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

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 MCQS

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

Mark the correct alternative in each of the following:

Question 1.

Which one of the following is not a measure of central value?

(a) Mean

(b) Range

(c) Median

(d) Mode

Solution:

Range (b)

Question 2.

The mean of n observations is \(\overline { X } \) . If k is added to each observation, then the new mean is

(a) \(\overline { X } \)

(b) \(\overline { X } \) + k

(c) \(\overline { X } \) – k

(d) k\(\overline { X } \)

Solution:

Mean of n observation = \(\overline { X } \)

By adding k to each observation the new mean will be \(\overline { X } \) + k (b)

Question 3.

The mean of n observations is \(\overline { X } \) . If each observation is multiplied by k, the mean of new observations is

(a) k\(\overline { X } \)

(b) \(\frac { \overline { X } }{ k } \)

(c) \(\overline { X } \) + k

(d) \(\overline { X } \) – k

Solution:

Mean of n observations = \(\overline { X } \)

By multiplying each observation by k,

the new mean = k\(\overline { X } \) (a)

Question 4.

The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is

(a) 98

(b) 99

(c) 100

(d) 101

Solution:

Mean of 7 numbers = 81

Total = 7 x 81 = 567

By discarding one number, the mean of the remaining 7 – 1 = 6 numbers = 78

Total = 6 x 78 = 468

Discarded number = 567 – 468 = 99 (b)

Question 5.

For which set of numbers do the mean, median and mode all have the same value?

(a) 2, 2, 2, 2, 4

(b) 1, 3, 3, 3, 5

(c) 1, 1, 2, 5, 6

(d) 1, 1, 1, 2, 5

Solution:

a) In set 2, 2, 2, 2, 4

Mode = 3 as it come in maximum times

This set has mean, median and mode same (b)

Question 6.

For the set of numbers 2, 2, 4, 5 and 12, which of the following statements is true?

(a) Mean = Median

(b) Mean > Mode

(c) Mean < Mode

(d) Mode = Median

Solution:

The given set is 2, 2, 4, 5, 12

Question 7.

If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value of x is

(a) 10

(b) 12

(c) 14

(d) 16

Solution:

Arithmetic mean of 7, 5, 13, x, 9 is 10

Question 8.

If the mean of five observations x, x + 2, x + 4, x + 6, x + 8, is 11, then the mean of first three observations is

(a) 9

(b) 11

(c) 13

(d) none of these

Solution:

Mean = 11

But mean of x, x + 2, x + 4, x+ 6, x + 8

Question 9.

Mode is

(a) least frequent value

(b) middle most value

(c) most frequent value

(d) none of these

Solution:

Mode is most frequent value (c)

Question 10.

The following is the data of wages per day: 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8 The mode of the data is

(a) 7

(b) 5

(c) 8

(d) 10

Solution:

Wages per day

5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8

=> 4, 5, 5, 5, 5, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10

Here 8 comes in maximum times

Mode = 8 (c)

Question 11.

The median of the following data :

is ,

(a) 0

(b) -1.5

(c) 2

(d) 3.5

Solution:

Arranging in ascending order,

-3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5, 6, 6, 6

Question 12.

The algebraic sum of the deviations of a set of n values from their mean is

(a) 0

(b) n – 1

(c) n

(d) n + 1

Solution:

The algebraic sum of deviation of a set of n values from that mean

Question 13.

A, B, C are three sets of values of X:

A : 2, 3, 7, 1, 3, 2, 3

B: 7, 5, 9, 12, 5, 3, 8

C: 4, 4, 11, 7 ,2, 3, 4

Which one of the following statements is

correct?

(a) Mean of A = Mode of C

(b) Mean of C = Median of B

(c) Median of B = Mode of A

(d) Mean, Median and Mode of A are equal.

Solution:

Arranging the sets in ascending order

A{2, 3, 7, 1,3,2,3)

= {1, 2, 2, 3, 3, 3, 7)

B = {7, 5, 9, 12, 5, 3, 8)

= {3, 5, 5, 7, 8, 9, 12)

C = {4, 4, 11,7,2,3,4)

= {2, 3, 4, 4, 4, 7, 11)

Mode = 5 {as it comes max times}

(c) Mean of set C = \(\\ \frac { 2+3+4+4+4+7+11 }{ 7 } \)

= \(\\ \frac { 35 }{ 7 } \) = 5

Median = \(\\ \frac { 7+1 }{ 2 } \) th =\(\\ \frac { 8 }{ 2 } \) =4th term = 4

Mode =4 {as it comes max times}

In set A,mean = median = mode = 3 (d)

Question 14.

The empirical relation between mean, mode and median is

(a) Mode = 3 Median — 2 Mean

(b) Mode 2 Median — 3 Mean

(c) Median 3 Mode — 2 Mean

(d) Mean = 3 Median —2 Mode

Solution:

The empirical relations between mean, mode

and median is

Mode = 3 Median — 2 Mean (a)

Question 15.

The mean of a, b, c, d and e is 28. If the mean of a, c, and e is 24, what is the mean of b and d?

(a) 31

(b) 32

(c) 33

(d) 34

Solution:

Mean of a, b, c, d and e = 28

Total of a, b, c, d and e = 28 x 5 = 140

Mean of a, c and e is = 24

Total of a, c, e = 24 x 3 = 72

Total of b and d = 140 – 72 = 68

Mean = \(\\ \frac { 68 }{ 2 } \) = 34 (d)

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