RS Aggarwal MCQs: Mean, Median and Mode of Ungrouped Data | Mathematics (Maths) Class 9 PDF Download

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Q u e s t i o n : 6 4
If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is
a
5
b
6
c
7
d
8
S o l u t i o n :
c
7
Mean of 5 observations = 11
We know:
Mean =
Sum of all observations
Total number of observations 
? 11 =
x+x+2+x+4+x+6+x+8
5
? 11 =
5x+20
5
? 5x +20 = 55 ? 5x = 35 ? x = 7
Q u e s t i o n : 6 5
If the mean of x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
a
10
1
3
b
10
2
3
c
11
1
3
d
11
2
3
S o l u t i o n :
c
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Q u e s t i o n : 6 4
If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is
a
5
b
6
c
7
d
8
S o l u t i o n :
c
7
Mean of 5 observations = 11
We know:
Mean =
Sum of all observations
Total number of observations 
? 11 =
x+x+2+x+4+x+6+x+8
5
? 11 =
5x+20
5
? 5x +20 = 55 ? 5x = 35 ? x = 7
Q u e s t i o n : 6 5
If the mean of x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
a
10
1
3
b
10
2
3
c
11
1
3
d
11
2
3
S o l u t i o n :
c
11
1
3
Mean of 5 observations = 9
We know:
Mean =
Sum of observations
Total number of observations
? 9 =
x+x+3+x+5+x+7+x+10
5
? 9 =
5x+25
5
? 5x +25 = 45 ? 5x = 20 ? x = 4Therefore, the last three observations are 4 +5 , 4 +7 and 4 +10 , i. e. , 9, 11 and
Q u e s t i o n : 6 6
If x
is the mean of x
1
, x
2
, x
3
, . . . , x
n
,
then ?
n
i=1
x
i
 - x = ?
a
-1
b 0
c 1
d n - 1
S o l u t i o n :
b
0
If x  is the mean of x
1
, x
2
, x
3
, x
4,...
x
n,
 then we have: ?
n
i=1
x
i
= x Or, ?
n
i=1
x
i
-x = 0
Q u e s t i o n : 6 7
If each observation of the data is increased by 8, then their mean
a
remains the same
b
is decreased by 8
c
is increased by 5
d
becomes 8 times the original mean
S o l u t i o n :
b
is decreased by 8
Let the numbers be x
1
, x
2,
...x
n.
Hence, mean = 
x 1 +x 2 +....+x n
n
Now the new numbers after decreasing every number by 8 : (x
1
-8) , (x
2
-8)...,(x
n
-8)
New Mean = 
x 1 -8 + x 2 -8 +....+ x n -8
n
           = 
x 1 +x 2 +....+x n -8n
n
           = 
x 1 +x 2 +....+x n
n
-8 ? New mean = mean - 8
          
Hence, mean is decreased by 8.
Q u e s t i o n : 6 8
The mean weight of six boys in a groups is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. The weight of the 6th boy is
a
52 kg
b
52.8 kg
c
53 kg
d
47 kg
S o l u t i o n :
c
53 kg
Mean weight of six boys = 48 kg
Let the weight of the 6th boy be x kg.
We know: Mean =
Sum of all observations 
Total number of observations 
=
51+45+49+46+44+x
6
=
235+x
6
Given: Mean = 48 kg ?
235+x
6
= 48 ? 235 +x = 288 ? x = 53Hence, the weight of the 6th boy is 53 kg.
Q u e s t i o n : 6 9
The mean of the marks scored by 50 students was found to be 39. Latter on it was discovered that a core of 43 was misread as 23. The correct mean is
a
38.6
b
39.4
c
39.8
d
39.2
( ) ( ) ( )
( )
( ) ( ) ( )
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Q u e s t i o n : 6 4
If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is
a
5
b
6
c
7
d
8
S o l u t i o n :
c
7
Mean of 5 observations = 11
We know:
Mean =
Sum of all observations
Total number of observations 
? 11 =
x+x+2+x+4+x+6+x+8
5
? 11 =
5x+20
5
? 5x +20 = 55 ? 5x = 35 ? x = 7
Q u e s t i o n : 6 5
If the mean of x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
a
10
1
3
b
10
2
3
c
11
1
3
d
11
2
3
S o l u t i o n :
c
11
1
3
Mean of 5 observations = 9
We know:
Mean =
Sum of observations
Total number of observations
? 9 =
x+x+3+x+5+x+7+x+10
5
? 9 =
5x+25
5
? 5x +25 = 45 ? 5x = 20 ? x = 4Therefore, the last three observations are 4 +5 , 4 +7 and 4 +10 , i. e. , 9, 11 and
Q u e s t i o n : 6 6
If x
is the mean of x
1
, x
2
, x
3
, . . . , x
n
,
then ?
n
i=1
x
i
 - x = ?
a
-1
b 0
c 1
d n - 1
S o l u t i o n :
b
0
If x  is the mean of x
1
, x
2
, x
3
, x
4,...
x
n,
 then we have: ?
n
i=1
x
i
= x Or, ?
n
i=1
x
i
-x = 0
Q u e s t i o n : 6 7
If each observation of the data is increased by 8, then their mean
a
remains the same
b
is decreased by 8
c
is increased by 5
d
becomes 8 times the original mean
S o l u t i o n :
b
is decreased by 8
Let the numbers be x
1
, x
2,
...x
n.
Hence, mean = 
x 1 +x 2 +....+x n
n
Now the new numbers after decreasing every number by 8 : (x
1
-8) , (x
2
-8)...,(x
n
-8)
New Mean = 
x 1 -8 + x 2 -8 +....+ x n -8
n
           = 
x 1 +x 2 +....+x n -8n
n
           = 
x 1 +x 2 +....+x n
n
-8 ? New mean = mean - 8
          
Hence, mean is decreased by 8.
Q u e s t i o n : 6 8
The mean weight of six boys in a groups is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. The weight of the 6th boy is
a
52 kg
b
52.8 kg
c
53 kg
d
47 kg
S o l u t i o n :
c
53 kg
Mean weight of six boys = 48 kg
Let the weight of the 6th boy be x kg.
We know: Mean =
Sum of all observations 
Total number of observations 
=
51+45+49+46+44+x
6
=
235+x
6
Given: Mean = 48 kg ?
235+x
6
= 48 ? 235 +x = 288 ? x = 53Hence, the weight of the 6th boy is 53 kg.
Q u e s t i o n : 6 9
The mean of the marks scored by 50 students was found to be 39. Latter on it was discovered that a core of 43 was misread as 23. The correct mean is
a
38.6
b
39.4
c
39.8
d
39.2
( ) ( ) ( )
( )
( ) ( ) ( )
S o l u t i o n :
b
39.4
Mean of the marks scored by 50 students = 39
Sum of the marks scored by 50 students = (39 ×50) = 1950
Correct sum = (1950 + 43 -
23) = 1970
? Mean =
1970
50
= 39. 4
Q u e s t i o n : 7 0
The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. The correct mean is
a
64.86
b
65.31
c
64.91
d
64.61
S o l u t i o n :
c
64.91
Mean of 100 items = 64
Sum of 100 items = 64 ×100 = 6400
Correct sum = (6400 + 36 + 90 -
26 -
9) = 6491
Correct mean =
6491
100
= 64. 91
Q u e s t i o n : 7 1
The mean of 100 observations is 50. If one of the observations 50 is replaced by 150, the resulting mean will be
a
50.5
b
51
c
51.5
d
52
S o l u t i o n :
b
51
Mean of 100 observations = 50
Sum of 100 observations = 100 ×50 = 5000
It is given that one of the observations, 50, is replaced by 150.
? New sum = (5000 -
50 + 150) = 5100
And,
Resulting mean =
5100
100
= 51
Q u e s t i o n : 7 2
Let x
be the mean of x
1
, x
2
, . . . , x
n
and y
be the mean of y
1
, y
2
, . . . , y
n
.
If z
is the mean of x
1
, x
2
, . . . , x
n
, y
1
, y
2
, . . . , y
n
,
then z = ?
a
(x + y)
b
1
2
x + y
c
1
n
x + y
d
1
2n
x + y
S o l u t i o n :
b
1
2
x + y
( )
( )
( )
( )
( ) ( )
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Q u e s t i o n : 6 4
If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is
a
5
b
6
c
7
d
8
S o l u t i o n :
c
7
Mean of 5 observations = 11
We know:
Mean =
Sum of all observations
Total number of observations 
? 11 =
x+x+2+x+4+x+6+x+8
5
? 11 =
5x+20
5
? 5x +20 = 55 ? 5x = 35 ? x = 7
Q u e s t i o n : 6 5
If the mean of x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
a
10
1
3
b
10
2
3
c
11
1
3
d
11
2
3
S o l u t i o n :
c
11
1
3
Mean of 5 observations = 9
We know:
Mean =
Sum of observations
Total number of observations
? 9 =
x+x+3+x+5+x+7+x+10
5
? 9 =
5x+25
5
? 5x +25 = 45 ? 5x = 20 ? x = 4Therefore, the last three observations are 4 +5 , 4 +7 and 4 +10 , i. e. , 9, 11 and
Q u e s t i o n : 6 6
If x
is the mean of x
1
, x
2
, x
3
, . . . , x
n
,
then ?
n
i=1
x
i
 - x = ?
a
-1
b 0
c 1
d n - 1
S o l u t i o n :
b
0
If x  is the mean of x
1
, x
2
, x
3
, x
4,...
x
n,
 then we have: ?
n
i=1
x
i
= x Or, ?
n
i=1
x
i
-x = 0
Q u e s t i o n : 6 7
If each observation of the data is increased by 8, then their mean
a
remains the same
b
is decreased by 8
c
is increased by 5
d
becomes 8 times the original mean
S o l u t i o n :
b
is decreased by 8
Let the numbers be x
1
, x
2,
...x
n.
Hence, mean = 
x 1 +x 2 +....+x n
n
Now the new numbers after decreasing every number by 8 : (x
1
-8) , (x
2
-8)...,(x
n
-8)
New Mean = 
x 1 -8 + x 2 -8 +....+ x n -8
n
           = 
x 1 +x 2 +....+x n -8n
n
           = 
x 1 +x 2 +....+x n
n
-8 ? New mean = mean - 8
          
Hence, mean is decreased by 8.
Q u e s t i o n : 6 8
The mean weight of six boys in a groups is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. The weight of the 6th boy is
a
52 kg
b
52.8 kg
c
53 kg
d
47 kg
S o l u t i o n :
c
53 kg
Mean weight of six boys = 48 kg
Let the weight of the 6th boy be x kg.
We know: Mean =
Sum of all observations 
Total number of observations 
=
51+45+49+46+44+x
6
=
235+x
6
Given: Mean = 48 kg ?
235+x
6
= 48 ? 235 +x = 288 ? x = 53Hence, the weight of the 6th boy is 53 kg.
Q u e s t i o n : 6 9
The mean of the marks scored by 50 students was found to be 39. Latter on it was discovered that a core of 43 was misread as 23. The correct mean is
a
38.6
b
39.4
c
39.8
d
39.2
( ) ( ) ( )
( )
( ) ( ) ( )
S o l u t i o n :
b
39.4
Mean of the marks scored by 50 students = 39
Sum of the marks scored by 50 students = (39 ×50) = 1950
Correct sum = (1950 + 43 -
23) = 1970
? Mean =
1970
50
= 39. 4
Q u e s t i o n : 7 0
The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. The correct mean is
a
64.86
b
65.31
c
64.91
d
64.61
S o l u t i o n :
c
64.91
Mean of 100 items = 64
Sum of 100 items = 64 ×100 = 6400
Correct sum = (6400 + 36 + 90 -
26 -
9) = 6491
Correct mean =
6491
100
= 64. 91
Q u e s t i o n : 7 1
The mean of 100 observations is 50. If one of the observations 50 is replaced by 150, the resulting mean will be
a
50.5
b
51
c
51.5
d
52
S o l u t i o n :
b
51
Mean of 100 observations = 50
Sum of 100 observations = 100 ×50 = 5000
It is given that one of the observations, 50, is replaced by 150.
? New sum = (5000 -
50 + 150) = 5100
And,
Resulting mean =
5100
100
= 51
Q u e s t i o n : 7 2
Let x
be the mean of x
1
, x
2
, . . . , x
n
and y
be the mean of y
1
, y
2
, . . . , y
n
.
If z
is the mean of x
1
, x
2
, . . . , x
n
, y
1
, y
2
, . . . , y
n
,
then z = ?
a
(x + y)
b
1
2
x + y
c
1
n
x + y
d
1
2n
x + y
S o l u t i o n :
b
1
2
x + y
( )
( )
( )
( )
( ) ( )
¯ z =
x 1 +x 2 +...+x n + y 1 +y 2 +...+y n
2n
Given: ¯ x = 
x 1 +x 2 +.....x n
n
? x
1
+x
2
+. . . . . . +x
n
= n ¯ xand
¯
y =
y 1 +y 2 +......+y n
n
? y
1
+y
2
+. . . . . . +y
n
= n ¯ y ? ¯ z = 
n ¯ x + n ¯ y
2n
          =
1
2 ¯ x + ¯ y
Q u e s t i o n : 7 3
If x
is the mean of x
1
, x
2
, . . . , x
n
,
then for a ? 0
, the mean of ax
1,
 ax
2
, . . . , ax
n
, 
x 1
a
,
x 2
a
, . . . ,
x n
a
is
a
a +
1
a
x
b
a +
1
a
x
2
c
a +
1
a
x
n
d
a+
1
a
x
2n
S o l u t i o n :
b
a +
1
a
x
2
Required mean =
ax 1 +ax 2 +...+ax n +
x 1
a
+
x 2
a
+...+
x n
a
2n
                             =
1
2
a x 1 +x 2 +...+x n
n
+
1
a
x 1 +x 2 +...+x n
n
                                                          =
1
2
a ¯ x +
1
a ¯ x                                
¯
 x =
x 1 +x 2 +...+x n
n
                       
Q u e s t i o n : 7 4
If x
1
, x
2
, . . . , x
n
are the means of n groups with n
1
, n
2
, . . . , n
n
number of observations respectively, then the mean x
of all the groups taken together is
a
?
n
i=1
n
i
x
i
b
?
n
i=1
n 2
n
i
x
i
c
n
?
i=1n i x i
n
?
i=1n i
d
n
?
i=1n i x i
2n
S o l u t i o n :
c 
n
?
i=1n i x i
n
?
i=1n i
Sum of all terms = n
1
¯ x
1
+n
2
¯ x
2
+. . . n
n
¯ x
n
Number of terms = n
1
+n
2
+. . . n
n
? Mean =
n
?
i=1n i
¯ x i
n
?
i=1n i
( ) ( )
( )
( )
( )
( )
( )
( )
( ) ( )
{
( ) ( )
}
{ } ( )
( )
( )
Page 5


       
                                
   
      
      
           
     
     
   
      
      
         
     
                                                                                
     
                                     
      
          
     
    
      
      
        
      
   
       
Q u e s t i o n : 6 4
If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is
a
5
b
6
c
7
d
8
S o l u t i o n :
c
7
Mean of 5 observations = 11
We know:
Mean =
Sum of all observations
Total number of observations 
? 11 =
x+x+2+x+4+x+6+x+8
5
? 11 =
5x+20
5
? 5x +20 = 55 ? 5x = 35 ? x = 7
Q u e s t i o n : 6 5
If the mean of x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
a
10
1
3
b
10
2
3
c
11
1
3
d
11
2
3
S o l u t i o n :
c
11
1
3
Mean of 5 observations = 9
We know:
Mean =
Sum of observations
Total number of observations
? 9 =
x+x+3+x+5+x+7+x+10
5
? 9 =
5x+25
5
? 5x +25 = 45 ? 5x = 20 ? x = 4Therefore, the last three observations are 4 +5 , 4 +7 and 4 +10 , i. e. , 9, 11 and
Q u e s t i o n : 6 6
If x
is the mean of x
1
, x
2
, x
3
, . . . , x
n
,
then ?
n
i=1
x
i
 - x = ?
a
-1
b 0
c 1
d n - 1
S o l u t i o n :
b
0
If x  is the mean of x
1
, x
2
, x
3
, x
4,...
x
n,
 then we have: ?
n
i=1
x
i
= x Or, ?
n
i=1
x
i
-x = 0
Q u e s t i o n : 6 7
If each observation of the data is increased by 8, then their mean
a
remains the same
b
is decreased by 8
c
is increased by 5
d
becomes 8 times the original mean
S o l u t i o n :
b
is decreased by 8
Let the numbers be x
1
, x
2,
...x
n.
Hence, mean = 
x 1 +x 2 +....+x n
n
Now the new numbers after decreasing every number by 8 : (x
1
-8) , (x
2
-8)...,(x
n
-8)
New Mean = 
x 1 -8 + x 2 -8 +....+ x n -8
n
           = 
x 1 +x 2 +....+x n -8n
n
           = 
x 1 +x 2 +....+x n
n
-8 ? New mean = mean - 8
          
Hence, mean is decreased by 8.
Q u e s t i o n : 6 8
The mean weight of six boys in a groups is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. The weight of the 6th boy is
a
52 kg
b
52.8 kg
c
53 kg
d
47 kg
S o l u t i o n :
c
53 kg
Mean weight of six boys = 48 kg
Let the weight of the 6th boy be x kg.
We know: Mean =
Sum of all observations 
Total number of observations 
=
51+45+49+46+44+x
6
=
235+x
6
Given: Mean = 48 kg ?
235+x
6
= 48 ? 235 +x = 288 ? x = 53Hence, the weight of the 6th boy is 53 kg.
Q u e s t i o n : 6 9
The mean of the marks scored by 50 students was found to be 39. Latter on it was discovered that a core of 43 was misread as 23. The correct mean is
a
38.6
b
39.4
c
39.8
d
39.2
( ) ( ) ( )
( )
( ) ( ) ( )
S o l u t i o n :
b
39.4
Mean of the marks scored by 50 students = 39
Sum of the marks scored by 50 students = (39 ×50) = 1950
Correct sum = (1950 + 43 -
23) = 1970
? Mean =
1970
50
= 39. 4
Q u e s t i o n : 7 0
The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. The correct mean is
a
64.86
b
65.31
c
64.91
d
64.61
S o l u t i o n :
c
64.91
Mean of 100 items = 64
Sum of 100 items = 64 ×100 = 6400
Correct sum = (6400 + 36 + 90 -
26 -
9) = 6491
Correct mean =
6491
100
= 64. 91
Q u e s t i o n : 7 1
The mean of 100 observations is 50. If one of the observations 50 is replaced by 150, the resulting mean will be
a
50.5
b
51
c
51.5
d
52
S o l u t i o n :
b
51
Mean of 100 observations = 50
Sum of 100 observations = 100 ×50 = 5000
It is given that one of the observations, 50, is replaced by 150.
? New sum = (5000 -
50 + 150) = 5100
And,
Resulting mean =
5100
100
= 51
Q u e s t i o n : 7 2
Let x
be the mean of x
1
, x
2
, . . . , x
n
and y
be the mean of y
1
, y
2
, . . . , y
n
.
If z
is the mean of x
1
, x
2
, . . . , x
n
, y
1
, y
2
, . . . , y
n
,
then z = ?
a
(x + y)
b
1
2
x + y
c
1
n
x + y
d
1
2n
x + y
S o l u t i o n :
b
1
2
x + y
( )
( )
( )
( )
( ) ( )
¯ z =
x 1 +x 2 +...+x n + y 1 +y 2 +...+y n
2n
Given: ¯ x = 
x 1 +x 2 +.....x n
n
? x
1
+x
2
+. . . . . . +x
n
= n ¯ xand
¯
y =
y 1 +y 2 +......+y n
n
? y
1
+y
2
+. . . . . . +y
n
= n ¯ y ? ¯ z = 
n ¯ x + n ¯ y
2n
          =
1
2 ¯ x + ¯ y
Q u e s t i o n : 7 3
If x
is the mean of x
1
, x
2
, . . . , x
n
,
then for a ? 0
, the mean of ax
1,
 ax
2
, . . . , ax
n
, 
x 1
a
,
x 2
a
, . . . ,
x n
a
is
a
a +
1
a
x
b
a +
1
a
x
2
c
a +
1
a
x
n
d
a+
1
a
x
2n
S o l u t i o n :
b
a +
1
a
x
2
Required mean =
ax 1 +ax 2 +...+ax n +
x 1
a
+
x 2
a
+...+
x n
a
2n
                             =
1
2
a x 1 +x 2 +...+x n
n
+
1
a
x 1 +x 2 +...+x n
n
                                                          =
1
2
a ¯ x +
1
a ¯ x                                
¯
 x =
x 1 +x 2 +...+x n
n
                       
Q u e s t i o n : 7 4
If x
1
, x
2
, . . . , x
n
are the means of n groups with n
1
, n
2
, . . . , n
n
number of observations respectively, then the mean x
of all the groups taken together is
a
?
n
i=1
n
i
x
i
b
?
n
i=1
n 2
n
i
x
i
c
n
?
i=1n i x i
n
?
i=1n i
d
n
?
i=1n i x i
2n
S o l u t i o n :
c 
n
?
i=1n i x i
n
?
i=1n i
Sum of all terms = n
1
¯ x
1
+n
2
¯ x
2
+. . . n
n
¯ x
n
Number of terms = n
1
+n
2
+. . . n
n
? Mean =
n
?
i=1n i
¯ x i
n
?
i=1n i
( ) ( )
( )
( )
( )
( )
( )
( )
( ) ( )
{
( ) ( )
}
{ } ( )
( )
( )
Q u e s t i o n : 7 5
The mean of the following data is 8.
x 3 5 7 9 11 13
y 6 8 15 p 8 4
The value of p is
a
23
b
24
c
25
d
21
S o l u t i o n :
c
25
 
x y
x
× y
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
Total 41 + p 303 + 9p
Now, Mean =
303+9p
41+p
Given: Mean = 8 ?
303+9p
41+p
= 8 ? 303 +9p = 328 +8p ? p = 25
Q u e s t i o n : 7 6
The runs scored by 11 members of a cricket team are
15, 34, 56, 27, 43, 29, 31, 13, 50, 20, 0
The median score is
a
27
b
29
c
31
d
20
S o l u t i o n :
b
29
Arranging the weight of 10 students in ascending order, we have:
0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56
Here, n is 11, which is an odd number.
Thus, we have:
Median = Value of 
n+1
2
th observation Median score = Value of 
11+1
2
th term                    = Value of 6th term                     = 29
Q u e s t i o n : 7 7
The weight of 10 students inkgs
are
55, 40, 35, 52, 60, 38, 36, 45, 31, 44
The median weight is
a
40 kg
b
41 kg
c
42 kg
d
44 kg
S o l u t i o n :
c
42 kg
Arranging the numbers in ascending order, we have:
31, 35, 36, 38, 40, 44, 45, 52, 55, 60
Here, n is 10, which is an even number.
Thus, we have:
Median = Mean of 
n
2
th observation & 
n
2
+1 th observationMedian weight = Mean of the weights of 
10
2
th student & 
10
2
+1 th student                     = Mean of the weights of 5th student
Q u e s t i o n : 7 8
The median of the numbers 4, 4, 5, 7, 6, 7, 7, 12, 3 is
a
4
b
5
c
6
d
( ) ( )
( ) ( ) ( ) ( )