Class 10 SELINA Solutions Maths Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)

Ex. 24(A)

Ex. 24(B)

Ex. 24(C)

Ex. 24(D)

Ex. 24(E)

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(A)

Solution 1

(i)

(ii)

Solution 2

(a) Here n = 9

(b)

If marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63

Solution 3

Numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

Here n = 10

Solution 4

(a) The mean of 7, 11, 6, 5 and 6

(b)

If we subtract 2 from each number, then the mean will be 7-2 = 5

Solution 5

No. of terms = 5

Mean = 8

Sum of numbers = 8 x 5 = 40 .(i)

But, sum of numbers = 6+4+7+a+10 = 27+a ..(ii)

From (i) and (ii)

27+a = 40

a = 13

Solution 6

No. of terms = 5 and mean = 8

Sum of numbers = 5 x 8 = 40 ..(i)

but sum of numbers = 6+y+7+x+14 = 27+y+x .(ii)

from (i) and (ii)

27 + y + x = 40

x + y = 13

y = 13 - x

Solution 7

Age in yrs x_i	Frequency (f_i)	f_ix_i
12	2	24
13	4	52
14	6	84
15	9	135
16	8	128
17	7	119
18	4	72
Total	40	614

Solution 8

No. of terms = 10

Mean = 69.5

Sum of the numbers = 69.5 x 10 = 695 ..........(i)

But sum of numbers = 72+70+x+62+ 50+71+90+64+58+82

= 619 + x ......(ii)

from (i) and (ii)

619 + x = 695

x = 76

Solution 9

Height (cm) x_i	No. of Plants f_i	f_ix_i
50	2	100
55	4	220
58	10	580
60	f	60f
65	5	325
70	4	280
71	3	213
Total	28+f	1718 + 60f

Mean = 60.95

Solution 10

Wages (Rs/day) (x)	No. of Workers (f)	fx
50	2	100
60	4	240
70	8	560
80	12	960
90	10	900
100	6	600
Total	42	3360

(i) Mean remains the same if the number of workers in each category is doubled.

Mean = 80

(ii) Mean will be increased by 60% if the wages per day per worker is increased by 60%

New mean =

(iii) No change in the mean if the number of workers is doubled but if wages per worker is reduced by 40%, then

New mean =

Solution 11

No. of matches (x)	No. of boxes (f)	fx
35	6	210
36	10	360
37	18	666
38	25	950
39	21	819
40	12	480
41	8	328
Total	100	3813

(i)

(ii) In the second case,

New mean = 39 matches

Total contents = 39 x 100 = 3900

But total number of matches already given = 3813

Number of new matches to be added = 3900 - 3813 = 87

Solution 12

Solution 13

Solution 14

Solution 15

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(B)

Solution 1

Age in years C.I.	x_i	Number of students (f_i)	x_if_i
16 - 18	17	2	34
18 - 20	19	7	133
20 - 22	21	21	441
22 - 24	23	17	391
24 - 26	25	3	75
Total		50	1074

Solution 2

(i) Direct Method

Weekly Wages (Rs)	Mid-Value x_i	No. of Workers (f_i)	f_ix_i
50-55	52.5	5	262.5
55-60	57.5	20	1150.0
60-65	62.5	10	625.0
65-70	67.5	10	675.0
70-75	72.5	9	652.5
75-80	77.5	6	465.0
80-85	82.5	12	990.0
85-90	87.5	8	700.0
Total		80	5520.00

(ii) Short - cut method

Weekly wages (Rs)	No. of workers (f_i)	Mid-value x_i	A = 72.5 d_i=x-A	f_id_i
50-55	5	52.5	-20	-100
55-60	20	57.5	-15	-300
60-65	10	62.5	-10	-100
65-70	10	67.5	-5	-50
70-75	9	A=72.5	0	0
75-80	6	77.5	5	30
80-85	12	82.5	10	120
85-90	8	87.5	15	120
Total	80			-280

Solution 3

(i) Short - cut method

Marks	No. of boys (f_i)	Mid-value x_i	A = 65 d_i=x-A	f_id_i
30 - 40	10	35	-30	-300
40 - 50	12	45	-20	-240
50 - 60	14	55	-10	-140
60 - 70	12	A = 65	0	0
70 - 80	9	75	10	90
80 - 90	7	85	20	140
90 - 100	6	95	30	180
Total	70			-270

(ii) Step - deviation method

Marks	No. of boys (f_i)	Mid-value x_i	A = 65	f_iu_i
30 - 40	10	35	-3	-30
40 - 50	12	45	-2	-24
50 - 60	14	55	-1	-14
60 - 70	12	A = 65	0	0
70 - 80	9	75	1	9
80 - 90	7	85	2	14
90 - 100	6	95	3	18
Total	70			-27

Here A = 65 and h = 10

Solution 4

C. I.	Frequency (f_i)	Mid-value x_i	A = 87.50	f_iu_i
63 - 70	9	66.50	-3	-27
70 - 77	13	73.50	-2	-26
77 - 84	27	80.50	-1	-27
84 - 91	38	A = 87.50	0	0
91 - 98	32	94.50	1	32
98 - 105	16	101.50	2	32
105 - 112	15	108.50	3	45
Total	150			29

Here A = 87.50 and h = 7

Solution 5

C. I.	frequency	Mid-value (x_i)	f_ix_i
0-10	8	5	40
10-20	22	15	330
20-30	31	25	775
30-40	f	35	35f
40-50	2	45	90
Total	63+f		1235+35f

Solution 6

Let the assumed mean A= 72.5

C.I	f_i	Mid value (x_i)	d_i=x_i -; A	f_id_i
50-55	5	52.5	-20	-100
55-60	20	57.5	-15	-300
60-65	10	62.5	-10	-100
65-70	10	67.5	-5	-50
70-75	9	72.5	0	0
75-80	6	77.5	5	30
80-85	12	82.5	10	120
85-90	8	87.5	15	120
Total	80		-280

Solution 7

C.I.	Frequency	Mid value x	fx
15-25	10	20	200
25-35	20	30	600
35-45	25	40	1000
45-55	15	50	750
55-65	5	60	300
Total	75		2850

Solution 8

Class	Frequency (f)	Mid Value (x)	fx
0 - 20	7	10	70
20 - 40	p	30	30p
40 - 60	10	50	500
60 - 80	9	70	630
80 - 100	13	90	1170
Total	39 + p		2370 + 30p

Here mean = 54 ..(ii)

from (i) and (ii)

Solution 9

Class	Freq (f)	Mid value	fx
0-20	5	10	50
20-40	f₁	30	30f₁
40-60	10	50	500
60-80	f₂	70	70f₂
80-100	7	90	630
100-120	8	110	880
Total	30+f₁+f₂		2060+30f₁+70f₂

Now, and

from (i)

using (i) and (ii)

Solution 10

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(C)

Solution 1

Arranging the given data in descending order:

8, 7, 6, 5, 4, 3, 3, 1, 0

The middle term is 4 which is the 5^th term.

Median = 4

Solution 2

Arranging the given data in descending order:

28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5

The middle terms are 24 and 24, 5^th and 6^th terms

Solution 3

Arranging in ascending order:

22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37

(i) Middle term is 10^th term i.e. 29

Median = 29

(ii) Lower quartile =

(iii) Upper quartile =

(iv) Interquartile range = q₃ - q₁ =35 - 26 = 9

Solution 4

Arrange in ascending order:

0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95

(i) Median is the mean of 8^th and 9^th term

(ii) Upper quartile =

(iii) Interquartile range =

Solution 5

Age (in years)	Frequency	Cumulative Frequency
11	2	2
12	4	6
13	6	12
14	10	22
15	8	30
16	7	37

Number of terms = 37

Median =

Median = 14

Solution 6

Weight (kg) x	no. of boys f	cumulative frequency
37	10	10
38	14	24
39	18	42
40	12	54
41	6	60

Number of terms = 60

(i) median = the mean of the 30^th and the 31^st terms

(ii) lower quartile (Q₁) =

(iii) upper quartile (Q₃) =

(iv) Interquartile range = Q₃ - Q₁= 40 - 38 = 2

Solution 7

Class	Frequency	Cumulative Frequency
0-10	4	4
10-20	9	13
20-30	15	28
30-40	14	42
40-50	8	50

Number of terms = 50

Through mark of 25.5 on the y-axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis, which meets x-axis at B.

The value of B is the median which is 28.

Solution 8

Weight (kg)	No. of boys	Cumulative Frequency
10-15	11	11
15-20	25	36
20-25	12	48
25-30	5	53
30-35	2	55

Number of terms = 55

Through mark of 28 on the y-axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis, which meets x-axis at B.

The value of B is the median which is 18.4 kg

Solution 9

Marks (less than)	Cumulative frequency
10	5
20	24
30	37
40	40
50	42
60	48
70	70
80	77
90	79
100	80

Number of terms = 80

\Median = 40^th term.

(i) Median = Through 40^th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

Value of B is the median = 40

(ii) Lower quartile (Q₁) = 20^th term = 18

(iii) Upper Quartile (Q₃) = 60^th term = 66

Solution 10

Height (in cm)	No. of pupils	Cumulative Frequency
121 - 130	12	12
131 - 140	16	28
141 - 150	30	58
151 - 160	20	78
161 - 170	14	92
171 - 180	8	100

Number of terms = 100

Through 50^th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

Value of B is the median = 148

Median height = 148cm

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(D)

Solution 1

(i) Mode = 7

Since 7 occurs 4 times

(ii) Mode = 11

Since it occurs 4 times

Solution 2

Mode is 122 cm because it occur maximum number of times. i.e. frequency is 18.

Solution 3

Mode is in 20-30, because in this class there are 20 frequencies.

Solution 4

Mode is in 30-35 because it has the maximum frequency.

Solution 5

which is 5.

Mode = 5 because it occurs maximum number of times.

Solution 6

Arranging the given data in ascending order:

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

(i) Mode = 16 as it occurs maximum number of times.

(ii)

(iii)Total marks = 7+10+12+12+14+15+16+16+16+17+19 =

154

(iv)

Solution 7

(i)

(ii) Median = mean of 8^th and 9^th term

(iii) Mode = 5 as it occurs maximum number of times.

Solution 8

Score x	No. of shots f	fx
0	0	0
1	3	3
2	6	12
3	4	12
4	7	28
5	5	25
Total	25	80

(i) Modal score = 4 as it has maximum frequency 7.

(ii)

(iii) Total score = 80

(iv)

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(E)

Solution 1

Taking Height of student along x-axis and cumulative frequency along y-axis we will draw an ogive.

(i)

Through mark for 80, draw a parallel line to x-axis which meets the curve; then from the curve draw a vertical line which meets the x-axis at the mark of 157.5.

(ii)Since, number of terms = 160

(iii)Through mark for 172 on x-axis, draw a vertical line which meets the curve; then from the curve draw a horizontal line which meets the y-axis at the mark of 145.

The number of students whose height is above 172 cm

= 160 - 144 = 16

Solution 2

Marks	No. of students	Cumulative frequency
9.5 - 19.5	14	14
19.5 - 29.5	16	30
29.5 - 39.5	22	52
39.5 - 49.5	26	78
49.5 - 59.5	18	96
59.5 - 69.5	11	107
69.5 - 79.5	6	113
79.5 - 89.5	4	117
89.5 - 99.5	3	120

Scale:

1cm = 10 marks on X axis

1cm = 20 students on Y axis

(i)

Through mark 60, draw a parallel line to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 43

(ii) Total marks = 100

75% of total marks = marks

The number of students getting more than 75% marks = 120 - 111 = 9 students.

Solution 3

Mean of 1, 7, 5, 3, 4 and 4 =

m=4

Now, mean of 3, 2, 4, 2, 3, 3 and p = m-1 = 4-1 = 3

Therefore, 17+p = 3 x n …. Where n = 7

17+p = 21

p = 4

Arranging in ascending order:

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

Mean = 4^th term = 3

Therefore, q = 3

Solution 4

Date	Number	C.f.
1	5	5
2	12	17
3	20	37
4	27	64
5	46	110
6	30	140
7	31	171
8	18	189
9	11	200
10	5	205
11	0	205
12	1	206

(i) Mode = 5^th July as it has maximum frequencies.

(ii) Total number of terms = 206

Upper quartile =

Lower quartile =

Solution 5

We plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100) to get the curve as follows:

At y = 50, affix A.

Through A, draw a horizontal line meeting the curve at B.

Through B, a vertical line is drawn which meets OX at M.

OM = 17.6 units

Hence, median income = 17.6 thousands

Solution 6

Arranging the terms in ascending order:

2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20

Number of terms = 20

(i)

(ii)

(iii) Mode = 15 as it has maximum frequencies i.e. 3

Solution 7

Marks	No. of students	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

(i)

Through mark 60.5, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 43

(ii) Number of students who obtained up to 75% marks in the test = 110

Number of students who obtained more than 75% marks in the test = 120 - 110 = 10

(iii) Number of students who obtained less than 40% marks in the test = 52 (from the graph; x=40, y=52)

(iv) Lower quartile = Q₁ =

Solution 8

Weight	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

(i) Number of students weighing more than 55 kg = 200-44 = 156

Therefore, percentage of students weighing 55 kg or more

(ii) 30% of students =

Heaviest 60students in weight = 9 + 21 + 30 = 60

weight = 65 kg ( from table)

(iii) (a) underweight students when 55.70 kg is standard = 46 (approx) from graph

(b) overweight students when 55.70 kg is standard = 200- 55.70 = 154 (approx) from graph

Solution 9

Marks obtained(x)	No. of students (f)	c.f.	fx
5	3	3	15
6	9	12	54
7	6	18	42
8	4	22	32
9	2	24	18
10	1	25	10
Total	25		171

Number of terms = 25

(i) Mean =

(ii)

(iii) Mode = 6 as it has maximum frequencies i.e. 6

Solution 10

C.I.	Frequency(f)	Mid value (x)	fx
10-20	5	15	75
20-30	3	25	75
30-40	f	35	35f
40-50	7	45	315
50-60	2	55	110
60-70	6	65	390
70-80	13	75	975
Total	36+f		1940+35f

Solution 11

Monthly Income (thousands)	No. of employees (f)	Cumulative frequency
6-7	20	20
7-8	45	65
8-9	65	130
9-10	95	225
10-11	60	285
11-12	30	315
12-13	5	320
Total	320

Number of employees = 320

(i)

Through mark 160, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = Rs 9.3 thousands

(ii) The number of employees with income below Rs 8500 = 95 (approx from the graph)

(iii) Number of employees with income below Rs 11500 = 305 (approx from the graph)

Therefore number of employees (senior employees) = 320-305 =15

(iv) Upper quartile =

Solution 12

(i)Draw the histogram

(ii) In the highest rectangle which represents modal class draw two lines AC and BD intersecting at P.

(iii) From P, draw a perpendicular to x-axis meeting at Q.

(iv) Value of Q is the mode = 82 (approx)

Solution 13

Marks	No. of students	Cumulative frequency
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

Number of students = 200

(i)

Through mark 100, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 57 marks (approx)

(ii) The number of students who failed (if minimum marks required to pass is 40)= 46 (approx from the graph)

(iii) The number of students who secured grade one in the examination = 200 - 188 = 12 (approx from the graph)

Solution 14

Solution 15

Since the frequency for x = 14 is maximum.

So Mode = 14.

According to the table it can be observed that the value of x from the 13^th term to the 17^th term is 13.

So the median = 13.

Solution 16

Solution 17

Solution 18

Histogram is as follows:

In the highest rectangle which represents modal class draw two lines AC and BD intersecting at E.

From E, draw a perpendicular to x-axis meeting at L.

Value of L is the mode. Hence, mode = 21.5

Solution 19

Marks	Number of students (Frequency)	Cumulative Frequency
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

The ogive is as follows:

Solution 20

Solution 21

Here the number of observations i. e is 10, which is even.'

So, the given data is 13, 35, 43, 46, 46, 50, 55, 61, 71, 80.

In the given data, 46 occurs most frequently.

∴ Mode = 46

Solution 22

The cumulative frequency table of the given distribution is as follows:

Wages (Rs.)	Upper limit	No. of workers	C.f.
400-450	450	2	2
450-500	500	6	8
500-550	550	12	20
550-600	600	18	38
600-650	650	24	62
650-700	700	13	75
700-750	750	5	80

The ogive is as follows:

Number of workers = n = 80

1) Median = term = 40^th term, draw a horizontal line which meets the curve at point A.

Draw vertical line parallel to y axis from A to meet x axis at B.

The value of point B is 605.

2) Lower quartile (Q1)= term=20^th term = 550

3) Through mark of point 625 on x axis draw a vertical line which meets the graph at point C Then through point C, draw a horizontal line which meets the y axis at the mark of 50.

Thus, the number of workers that earn more than Rs. 625 daily = 80 - 50 = 30

Solution 23

i. The frequency distribution table is as follows:

Class interval	Frequency
0-10	2
10- 20	5
20-30	8
30-40	4
40-50	6

ii.

Class interval	Frequency (f)	Mean value (x)	fx
0-10	2	5	10
10- 20	5	15	75
20-30	8	25	200
30-40	4	35	140
40-50	6	45	270
	Sf = 25		Sf = 695

iii. Here the maximum frequency is 8 which is corresponding to class 20 - 30.

Hence, the modal class is 20 - 30.

Class 10 SELINA Solutions Maths Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(A)

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(B)

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(C)

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(D)

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(E)

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

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