Recalling the concepts learnt from the previous chapter, this exercise contains a mixture of problems based on finding the mean, median, mode and drawing ogives for suitable questions. If students are searching for a resource to clarify their doubts, then stop here at the Selina Solutions for Class 10 Maths as these solutions are created by experts at BYJU’S, keeping in mind the latest ICSE patterns. These exercise solutions are available in the Concise Selina Solutions for Class 10 Maths Chapter 24 Measures of Central Tendency Exercise 24(E) PDF, provided in the link mentioned below.
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1. The following distribution represents the height of 160 students of a school.
Height (in cm) 
No. of Students 
140 – 145 
12 
145 – 150 
20 
150 – 155 
30 
155 – 160 
38 
160 – 165 
24 
165 – 170 
16 
170 – 175 
12 
175 – 180 
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 interquartile range.
iii. The number of students whose height is above 172 cm.
Solution:
Height (in cm) 
No. of Students 
Cumulative frequency 
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 
N = 160 
Now, let’s draw an ogive taking height of student along xaxis and cumulative frequency along yaxis.
(i) So,
Median = 160/2 = 80^{th} term
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) As, the number of terms = 160
Lower quartile (Q_{1}) = (160/4) = 40^{th} term = 152
Upper quartile (Q_{3}) = (3 x 160/4) = 120^{th} term = 164
Inner Quartile range = Q_{3} – Q_{1}
= 164 – 152
= 12
(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.
Now,
The number of students whose height is above 172 cm
= 160 – 144 = 16
2. Draw ogive for the data given below and from the graph determine: (i) the median marks.
(ii) the number of students who obtained more than 75% marks.
Marks 
10 – 19 
20 29 
30 – 39 
40 – 49 
50 – 59 
60 – 69 
70 – 79 
80 – 89 
90 – 99 
No. of students 
14 
16 
22 
26 
18 
11 
6 
4 
3 
Solution:
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) So, the median = 120/ 2 = 60^{th} term
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 = 75/100 x 100 = 75 marks
Hence, the number of students getting more than 75% marks = 120 – 111 = 9 students.
3. The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m – 1 and median q. Find p and q.
Solution:
Mean of 1, 7, 5, 3, 4 and 4 = (1 + 7 + 5 + 3 + 4 + 4)/ 6 = 24/6 = 4
So, m = 4
Now, given that
The mean of 3, 2, 4, 2, 3, 3 and p = m 1 = 4 – 1 = 3
Thus, 17 + p = 3 x n …. ,where n = 7
17 + p = 21
p = 4
Arranging the terms in ascending order, we have:
2, 2, 3, 3, 3, 3, 4, 4
Mean = 4^{th} term = 3
Hence, q = 3
4. In a malaria epidemic, the number of cases diagnosed were as follows:
Date (July) 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
Number 
5 
12 
20 
27 
46 
30 
31 
18 
11 
5 
0 
1 
On what days do the mode and upper and lower quartiles occur?
Solution:
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 = 206 x (3/4) = 154.5^{th} = 7^{th} July
Lower quartile = 206 x (1/4) = 51.5^{th} = 4^{th} July
5. The income of the parents of 100 students in a class in a certain university are tabulated below.
Income (in thousand Rs) 
0 – 8 
8 – 16 
16 – 24 
24 – 32 
32 – 40 
No. of students 
8 
35 
35 
14 
8 
(i) Draw a cumulative frequency curve to estimate the median income.
(ii) If 15% of the students are given freeships on the basis of the basis of the income of their parents, find the annual income of parents, below which the freeships will be awarded.
(iii) Calculate the Arithmetic mean.
Solution:
(i) Cumulative Frequency Curve
Income (in thousand Rs.) 
No. of students f 
Cumulative Frequency 
Class mark x 
fx 
0 – 8 
8 
8 
4 
32 
8 – 16 
35 
43 
12 
420 
16 – 24 
35 
78 
20 
700 
24 – 32 
14 
92 
28 
392 
32 – 40 
8 
100 
36 
288 
∑f_{ }= 100 
∑ fx = 1832 
We plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100) to get the curve as follows:
Here, N = 100
N/2 = 50
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
(ii) 15% of 100 students = (15 x 100)/ 100 = 15
From c.f. 15, draw a horizontal line which intersects the curve at P.
From P, draw a perpendicular to x – axis meeting it at Q which is equal to 9.6
Thus, freeship will be awarded to students provided annual income of their parents is upto 9.6 thousands.
(ii) Mean = ∑ fx/ ∑ f = 1832/100 = 18.32
6. The marks of 20 students in a test were as follows:
2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20.
Calculate:
(i) the mean (ii) the median (iii) the mode
Solution:
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
∑ x = 2 + 6 + 8 + 9 + 11 + 11 + 12 + 13 + 13 + 14 + 14 + 15 + 15 + 15 + 15 + 16 + 16 + 18 + 19 + 20 = 257
(i) Mean = ∑x/ ∑n = 257/ 20 = 12.85
(ii) Median = (10^{th} term + 11^{th} term)/ 2 = (13 + 14)/ 2 = 27/ 2 = 13.5
(iii) Mode = 15 since it has maximum frequencies i.e. 3
7. The marks obtained by 120 students in a mathematics test is given below:
Marks 
No. of students 
010 
5 
1020 
9 
2030 
16 
3040 
22 
4050 
26 
5060 
18 
6070 
11 
7080 
6 
8090 
4 
90100 
3 
Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate:
(i) the median
(ii) the number of students who obtained more than 75% in test.
(iii) the number of students who did not pass in the test if the pass percentage was 40.
(iv) the lower quartile
Solution:
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) Median = (120 + 1)/ 2 = 60.5^{th} term
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.
Then, 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} = 120 x (1/4) = 30^{th} term = 30
8. Using a graph paper, draw an ogive for the following distribution which shows a record of the width in kilograms of 200 students.
Weight 
Frequency 
40 – 45 
5 
45 – 50 
17 
50 – 55 
22 
55 – 60 
45 
60 – 65 
51 
65 – 70 
31 
70 – 75 
20 
75 – 80 
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 student fall
(iii) The number of students who are (a) underweight (b) overweight, if 55.70 kg is considered as standard weight.
Solution:
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) The number of students weighing more than 55 kg = 200 – 44 = 156
Thus, the percentage of students weighing 55 kg or more = (156/200) x 100 = 78 %
(ii) 30% of students = (30 x 200)/ 100 = 60
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
9. 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:
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 = 171/25 = 6.84
(ii) Median = (25 + 1)/ 2 ^{th} = 13^{th} term = 7
(iii) Mode = 6 since it has the maximum frequency i.e. 6
10. The mean of the following distribution is 52 and the frequency of class interval 30 – 40 is ‘f’. Find f.
Class Interval 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
60 – 70 
70 – 80 
Frequency 
5 
3 
f 
7 
2 
6 
13 
Solution:
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 
Mean = ∑ fx/ ∑ f = (1940 + 35f)/ (36 + f) …… (i)
But, given mean = 52 …. (ii)
From (i) and (ii), we have
(1940 + 35f)/ (36 + f) = 52
1940 + 35f = 1872 + 52f
17f = 68
Thus, f = 4
11. The monthly income of a group of 320 employees in a company is given below:
Monthly Income (thousands) 
No. of employees 
6 – 7 
20 
7 – 8 
45 
8 – 9 
65 
9 – 10 
95 
10 – 11 
60 
11 – 12 
30 
12 – 13 
5 
Draw an ogive of the given distribution on a graph paper 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) number of employees whose income is below Rs 8500.
(iii) if salary of a senior employee is above Rs 11,500, find the number of senior employees in the company.
(iv) the upper quartile.
Solution:
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) Median = 320/2 = 160^{th} term
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 8,500 = 95 (approx from the graph)
(iii) Number of employees with income below Rs 11,500 = 305 (approx from the graph)
Thus, the number of employees (senior employees) = 320 – 305 = 15
(iv) Upper quartile = Q_{3} = 320 x (3/4) = 240^{th} term = 10.3 thousands = Rs 10,300