / number of observations

Mean = (5.7+6.6+7.2+9.3+6.2)/5

= 35/5 = 7

Hence the mean of the given observations is 7.

(b) Given weight of babies are 3, 3.2, 3.4, 3.5, 4, 3.6, 4.1, 3.2

Number of observations = 8

Mean = sum of observations / number of observations

Mean = (3+3.2+3.4+3.5+4+3.6+4.1+3.2)/8

= 28/8 = 3.5 kg

Hence the mean of the weight of babies is 3.5 kg.

2.The marks obtained by 15 students in a class test are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20 find
(i) the mean of their marks.
(ii) the mean of their marks when the marks of each student are increased by 4.
(iii) the mean of their marks when 2 marks are deducted from the marks of each student.
(iv) the mean of their marks when the marks of each student are doubled.

Solution:

(i) Marks obtained by students are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20.

Number of students = 15

Mean = sum of observations / number of observations

= 12+14+07+09+23+11+08+13+11+19+16+24+17+03+20

= 207/15

= 13.8

Hence the mean of their marks is 13.8.

(ii) If mark of each student is increased by 4, total increased marks = 4×15 = 60

Total increase in sum of marks = 207+60 = 267

mean = sum of marks/number of students

mean = 267/15 = 17.8

Hence the mean is 17.8.

(iii) If mark of each student is deducted by 2, total deducted marks = 2×15 = 30

Total decrease in sum of marks = 207-30 = 177

mean = sum of marks/number of students

mean = 177/15 = 11.8

Hence the mean is 11.8.

(iv) If mark of each student is doubled , then new sum of marks = 2×207 = 414

mean = new sum of marks/number of students

mean = 414/15 = 27.6

Hence the mean is 27.6.

3. (a) The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.
(b) The mean of 9 variates is 11. If eight of them are 7, 12, 9, 14, 21, 3, 8 and 15 find the 9th variate.

Solution:

(a)Given observations are 6, y, 7, x, 14.

Mean = 8

Number of observations = 5

Mean = Sum of observations/number of observations

8 = (6+y+7+x+14)/5

40 = 27+x+y

40-27 = x+y

13 = x+y

y = 13-x

Hence the answer is y = 13-x.

(b)Given mean = 11

Number of variates = 9

Variates are 7, 12, 9, 14, 21, 3, 8 ,15

Let the 9th variate be x.

Sum of variates = 7+12+9+14+21+3+8+15+x

= 89+x

Mean = Sum of variates/number of variates

11 = (89+x)/9

11×9 = 89+x

99 = 89+x

x = 99-89 = 10

Hence the 9th variate is 10.

4. (a) The mean age of 33 students of a class is 13 years. If one girl leaves the class, the mean becomes ML Aggarwal Sol Class 10 Maths chapter 21-1   years. What is the age of the girl ?
(b) In a class test, the mean of marks scored by a class of 40 students was calculated as 18.2. Later on, it was detected that marks of one student was wrongly copied as 21 instead of 29. Find the correct mean.

Solution:

(a)Given mean age = 13

Number of students = 33

Sum of ages = mean ×number of students

= 13×33

= 429

After a girl leaves, the mean of 32 students becomes
ML Aggarwal Sol Class 10 Maths chapter 21-2= 207/16

Now sum of ages = 32×207/16

= 414

So the age of the girl who left = 429-414 = 15 years.

Hence the age of the girl who left is 15 years.

(b)Mean of marks = 18.2

Number of students = 40

Total marks of 40 students = 40×18.2 = 728

Difference of marks when copied wrongly = 29-21 = 8

So total marks = 728+8 = 736

mean = 736/40

= 18.4

Hence the correct mean is 18.4.

5. Find the mean of 25 given numbers when the mean of 10 of them is 13 and the mean of the remaining numbers is 18.

Solution:

Mean of 10 numbers = 13

Sum of numbers = 13×10 = 130

Mean of remaining 15 numbers = 18

Sum of numbers = 15×18 = 270

Sum of all numbers = 130+270 = 400

Mean = sum of numbers/25 = 400/25 = 16

Hence the mean of 25 numbers is 16.

6. Find the mean of the following distribution:

Number

5

10

15

20

25

30

35

Frequency

1

2

5

6

3

2

1

Solution:

Number (x)

Frequency (f)

fx

5

1

5×1 = 5

10

2

10×2 = 20

15

5

15×5 = 75

20

6

20×6 = 120

25

3

25×3 = 75

30

2

30×2 = 60

35

1

35×1 = 35

Total

Ʃf = 20

Ʃfx = 390

Mean = Ʃfx/Ʃf

= 390/20 = 19.5

Hence the mean is 19.5.

7. The contents of 100 match boxes were checked to determine the number of matches they contained

No. of matches

35

36

37

38

39

40

41

No. of boxes

6

10

18

25

21

12

8

(i) Calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to; bring the mean up to exactly 39 matches. (1997)

Solution:

(i)

No. of matches (x)

Number of boxes (f)

fx

35

6

35×6 = 210

36

10

36×10 = 360

37

18

37×18 = 666

38

25

38×25 = 950

39

21

39×21 = 819

40

12

40×12 = 480

41

8

41×8 = 328

Total

Ʃf = 100

Ʃfx = 3813

Mean = Ʃfx/Ʃf

= 3813/100

= 38.13

= 38.1

Hence the mean is 38.1.

(ii)New mean = 39

Ʃfx = 39×100 = 3900

So number of extra matches to be added = 3900-3813 = 87

Hence the number of extra matches to be added is 87.

8. Calculate the mean for the following distribution :

Pocket money (in Rs)

60

70

80

90

100

110

120

No. of students

2

6

13

22

24

10

3

Solution:

Pocket money in Rs (x)

Number of students (f)

fx

60

2

60×2 = 120

70

6

70×6 = 420

80

13

80×13 = 1040

90

22

90×22 = 1980

100

24

100×24 = 2400

110

10

110×10 = 1100

120

3

120×3 = 360

Total

Ʃf = 80

Ʃfx = 7420

Mean = Ʃfx/Ʃf

= 7420/80

= 92.75

Hence the mean is 92.75.

9. Six coins were tossed 1000 times, and at each toss the number of heads were counted and the results were recorded as under :

No. of heads

6

5

4

3

2

1

0

No. of tosses

20

25

160

283

338

140

34

Calculate the mean for this distribution.

Solution:

No. of heads (x)

No. of tosses (f)

fx

6

20

6×20 = 120

5

25

5×25 = 125

4

160

4×160 = 640

3

283

3×283 = 849

2

338

2×338 = 676

1

140

1×140 =140

0

34

0×34 = 0

Total

Ʃf = 1000

Ʃfx = 2550

Mean = Ʃfx/Ʃf

= 2550/1000

= 2.55

Hence the mean is 2.55.

10. Find the mean for the following distribution.

Numbers

60

61

62

63

64

65

66

Cumulative frequency

8

18

33

40

49

55

60

Solution:

Numbers

(x)

Cumulative frequency

Frequency (f)

fx

60

8

8

60×8 = 480

61

18

18-8 = 10

61×10 = 610

62

33

33-18 = 15

62×15 = 930

63

40

40-33 = 7

63×7 = 441

64

49

49-40 = 9

64×9 = 576

65

55

55-49 = 6

65×6 = 390

66

60

60-55 = 5

66×5 = 330

Total

Ʃf = 60

Ʃfx = 3757

Mean = Ʃfx/Ʃf

= 3757/60

= 62.616

= 62.62

Hence the mean is 62.62.

11.

Category

A

B

C

D

E

F

G

Wages (in Rs) per day

50

60

70

80

90

100

110

No. of workers

2

4

8

12

10

6

8

(i) Calculate the mean wage correct to the nearest rupee (1995)
(ii) If the number of workers in each category is doubled, what would be the new mean wage ?

Solution:

Category

Wages in Rs. (x)

No. of workers f

fx

A

50

2

100

B

60

4

240

C

70

8

560

D

80

12

960

E

90

10

900

F

100

6

600

G

110

8

880

Total

Ʃf = 50

Ʃfx = 4240

Mean = Ʃfx/Ʃf

= 4240/50

= 84.8

= 85

Hence the mean is 85.

(ii)If number of workers is doubled, then total number of workers = 50×2 = 100

So wages will be doubled.

Total wages = 4240×2 = 8480

Mean = Ʃfx/Ʃf

= 8480/100

= 84.8

= 85

Hence the mean is 85.

12.If the mean of the following distribution is 7.5, find the missing frequency ” f “.

Variate

5

6

7

8

9

10

11

12

Frequency

20

17

f

10

8

6

7

6

Solution:

Variate (x)

Frequency (f)

fx

5

20

100

6

17

102

7

f

7f

8

10

80

9

8

72

10

6

60

11

7

77

12

6

72

Total

Ʃf = 74+f

Ʃfx = 563+7f

Given mean = 7.5

Mean = Ʃfx/Ʃf

7.5= (563+f)/(74+7f)

7.5×(74+f) = 563+7f

555+7.5f = 563+7f

7.5f-7f = 563-555

0.5f = 8

f = 8/0.5 = 16

Hence the value of missing frequency f is 16.

13. Find the value of the missing variate for the following distribution whose mean is 10

Variate (xi)

5

7

9

11

_

15

20

Frequency (fi)

4

4

4

7

3

2

1

Solution:

Let the missing variate be x.

Variate (xi)

Frequency (fi)

fixi

5

4

20

7

4

28

9

4

36

11

7

77

x

3

3x

15

2

30

20

1

20

Total

Ʃfi =25

Ʃfixi = 211+3x

Given mean = 10

Mean = Ʃfixi/Ʃfi

10= (211+3x)/25

10×25 = 211+3x

250 = 211+3x

250-211 = 3x

39 = 3x

x = 39/3 = 13

Hence the missing variate is 13.

14. Marks obtained by 40 students in a short assessment are given below, where a and b are two missing data.

Marks

5

6

7

8

9

No. of students

6

a

16

13

b

If the mean of the distribution is 7.2, find a and b.

Solution:

Marks (x)

No. of students (f)

fx

5

6

30

6

a

6a

7

16

112

8

13

104

9

b

9b

Total

Ʃf = 35+a+b

Ʃfx = 246+6a+9b

Given number of students = 40

Ʃf = 35+a+b = 40

a+b = 40-35 = 5

a = 5-b ……(i)

Mean = Ʃfx/Ʃf

Given mean = 7.2

( 246+6a+9b) /40 = 7.2

( 246+6a+9b) = 40×7.2

( 246+6a+9b) = 288

6a+9b = 288-246

6a+9b = 288-246

6a+9b = 42

2a+3b = 14 …..(ii)

Substitute (i) in (ii)

2(5-b)+3b = 14

10-2b+3b = 14

10+b = 14

b = 14-10 = 4

a = 5-4 = 1

Hence the value of a and b is 1 and 4 respectively.

15. Find the mean of the following distribution.

Class interval

0-10

10-20

20-30

30-40

40-50

Frequency

10

6

8

12

5

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Class interval

Frequency fi

Class mark xi

fixi

0-10

10

5

50

10-20

6

15

90

20-30

8

25

200

30-40

12

35

420

40-50

5

45

225

Total

Ʃfi = 41

Ʃfixi = 985

Mean = Ʃfixi/ Ʃfi

= 985/41

= 24.024

= 24.02 (approx)

Hence the mean of the distribution is 24.02.

16. Calculate the mean of the following distribution:

Class interval

0-10

10-20

20-30

30-40

40-50

50-60

Frequency

8

5

12

35

24

16

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Class interval

Frequency fi

Class mark xi

fixi

0-10

8

5

40

10-20

5

15

75

20-30

12

25

300

30-40

35

35

1225

40-50

24

45

1080

50-60

16

55

880

Total

Ʃfi = 100

Ʃfixi = 3600

Mean = Ʃfixi/ Ʃfi

= 3600/100

= 36

Hence the mean of the distribution is 36.

17. Calculate the mean of the following distribution using step deviation method:

Marks

0-10

10-20

20-30

30-40

40-50

50-60

No. of students

10

9

25

30

16

10

Solution:

Class mark (xi) = (upper limit + lower limit)/2

Let assumed mean (A) = 25

Class size (h) = 10

Class Interval

No. of students (fi)

Class mark (xi)

di = xi – A

ui = di/h

fiui 

0-10

10

5

-20

-2

-20

10-20

9

15

-10

-1

-9

20-30

25

25

0

0

0

30-40

30

35

10

1

30

40-50

16

45

20

2

32

50-60

10

55

30

3

30

Total

∑fi = 100

∑fiui =63


By step deviation method, Mean = x̄ = A+h∑fiui /∑fi

 = 25+10(63/100)

= 25+10×0.63

= 25+6.3

=31.3

Hence the mean of the distribution is 31.3.

18. Find the mean of the following frequency distribution:

Class intervals

0-50

50-100

100-150

150-200

200-250

250-300

frequency

4

8

16

13

6

3

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Class interval

Frequency fi

Class mark xi

fixi

0-50

4

25

100

50-100

8

75

600

100-150

16

125

2000

150-200

13

175

2275

200-250

6

225

1350

250-300

3

275

825

Total

Ʃfi = 50

Ʃfixi = 7150

Mean = Ʃfixi/ Ʃfi

= 7150/50

=143

Hence the mean of the distribution is 143.

19. The following table gives the daily wages of workers in a factory:

Wages in Rs.

45-50

50-55

55-60

60-65

65-70

70-75

75-80

No. of workers

5

8

30

25

14

12

6

Calculate their mean by short cut method.

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Assumed mean, A = 62.5

Wages in Rs.

No. of workers (fi)

Class mark (xi)

di = xi – A

fidi 

45-50

5

47.5

-15

-75

50-55

8

52.5

-10

-80

55-60

30

57.5

-5

-150

60-65

25

62.5

0

0

65-70

14

67.5

5

70

70-75

12

72.5

10

120

75-80

6

77.5

15

90

Total

∑fi = 100

∑fidi = -25

By short cut method, Mean = x̄ = A+∑fidi /∑fi

= 62.5+-25/100

= 62.5-0.25

= 62.25

Hence the mean of the distribution is Rs.62.25.

20. Calculate the mean of the distribution given below using the short cut method.

Marks

11-20

21-30

31-40

41-50

51-60

61-70

71-80

No. of students

2

6

10

12

9

7

4

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Assumed mean, A = 45.5

Marks

No. of students (fi)

Class mark (xi)

di = xi – A

fidi 

11-20

2

15.5

-30

-60

21-30

6

25.5

-20

-120

31-40

10

35.5

-10

-100

41-50

12

45.5

0

0

51-60

9

55.5

10

90

61-70

7

65.5

20

140

71-80

4

75.5

30

120

Total

∑fi = 50

∑fidi = 70

By short cut method, Mean = x̄ = A+∑fidi /∑fi

= 45.5+70/50

= 45.5+1.4

= 46.9

Hence the mean of the distribution is Rs.46.9.

21. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

No. of days

0-6

6-10

10-14

14-20

20-28

28-38

38-40

No. of students

11

10

7

4

4

3

1

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

No. of days

Frequency fi

Class mark xi

fixi

0-6

11

3

33

6-10

10

8

80

10-14

7

12

84

14-20

4

17

68

20-28

4

24

96

28-38

3

33

99

38-40

1

39

39

Total

Ʃfi = 40

Ʃfixi = 499

Mean = Ʃfixi/ Ʃfi

= 499/40

=12.475

Hence the mean number of days a student was absent is 12.475.

22. The mean of the following distribution is 23.4. Find the value of p.

Class intervals

0-8

8-16

16-24

24-32

32-40

40-48

Frequency

5

3

10

P

4

2

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Class intervals

Frequency fi

Class mark xi

fixi

0-8

5

4

20

8-16

3

12

36

16-24

10

20

200

24-32

P

28

28P

32-40

4

36

144

40-48

2

44

88

Total

Ʃfi = 24+P

Ʃfixi = 488+28P

Given mean = 23.4

Mean = Ʃfixi/ Ʃfi

23.4 = (488+28P)/(24+P)

23.4×(24+P) = 488+28P

561.6+23.4P = 488+28P

561.6-488 = 28P -23.4P

73.6 = 4.6 P

P = 73.6/4.6 = 16

Hence the value of P is 16.

23. The following distribution shows the daily pocket allowance for children of a locality. The mean pocket allowance is Rs. 18. Find the value of f.

Daily pocket allowance in Rs.

11-13

13-15

15-17

17-19

19-21

21-23

23-25

No. of children

3

6

9

13

f

5

4

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Daily pocket allowance in Rs.

No. of children fi

Class mark xi

fixi

11-13

3

12

36

13-15

6

14

84

15-17

9

16

144

17-19

13

18

234

19-21

f

20

20f

21-23

5

22

110

23-25

4

24

96

Total

Ʃfi = 40+f

Ʃfixi = 704+20f

Given mean = 18

Mean = Ʃfixi/ Ʃfi

18 = (704+20f)/( 40+f)

18×(40+f) = 704+20f

720 +18f = 704+20f

720-704 = 20f-18f

16 = 2f

f = 16/2 = 8

Hence the value of f is 8.

24. The mean of the following distribution is 50 and the sum of all the frequencies is 120. Find the values of p and q.

Class intervals

0-20

20-40

40-60

60-80

80-100

Frequency

17

P

32

q

19

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Class intervals

Frequency fi

Class mark xi

fixi

0-20

17

10

170

20-40

P

30

30P

40-60

32

50

1600

60-80

q

70

70q

80-100

19

90

1710

Total

Ʃ fi = 68+P+q

Ʃfixi = 3480+30P+70q

Given sum of all frequencies, Ʃ fi = 120

68+P+q = 120

P+q = 120-68 = 52

P+q = 52

P = 52-q …(i)

Given mean = 50

Mean = Ʃfixi / Ʃ fi

50 = (3480+30P+70q)/120

50×120 = 3480+30P+70q

6000 = 3480+30P+70q

6000- 3480 = 30P+70q

2520 = 30P+70q

252 = 3P+7q …(ii)

Substitute (i) in (ii)

252 = 3(52-q)+7q

252 = 156-3q+7q

252-156 = 4q

4q = 96

q = 96/4 = 24

P = 52-24 = 28

Hence the value of P and q is 28 and 24 respectively.

25.The mean of the following frequency distribution is 57.6 and the sum of all the frequencies is 50. Find the values of p and q.

Class intervals

0-20

20-40

40-60

60-80

80-100

100-120

Frequency

7

P

12

q

8

5

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Class intervals

Frequency fi

Class mark xi

fixi

0-20

7

10

70

20-40

P

30

30P

40-60

12

50

600

60-80

q

70

70q

80-100

8

90

720

100-120

5

110

550

Total

Ʃ fi = 32+P+q

Ʃfixi = 1940+30P+70q

Given sum of all frequencies, Ʃ fi = 50

32+P+q = 50

P+q = 50-32= 18

P+q = 18

P = 18-q …(i)

Given mean = 57.6

Mean = Ʃfixi / Ʃ fi

57.6 = (1940+30P+70q)/50

57.6 ×50 = 1940+30P+70q

2880= 1940+30P+70q

2880- 1940 = 30P+70q

940 = 30P+70q

94 = 3P+7q …(ii)

Substitute (i) in (ii)

94 = 3(18-q)+7q

94 = 54-3q+7q

94-54= 4q

40 = 4q

q = 40/4 = 10

P = 18-10 = 8

Hence the value of P and q is 8 and 10 respectively.

26. The following table gives the life time in days of 100 electricity tubes of a certain make :

Life time in days

No. of tubes

Less than 50

8

Less than 100

23

Less than 150

55

Less than 200

81

Less than 250

93

Less than 300

100

Find the mean life time of electricity tubes.

Solution:

Class mark (xi) = (upper limit + lower limit)/2

Let assumed mean (A) = 175

Class size (h) = 50

Class Interval

No. of tubes (cf)

Class mark (xi)

di = xi – A

ui = di/h

Frequency (fi)

fiui 

0-50

8

25

-150

-3

8

-24

50-100

23

75

-100

-2

15

-30

100-150

55

125

-50

-1

32

-32

150-200

81

175

0

0

26

0

200-250

93

225

50

1

12

12

250-300

100

275

100

2

7

14

Total

∑fi = 100

∑fiui =-60


By step deviation method, Mean = x̄ = A+h∑fiui /∑fi

 = 175+50(-60/100)

= 175+50×-0.60

= 175-30

= 145

Hence the mean of the electricity tubes is 145.

27. Using the information given in the adjoining histogram, calculate the mean correct to one decimal place.

ML Aggarwal Sol Class 10 Maths chapter 21-3

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Assumed mean, A = 45

Class interval

frequency (fi)

Class mark (xi)

di = xi – A

fidi 

20-30

3

25

-20

-60

30-40

5

35

-10

-50

40-50

12

45

0

0

50-60

9

55

10

90

60-70

4

65

20

80

∑fi = 33

∑fidi = 60

By short cut method, Mean = x̄ = A+∑fidi /∑fi

= 45+60/33

= 45+1.81

= 46.81

= 46.8 [corrected to one decimal place]

Hence the mean is Rs.46.8.

Exercise 21.2

1. A student scored the following marks in 11 questions of a question paper : 3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7 Find the median marks.

Solution:

Arranging the data in the ascending order

1,2,2,3,4,5,5,6,7,7,8

Here number of terms, n = 11

Here n is odd.

So median = [(n+1)/2 ]th observation

= (11+1)/2

= 12/2

= 6th observation

Here 6th observation is 5.

Hence the median is 5.

2. (a) Find the median of the following set of numbers : 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7 (1990)
(b)For the following set of numbers, find the median: 10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15.

Solution:

(a) Arranging the numbers in ascending order :

0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9

Here, n = 12 which is even

Median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (12/2 th term + ((12/2)+1)th term)

= ½ (6 th term + (6+1)th term)

= ½ (6 th term + 7th term)

= ½ (4+5)

= 9/2

= 4.5

Hence the median is 4.5.

(b) Arranging the numbers in ascending order :

3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81

Here, n = 12 which is even

Median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (12/2 th term + ((12/2)+1)th term)

= ½ (6 th term + (6+1)th term)

= ½ (6 th term + 7th term)

= ½ (15+17)

= ½ ×32

= 16

Hence the median is 16.

3. Calculate the mean and the median of the numbers : 2, 1, 0, 3, 1, 2, 3, 4, 3, 5

Solution:

Arranging the numbers in ascending order :

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

Here, n = 10 which is even

Median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (10/2 th term + ((10/2)+1)th term)

= ½ (5 th term + (5+1)th term)

= ½ (5 th term + 6th term)

= ½ (2+3)

= ½ ×5

= 2.5

Hence the median is 2.5.

Mean = sum of the observations/ number of observations

= Ʃxi/n

= (0+1+1+2+2+3+3+3+4+5)/10

= 24/10

= 2.4

Hence the mean is 2.4.

4.The median of the observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.

Solution:

Observation are as follows :

11, 12, 14, (x-2), (x+4), (x+9), 32, 38, 47

n = 9

Here n is odd. So median = ((n+1)/2)th term

= (9+1)/2 )th term

= 5th term

= x+4

Given median = 24

x+4 = 24

x = 24 -4 = 20

Sum of observations = 11+12+14+(x-2)+(x+4)+(x+9)+32+38+47

= 165+3x

Substitute x = 20

Sum of observations = 165+3×20

= 165+60

= 225

Mean = Sum of observations /number of observations

= 225/9 = 25

Hence the value of x is 20 and mean is 25.

5.The mean of the numbers 1, 7, 5, 3, 4, 4, is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m-1 and median q. Find
(i) p
(ii) q
(iii) the mean of p and q.

Solution:

(i) Mean of 1, 7, 5, 3, 4, 4 is m.

Here n = 6

Mean, m = (1+7+5+3+4+4)/6

m = 24/6

m = 4


Given the numbers 3, 2, 4, 2, 3, 3, p have mean m-1.

So m-1 = (3+2+4+2+3+3+p)/7

4-1 = (17+p)/7

3 = (17+p)/7

3×7 = 17+p

21 = 17+p

p = 21-17

p = 4

Hence the value of p is 4.

(ii) Given the numbers have median q.

Arranging them in ascending order

2, 2, 3, 3, 3, 4, 4

Here n = 7 which is odd

So median = ((n+1)/2)th term

q = ((7+1)/2 )th term

q = (8/2 )th term

q = 4th term

q = 3

So value of q is 3.

(iii)mean of p and q = (p+q)/2

= (4+3)/2

= 7/2

= 3.5

Hence the mean of p and q is 3.5.

6. Find the median for the following distribution:

Wages per day in Rs.

38

45

48

55

62

65

No. of workers

14

8

7

10

6

2

Solution:

We write the distribution in cumulative frequency table.

Wages per day in Rs.

No. of workers (f)

Cumulative frequency

38

14

14

45

8

22

48

7

29

55

10

39

62

6

45

65

2

47

Here total number of observations, n = 47 which is odd.

So median =(( n+1)/2)th term

= ((47+1)/2 )th term

= (48/2 )th term

= 24th term

= 48 [Since 23rd to 29th observation is 48]

Hence the median is 48.

7. Find the median for the following distribution.

Marks

35

45

50

64

70

72

No. of students

3

5

8

10

5

5

Solution:

We write the distribution in cumulative frequency table.

Marks

No. of students (f)

Cumulative frequency

35

3

3

45

5

8

50

8

16

64

10

26

70

5

31

72

5

36

Here total number of observations, n = 36 which is even.

Median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (36/2 th term + ((36/2)+1)th term)

= ½ (18 th term + (18+1)th term)

= ½ (18 th term + 19th term)

= ½ (64+64) [Since 17th to 26th observation is 64]

= ½ ×128

= 64

Hence the median is 64.

8.Marks obtained by 70 students are given below :

Marks

20

70

50

60

75

90

40

No. of students

8

12

18

6

9

5

12

Calculate the median marks.

Solution:

We write the marks in ascending order in cumulative frequency table.

Marks

No. of students (f)

Cumulative frequency

20

8

8

40

12

20

50

18

38

60

6

44

70

12

56

75

9

65

90

5

70

Here total number of observations, n = 70 which is even.

Median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (70/2 th term + ((70/2)+1)th term)

= ½ (35 th term + (35+1)th term)

= ½ (35 th term + 36th term)

= ½ (50+50) [Since all observations from 21st to 38th are 50]

= ½ ×100

= 50

Hence the median is 50.

9. Calculate the mean and the median for the following distribution :

Number

5

10

15

20

25

30

35

Frequency

1

2

5

6

3

2

1

Solution:

We write the numbers in cumulative frequency table.

Marks (x)

No. of students (f)

Cumulative frequency

fx

5

1

1

5

10

2

3

20

15

5

8

75

20

6

14

120

25

3

17

75

30

2

19

60

35

1

20

35

Total

Ʃf = 20

Ʃfx = 390

Mean = Ʃfx/Ʃf

= 390/20

= 19.5

Hence the mean is 19.5.

Here number of observations, n = 20 which is even.

So median = = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (20/2 th term + ((20/2)+1)th term)

= ½ (10 th term + (10+1)th term)

= ½ (10 th term + 11th term)

= ½ (20+20) [Since all observations from 9th to 14th are 20]

= ½ ×140

= 20

Hence the median is 20.

10. The daily wages in (rupees of) 19 workers are

41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35.

find :
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range

Solution:

Arranging the observations in ascending order

21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53

Here n = 19 which is odd.

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

= (19+1)/2

= 20/2

= 10th term

= 31

Hence the median is 31.

(ii) Lower quartile, Q1 = ((n+1)/4) th term

= (19+1)/4

= 20/4

= 5 th term

= 27

Hence the lower quartile is 27.

(iii)Upper quartile, Q3 = (3(n+1)/4) th term

= (3×(19+1)/4) th term

= (3×(20/4)) th term

= (3×5) th term

= 15 th term

= 41

Hence the upper quartile is 41.

(iv)Interquartile range = Q3-Q1

= 41-27

= 14

Hence the Interquartile range is 14.

11.From the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range

Variate

15

18

20

22

25

27

30

Frequency

4

6

8

9

7

8

6

Solution:

We write the variates in cumulative frequency table.

Variate

Frequency (f)

Cumulative frequency

15

4

4

18

6

10

20

8

18

22

9

27

25

7

34

27

8

42

30

6

48

(i) Here number of observations, n = 48 which is even.

So median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (48/2 th term + ((48/2)+1)th term)

= ½ (24 th term + (24+1)th term)

= ½ (24 th term + 25th term)

= ½ (22+22) [Since all observations from 19th to 27th are 22]

= ½ ×44

= 22

Hence the median is 22.

(ii) Lower quartile, Q1 = (n/4) th term

= (48)/4

= 12 th term

= 20

Hence the lower quartile is 20.

(iii)Upper quartile, Q3 = (3n/4) th term

= (3×48/4) th term

= (3×12)th term

= 36 th term

= 27

Hence the upper quartile is 27.

(iv)Interquartile range = Q3-Q1

= 27-20

= 7

Hence the Interquartile range is 7.

12. For the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile

Variate

25

31

34

40

45

48

50

60

Frequency

3

8

10

15

10

9

6

2

Solution:

We write the variates in cumulative frequency table.

Variate

Frequency (f)

Cumulative frequency

25

3

3

31

8

11

34

10

21

40

15

36

45

10

46

48

9

55

50

6

61

60

2

63

(i) Here number of observations, n = 63 which is odd.

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

= (63+1)/2

= 64/2

= 32th term

= 40

Hence the median is 40.

(ii) Lower quartile, Q1 = ((n+1)/4) th term

= (63+1)/4

= 64/4

= 16 th term

= 34

Hence the lower quartile is 34.

(iii)Upper quartile, Q3 = (3(n+1)/4) th term

= (3×(63+1)/4) th term

= (3×(64/4)) th term

= (3×16) th term

= 48 th term

= 48

Hence the upper quartile is 48.

Exercise 21.3

1.Find the mode of the following sets of numbers ;
(i) 3, 2, 0, 1, 2, 3, 5, 3
(ii) 5, 7, 6, 8, 9, 0, 6, 8, 1, 8
(iii) 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7

Solution:

Mode is the number which appears most often in a set of numbers.

(i)Given set is 3, 2, 0, 1, 2, 3, 5, 3.

In this set, 3 occurs maximum number of times.

Hence the mode is 3.

(ii) Given set is 5, 7, 6, 8, 9, 0, 6, 8, 1, 8.

In this set, 8 occurs maximum number of times.

Hence the mode is 8.

(iii) Given set is 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7.

In this set, 5 occurs maximum number of times.

Hence the mode is 5.

2. Calculate the mean, the median and the mode of the numbers : 3, 2, 6, 3, 3, 1, 1, 2

Solution:

We arrange given data in ascending order 1, 1, 2, 2, 3, 3, 3, 6

Mean = Ʃxi/n

= (1+1+2+2+3+3+3+6)/8

= 21/8

= 2.625

Hence the mean is 2.625.

Here number of observations, n = 8 which is even.

So median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (8/2 th term + ((8/2)+1)th term)

= ½ (4 th term + (4+1)th term)

= ½ (4 th term + 5th term)

= ½ (2+3)

= ½ ×5

= 2.5

Hence the median is 2.5.

In the given set, 3 occurs maximum number of times.

Hence the mode is 3.

3. Find the mean, median and mode of the following distribution : 8, 10, 7, 6, 10, 11, 6, 13, 10

Solution:

We arrange given data in ascending order 6, 6, 7, 8, 10, 10, 10, 11, 13

Mean = Ʃxi/n

= (6+6+7+8+10+10+10+11+13)/9

= 81/9

= 9

Hence the mean is 9.

Here number of observations, n = 9 which is odd.

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

= (9+1)/2

= 10/2

= 5th term

= 10

Hence the median is 10.

In the given set, 10 occurs maximum number of times.

Hence the mode is 10.

4. Calculate the mean, the median and the mode of the following numbers : 3, 1, 5, 6, 3, 4, 5, 3, 7, 2

Solution:

We arrange given data in ascending order 1, 2, 3, 3, 3, 4, 5, 5, 6, 7

Mean = Ʃxi/n

= (1+2+3+3+3+4+5+5+6+7)/10

= 39/10

= 3.9

Here number of observations, n = 10 which is even.

So median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (10/2 th term + ((10/2)+1)th term)

= ½ (5 th term + (5+1)th term)

= ½ (5 th term + 6th term)

= ½ (3+4)

= ½ ×7

= 3.5

Hence the median is 3.5.

In the given set, 3 occurs maximum number of times.

Hence the mode is 3.

5. The marks of 10 students of a class in an examination arranged in ascending order are as follows: 13, 35, 43, 46, x, x +4, 55, 61,71, 80
If the median marks is 48, find the value of x. Hence, find the mode of the given data. (2017)

Solution:

Given data in ascending order: 13, 35, 43, 46, x, x +4, 55, 61,71, 80

Given median = 48

Number of observations, n = 10 which is even.

median = ½ ( n/2 th term + ((n/2)+1)th term)

48 = ½ (10/2 th term + ((10/2)+1)th term)

48 = ½ (5 th term + (5+1)th term)

48 = ½ (5 th term + 6th term)

48 = ½ (x+x+4)

48 = ½ ×(2x+4)

48 =x+2

x = 48-2 = 46

x+4 = 46+4 = 50

So the distribution becomes

13, 35, 43, 46, 46, 50, 55, 61,71, 80

Here 46 occurs maximum number of times.

Hence the mode is 46.

6.A boy scored the following marks in various class tests during a term each test being marked out of 20: 15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16
(i) What are his modal marks ?
(ii) What are his median marks ?
(iii) What are his mean marks ?

Solution:

(i)We arrange given marks in ascending order

7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19

16 appears maximum number of times.

Hence his modal mark is 16.

(ii)Here number of observations, n = 11 which is odd.

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

= (11+1)/2

= 12/2

= 6th term

= 15

Hence the median is 15.

(iii) Mean = Ʃxi/n

= 7+ 10+12+12+14+ 15+16+16+16+ 17+ 19

= 154/11

= 14

Hence the mean is 14.

7. Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks : 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8

Solution:

Given data is 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8

Number of observations, n = 16

Mean = Ʃxi/n

= (0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8)/16

= 64/16

= 4

Hence the mean is 4.

Here n = 16 which is even.

So median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (16/2 th term + ((16/2)+1)th term)

= ½ (8 th term + (8+1)th term)

= ½ (8 th term + 9th term)

= ½ (4+5)

= 9/2

= 4.5

Hence the median is 4.5.

Here 5 appears maximum number of times.

Hence mode is 5.

8. Find the mode and median of the following frequency distribution :

x

10

11

12

13

14

15

f

1

4

7

5

9

3

Solution:

We write the data in cumulative frequency table.

x

Frequency (f)

Cumulative frequency

10

1

1

11

4

5

12

7

12

13

5

17

14

9

26

15

3

29

Here number of observations, n = 29 which is odd.

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

= (29+1)/2

= 30/2

= 15th term

= 13

Hence the median is 13.

Here the frequency corresponding to 14 is maximum.

Hence the mode is 14.

9. The marks obtained by 30 students in a class assessment of 5 marks is given below:

Marks

0

1

2

3

4

5

No. of students

1

3

6

10

5

5

Calculate the mean, median and mode of the above distribution.

Solution:

We write the data in cumulative frequency table.

Marks x

Frequency (f)

Cumulative frequency

fx

0

1

1

0

1

3

4

3

2

6

10

12

3

10

20

30

4

5

25

20

5

5

30

25

Total

Ʃf = 30

Ʃfx = 90

Mean = Ʃfx/Ʃf

= 90/30

= 3

Hence the mean is 3.

Here number of observations, n = 30 which is even.

So median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (30/2 th term + ((30/2)+1)th term)

= ½ (15 th term + (15+1)th term)

= ½ (15 th term + 16th term)

= ½ (3+3)

= 6/2

= 3

Hence the median is 3.

Here the mark 3 occurs maximum number of times.

Hence the mode is 3.

10. The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Marks obtained

5

6

7

8

9

10

No. of students

3

9

6

4

2

1

Solution:

We write the marks in cumulative frequency table.

Marks x

Frequency (f)

fx

Cumulative frequency

5

3

15

3

6

9

54

12

7

6

42

18

8

4

32

22

9

2

18

24

10

1

10

25

Total

Ʃf = 25

Ʃfx = 171

Mean = Ʃfx/Ʃf

= 171/25

= 6.84

Hence the mean is 6.84.

Here number of observation, n = 25 which is odd.

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

= (25+1)/2

= 26/2

= 13th term

= 7

Hence the median is 7.

Here the frequency corresponding to 6 is maximum.

Hence the mode is 6.

11. At a shooting competition, the scores of a competitor were as given below :

Score

0

1

2

3

4

5

No. of shots

0

3

6

4

7

5

(i) What was his modal score ?
(ii) What was his median score ?
(iii) What was his total score ?
(iv) What was his mean ?

Solution:

We write the marks in cumulative frequency table.

Score x

No. of shots (f)

fx

Cumulative frequency

0

0

0

0

1

3

3

3

2

6

12

9

3

4

12

13

4

7

28

20

5

5

25

25

Total

Ʃf = 25

Ʃfx = 80

(i) Here the frequency corresponding to 4 is maximum. 4 occurs 7 times.

Hence his modal score is 4.

(ii)Here number of observation, n = 25 which is odd.

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

= (25+1)/2

= 26/2

= 13th term

= 3

Hence his median score is 3.

(iii)Total score = Ʃfx = 80

Hence his total score is 80.

(iv)Mean = Ʃfx/Ʃf

= 80/25

= 3.2

Hence his mean score is 3.2.

12. (i) Using step-deviation method, calculate the mean marks of the following distribution.
(ii) State the modal class.

Class interval

50-55

55-60

60-65

65-70

70-75

75-80

80-85

85-90

Frequency

5

20

10

10

9

6

12

8

Solution:

(i) Class mark (xi) = (upper limit + lower limit)/2

Let assumed mean (A) = 67.5

Class size (h) = 5

Class Interval

Frequency (fi)

Class mark (xi)

di = xi – A

ui = di/h

fiui 

50-55

5

52.5

-15

-3

-15

55-60

20

57.5

-10

-2

-40

60-65

10

62.5

5

-1

-10

65-70

10

67.5

10

0

0

70-75

9

72.5

15

1

9

75-80

6

77.5

20

2

12

80-85

12

82.5

25

3

36

85-90

8

87.5

30

4

32

Total

∑fi = 80

∑fiui = 24


By step deviation method, Mean = x̄ = A+h∑fiui /∑fi

 = 67.5+5(24/80)

= 67.5+5×0.3

= 67.5+1.5

= 69

Hence the mean of the distribution is 69.

(ii) Modal class is the class with highest frequency.

Here the modal class is 55-60.

13. The following table gives the weekly wages (in Rs.) of workers in a factory :

Weekly wages (in Rs)

50-55

55-60

60-65

65-70

70-75

75-80

80-85

85-90

No. of workers

5

20

10

10

9

6

12

8

Calculate:
(i) The mean.
(ii) the modal class
(iii) the number of workers getting weekly wages below Rs. 80.
(iv) the number of workers getting Rs. 65 or more but less than Rs. 85 as weekly wages.

Solution:

We write the given data in cumulative frequency table.

Class Interval

Frequency (fi)

Class mark (xi)

Cumulative frequency

fixi 

50-55

5

52.5

5

262.5

55-60

20

57.5

25

1150

60-65

10

62.5

35

625

65-70

10

67.5

45

675

70-75

9

72.5

54

652.5

75-80

6

77.5

60

465

80-85

12

82.5

72

990

85-90

8

87.5

80

700

Total

∑fi = 80

∑fixi = 5520

(i) Mean = ∑fixi /∑fi

= 5520/80

= 69

Hence the mean is 69.

(ii) Modal class is the class with highest frequency.

Here the modal class is 55-60.

(iii) The number of workers getting weekly wages below Rs. 80 is 60.

[Check the cumulative frequency column and class interval column. 60 workers get below Rs. 80]

(iv) The number of workers getting Rs. 65 or more but less than Rs. 85 as weekly wages = 72-35 = 37

[Check the cumulative frequency column and class interval column. ]


Exercise 21.4

1. Draw a histogram for the following frequency distribution and find the mode from the graph :

Class

0-5

5-10

10-15

15-20

20-25

25-30

Frequency

2

5

18

14

8

5

Solution:

Construct histogram using given data.

Class

0-5

5-10

10-15

15-20

20-25

25-30

Frequency

2

5

18

14

8

5

Represent class on X-axis and frequency on Y-axis.

ML Aggarwal Sol Class 10 Maths chapter 21-4

In the highest rectangle, draw two straight lines AC and BD.

P is the point of intersection.

Draw a vertical line through P to meet the X-axis at M.

The abscissa of M is 14.

Hence the mode is 14.

2. Find the modal height of the following distribution by drawing a histogram :

Height (in cm)

140-150

150-160

160-170

170-180

180-190

No. of students

7

6

4

10

2

Solution:

Construct histogram using given data.

Height (in cm)

140-150

150-160

160-170

170-180

180-190

No. of students

7

6

4

10

2

Represent height on X-axis and number of students on Y-axis.

Take scale: X axis : 2 cm = 10 (class interval)

Y axis : 1 cm = 1 (frequency)

ML Aggarwal Sol Class 10 Maths chapter 21-5

In the highest rectangle, draw two straight lines AC and BD.

P is the point of intersection.

Draw a vertical line through P to meet the X-axis at M.

The abscissa of M is 174.

Hence the mode is 174.

3. A Mathematics aptitude test of 50 students was recorded as follows :

Marks

50-60

60-70

70-80

80-90

90-100

No. of students

4

8

14

19

5

Draw a histogram for the above data using a graph paper and locate the mode. (2011)

Solution:

Construct histogram using given data.

Marks

50-60

60-70

70-80

80-90

90-100

No. of students

4

8

14

19

5

Represent marks on X-axis and number of students on Y-axis.

Take scale: X axis : 2 cm = 10 (class interval)

Y axis : 1 cm = 1 (frequency)

ML Aggarwal Sol Class 10 Maths chapter 21-6

In the highest rectangle, draw two straight lines AC and BD.

P is the point of intersection.

Draw a vertical line through P to meet the X-axis at M.

The abscissa of M is 82.5.

Hence the mode is 82.5.

4. Draw a histogram and estimate the mode for the following frequency distribution :

Classes

0-10

10-20

20-30

30-40

40-50

50-60

Frequency

2

8

10

5

4

3

Solution:

Construct histogram using given data.

Classes

0-10

10-20

20-30

30-40

40-50

50-60

Frequency

2

8

10

5

4

3

Represent classes on X-axis and frequency on Y-axis.

Take scale: X axis : 2 cm = 10 (class interval)

Y axis : 1 cm = 1 (frequency)

ML Aggarwal Sol Class 10 Maths chapter 21-7

In the highest rectangle, draw two straight lines AC and BD.

P is the point of intersection.

Draw a vertical line through P to meet the X-axis at M.

The abscissa of M is 23.

Hence the mode is 23.

5. IQ of 50 students was recorded as follows.

IQ score

80-90

90-100

100-110

110-120

120-130

130-140

No. of students

6

9

16

13

4

2

Draw a histogram for the above data and estimate the mode.

Solution:

Construct histogram using given data.

Represent classes on X-axis and frequency on Y-axis.

Take scale: X axis : 1 cm = 10 (class interval)

Y axis : 1 cm = 1 (frequency)

ML Aggarwal Sol Class 10 Maths chapter 21-8

In the highest rectangle, draw two straight lines AB and CD.

M is the point of intersection.

Draw a vertical line through M to meet the X-axis at L.

The abscissa of L is 107.

Hence the mode is 107.

6. Use a graph paper for this question. The daily pocket expenses of 200 students in a school are given below:

Pocket expenses (in Rs)

Number of students

(Frequency )

0-5

10

5-10

14

10-15

28

15-20

42

20-25

50

25-30

30

30-35

14

35-40

12

Draw a histogram representing the above distribution and estimate the mode from the graph.

Solution:

Construct histogram using given data.

Represent classes on X-axis and frequency on Y-axis.

Take scale: X axis : 2 cm = 5 (class interval)

Y axis : 1 cm = 5 (frequency)

ML Aggarwal Sol Class 10 Maths chapter 21-9

In the highest rectangle, draw two straight lines AC and BD.

P is the point of intersection.

Draw a vertical line through P to meet the X-axis at M.

The abscissa of M is 21.

Hence the mode is 21.

7. Draw a histogram for the following distribution :

Wt. in kg

40-44

45-49

50-54

55-59

60-64

65-69

No. of students

2

8

12

10

6

4

Hence estimate the modal weight.

Solution:

The given distribution is not continuous.

Adjustment factor = (45-44)/2 = ½ = 0.5

We subtract 0.5 from lower limit of the class interval and add 0.5 to upper limit.

So the new table in continuous form is given below.

Weight in kg

Number of students

(Frequency )

39.5-44.5

2

44.5-49.5

8

49.5-54.5

12

54.5-59.5

10

59.5-64.5

6

64.5-69.5

4

Construct histogram using given data.

Represent weight on X-axis and no. of students on Y-axis.

Take scale: X axis : 2 cm = 5 (class interval)

Y axis : 1 cm = 1 (frequency)

ML Aggarwal Sol Class 10 Maths chapter 21-10

In the highest rectangle, draw two straight lines AC and BD.

P is the point of intersection.

Draw a vertical line through P to meet the X-axis at M.

The abscissa of M is 52.75.

Hence the mode is 52.75.

8. Find the mode of the following distribution by drawing a histogram

Mid value

12

18

24

30

36

42

48

Frequency

20

12

8

24

16

8

12

Also state the modal class.

Solution:

Mid value

Frequency

12

20

18

12

24

8

30

24

36

16

42

8

48

12

Here mid value and frequency is given.

We can find the class size, h by subtracting second mid value from first mid value.

h = 18-12 = 6

So to find the lower limit of class interval, we subtract h/2 to the mid value.

To find the upper limit of class interval, we add h/2 to the mid value.

Here h/2 = 6/2 = 3

So lower limit = 12-3 = 9

Upper limit = 12+3 = 15

So the class interval is 9-15

Likewise we find the class interval of other values.

Mid value

Class interval

Frequency

12

9-15

20

18

15-21

12

24

21-27

8

30

27-33

24

36

33-39

16

42

39-45

8

48

45-51

12

Construct histogram using given data.

Take scale: X axis : 2 cm = 6 (class interval)

Y axis : 1 cm = 2 (frequency)

ML Aggarwal Sol Class 10 Maths chapter 21-11

In the highest rectangle, draw two straight lines AB and CD.

M is the point of intersection.

Draw a vertical line through M to meet the X-axis at L.

The abscissa of L is 30.5.

Hence the mode is 30.5.

Modal class is the class with highest frequency.

Hence the modal class is 27-33.

Exercise 21.5

1.Draw an ogive for the following frequency distribution: 

Height ( in cm )

150-160

160-170

170-180

180-190

190-200

No. of students

8

3

4

10

2

Solution:

We write the given data in cumulative frequency table.

Height in cm

No of students

Cumulative frequency

150-160

8

8

160-170

3

11

170-180

4

15

180-190

10

25

190-200

2

27

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis).

Plot the points (160, 8), (170, 11), (180, 15), (190, 25) and (200, 27) on the graph.

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-12

2. Draw an ogive for the following data: 

Class intervals

1-10

11-20

21-30

31-40

41-50

51-60

Frequency

3

5

8

7

6

2

Solution:

The given distribution is not continuous.

Adjustment factor = (11-10)/2 = ½ = 0.5

We subtract 0.5 from lower limit of the class interval and add 0.5 to upper limit.

So the new table in continuous form is given below.

We write the given data in cumulative frequency table.

Class intervals

frequency

Cumulative frequency

0.5-10.5

3

3

10.5-20.5

5

8

20.5-30.5

8

16

30.5-40.5

7

23

40.5-50.5

6

29

50.5-60.5

2

31

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis).

Plot the points (10.5, 3), (20.5, 8), (30.5, 16), (40.5, 23) , (50.5, 29) and (60.5, 31) on the graph.

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-13

3. Draw a cumulative frequency curve for the following data: 

Marks obtained

24-29

29-34

34-39

39-44

44-49

49-54

54-59

No. of students

1

2

5

6

4

3

2

Solution:

We write the given data in cumulative frequency table.

Marks obtained

No of students

Cumulative frequency

24-29

1

1

29-34

2

3

34-39

5

8

39-44

6

14

44-49

4

18

49-54

3

21

54-59

2

23

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis).

Plot the points (29, 1), (34, 3), (39, 8), (44, 14), (49, 18) , (54, 21)  and (59, 23) on the graph.

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-14

Exercise 21.6

1. The following table shows the distribution of the heights of a group of a factory workers.

Height ( in cm )

150-155

155-160

160-165

165-170

170-175

175-180

180-185

No. of workers

6

12

18

20

13

8

6

(i) Determine the cumulative frequencies.

(ii) Draw the cumulative frequency curve on a graph paper. Use 2 cm = 5 cm height on one axis and 2 cm = 10 workers on the other.

(iii) From your graph, write down the median height in cm.

Solution:

(i) We write the given data in cumulative frequency table.

Height in cm

No of workers f

Cumulative frequency

150-155

6

6

155-160

12

18

160-165

18

36

165-170

20

56

170-175

13

69

175-180

8

77

180-185

6

83

(ii)Plot the points (155, 6), (160, 18), (165, 36), (170, 56), (175, 69), (180, 77) and (185, 83) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-15

(iii)Here n = 83 , which is odd.

So median = ((n+1)/2)th observation

= ((83+1)/2)th observation

= (84/2)th observation

= 42th observation

Take a point A(42) on Y-axis. From A, draw a horizontal line parallel to X-axis meeting the curve at B. From B draw a line perpendicular on the x-axis which meets it at C. 

C is the median which is 166.5 cm.

2. Using the data given below construct the cumulative frequency table and draw the-Ogive. From the ogive determine the median. 

Marks

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

No. of students

3

8

12

14

10

6

5

2

Solution:

We write the given data in cumulative frequency table.

Marks

No of students f

Cumulative frequency c.f

0-10

3

3

10-20

8

11

20-30

12

23

30-40

14

37

40-50

10

47

50-60

6

53

60-70

5

58

70-80

2

60

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),

Plot the points (10, 3), (20, 11), (30, 23), (40, 37), (50, 47), (60, 53), (70, 58) and (80, 60) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-16

Here number of observations, n = 60 which is even.

So median = ( n/2) th term

= (60/2 th term

= 30 th term

Mark a point A(30) on Y-axis. From A, draw a horizontal line parallel to X-axis meeting the curve at P. From P draw a line perpendicular on the x-axis which meets it at Q. 

Q is the median .

Q = 35

Hence the median is 35 .

3. Use graph paper for this question. The following table shows the weights in gm of a sample of 100 potatoes taken from a large consignment: 

Weight (gm)

50-60

60-70

70-80

80-90

90-100

100-110

110-120

120-130

Frequency

8

10

12

16

18

14

12

10

(i) Calculate the cumulative frequencies. 

(ii) Draw the cumulative frequency curve and from it determine the median weight of the potatoes. (1996)

Solution:

(i)We write the given data in cumulative frequency table.

Marks

frequency f

Cumulative frequency c.f

50-60

8

8

60-70

10

18

70-80

12

30

80-90

16

46

90-100

18

64

100-110

14

78

110-120

12

90

120-130

10

100

(ii)To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),

Plot the points (60, 8), (70, 18), (80, 30), (90, 46), (100, 64), (110, 78), (120, 90) and (130, 100) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-17

Here n = 100 which is even.

So median = ( n/2 th term)

= (100/2 th term)

= (50 th term)

Now mark a point A (50) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at P. From P, draw a perpendicular on x-axis meeting it at Q. 

Q is the median.

Q = 93 gm.

Hence the median is 93.

4. Attempt this question on graph paper. 

Age( yrs)

5-15

15-25

25-35

35-45

45-55

55-65

65-75

No. of casualities due to accidents

6

10

15

13

24

8

7

(i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10 years, on one axis and 2 cm = 10 casualties on the other.

 (ii) From your graph determine

(1) the median and (2) the upper quartile

Solution:

(i)We write the given data in cumulative frequency table.

Age (yrs)

No of casualities due to accidents f

Cumulative frequency c.f

5-15

6

6

15-25

10

16

25-35

15

31

35-45

13

44

45-55

24

68

55-65

8

76

65-75

7

83

(ii)To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),

Plot the points (15, 6), (25, 16), (35, 31), (45, 44), (55, 68), (65, 76) and (75, 83) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-18

(ii)(1). Here n = 83, which is odd.

So median = (n+1)/2)th term

= ((83+1)/2)

= 84/2

= 42

Now mark a point A (42) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at P. From P, draw a perpendicular on x-axis meeting it at Q. 

Q is the median.

Q = 43

Hence the median is 43.

(ii)(2). Upper quartile = (3(n+1)/4)

= (3×(83+1)/4)

= (3×(84)/4)

= 63

Now mark a point B (63) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at L. From L, draw a perpendicular on x-axis meeting it at M. 

M = 52

Hence the upper quartile is 52.

5. The weight of 50 workers is given below: 

Weight in kg

50-60

60-70

70-80

80-90

90-100

100-110

110-120

No. of workers

4

7

11

14

6

5

3

Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis , and 2 cm = 5 workers along the other axis.

Use a graph to estimate the following:

 (i) the upper and lower quartiles. 

(ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight. (2015)

Solution:

We write the given data in cumulative frequency table.

Weight in kg

No of workers f

Cumulative frequency c.f

50-60

4

4

60-70

7

11

70-80

11

22

80-90

14

36

90-100

6

42

100-110

5

47

110-120

3

50

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),

Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47) and (120, 50) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-19

(i)Here n = 50, which is even.

Upper quartile = 3n/4

= 3×50/4

= 150/4

= 37.5

Now mark a point A (37.5) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at B. From B, draw a perpendicular on x-axis meeting it at C. 

C = 92.5

Hence the upper quartile is 92.5 kg.

Lower quartile, Q1 = (n/4) th term

= 50/4

= 12.5

Now mark a point D(12.5) on the Y-axis and from D draw a line parallel to X-axis meeting the curve at E. From E, draw a perpendicular on x-axis meeting it at F. 

F = 72

Hence the lower quartile is 72 kg.

(ii) Mark on the graph point P which is 95 kg on X axis.

Through P draw a vertical line to meet the ogive at Q. Through Q, draw a horizontal line to meet y-axis at R. 

The ordinate of point R represents 40 workers on the y-axis .

 The number of workers who are 95 kg and above = Total number of workers – number of workers of weight less than 95 kg = 50-40 = 10

6. The table shows the distribution of scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the x-axis and 2 cm = 20 shooters on the y-axis)

Scores

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

No. of shooters

9

13

20

26

30

22

15

10

8

7

Use your graph to estimate the following:

 (i) The median. 

(ii) The interquartile range. 

(iii) The number of shooters who obtained a score of more than 85%.

Solution:

We write the given data in cumulative frequency table.

Scores

No of shooters f

Cumulative frequency c.f

0-10

9

9

10-20

13

22

20-30

20

42

30-40

26

68

40-50

30

98

50-60

22

120

60-70

15

135

70-80

10

145

80-90

8

153

90-100

7

160

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),

Plot the points (10, 9), (20, 22), (30, 42), (40, 68), (50, 98), (60, 120), (70, 135), (80, 145), (90, 153) and (100, 160) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-20

(i)Here n = 160, which is even.

So median = n/2 = 80

Now mark a point A(80) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at B. From P, draw a perpendicular on x-axis meeting it at C. 

C is the median.

C = 44

(ii) lower quartile, Q1 = (n/4)th term

= 160/4

= 40

Now mark a point D(40) on the Y-axis and from that point draw a line parallel to X-axis meeting the curve at E. From E, draw a perpendicular on x-axis meeting it at F. 

F = 29

So Q1= 29

Upper quartile, Q3 = (3n/4)th term

= 3×160/4

= 3×40

= 120

Mark a point P(120) on the Y-axis and from that point draw a line parallel to X-axis meeting the curve at Q. From Q, draw a perpendicular on x-axis meeting it at R. 

R = 60

So Q3 = 60

Inter quartile range = Q3– Q1

= 60-29

= 31

Hence the Inter quartile range is 31.

(iii) Mark a point Z(85) on the X axis.

From Z on X-axis, draw a perpendicular to it meeting the curve at Y. From Y, draw a line parallel to X-axis meeting Y-axis at X. X is the required point which is 150.

 Number of shooters getting more than 85% scores = Total number of shooters – number of shooters who got till 85% = 160-150 = 10.

Hence the number of shooters getting more than 85% scores is 10.

7. The daily wages of 80 workers in a project are given below 

Wages in Rs

400-450

450-500

500-550

550-600

600-650

650-700

700-750

No. of workers

2

6

12

18

24

13

5

Use a graph paper to draw an ogive for the above distribution. ( a scale of 2 cm = Rs 50 on x- axis and 2 cm = 10 workers on y-axis). your ogive to estimate

 (i) the median wage of the workers. 

(ii) the lower quartile wage of the workers. 

(iii) the number of workers who earn more than Rs 625 daily. (2017)

Solution:

We write the given data in cumulative frequency table.

Wages in Rs.

No of workers f

Cumulative frequency c.f

400-450

2

2

450-500

6

8

500-550

12

20

550-600

18

38

600-650

24

62

650-700

13

75

700-750

5

80

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),

Plot the points (450, 9), (500, 22), (550, 42), (600, 68), (650, 98), (700, 120) and (750, 135) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-21

(i)Here n = 80.

Median = (n/2)th term

= 80/2

= 40th term

Mark a point (40) on Y axis. Draw a line from that point parallel to X axis. Let it meet the curve at A.

Draw a perpendicular from A to meet X axis at B.

The value of B is 604.

Hence the median is 604.

(ii)Lower quartile, Q1 = (n/4)th term

= 80/4

= 20th term

= 550 [from graph]

(iii) Draw a vertical line through the point 625 on X axis. which meets the graph at point C. From C, draw a horizontal line which meets the y-axis at the mark of 50. 

Thus, number of workers that earn more Rs 625 daily = Total no. of workers – no. of workers who earn upto 625

= 80-50 = 30

8. Marks obtained by 200 students in an examination are given below

marks

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

No. of students

5

11

10

20

28

37

40

29

14

6

Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis.

Using the graph, determine 

(i) The median marks.

 (ii) The number of students who failed if minimum marks required to pass is 40. 

(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.

Solution:

We write the given data in cumulative frequency table.

Marks

No of students f

Cumulative frequency c.f

0-10

5

5

10-20

11

16

20-30

10

26

30-40

20

46

40-50

28

74

50-60

37

111

60-70

40

151

70-80

29

180

80-90

14

194

90-100

6

200

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),

Plot the points (10, 5), (20, 16), (30, 26), (40, 46), (50, 74), (60, 111) , (70, 151) (80, 180) (90, 194) and (100, 200) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-22

(i)Here n= 200

Median = (n/2)th term

= 200/2

= 100th term

= 57 [from graph]

(ii)number of students failed if minimum marks required to pass is 40 = 44 [from graph]

(iii)Number of students who got grade 1 = number of students who scored 85 and more

= 200-188

= 12[From graph]

9. The monthly income of a group of 320 employees in a company is given below 

Monthly income

No. of employees

6000-7000

20

7000-8000

45

8000-9000

65

9000-10000

95

10000-11000

60

11000-12000

30

12000-13000

5

Draw an ogive of the given distribution on a graph sheet taking 2 cm = Rs. 1000 on one axis and 2 cm = 50 employees on the other axis.

From the graph determine

 (i) the median wage. 

(ii) the number of employees whose income is below Rs. 8500.

 (iii) If the salary of a senior employee is above Rs. 11500, find the number of senior employees in the company. 

(iv) the upper quartile.

Solution:

We write the given data in cumulative frequency table.

Monthly income

No. of employees

Cumulative frequency c.f

6000-7000

20

20

7000-8000

45

65

8000-9000

65

130

9000-10000

95

225

10000-11000

60

285

11000-12000

30

315

12000-13000

5

320

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),

Plot the points (7000, 20), (8000, 65), (9000, 130), (10000, 225), (11000, 285), (12000, 315) and (13000, 320) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-23

(i)here n = 320

Median = (n/2)th term

= 320/2

= 160th term

Mark the point A(160) on Y axis.

Draw a line parallel to x axis from that point.

Let it meet the curve at P.

Draw a perpendicular from P to X axis which meets at M.

M is the median.

Here median is 9300. [from graph]

(ii)Mark the point B(8500) on X axis.

Draw a line parallel to Y axis which meets curve at Q.

From Q draw a line parallel to X axis which meets Y axis at N.

N = 98

number of employees whose income is below 8500 = 98

(iii) Mark the point C(11500) on the x-axis.

Draw a line perpendicular to x-axis meeting the curve at R. 

From R, draw a line parallel to x-axis meeting y-axis at L which is 300 

No. of employees getting more than Rs. 11500 = 320-300 = 20

(iv)upper quartile = 3n/4

= 3×320/4

= 240

Mark the point T(240) on Y axis.

From that point on y-axis, draw a line perpendicular on the x-axis which meets the curve at S.

 From S, draw a perpendicular on x-axis meeting it at U, which is 10250. 

Hence upper quartile is 10250.

10. Using a graph paper, draw an ogive for the following distribution which shows a record of the weight in kilograms of 200 students 

Weight

40-45

45-50

50-55

55-60

60-65

65-70

70-75

75-80

Frequency

5

17

22

45

51

31

20

9

Use your ogive to estimate the following: 

(i) The percentage of students weighing 55 kg or more. 

(ii) The weight above which the heaviest 30% of the students fall, 

(iii) The number of students who are : 

1. under-weight and 

2. over-weight, if 55.70 kg is considered as standard weight.

Solution:

We write the given data in cumulative frequency table.

Weight

Frequency

Cumulative frequency c.f

40-45

5

5

45-50

17

22

50-55

22

44

55-60

45

89

60-65

51

140

65-70

31

171

70-75

20

191

75-80

9

200

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),

Plot the points (45, 5), (50, 22), (55, 44), (60, 89), (65, 140), (70, 171), (75, 191) and (80, 200) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-24

(i)Total number of students = 200

The number of students weighing 55 kg or more = 200-44 = 156  [From the graph]

Percentage = (156/200)×100

= 156/2

= 78%.

(ii)30% of 200 = (30/100)×200

= 30×2

= 60

No of heaviest students = 31+20+9 = 60

60 students fall above 65 kg.

(iii)If 55.70 kg is the standard weight,

No. of students who are under weight = 47 [from graph]

No. of students who are overweight = 200-47 = 153

11. The marks obtained by 100 students in a Mathematics test are given below

Marks

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

No. of students

3

7

12

17

23

14

9

6

5

4

Draw an ogive on a graph sheet and from it determine the : 

(i) median 

(ii) lower quartile 

(iii) number of students who obtained more than 85% marks in the test. 

(iv) number of students who did not pass in the test if the pass percentage was 35.

Solution:

We write the given data in cumulative frequency table.

Marks

No. of students

Cumulative frequency c.f

0-10

3

3

10-20

7

10

20-30

12

22

30-40

17

39

40-50

23

62

50-60

14

76

60-70

9

85

70-80

6

91

80-90

5

96

90-100

4

100

Plot the points (10, 3), (20, 10), (30, 22), (40, 39), (50, 62), (60, 76), (70, 85), (80, 91), (90, 96) and (100, 100) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-25

(i)Here n = 100

Median = n/2

= 50 th term

Mark a point A(50) on Y axis. Draw a line parallel to X axis from A.

Let it meet the curve at B. From B draw a perpendicular which meets X axis at C.

The point C is 45.

Hence median is 45.

(ii)Lower quartile = n/4

= 100/4

= 25th term

Mark a point P (25) on Y axis. Draw a line parallel to X axis from that point.

Let it meet the curve at Q. From that point draw a perpendicular which meets X axis at R.

The point R is 32.

Hence lower quartile is 32.

(iii)no. of students who obtained more than 85% = 100-94 = 6 [from graph]

(iv)No of students who failed if 35% is the pass percentage = 25 [from graph]

12. The marks obtained by 120 students in a Mathematics test are-given below

Marks

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

No. of students

5

9

16

22

26

18

11

6

4

3

Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for ogive to estimate the following: 

(i) the median 

(ii) the lower quartile 

(iii) the number of students who obtained more than 75% marks in the test.

(iv) the number of students who did not pass in the test if the pass percentage was 40. (2002)

Solution:

We write the given data in cumulative frequency table.

Marks

No. of students

Cumulative frequency c.f

0-10

5

5

10-20

9

14

20-30

16

30

30-40

22

52

40-50

26

78

50-60

18

96

60-70

11

107

70-80

6

113

80-90

4

117

90-100

3

120

Plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117) and (100, 120) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-26

(i)Here n = 120

Median = (n/2)th term

= 120/2

= 60th term

Mark point A(60) on Y axis. Draw a line parallel to X axis from A.

Let it meet the curve at B. Draw a straight line from B to X axis which meets at C.

C = 50

Hence median is 50.

(ii)Lower quartile = (n/4)th term

= 120/4

= 30th term

Mark a point P (30) on Y axis. Draw a line parallel to X axis from that point.

Let it meet the curve at Q. From that point draw a perpendicular which meets X axis at R.

The point R is 30.

Hence lower quartile is 30.

(iii)Mark a point U(75) on X axis.

Draw a line parallel to Y axis which meets curve at T.

From T, draw a line parallel to X axis to meet Y axis at S.

S = 110

No. of students who obtained more than 75% = 120-110 = 10

(iv) Mark a point Z(40) on X axis.

Draw a line parallel to Y axis which meets curve at Y.

From Y, draw a line parallel to X axis to meet Y axis at X.

X = 52

No of students who failed if 40% is the pass percentage is 52.

13. The following distribution represents the height of 160 students of a school. 

Height

140-145

145-150

150-155

155-160

160-165

165-170

170-175

175-180

No. of students

12

20

30

38

24

16

12

8

Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis.

Using the graph, determine :

 (i)The median height. 

(ii)The inter quartile range. 

(iii) The number of students whose height is above 172 cm.

Solution:

We write the given data in cumulative frequency table.

Height

No. of students

Cumulative frequency c.f

140-145

12

12

145-150

20

32

150-155

30

62

155-160

38

100

160-165

24

124

165-170

16

140

170-175

12

152

175-180

8

160

Plot the points (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152), and (180, 160) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-27

(i)Here n = 160

Median = (n/2)th term

= 160/2

= 80th term

Mark point A(80) on Y axis. Draw a line parallel to X axis from A.

Let it meet the curve at B. Draw a straight line from B to X axis which meets at C.

C = 157.5

Hence median is 157.5.

(ii) Lower quartile , Q1 = (n/4)th term

= 160/4

= 40th term

Proceeding in the same way mentioned in (i),

we get lower quartile = 152 [Point R]

Upper quartile, Q3 = 3n/4

= 3×160/4

= 3×40

= 120th term

Proceeding in the same way mentioned in (i),

we get upper quartile = 164 [Point Z]

Interquartile range = Q3-Q1

= 164-152

= 12

(iii)Mark a point O(172 ) on X axis. Draw a line parallel to Y axis from O.

Let it meet the curve at N. Draw a straight line from N to Y axis which meets at M.

M = 144

Hence number of students whose height is more than 172 cm is 160-144 = 16

14. 100 pupils in a school have heights as tabulated below

Height in cm

121-130

131-140

141-150

151-160

161-170

171-180

No. of pupils

12

16

30

20

14

8

Draw the ogive for the above data and from it determine the median (use graph paper).

 Solution:

We write the given data in cumulative frequency table (in continuous distribution):

Height

No. of students

Cumulative frequency c.f

120.5-130.5

12

12

130.5-140.5

16

28

140.5-150.5

30

58

150.5-160.5

20

78

160.5-170.5

14

92

170.5-180.5

8

100

Plot the points (130.5, 12), (140.5, 28), (150.5, 58), (160.5, 78), (170.5, 92) and (180.5, 100) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-28

Here n = 100

Median = (n/2)th term

= 100/2

= 50th term

Mark point A(50) on Y axis. Draw a line parallel to X axis from A.

Let it meet the curve at P. Draw a straight line from P to X axis which meets at Q.

Q = 147.5

Hence median is 147.5.

Chapter Test

1. Arun scored 36 marks in English, 44 marks in Civics, 75 marks in Mathematics and x marks in Science. If he has scored an average of 50 marks, find x.

Solution:

Marks scored in English = 36

Marks scored in Civics = 44

Marks scored in Mathematics = 75

Marks scored in Science = x

No. of subjects = 4

Average marks = sum of marks / No. of subjects = 50 [Given]

(36+44+75+x)/4 = 50

155+x = 4×50

155+x = 200

x = 200-155

x = 45

Hence the value of x is 45.

2. The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of new set of 20 numbers.

Solution:

Given the mean of 20 numbers = 18

Sum of numbers = 18×20 = 360

If 3 is added to each of first 10 numbers, then new sum = (3×10)+360

= 30+360

= 390

New mean = 390/20

= 19.5

Hence the mean of new set of 20 numbers is 19.5.

3. The average height of 30 students is 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for computation of mean. Find the correct mean.

Solution:

Average height of 30 students = 150 cm

So sum of height = 150×30 = 4500

Difference between correct value and wrong value = 165-135 = 30

So actual sum = 4500+30 = 4530

So actual mean = 4530/30 = 31

Hence the correct mean 31.

4. There are 50 students in a class of which 40 are boys and the rest girls. The average weight of the students in the class is 44 kg and average weight of the girls is 40 kg. Find the average weight of boys.

Solution:

Total number of students = 50

No. of boys = 40

No. of girls = 50-40 = 10

Average weight of 50 students = 44 kg

So sum of weight = 44×50 = 2200 kg

Average weight of girls = 40 kg

So sum of weight of girls = 40×10 = 400 kg

Total weight of boys = 2200-400 = 1800 kg

Average weight of boys = 1800/40 = 45 kg

Hence the average weight of boys is 45 kg.

5. The contents of 50 boxes of matches were counted giving the following results.

No. of matches

41

42

43

44

45

46

No. of boxes

5

8

13

12

7

5

Calculate the mean number of matches per box.

Solution:

No. o f matches x

No. of boxes f

fx

41

5

205

42

8

336

43

13

559

44

12

528

45

7

315

46

5

230

Total

Ʃf = 50

Ʃfx =2173

Mean = Ʃfx/Ʃf

= 2173/50

= 43.46

Hence the mean is 43.46.

6. The heights of 50 children were measured (correct to the nearest cm) giving the following results

Height (in cm)

65

66

67

68

69

70

71

72

73

No. of children

1

4

5

7

11

10

6

4

2

Calculate the mean height for this distribution correct to one place of decimal. 

Solution:

Height x

No. of children f

fx

65

1

65

66

4

264

67

5

335

68

7

476

69

11

759

70

10

700

71

6

426

72

4

288

73

2

146

Total

Ʃf = 50

Ʃfx = 3459

Mean = Ʃfx/Ʃf

= 3459/50

= 69.18

= 69.2 [corrected to one decimal place]

Hence the mean is 69.2.

7. Find the value of p for the following distribution whose mean is 20.6.

Variate (xi)

10

15

20

25

35

Frequency (fi)

3

10

p

7

5

Solution:

Variate (xi)

Frequency (fi)

fx

10

3

30

15

10

150

20

p

20p

25

7

175

35

5

175

Total

Ʃfi = 25+p

Ʃfi xi = 530+20p

Mean = Ʃfx/Ʃf

20.6 = (530+20p )/(25+p) [Given mean = 20.6]

20.6(25+p) = (530+20p)

515 + 20.6p = 530+20p

20.6p-20p = 530-515

0.6p = 15

p = 15/0.6

p = 25

Hence the value of p is 25.

8. Find the value of p if the mean of the following distribution is 18. 

Variate (xi)

13

15

17

19

20+p

23

Frequency (fi)

8

2

3

4

5p

6

Solution:

Variate (xi)

Frequency (fi)

fi xi

13

8

104

15

2

30

17

3

51

19

4

76

20+p

5p

5p2+100p

23

6

138

Total

Ʃfi = 23+5p

Ʃfi xi = 399+5p2+100p

Mean = Ʃfi xi / Ʃfi

18 = (399+5p2+100p)/( 23+5p) [Given mean = 18]

18(23+5p) = 399+5p2+100p

414 + 90p = 399+5p2+100p

5p2+100p-90p+399-414 = 0

5p2+10p-15 = 0

Dividing by 5, we get

p2+2p-3 = 0

(p-1)(p+3) = 0

p-1 = 0 or p+3 = 0

p = 1 or p = -3

p cannot be negative.

So p = 1

Hence the value of p is 1.

9. Find the mean age in years from the frequency distribution given below:

Age in years

25-29

30-34

35-39

40-44

45-49

50-54

55-59

No. of persons

4

14

22

16

6

5

3

Solution:

The given distribution is not continuous.

Adjustment factor = (30-29)/2 = ½ = 0.5

We subtract 0.5 from lower limit of the class interval and add 0.5 to upper limit.

So the new table in continuous form is given below.

Age

Mid value xi

No. of persons fi

fixi

24.5-29.5

27

4

108

29.5-34.5

32

14

448

34.5-39.5

37

22

814

39.5-44.5

42

16

672

44.5-49.5

47

6

282

49.5-54.5

52

5

260

54.5-59.5

57

3

171

Total

Ʃfi = 70

Ʃ fixi = 2755

Mean = Ʃfi xi / Ʃfi

= 2755/70

= 39.357

= 39.36

Hence the mean age is 39.36 years.

10. Calculate the Arithmetic mean, correct to one decimal place, for the following frequency

Marks

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

Students

2

4

5

16

20

10

6

8

4

Solution:

Class mark, xi = (upper class limit + lower class limit)/2

Marks

Students fi

Class mark xi

fixi

10-20

2

15

30

20-30

4

25

100

30-40

5

35

175

40-50

16

45

720

50-60

20

55

1100

60-70

10

65

650

70-80

6

75

450

80-90

8

85

680

90-100

4

95

380

Total

Ʃfi = 75

Ʃfixi = 4285

Mean = Ʃfi xi / Ʃfi

= 4285/75

= 57.133

= 57.1

Hence the mean is 57.1.

11. The mean of the following frequency distribution is 62.8. Find the value of p. 

Class

0-20

20-40

40-60

60-80

80-100

100-120

Frequency

5

8

p

12

7

8

Solution:

Class

Frequency fi

Class mark xi

fixi

0-20

5

10

50

20-40

8

30

240

40-60

p

50

50p

60-80

12

70

840

80-100

7

90

630

100-120

8

110

880

Total

Ʃfi = 40+p

Ʃfixi = 2640+50p

Mean = Ʃfi xi / Ʃfi

62.8 = (2640+50p)/( 40+p) [Given mean = 62.8]

62.8(40+p) = 2640+50p

2512+62.8p = 2640+50p

62.8p-50p = 2640-2512

12.8p = 128

p = 128/12.8

p = 10

Hence the value of p is 10.

12. The daily expenditure of 100 families are given below. Calculate f1, and f2, if the mean daily expenditure is Rs 188. 

Expenditure in Rs

140-160

160-180

180-200

200-220

220-240

No. of families

5

25

f1

f2

5

Solution:

Given mean = 188

Class

Frequency fi

Class mark xi

fixi

140-160

5

150

750

160-180

25

170

4250

180-200

f1

190

190f1

200-220

f2

210

210f2

220-240

5

230

1150

Total

Ʃfi = 35+f1+f2 = 100

Ʃfixi = 6150+190f1+210f2

Given no. of families = 100

So 35+f1+f2 = 100

f1+f2 = 100-35 = 65

f1 = 65-f2 ..(i)

Mean = Ʃfi xi / Ʃfi

188 = (6150+190f1+210f2)/100 [Given mean = 188]

188(100) = 6150+190f1+210f2

18800 = 6150+190f1+210f2

18800-6150 = 190f1+210f2

12650 = 190f1+210f2 ..(ii)

Substitute (i) in (ii)

12650 = 190(65-f2)+210f2

12650 = 12350-190f2+210f2

12650-12350 = -190f2+210f2

300 = 20f2

f2 = 300/20 = 15

Put f2 in (i)

f1 = 65-15

f1 = 50

Hence the value of f1 and f2 is 50 and 15 respectively.

13. The measures of the diameter of the heads of 150 screw is given in the following table. If the mean diameter of the heads of the screws is 51.2 mm, find the values of p and q .

Diameter in mm

32-36

37-41

42-46

47-51

52-56

57-61

62-66

No. of screws

15

17

p

25

q

20

30

Solution:

Given mean = 51.2 mm

The given distribution is not continuous.

Adjustment factor = (37-36)/2 = ½ = 0.5

We subtract 0.5 from lower limit of the class interval and add 0.5 to upper limit.

So the new table in continuous form is given below.

Diameter in mm

Mid value xi

No. of screws fi

fixi

31.5-36.5

34

15

510

36.5-41.5

39

17

663

41.5-46.5

44

p

44p

46.5-51.5

49

25

1225

51.5-56.5

54

q

54q

56.5-61.5

59

20

1180

61.5-66.5

64

30

1920

Total

Ʃfi = 107+p+q

Ʃ fixi = 5498+44p+54q

Given No. of screws = 150

107+p+q = 150

p = 150-107-q

p = 43-q ..(i)

Mean = Ʃfi xi / Ʃfi

51.2 = (5498+44p+54q)/150

51.2×150 = 5498+44p+54q

7680 = 5498+44(43-q)+54q

7680 = 5498+1892-44q+54q

7680 -5498-1892 = -44q+54q

290 = 10q

q = 290/10 = 29

Put q in (i)

p = 43-29

p = 14

Hence the value of p and q is 14 and 29 respectively.

14. The median of the following numbers, arranged in ascending order is 25. Find x.

11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46

Solution:

Here n = 10, which is even

Median = 25

So Median = ½ ( n/2 th term + ((n/2)+1)th term)

25 = ½ (( 10/2 )th term + (10/2)+1)th term)

25 = ½ (( 5 )th term + (6)th term)

25 = ½ (x+2 + x+4)

25 = ½ (2x+6)

2x+6 = 25×2

2x+6 = 50

2x = 50-6 = 44

x = 44/2 = 22

Hence the value of x is 22.

15. If the median of 5, 9, 11, 3, 4, x, 8 is 6, find the value of x.

Solution:

Arranging numbers in ascending order

3,4,5,x,8,9,11

Here n = 7 which is odd

Given median = 6

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

6 = ((7+1)/2)th term

6 = ((8/2)th term

6 = 4th term

6 = x

Hence the value of x is 6.

16. Find the median of: 17, 26, 60, 45, 33, 32, 29, 34, 56 .

If 26 is replaced by 62, find the new median.

Solution:

Arranging numbers in ascending order

17,26,29,32,33,34,45,56,60

Here n = 9 which is odd.

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

Median = ((9+1)/2)th term

Median = (10/2)th term

Median = 5th term

Median = 33

Hence median is 33.

If 26 is replaced by 62, new set of numbers in ascending order is shown below.

17,29,32,33,34,45,56,60,62

Here n = 9 which is odd.

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

Median = ((9+1)/2)th term

Median = (10/2)th term

Median = 5th term

Median = 34

Hence median is 34.

17. The marks scored by 16 students in a class test are : 3, 6, 8, 13, 15, 5, 21, 23, 17, 10, 9, 1, 20, 21, 18, 12 

Find 

(i) the median 

(ii) lower quartile 

(iii) upper quartile

Solution:

Arranging data in ascending order

1,3,5,6,8,9,10,12,13,15,17,18,20,21,21,23

Here n = 16 which is even

(i) So median = ½ ( n/2 th term + ((n/2)+1)th term)

= ½ (16/2 th term + ((16/2)+1)th term)

= ½ (8 th term + (8+1)th term)

= ½ (8 th term + 9th term)

= ½ (12+13)

= ½ ×25

= 12.5

Hence the median is 12.5.

(ii) Lower quartile, Q1 = (n/4) th term

= (16)/4

= 4 th term

= 6

Hence the lower quartile is 6.

(iii)Upper quartile, Q3 = (3n/4) th term

= (3×16/4) th term

= (3×4)th term

= 12 th term

= 18

Hence the upper quartile is 18.

18. Find the median and mode for the set of numbers : 2, 2, 3, 5, 5, 5, 6, 8, 9

Solution:

Here n = 9 which is odd.

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

Median = ((9+1)/2)th term

Median = (10/2)th term

Median = 5 th term

Median = 5

Mode is the number which appears most often in a set of numbers.

Here 5 occurs maximum number of times.

So mode is 5.

19. Calculate the mean, the median and the mode of the following distribution.

Age in years

12

13

14

15

16

17

18

No. of students

2

3

5

6

4

3

2

Solution:

Age in years xi

No. of students fi

Cumulative frequency

fixi

12

2

2

24

13

3

5

39

14

5

10

70

15

6

16

90

16

4

20

64

17

3

23

51

18

2

25

36

Total

Ʃfi = 25

Ʃ fixi = 374

Mean = Ʃ fixi/ Ʃfi

= 374/25

= 14.96

Hence the mean is 14.96.

Here n = 25 which is odd.

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

Median = ((25+1)/2)th term

Median = (26/2)th term

Median = 13 th term

Median = 15

Hence the median is 15.

Here 15 occurs maximum number of times. i.e., 6 times.

Hence the mode is 15.

20. The daily wages of 30 employees in an establishment are distributed as follows:

Daily wages in Rs

0-10

10-20

20-30

30-40

40-50

50-60

No. of employees

1

8

10

5

4

2

Estimate the modal daily wages for this distribution by a graphical method.

Solution:

Daily wages in Rs.

No. of employees

0-10

1

10-20

8

20-30

10

30-40

5

40-50

4

50-60

2

Taking daily wages on x-axis and No. of employees on the y-axis and draw a histogram as shown below.

ML Aggarwal Sol Class 10 Maths chapter 21-29

Join AB and CD intersecting each other at M. From M draw ML perpendicular to x-axis.

L is the mode 

Here Mode = Rs 23

Hence the mode is Rs. 23.

21. Using the data given below, construct the cumulative frequency table and draw the ogive.

From the ogive, estimate :

 (i) the median 

(ii) the inter quartile range. 

Marks

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

Frequency

3

8

12

14

10

6

5

2

Also state the median class

Solution:

Arranging the data in cumulative frequency table.

Marks

Frequency

Cumulative frequency

0-10

3

3

10-20

8

11

20-30

12

23

30-40

14

37

40-50

10

47

50-60

6

53

60-70

5

58

70-80

2

60

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis).

Plot the points (10, 3), (20, 11), (30, 23), (40, 37), (50, 47), (60, 53), (70, 58) and (80, 60) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-30

(i)Here number of observations, n = 60 which is even.

So median = ( n/2) th term

= (60/2) th term

= 30 th term

Mark a point A(30) on Y-axis. From A, draw a horizontal line parallel to X-axis meeting the curve at P. From P draw a line perpendicular to the x-axis which meets it at Q. 

Q is the median .

Q = 35

Hence the median is 35 .

(ii) Lower quartile, Q1 = n/4 = 60/4 = 15th term

Upper quartile, Q3 = 3n/4 = 3×60/4 = 45th term

Mark a point B(15) and C(45) on Y-axis. From B and C, draw a horizontal line parallel to X-axis meeting the curve at L and M respectively. From L and M, draw lines perpendicular to the x-axis which meets it at E and F respectively. 

E is the lower quartile .

E = 22.3

F is the upper quartile.

F = 47

Inter quartile range = Q3-Q1

= 47-22.3

= 24.7

Hence interquartile range is 24.7.

22. Draw a cumulative frequency curve for the following data : 

Marks obtained

0-10

10-20

20-30

30-40

40-50

No. of students

8

10

22

40

20

Hence determine:

 (i) the median 

(ii) the pass marks if 85% of the students pass. 

(iii) the marks which 45% of the students exceed.

Solution:

Arranging the data in cumulative frequency table.

Marks obtained

No. of students f

Cumulative frequency

0-10

8

8

10-20

10

18

20-30

22

40

30-40

40

80

40-50

20

100

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis).

Plot the points (10, 8), (20, 18), (30, 40), (40, 80), and(50,100) on the graph. 

Join the points with the free hand. We get an ogive as shown: 

ML Aggarwal Sol Class 10 Maths chapter 21-31

(i) Here number of observations, n = 100 which is even.

So median = ( n/2) th term

= (100/2) th term

= 50 th term

Mark a point A(50) on Y-axis. From A, draw a horizontal line parallel to X-axis meeting the curve at P. From P, draw a line perpendicular to the x-axis which meets it at Q. 

Q is the median .

Q = 32.5

Hence the median is 32.5 .

(ii)Total number of students = 100

85% of 100 = 85

Remaining number of students = 100-85 = 15

Mark a point B(15) on Y axis. From B, draw a horizontal line parallel to X-axis meeting the curve at L. From L, draw a line perpendicular to the x-axis which meets it at M. 

Here M = 18

The pass marks will be 18 if 85% of students passed.

(iii) Total number of students = 100

45% of 100 = 45

Remaining number of students = 100-45 = 55

Mark a point C(55) on Y axis. From C, draw a horizontal line parallel to X-axis meeting the curve at E. From E, draw a line perpendicular to the x-axis which meets it at F. 

Here F = 34

Hence marks which 45% of students exceeds is 34 marks.

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