RS Aggarwal MCQs: Presentation of Data in Tabular Form | Mathematics (Maths) Class 9 PDF Download

 Page 1


                        
 
 
 
 
 
 
  
  
     
    
  
     
 
        
        
        
         
 
     
       
 
 
Question:21
The range of the data
12, 25, 15, 18, 17, 20, 22, 6, 16, 11, 8, 19, 10, 30, 20, 32 is
a
10
b
15
c
18
d
26
Solution:
d
26
We have:
Maximum value = 32
Minimum value = 6
We know:
Range = Maximum value -
Minimum value 
          =32 -
Page 2


                        
 
 
 
 
 
 
  
  
     
    
  
     
 
        
        
        
         
 
     
       
 
 
Question:21
The range of the data
12, 25, 15, 18, 17, 20, 22, 6, 16, 11, 8, 19, 10, 30, 20, 32 is
a
10
b
15
c
18
d
26
Solution:
d
26
We have:
Maximum value = 32
Minimum value = 6
We know:
Range = Maximum value -
Minimum value 
          =32 -
6
          =26
Question:22
The class mark of the class 100-120 is
a
100
b
110
c
115
d
120
Solution:
b
110
Class mark = 
Upper limit+Lower limit
2
=
120+100
2
= 110
Question:23
In the class intervals 10-20, 20-30, the number 20 is included in
a
10-20
b
20-30
c
in each of 10-20 and 20-30
d
in none of 10-20 and 20-30
Solution:
b
20-30
This is the continuous form of frequency distribution. Here, the upper limit of each class is excluded, while the lower limit is
included. So, the number 20 is included in the class interval 20-30.
Question:24
The class marks of a frequency distribution are 15, 20, 25, 30, .. . The class corresponding to the class marks 20 is
a
12.5-17.5
b 17.5-22.5
c 18.5-21.5
d 19.5-20.5
Solution:
b
17.5 -
22.5
We are given frequency distribution 15, 20, 25, 30,...
Class size = 20 -
15 = 5
Class marks = 20
Now,
Lower limit = 20 -
5
2
=
35
2
= 17. 5Upper limit = 20 +
5
2
=
45
2
= 22. 5
Thus, the required class is 17.5 -
22.5.
Question:25
( ) ( )
Page 3


                        
 
 
 
 
 
 
  
  
     
    
  
     
 
        
        
        
         
 
     
       
 
 
Question:21
The range of the data
12, 25, 15, 18, 17, 20, 22, 6, 16, 11, 8, 19, 10, 30, 20, 32 is
a
10
b
15
c
18
d
26
Solution:
d
26
We have:
Maximum value = 32
Minimum value = 6
We know:
Range = Maximum value -
Minimum value 
          =32 -
6
          =26
Question:22
The class mark of the class 100-120 is
a
100
b
110
c
115
d
120
Solution:
b
110
Class mark = 
Upper limit+Lower limit
2
=
120+100
2
= 110
Question:23
In the class intervals 10-20, 20-30, the number 20 is included in
a
10-20
b
20-30
c
in each of 10-20 and 20-30
d
in none of 10-20 and 20-30
Solution:
b
20-30
This is the continuous form of frequency distribution. Here, the upper limit of each class is excluded, while the lower limit is
included. So, the number 20 is included in the class interval 20-30.
Question:24
The class marks of a frequency distribution are 15, 20, 25, 30, .. . The class corresponding to the class marks 20 is
a
12.5-17.5
b 17.5-22.5
c 18.5-21.5
d 19.5-20.5
Solution:
b
17.5 -
22.5
We are given frequency distribution 15, 20, 25, 30,...
Class size = 20 -
15 = 5
Class marks = 20
Now,
Lower limit = 20 -
5
2
=
35
2
= 17. 5Upper limit = 20 +
5
2
=
45
2
= 22. 5
Thus, the required class is 17.5 -
22.5.
Question:25
( ) ( )
In a frequency distribution, the mid-value of a class is 10 and width of each class is 6. The lower limit of the class is
a
6
b
7
c
8
d
12
Solution:
b
7
Given:
Mid value of the class = 10
Width of each class = 6
Now,
Let the lower limit be x. 
We know:
Upper limit = Lower limit + Class size
                  = x + 6
Also,
Mid value =
x+x+6
2
=
2x+6
2
= x +3 ? x +3 = 10 ? x = 7
Thus, the lower limit is 7.
Question:26
The mid-value of a class interval is 42 and the class size is 10. The lower and upper limits are
a
37-47
b 37.5-47.5
c 36.5-47.5
d 36.5-46.5
Solution:
a
37–47
Let the lower limit be x. 
Here,
Class size = 10
? Upper limit = Class size + Lower limit
Upper limit = (x + 10)
Mid value of the class interval = 42
?
x+x+10
2
= 42 ?
2x+10
2
= 42 ? 2x +10 = 84 ? 2x = 74 ? x = 37Thus, we have: Lower limit = 37 Upper limit = 37 +10 = 47
Question:27
Let m be the midpoint and u be the upper class limit of a class in a continuous frequency distribution. The lower class limit of
the class is
a
2m - u
b 2m + u
c m - u
d m + u
Solution:
a
2m -
u
Given:
Mid value = m
Page 4


                        
 
 
 
 
 
 
  
  
     
    
  
     
 
        
        
        
         
 
     
       
 
 
Question:21
The range of the data
12, 25, 15, 18, 17, 20, 22, 6, 16, 11, 8, 19, 10, 30, 20, 32 is
a
10
b
15
c
18
d
26
Solution:
d
26
We have:
Maximum value = 32
Minimum value = 6
We know:
Range = Maximum value -
Minimum value 
          =32 -
6
          =26
Question:22
The class mark of the class 100-120 is
a
100
b
110
c
115
d
120
Solution:
b
110
Class mark = 
Upper limit+Lower limit
2
=
120+100
2
= 110
Question:23
In the class intervals 10-20, 20-30, the number 20 is included in
a
10-20
b
20-30
c
in each of 10-20 and 20-30
d
in none of 10-20 and 20-30
Solution:
b
20-30
This is the continuous form of frequency distribution. Here, the upper limit of each class is excluded, while the lower limit is
included. So, the number 20 is included in the class interval 20-30.
Question:24
The class marks of a frequency distribution are 15, 20, 25, 30, .. . The class corresponding to the class marks 20 is
a
12.5-17.5
b 17.5-22.5
c 18.5-21.5
d 19.5-20.5
Solution:
b
17.5 -
22.5
We are given frequency distribution 15, 20, 25, 30,...
Class size = 20 -
15 = 5
Class marks = 20
Now,
Lower limit = 20 -
5
2
=
35
2
= 17. 5Upper limit = 20 +
5
2
=
45
2
= 22. 5
Thus, the required class is 17.5 -
22.5.
Question:25
( ) ( )
In a frequency distribution, the mid-value of a class is 10 and width of each class is 6. The lower limit of the class is
a
6
b
7
c
8
d
12
Solution:
b
7
Given:
Mid value of the class = 10
Width of each class = 6
Now,
Let the lower limit be x. 
We know:
Upper limit = Lower limit + Class size
                  = x + 6
Also,
Mid value =
x+x+6
2
=
2x+6
2
= x +3 ? x +3 = 10 ? x = 7
Thus, the lower limit is 7.
Question:26
The mid-value of a class interval is 42 and the class size is 10. The lower and upper limits are
a
37-47
b 37.5-47.5
c 36.5-47.5
d 36.5-46.5
Solution:
a
37–47
Let the lower limit be x. 
Here,
Class size = 10
? Upper limit = Class size + Lower limit
Upper limit = (x + 10)
Mid value of the class interval = 42
?
x+x+10
2
= 42 ?
2x+10
2
= 42 ? 2x +10 = 84 ? 2x = 74 ? x = 37Thus, we have: Lower limit = 37 Upper limit = 37 +10 = 47
Question:27
Let m be the midpoint and u be the upper class limit of a class in a continuous frequency distribution. The lower class limit of
the class is
a
2m - u
b 2m + u
c m - u
d m + u
Solution:
a
2m -
u
Given:
Mid value = m
Upper limit = u
We know:
Lower limit+Upper limit
2
= Mid value ?
Lower limit +u
2
= m ? Lower limit +u = 2m ? Lower limit = 2m -u
Question:28
The width of each of the five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is
10. The upper class limit of the highest class is
a
45
b
25
c
35
d
40
Solution:
c
35
We have:
Class width = 5
Lower class limit of the lowest class = 10
Now,
Upper class limit of the highest class = 10 + 5 ×
5 = 35
Question:29
Let L be the lower class boundary of a class in a frequency distribution and m be the midpoint of the class. Which one of the
following is the upper class boundary of the class?
a
m +
(m + L)
2
b
L + 
m + L
2
c
2m - L
d
m - 2L
Solution:
c
2m -
L
Mid value =
Lower limit+Upper limit
2
? m =
L+U
2
? U = 2m -L ? Upper class boundary of the class = 2m -L