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
(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 72 = 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_{i}x_{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_{i}x_{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_{i}f_{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) 
MidValue x_{i} 
No. of Workers (f_{i}) 
f_{i}x_{i} 
5055 
52.5 
5 
262.5 
5560 
57.5 
20 
1150.0 
6065 
62.5 
10 
625.0 
6570 
67.5 
10 
675.0 
7075 
72.5 
9 
652.5 
7580 
77.5 
6 
465.0 
8085 
82.5 
12 
990.0 
8590 
87.5 
8 
700.0 
Total 

80 
5520.00 
(ii) Short  cut method
Weekly wages (Rs) 
No. of workers (f_{i}) 
Midvalue x_{i} 
A = 72.5 d_{i}=xA 
f_{i}d_{i} 
5055 
5 
52.5 
20 
100 
5560 
20 
57.5 
15 
300 
6065 
10 
62.5 
10 
100 
6570 
10 
67.5 
5 
50 
7075 
9 
A=72.5 
0 
0 
7580 
6 
77.5 
5 
30 
8085 
12 
82.5 
10 
120 
8590 
8 
87.5 
15 
120 
Total 
80 


280 
Solution 3
(i) Short  cut method
Marks 
No. of boys (f_{i}) 
Midvalue x_{i} 
A = 65 d_{i}=xA 
f_{i}d_{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}) 
Midvalue x_{i} 
A = 65

f_{i}u_{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}) 
Midvalue x_{i} 
A = 87.50

f_{i}u_{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 
Midvalue (x_{i}) 
f_{i}x_{i} 
010 
8 
5 
40 
1020 
22 
15 
330 
2030 
31 
25 
775 
3040 
f 
35 
35f 
4050 
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_{i}d_{i} 
5055 
5 
52.5 
20 
100 
5560 
20 
57.5 
15 
300 
6065 
10 
62.5 
10 
100 
6570 
10 
67.5 
5 
50 
7075 
9 
72.5 
0 
0 
7580 
6 
77.5 
5 
30 
8085 
12 
82.5 
10 
120 
8590 
8 
87.5 
15 
120 
Total 
80 

280 

Solution 7
C.I. 
Frequency 
Mid value x 
fx 
1525 
10 
20 
200 
2535 
20 
30 
600 
3545 
25 
40 
1000 
4555 
15 
50 
750 
5565 
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 
020 
5 
10 
50 
2040 
f_{1} 
30 
30f_{1} 
4060 
10 
50 
500 
6080 
f_{2} 
70 
70f_{2} 
80100 
7 
90 
630 
100120 
8 
110 
880 
Total 
30+f_{1}+f_{2} 

2060+30f_{1}+70f_{2} 
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 = q_{3}  q_{1} =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 (Q_{1}) =
(iii) upper quartile (Q_{3}) =
(iv) Interquartile range = Q_{3}  Q_{1 }= 40  38 = 2
Solution 7
Class 
Frequency 
Cumulative Frequency 
010 
4 
4 
1020 
9 
13 
2030 
15 
28 
3040 
14 
42 
4050 
8 
50 
Number of terms = 50
Through mark of 25.5 on the yaxis, draw a line parallel to xaxis which meets the curve at A. From A, draw a perpendicular to xaxis, which meets xaxis at B.
The value of B is the median which is 28.
Solution 8
Weight (kg) 
No. of boys 
Cumulative Frequency 
1015 
11 
11 
1520 
25 
36 
2025 
12 
48 
2530 
5 
53 
3035 
2 
55 
Number of terms = 55
Through mark of 28 on the yaxis, draw a line parallel to xaxis which meets the curve at A. From A, draw a perpendicular to xaxis, which meets xaxis 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 xaxis which meets the curve at A. From A, draw a perpendicular to xaxis which meets it at B.
Value of B is the median = 40
(ii) Lower quartile (Q_{1}) = 20^{th} term = 18
(iii) Upper Quartile (Q_{3}) = 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 xaxis which meets the curve at A. From A, draw a perpendicular to xaxis 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 2030, because in this class there are 20 frequencies.
Solution 4
Mode is in 3035 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 xaxis and cumulative frequency along yaxis we will draw an ogive.
(i)
Through mark for 80, draw a parallel line to xaxis which meets the curve; then from the curve draw a vertical line which meets the xaxis at the mark of 157.5.
(ii)Since, number of terms = 160
(iii)Through mark for 172 on xaxis, draw a vertical line which meets the curve; then from the curve draw a horizontal line which meets the yaxis 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 xaxis which meets the curve at A. From A, draw a perpendicular to xaxis 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 = m1 = 41 = 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. 
010 
5 
5 
1020 
9 
14 
2030 
16 
30 
3040 
22 
52 
4050 
26 
78 
5060 
18 
96 
6070 
11 
107 
7080 
6 
113 
8090 
4 
117 
90100 
3 
120 
(i)
Through mark 60.5, draw a parallel line to xaxis which meets the curve at A, From A draw a perpendicular to xaxis 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 = Q_{1} =
Solution 8
Weight 
Frequency 
C. f. 
4045 
5 
5 
4550 
17 
22 
5055 
22 
44 
5560 
45 
89 
6065 
51 
140 
6570 
31 
171 
7075 
20 
191 
7580 
9 
200 
(i) Number of students weighing more than 55 kg = 20044 = 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 
1020 
5 
15 
75 
2030 
3 
25 
75 
3040 
f 
35 
35f 
4050 
7 
45 
315 
5060 
2 
55 
110 
6070 
6 
65 
390 
7080 
13 
75 
975 
Total 
36+f 
1940+35f 
Solution 11
Monthly Income (thousands) 
No. of employees (f) 
Cumulative frequency 
67 
20 
20 
78 
45 
65 
89 
65 
130 
910 
95 
225 
1011 
60 
285 
1112 
30 
315 
1213 
5 
320 
Total 
320 
Number of employees = 320
(i)
Through mark 160, draw a parallel line to xaxis which meets the curve at A, From A draw a perpendicular to xaxis 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) = 320305 =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 xaxis meeting at Q.
(iv) Value of Q is the mode = 82 (approx)
Solution 13
Marks 
No. of students

Cumulative frequency 
010 
5 
5 
1020 
11 
16 
2030 
10 
26 
3040 
20 
46 
4050 
28 
74 
5060 
37 
111 
6070 
40 
151 
7080 
29 
180 
8090 
14 
194 
90100 
6 
200 
Number of students = 200
(i)
Through mark 100, draw a parallel line to xaxis which meets the curve at A, From A draw a perpendicular to xaxis 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 xaxis meeting at L.
Value of L is the mode. Hence, mode = 21.5
Solution 19
Marks 
Number of students (Frequency) 
Cumulative Frequency 
010 
3 
3 
1020 
7 
10 
2030 
12 
22 
3040 
17 
39 
4050 
23 
62 
5060 
14 
76 
6070 
9 
85 
7080 
6 
91 
8090 
5 
96 
90100 
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. 
400450 
450 
2 
2 
450500 
500 
6 
8 
500550 
550 
12 
20 
550600 
600 
18 
38 
600650 
650 
24 
62 
650700 
700 
13 
75 
700750 
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 
010 
2 
10 20 
5 
2030 
8 
3040 
4 
4050 
6 
ii.
Class interval 
Frequency (f) 
Mean value (x) 
fx 
010 
2 
5 
10 
10 20 
5 
15 
75 
2030 
8 
25 
200 
3040 
4 
35 
140 
4050 
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.