Selina Textbook Solutions: Measures of Central Tendency - Mean | Mathematics Class 10 ICSE PDF Download

 Page 1


Measures of Central Tendency  
(Mean, Median, Quartiles & Mode) 
 
Exercise 24A 
Question 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: 
 
 
Question 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. 
Page 2


Measures of Central Tendency  
(Mean, Median, Quartiles & Mode) 
 
Exercise 24A 
Question 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: 
 
 
Question 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: 
 
Question 3. 
Find the mean of the natural numbers from 3 to 12. 
 
Solution: 
 
 
Question 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: 
 
 
Question 5. 
If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’ 
Page 3


Measures of Central Tendency  
(Mean, Median, Quartiles & Mode) 
 
Exercise 24A 
Question 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: 
 
 
Question 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: 
 
Question 3. 
Find the mean of the natural numbers from 3 to 12. 
 
Solution: 
 
 
Question 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: 
 
 
Question 5. 
If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’ 
 
Solution: 
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 
Question 6. 
The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’. 
 
Solution: 
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 
Question 7. 
 
Solution: 
 
Page 4


Measures of Central Tendency  
(Mean, Median, Quartiles & Mode) 
 
Exercise 24A 
Question 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: 
 
 
Question 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: 
 
Question 3. 
Find the mean of the natural numbers from 3 to 12. 
 
Solution: 
 
 
Question 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: 
 
 
Question 5. 
If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’ 
 
Solution: 
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 
Question 6. 
The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’. 
 
Solution: 
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 
Question 7. 
 
Solution: 
 
Question 8. 
If 69.5 is the mean of 72, 70, ‘x’, 62, 50, 71, 90, 64, 58 and 82, find the value of ‘x’. 
 
Solution: 
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 
Question 9. 
 
 
Solution: 
 
Page 5


Measures of Central Tendency  
(Mean, Median, Quartiles & Mode) 
 
Exercise 24A 
Question 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: 
 
 
Question 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: 
 
Question 3. 
Find the mean of the natural numbers from 3 to 12. 
 
Solution: 
 
 
Question 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: 
 
 
Question 5. 
If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’ 
 
Solution: 
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 
Question 6. 
The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’. 
 
Solution: 
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 
Question 7. 
 
Solution: 
 
Question 8. 
If 69.5 is the mean of 72, 70, ‘x’, 62, 50, 71, 90, 64, 58 and 82, find the value of ‘x’. 
 
Solution: 
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 
Question 9. 
 
 
Solution: 
 
Question 10. 
 
 
Solution: