# RD Sharma Solutions Class 9 Measures Of Central Tendency Exercise 24.3

## RD Sharma Solutions Class 9 Chapter 24 Exercise 24.3

Find the median of the following data :

Q1 . 83 , 37 , 70 , 29 , 45 , 63 , 41 , 70 , 34 , 54

SOLUTION :

Numbers are 83 , 37 , 70 , 29 , 45 , 63 , 41 , 70 , 34 , 54

Arranging the numbers in ascending order

29 , 34 , 37 , 41 , 45 , 54 , 63 , 70 , 70 , 83

n = 10(even)

$∴ median=\frac{\frac{n}{2}th \;value+\left ( \frac{n}{2} +1\right )th\;value}{2}$

$=\frac{\frac{10}{2}th \;value+\left ( \frac{10}{2} +1\right )th\;value}{2}$

$=\frac{5th \;value+6th\;value}{2}$

$=\frac{45+54}{2}$

$=\frac{99}{2}$ = 49.5

Q2 . 133 , 73 , 89 , 108 , 94 , 104 , 94 , 85 , 100 , 120

SOLUTION :

Numbers are 133 , 73 , 89 , 108 , 94 , 104 , 94 , 85 , 100 , 120

Arranging the numbers in ascending order

73 , 85 , 89 ,94 ,  94 , 100 , 104 , 108 , 120 , 133

n = 10(even)

$∴ median=\frac{\frac{n}{2}th \;value+\left ( \frac{n}{2} +1\right )th\;value}{2}$

$=\frac{\frac{10}{2}th \;value+\left ( \frac{10}{2} +1\right )th\;value}{2}$

$=\frac{5th \;value+6th\;value}{2}$

$=\frac{94+100}{2}$

$=\frac{194}{2}$ = 97

Q3 . 31 , 38 , 27 , 28 , 36 , 25 , 35 , 40

SOLUTION :

Numbers are 31 , 38 , 27 , 28 , 36 , 25 , 35 , 40

Arranging the numbers in ascending order

25 , 27 , 28 , 31 , 35 , 36 , 38 , 40

n = 8(even)

$∴ median=\frac{\frac{n}{2}th \;value+\left ( \frac{n}{2} +1\right )th\;value}{2}$

$=\frac{\frac{8}{2}th \;value+\left ( \frac{8}{2} +1\right )th\;value}{2}$

$=\frac{4th \;value+5th\;value}{2}$

$=\frac{31+35}{2}$

$=\frac{66}{2}$ = 33

Q4 . 15 , 6 , 16 , 8 , 22 , 21 , 9 , 18 , 25

SOLUTION :

Numbers are 15 , 6 , 16 , 8 , 22 , 21 , 9 , 18 , 25

Arranging the numbers in ascending order

6 , 8 , 9 , 15 , 16 , 21 , 22 , 25

n = 9 (odd)

$∴ Median= \left ( \frac{n+1}{2} \right )th\; value$

= $\left ( \frac{9+1}{2} \right )th\; value$

= $5th\; value$ = 16

Q5 . 41 , 43 , 127 , 99 , 71 , 92 , 71 , 58 , 57

SOLUTION :

Numbers are 41 , 43 , 127 , 99 , 71 , 92 , 71 , 58 , 57

Arranging the numbers in ascending order

41 , 43 , 57 , 58 , 71 , 71 , 92 , 99 , 127

n = 9 (odd)

$∴ Median= \left ( \frac{n+1}{2} \right )th\; value$

= $\left ( \frac{9+1}{2} \right )th\; value$

= $5th\; value$ = 71

Q6 . 25 , 34 , 31 , 23 , 22 , 26 , 35 , 29 , 20 , 32

SOLUTION :

Numbers are 25 , 34 , 31 , 23 , 22 , 26 , 35 , 29 , 20 , 32

Arranging the numbers in ascending order

20 , 22 , 23 , 25 , 26 , 29 , 31 , 32 , 34 , 35

n = 10(even)

$∴ median=\frac{\frac{n}{2}th \;value+\left ( \frac{n}{2} +1\right )th\;value}{2}$

$=\frac{\frac{10}{2}th \;value+\left ( \frac{10}{2} +1\right )th\;value}{2}$

$=\frac{5th \;value+6th\;value}{2}$

$=\frac{26+29}{2}$

$=\frac{55}{2}$ = 27.5

Q7 . 12 , 17 , 3 , 14 , 5 , 8 , 7 , 15

SOLUTION :

Numbers are 12 , 17 , 3 , 14 , 5 , 8 , 7 , 15

Arranging the numbers in ascending order

3 , 5 , 7 , 8 , 12 , 14 , 15 , 17

n = 8(even)

$∴ median=\frac{\frac{n}{2}th \;value+\left ( \frac{n}{2} +1\right )th\;value}{2}$

$=\frac{\frac{8}{2}th \;value+\left ( \frac{8}{2} +1\right )th\;value}{2}$

$=\frac{4th \;value+5th\;value}{2}$

$=\frac{8+12}{2}$

$=\frac{20}{2}$ = 10

Q8 . 92 , 35 , 67 , 85 , 72 , 81 , 56 , 51 , 42 , 69

SOLUTION :

Numbers are 92 , 35 , 67 , 85 , 72 , 81 , 56 , 51 , 42 , 69

Arranging the numbers in ascending order

35 , 42 , 51 , 56 , 67 , 69 , 72 , 81 , 85 , 92

n = 10(even)

$∴ median=\frac{\frac{n}{2}th \;value+\left ( \frac{n}{2} +1\right )th\;value}{2}$

$=\frac{\frac{10}{2}th \;value+\left ( \frac{10}{2} +1\right )th\;value}{2}$

$=\frac{5th \;value+6th\;value}{2}$

$=\frac{67+69}{2}$

$=\frac{136}{2}$ = 68

Q9 . Numbers 50 , 42 , 35 , 2x + 10 , 2x – 8 , 12 , 11 , 8 are written in descending order and their median is 25 , find x.

SOLUTION :

Given the number of observation , n = 8

$∴ median=\frac{\frac{n}{2}th \;value+\left ( \frac{n}{2} +1\right )th\;value}{2}$

$=\frac{\frac{8}{2}th \;value+\left ( \frac{8}{2} +1\right )th\;value}{2}$

$=\frac{4th \;value+5th\;value}{2}$

$=\frac{2x+10+2x-8}{2}$

= 2x +  1

Given median = 25

$∴ 2x+1=25$

$\Rightarrow 2x=24$

$\Rightarrow x=12$

Q10 . Find the median of the following observations : 46 , 64 , 87 , 41 , 58 , 77 , 35 , 90 , 55 , 92 , 33 .If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?

SOLUTION :

Given the numbers are  46 , 64 , 87 , 41 , 58 , 77 , 35 , 90 , 55 , 92 , 33

Arranging the numbers in ascending order

33 , 35 , 41 , 46 , 55 , 58 , 64 , 77 , 87 , 90 , 92

n = 11 (odd)

$∴ Median= \left ( \frac{n+1}{2} \right )th\; value$

= $\left ( \frac{11+1}{2} \right )th\; value$

= $6th\; value$ = 58

If 92 is replaced by 99 and 41 by 43

Then the new values are : 33 , 35 , 43 , 46 , 55 , 58 , 64 , 77 , 87 , 90 , 99

n = 11 (odd)

$∴ New Median= \left ( \frac{n+1}{2} \right )th\; value$

= $\left ( \frac{11+1}{2} \right )th\; value$

= $6th\; value$ = 58

Q11 . Find the median of the following data : 41 , 43 , 127 , 99 , 61 , 92 , 71 , 58 , 57 .If 58 is replaced by 85 , what will be the new median ?

SOLUTION :

Given the numbers are  41 , 43 , 127 , 99 , 61 , 92 , 71 , 58 , 57

Arranging the numbers in ascending order

41 , 43 , 57 , 58 , 61 , 71 , 92 , 99 , 127

n = 9 (odd)

$∴ New Median= \left ( \frac{n+1}{2} \right )th\; value$

= $\left ( \frac{9+1}{2} \right )th\; value$

= $5th\; value$ = 61

If 58 is replaced by 85

Then the new values be in order are : 41 , 43 , 57, 61 , 71 , 85 , 92 , 99 , 127

$∴ New Median= \left ( \frac{n+1}{2} \right )th\; value$

= $\left ( \frac{9+1}{2} \right )th\; value$

= $5th\; value$ = 71

Q12 . The weights (in kg ) of 15 students are : 31 , 35 , 27 , 29 , 32 , 43 , 37 , 41 , 34 , 28 , 36 , 44 , 45 , 42 , 30.Find the median . If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg , find the new median .

SOLUTION :

Given the numbers are  31 , 35 , 27 , 29 , 32 , 43 , 37 , 41 , 34 , 28 , 36 , 44 , 45 , 42 , 30

Arranging the numbers in ascending order

27 , 28 , 29 , 30 , 31 , 32 , 34 , 35 , 36 , 37 , 41 , 42 , 43 , 44 , 45.

n = 15 (odd)

$∴ New Median= \left ( \frac{n+1}{2} \right )th\; value$

= $\left ( \frac{15+1}{2} \right )th\; value$

= $8th\; value$ = 35 kg

If the weight 44 kg is replaced by 46 kg and 27 kg is replaced by 25 kg

Then the new values be in order are : 25 , 28 , 29 , 30 , 31 , 32 , 34 , 35 , 36 , 37 , 41 , 42 , 43 , 45 , 46

$∴ New Median= \left ( \frac{n+1}{2} \right )th\; value$

= $\left ( \frac{15+1}{2} \right )th\; value$

= $8th\; value$ = 35 kg

Q13 . The following observation s have been arranged in ascending order. If the median of the data is 63 , find the value of x : 29 , 32 , 48 , 50 , x , x + 2 , 72 , 78 , 84 , 95.

SOLUTION :

Total number of observation in the given data is 10 (even number) . So median of this data will be mean of $\frac{10}{2}$ i . e, 5th observation and $\frac{10}{2}+1$ i . e , 6th observation.

So , median of data = $\frac{5th\; observation+6th\;observation}{2}$

$\Rightarrow 63=\frac{x+x+2}{2}$

$\Rightarrow 63=\frac{2x+2}{2}$

$\Rightarrow 63=x+1$

$\Rightarrow x=62$

#### Practise This Question

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