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(A) Page No: 356
1. Find the mean of the following set of numbers:
(i) 6, 9, 11, 12 and 7
(ii) 11, 14, 23, 26, 10, 12, 18 and 6
Solution:
(i) By definition, we know
Mean = ∑x/ n
Here, n = 5
Thus,
Mean = (6 + 9 + 11 + 12 + 7)/ 5 = 45/5 = 9
(ii) By definition, we know
Mean = ∑x/ n
Here, n = 8
Thus,
Mean = (11 + 14 + 23 + 26 + 10 + 12 + 18 + 6)/ 8 = 120/8 = 15
2. Marks obtained (in mathematics) by 9 student are given below:
60, 67, 52, 76, 50, 51, 74, 45 and 56
(a) find the arithmetic mean
(b) if marks of each student be increased by 4; what will be the new value of arithmetic mean.
Solution:
(a) Mean = ∑x/ n
Here, n = 9
Thus,
Mean = (60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56)/ 9 = 531/9 = 59
(b) If the marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63
3. Find the mean of the natural numbers from 3 to 12.
Solution:
The numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Here n = 10
Mean = ∑x/ n
= (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)/ 10 = 75/10 = 7.5
4. (a) Find the mean of 7, 11, 6, 5, and 6
(b) If each number given in (a) is diminished by 2, find the new value of mean.
Solution:
(a) Mean = ∑x/ n , here n = 5
= (7 + 11 + 6 + 5 + 6)/ 5 = 35/5 = 7
(b) If 2 is subtracted from each number, then the mean will he changed as 7 – 2 = 5
5. If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’
Solution:
Given,
No. of terms (n) = 5
Mean = 8
Sum of all terms = 8 x 5 = 40 …… (i)
But, sum of numbers = 6 + 4 + 7 + a + 10 = 27 + a ….. (ii)
On equating (i) and (ii), we get
27 + a = 40
Thus, a = 13
6. The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’.
Solution:
Given,
No. of terms (n) = 5 and mean = 8
So, the sum of all terms = 5 x 8 = 40 ……. (i)
but sum of numbers = 6 + y + 7 + x + 14 = 27 + y + x ……. (ii)
On equating (i) and (ii), we get
27 + y + x = 40
x + y = 13
Hence, y = 13 – x
7. The ages of 40 students are given in the following table:
| Age( in yrs) | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| Frequency | 2 | 4 | 6 | 9 | 8 | 7 | 4 |
Find the arithmetic mean.
Solution:
| Age in yrs xi | Frequency (fi) | fixi |
| 12 | 2 | 24 |
| 13 | 4 | 52 |
| 14 | 6 | 84 |
| 15 | 9 | 135 |
| 16 | 8 | 128 |
| 17 | 7 | 119 |
| 18 | 4 | 72 |
| Total | 40 | 614 |
Mean = ∑fi xi/ ∑fi = 614/40 = 15.35
0 Comments
Please Comment