## RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS

Other Exercises

- RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1
- RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.2
- RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3
- RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4
- RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS
- RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency MCQS

Question 1.

If the ratio of mode and median of a certain data is 6 : 5, then find the ratio of its mean and median.

Solution:

We know that

Mode = 3 median – 2 mean…(i)

and \(\frac { mode }{ median } \) = \(\frac { 6 }{ 5 } \)

Mode = \(\frac { 6 }{ 5 } \)median

∴From (i), \(\frac { 6 }{ 5 } \) median = 3 median – 2 mean

=> 2 mean = 3 median – \(\frac { 6 }{ 5 } \)median

2 mean = \(\frac { 15-6 }{ 5 } \)median = \(\frac { 9 }{ 5 } \)median

\(\frac { mean }{ median } \) = \(\frac { 9 }{ 5X2 } \) = \(\frac { 9 }{ 10 } \)

∴Ratio = 9:10

Question 2.

If the mean of x + 2, 2x + 3, 3x + 4, 4x + 5 is x + 2, find x.

Solution:

Mean of x + 2, 2x + 3, 3x + 4, 4x + 5 = x + 2

=> \(\frac { x + 2+2x + 3+3x + 4+4x + 5 }{ 4 } \) = x + 2

=> 10x + 14 = 4x + 8

=> 10x – 4x = 8 – 14

=> 6x= – 6

∴ x = – 1

Question 3.

If the median of scores ,\(\frac { x }{ 2 } \), \(\frac { x }{ 3 } \), \(\frac { x }{ 4 } \), \(\frac { x }{ 5} \) and \(\frac { x }{ 6 } \) (where x > 0) is 6, then find the value \(\frac { x }{ 6 } \)

Solution:

\(\frac { x }{ 2 } \), \(\frac { x }{ 3 } \), \(\frac { x }{ 4 } \), \(\frac { x }{ 5} \), \(\frac { x }{ 6 } \)

Here n = 5

Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 5+1 }{ 2 } \) th

\(\frac { 6 }{ 2 } \) = 3rd term = \(\frac { x }{ 4 } \)

\(\frac { x }{ 4 } \) = 6 => x = 24

\(\frac { x }{ 6 } \) = \(\frac { 24 }{ 6 } \) = 4

∴Hence = \(\frac { x }{ 6 } \) = 4

Question 4.

If the mean of 2, 4, 6, 8, x, y is 5, then find the value of x + y.

Solution:

Mean of 2, 4, 6, 8, x, y is 5

\(\frac { 2+4+6+8+x+y }{ 6 } \) = 5

\(\frac { 20+x+y }{ 6 } \) = 5

=> 20 + (x +y) = 30

=> x + y = 30 – 20 = 10

∴x + y = 10

Question 5.

If the mode of scores 3, 4, 3, 5, 4, 6, 6, x is 4, find the value of x.

Solution:

Mode of 3, 4, 3, 5, 4, 6, 6, x is 4

∴ 4 comes in maximum times

But here ,

3 2

4 2

5 1

6 2

3, 4 and 6 are equal in number

∴ x must be 4 so that it becomes in maximum times

Question 6.

If the median of 33, 28, 20. 25, 34, x is 29. find the maximum possible value of x.

Solution:

Median of 33, 28, 20, 25, 34, x is 29

Now arranging in ascending order 20, 25, 28, x, 33, 34

Here n = 6

Median = \(\frac { 1 }{ 2 } \left[ \frac { 6 }{ 2 } th\quad term+\left( \frac { 6 }{ 2 } +1 \right) th\quad term \right] \)

29 = \(\frac { 1 }{ 2 } \) [3rd term + 4th term]

29 = \(\frac { 1 }{ 2 } \) [28+x]

58 = 28 + x

=> x = 58 – 28 = 30

∴Possible value of x = 30

Question 7.

If the median of the scores 1, 2, x, 4, 5 (where 1 <2 <x <4 <5) is 3, then find the mean of the scores.

Solution:

Scores are 1, 2, x, 4, 5 and median 3

Here n = 5 which is odd

Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 5+1 }{ 2 } \) = \(\frac { 6 }{ 2 } \) th

=> 3 = 3rd term = x

=> 3 = x

∴ x = 3

Mean of the score = \(\frac { 1+2+3+4+5 }{ 5 } \) = 3

Question 8.

If the ratio of mean and median of a certain data is 2 : 3, then find the ratio of its mode and mean.

Solution:

We know that mode = 3 median – 2 mean

\(\frac { mode }{ mean } \) = \(\frac { 5 }{ 2 } \)

Ratio in mode and mean = 5 : 2

Question 9.

The arithmetic mean and mode of a data are 24 and 12 respectively, then find the median of the data.

Solution:

Mean = 24

Mode = 12

We know that mode = 3 median – 2 mean

12 = 3 median – 2 x 24

12 = 3 median – 48

3 median 12 + 48 = 60

Median = \(\frac { 60 }{ 3 } \) = 20

Question 10.

If the difference of mode and median of a data is 24, then find the difference of median and mean.

Solution:

Mode – Median = 24

Mode = 24 + median

But mode = 3 median – 2 mean

3 median – 2 mean = 24 + median

3 median – median – 2 mean = 24

=> 2 median – 2 mean = 24

=> Median – Mean = 12 (Dividing by 2)

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS are helpful to complete your math homework.

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