Selina Solutions Class 10 Maths Chapter 24 Measures of Central Tendency
Recalling the concepts learned from the previous chapter, the practice consists of a mixture of problems based on extracting the mean, median, polymer, and pylon for appropriate questions. If students are looking for a resource to clarify their doubts, stop here at Selina Solutions of Class 10 Mathematics as these solutions have been created by experts from Mathematics PDF Solutions keeping in mind the latest ICSE patterns. These practice solutions are short Selina solutions for class 10 mathematics chapter 24 central trend practice measures pdf available in the following link.
Selina Solutions Concise Maths Class 10 Chapter 24 Measures of Central Tendency Exercise 24(C)
1. A student got the following marks in 9 questions of a question paper.
3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.
Solution:
Arranging the given data in descending order:
8, 7, 6, 5, 4, 3, 3, 1, 0
Clearly, the middle term is 4 which is the 5th term.
Hence, median = 4
2. The weights (in kg) of 10 students of a class are given below:
21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24.
Find the median of their weights.
Solution:
Arranging the given data in descending order:
28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5
It’s seen that,
The middle terms are 24 and 24, 5th and 6th terms
Thus,
Median = (24 + 24)/ 2 = 48/2 = 24
3. The marks obtained by 19 students of a class are given below:
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28. Find:
(i) median (ii) lower quartile
(iii) upper quartile (iv) interquartile range
Solution:
Arranging in ascending order:
22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37
(i) The middle term is 10th term i.e. 29
Hence, median = 29
(ii) Lower quartile
(iii) Upper quartile =
(iv) Interquartile range = q3 – q1 =35 – 26 = 9
4. From the following data, find:
(i) Median
(ii) Upper quartile
(iii) Inter-quartile range
25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83
Solution:
Arranging the given data in ascending order, we have:
0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95
(i) Median is the mean of 8th and 9th term
Thus, median = (40 + 45)/ 2 = 85/2 = 42.5
(ii) Upper quartile =
(iii) Interquartile range is given by,
q1 = 16th/4 term = 18; q3 = 65
Interquartile range = q3 – q1
Thus,
q3 – q1 = 65 – 18 = 47
5. The ages of 37 students in a class are given in the following table:
Age (in years) | 11 | 12 | 13 | 14 | 15 | 16 |
Frequency | 2 | 4 | 6 | 10 | 8 | 7 |
Find the median.
Solution:
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 (n) = 37
Median = 
And, the 19th term is 14
Therefore, the median = 14
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