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 TEST YOURSELF
Solution 1(a)
Correct option: (i)
Given A.P. is 2, 4, 6, 8, ….., 40
Here, first term, a = 2 and common difference, d = 2
Last term, 40 = 2 + (n – 1)(2)
38 = (n – 1)(2)
n = 20
Solution 1(b)
Correct option: (i) 12
Data arranged in ascending order,
8, 8, 9, 10, 12, 12, 12, 13, 15
Number of observations, n = 9 (odd)
Then, median =
Solution 1(c)
Correct option: (iv) 21
Data arranged in ascending order,
10, 12, 14, 16, 17, x
Number of observations, n = 6 (even)
Median =
Now, Mean = Median
Solution 1(d)
Correct option: (iii) 6
First six prime numbers are as follows:
2, 3, 5, 7, 11, 13
Number of observations, n = 6 (even)
Median =
Solution 1(e)
Correct option: (iii) 44
n = 80 (even)
From an ogive,
Interquartile range = Q_{3} – Q_{1} = 54 – 10 = 44
Solution 2
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 3
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 4
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 5
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 6
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 7
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 8
Solution 9
Solution 10
Solution 11
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 12
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.
Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(A)
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
(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 4
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 5
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 6
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 7
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 8
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 9
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 10
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 11
Solution 12
Solution 13
Solution 14
Solution 1(a)
Correct option: (i) 38
Let the fifth variable be x.
Then,
62 + x = 100
x = 38
Solution 1(b)
Correct option: (ii) 136 cm
Mean height of 5 students = 140 cm
Then, the total height of 5 students = 140 × 5 = 700 cm
Height of one student = 156 cm
Then, the total height of the remaining 4 students = 700 – 156 = 544
Hence, mean height of the remaining 4 students
Solution 1(c)
Correct option: (iii) 16 years
Mean age of 8 boys = 16 years
Then, sum of the ages of 8 boys = 16 × 8 = 128 years
Sum of the ages of two more boys = 18 + 14 = 32 years
Now, total age of 10 boys = 128 + 32 = 160 years
Hence, the resulting mean age
Solution 1(d)
Correct option: (iv) 18
Mean value of 15 numbers = 20
Then, the (incorrect) sum of values of 15 numbers = 20 × 15 = 300
Correct sum = 300 – 45 + 15 = 270
Therefore, correct mean
Solution 1(e)
Correct option: (iii)
Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(B)
Solution 2
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 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
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 7
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 8
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 9
Solution 1(a)
Correct option: (i) 41
C.I.  Class Mark (x)  f  fx 
3040  35  2  70 
4050  45  3  135 

 Σf_{}= 5  Σfx = 205 
Alternate Method:
The mean of the set of observations lies between the range of the given data.
Since, the range of observations is from 30 to 50, only option (i), that is, 41 lies between 30 and 50.
Hence, 41 is the mean.
Solution 1(b)
Correct option: (i) 17
Let the assumed mean, A = 15
Class width, I = 10
C.I.  f  x  u = (x – A)/i  f × u 
010  5  5  –1  –5 
1020  10  15  0  0 
2030  10  25  1  10 
 Σf = 25 

 Σfu = 5 
Solution 1(c)
Correct option: () Back answer is 30
x  f  fx 
10  10  100 
20  a  20a 
30  10  300 
Note: All four values of ‘a’ given in the options satisfy the equation.
In fact, a can take any value apart from this.
Solution 1(d)
Correct option: (iii) x_{1} – x_{2} = 0
x  f  fx 
10  x_{1}  10x_{1} 
20  20  400 
30  x_{2}  30x_{2} 
Solution 1(e)
Correct option: (ii) 47
C.I.  Class Mark (x)  f  fx 
4050  45  40  1800 
5060  55  10  550 

 Σf_{}= 50  Σfx = 2350 
Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(C)
Solution 2
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 3
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 4
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 5
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 6
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 7
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
Solution 8
Mode is in 2030, because in this class there are 20 frequencies.
Solution 9
Mode is in 3035 because it has the maximum frequency.
Solution 10
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 11
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)
Solution 1(a)
Correct option: (ii) 18
Arranging data in ascending order,
14, 15, 18, 21, 29
Number of observations, n = 5 (odd)
Then, median =
Solution 1(b)
Correct option: (iii) 5
Arranging data in ascending order,
0, 2, 3, 4, 6, 8, 11, 16
Number of observations, n = 8 (even)
Then, median =
Solution 1(c)
Correct option: (iv) 23
Data arranged in ascending order,
5, 15, 20, x, 28, 30, 35
Number of observations, n = 7 (odd)
Then, median =
Therefore, x = 23
Solution 1(d)
Correct option: (iii) 60 – 20
Data: 10, 20, 30, 40, 50, 60, 70 and 80
Number of observations, n = 8
Interquartile range = Q_{3} – Q_{1} = 60 – 20
Solution 1(e)
Correct option: (iii) 5060
Modal class is the class with the highest frequency.
Therefore, modal class is 5060.