# R.D. Sharma Solutions Class 10th: Ch 7 Statistics MCQ's

## Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math MCQ's

**Multiple Choice Questions**

1. Which of the following is not a measure of central tendency ?

(a) Median

(b) Median

(c) Mode

(d) Standard deviation

**Solution**

Hence , the correct option is c

2. The algebraic sum of the deviations of a frequency distribution from its mean is

(a) Always positive

(b) Always Negative

(c) 0

(d) A non – zero number

**Solution**

The algebraic sum of the deviation of a frequency distribution from its mean is zero .

Hence , the correct option is c.

(a) n+1/2

(b) n-1/2

(c) n/2

(d) n/2 + 1

**Solution**

= 1+2+3+…+n/n

= n(n+1)/2/n

= n+1/2

Hence , the correct option is a .

(a) Mode = 3 Mean – 2 Median

(b) Mode = 2 median – 3 Mean

(c) Mode = 3 Median – 2 Mean

(d) Mode = 3 Median + 2 Mean

**Solution**

The relation between mean, median and mode is

Mode = 3 Median – 2 Mean

Mode = 3 Median – 2 Mean

Hence, the correct option is c .

5. Which of the following cannot be determined graphically ?

(a) Mean

(b) Median

(c) Mode

(d) None of these

**Solution**

‘Mean’ cannot be determined by graphically.

Hence, the correct option is a .

6. The median of a given frequency distribution is found graphically with the help of

(b) Frequency curve

(c) Frequency Polygon

(d) Ogive

**Solution**

The median of a given frequency distribution is found graphically with the help of ‘ogive’ .

Hence , the correct option is d .

7. The mode of a frequency distribution can be determined graphically from

(a) Histogram

(b) Frequency polygon

(c) Ogive

(d) Frequency curve

The mode of a frequency distribution can be determined from ‘’Histogram’’ .

Hence, the correct option is a.

(a) Histogram

(b) Frequency polygon

(c) Ogive

(d) Frequency curve

**Solution**The mode of a frequency distribution can be determined from ‘’Histogram’’ .

Hence, the correct option is a.

8. Mode is

(a) Least frequency value

(b) Middle most value

(c) Most frequent value

(d) None of these

**Solution**

Mode is ‘’ Most frequent value’’ .

Hence , the correct option is c .

9. The mean of n observation is x̅ . If the first item is increased by 1, second by 2 and so on, then the new mean is

(a) x̅ + n(b) x̅ + n/2

(c) x̅ + n+1/2

(d) None of these

**Solution**

Let x

_{1}, x

_{2}, x

_{3}, …..,x

_{n}be the n observations .

Mean = x̅ = x

_{1}+x

_{2}+…+x

_{n}/n

⇒ x

_{1}+ x

_{2}+x

_{3}+ …+ x

_{n}= n x̅

If the first item is increased by 1, second by 2 and so on .

Then, the new observations are x

_{1}+ x

_{2}+ 2 , x

_{3}+ 3, …, x

_{n}+ n.

New mean = (x

_{1}+ 1) + (x

_{2}+2) + (x

_{3}+ 3) + … +(x

_{n}+n)/n

= x

_{1}+ x

_{2}+ x

_{3}+ … + x

_{n}+ (1+2+3+…+n)

= n x̅ + (n(n+1)/2)/n

= x̅ + n+1/2

Hence , the correct answer is c .

(a) Mode = 2 median – 3 mean

(b) Mode = 2 median + 3 mean

(c) Mode = 3 median – 2 mean

(d) Mode = 3 median + 2 mean

**Solution**

We have ,

Mode = 3 median – 2 mean

Hence , the correct option is c .

Variable (x) : 1 2 3 4 5

Frequency 4 5 y 1 2

**Solution**

Mean = Î£fx/ Î£f

2.6/1 = 28 + 3y/12 + y

31.2 + 2.6y = 28 + 3y

0.4y = 3.2

y = 3.2/0.4

y = 8

Hence, the correct option is b .

(a) Mode = 2 median – 3 mean

(b) Mode = median – 2 mean

(c) Mode = 2 median – mean

(d) Mode = 3 median – 2 mean

**Solution**

Mode = 3 median - 2 mean

Hence, the correct option is d .

_{i}/ f

_{i}, i = 1,2,…,n is given by

**Solution**

x̅ = Î£f

_{i}x

_{i}/Î£f

_{i}

Hence, the correct option is a .

(a) 1

(b) 2

(c) 6

(d) 4

**Solution**

The given observations are x, x+3, x+6, x+9, and x +12.

∴ Î£ x = 5x + 30, n = 5, x̅ = 10

Now,

x̅ = Î£x/n

10 = 5x + 30/5

50 = 5x + 30

5x = 20

x = 4

Hence, the correct option is d.

(a) 27

(b) 25

(c) 28

(d) 30

**Solution**

The given observations are 24, 25, 26, x + 2, x + 3, 30, 31, 34

Median = 27.5

Here, n = 8

Median = (n/2)

^{th}term + (n/2 + 1)

^{th}term / 2

27.5 = 4th term + 5 th term/2

27.5 = (x+2) + (x+3) / 2

27.5 = 2x + 5 / 2

2x + 5 = 55

2x = 50

x = 25

Hence , the correct option is b .

16. If the median of the data : 6,7,x-2, x,17,20 written ascending order is, 16 . Then x =

(a) 15

(b) 16

(c) 17

(d) 18

**Solution**

The given observations arranged in ascending order are

6,7,x-2,x,17,20

n = 6(even), median = 16

Median = (n/2)

^{th}term + (n/2 +1)

^{th}term/2

= 3rd term + 4th term/2

= x – 2 + x /2

= 2x – 2 / 2

= 16 = 2x – 2 / 2

= 2x – 2 = 32

x = 17

Hence the correct option is c.

17. The median of first 10 prime numbers is

(a) 11

(b) 12

(c) 13

(d) 14

**Solution**

First 10 prime numbers are 2 ,3 , 5 ,7 , 11 , 13 , 17, 19 , 23, 29 .

n = 10 (even)

Median = (n/2)th term + (n/2+1)th term

= 5th term + 6th term / 2

= 11 + 13 / 2

= 24/2

= 12

Hence, the correct option is b.

18. If the mode of the data : 64,60,48, x, 43, 34 is 43, then x + 3

(a) 44(b) 45

(c) 46

(d) 48

**Solution**

Now, this is possible only when x = 43. In this case, the frequency of the observation 43 would be 3.

Hence,

x + 3 = 46

Hence, the correct option is c .

19. If the mode of the data : 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =

(a) 15

(b) 16

(c) 17

(d) 19

**Solution**

This is possible only when x = 15. In this case, the frequency of 15 would be 3.

Hence , the correct answer is a .

20. The mean of 1,3,4,5,7,4 is m. The numbers 3,2,2,4,3,3,p have mean m – 1 and median q. Then, p+q =

(a) 4

(b) 5

(c) 6

(d) 7

(b) 5

(c) 6

(d) 7

**Solution**
1,3,4,5,7,4

Mean = 1+3+4+5+7+4/6

= 24/6

= 4

Consider the numbers 3,2,2,4,3,3,p.

Mean = 3+2+2+4+3+3+p / 7.

⇒ 7× (4 – 1) = 17 + p

⇒ 21 = 17 + p

⇒ p = 4

Arranging the numbers 3,2,2,4,3,3,4 in ascending order, we have

2,2,3,3,3,4,4

Median = (n+1/2)

q = (7+1/2)

= (8/2)

= 4

q = 3

so,

p+q = 4 + 3

= 7

Hence , the correct option is d.

21. If the mean of frequency distribution is 8.1 and Î£f

Mean = 1+3+4+5+7+4/6

= 24/6

= 4

Consider the numbers 3,2,2,4,3,3,p.

Mean = 3+2+2+4+3+3+p / 7.

⇒ 7× (4 – 1) = 17 + p

⇒ 21 = 17 + p

⇒ p = 4

Arranging the numbers 3,2,2,4,3,3,4 in ascending order, we have

2,2,3,3,3,4,4

Median = (n+1/2)

^{th}termq = (7+1/2)

^{th}term= (8/2)

^{th}term= 4

^{th}termq = 3

so,

p+q = 4 + 3

= 7

Hence , the correct option is d.

21. If the mean of frequency distribution is 8.1 and Î£f

_{i}x_{i}= 132 + 5k, Î£f_{i}= 20, then k =
(a) 3

(b) 4

(c) 5

(d) 6

Given:

Î£f

Then,

Mean = Î£f

8.1 = 132 + 5k / 20

162 = 132 + 5k

5k = 30

k = 6

Hence , the correct option is d .

22. If the mean of 6,7,x,8,y,14 is 9, then

(a) x+y = 21

(b) x+y = 19

(c) x-y = 19

(d) x-y = 21

The given observation are 6,7,x,8,y,14.

Mean = 9 (Given)

⇒ 6+7+x+8+y+14 / 6 = 9

⇒ 35 + x + y = 54

⇒ x + y = 54 – 35 = 19

Hence , the correct option is b .

(c) 5

(d) 6

**Solution**Given:

Î£f

_{i}x_{i}= 132 + 5k, Î£f_{i}= 20 and mean = 8.1Then,

Mean = Î£f

_{i}x_{i}/ Î£f_{i}8.1 = 132 + 5k / 20

162 = 132 + 5k

5k = 30

k = 6

Hence , the correct option is d .

22. If the mean of 6,7,x,8,y,14 is 9, then

(a) x+y = 21

(b) x+y = 19

(c) x-y = 19

(d) x-y = 21

**Solution**The given observation are 6,7,x,8,y,14.

Mean = 9 (Given)

⇒ 6+7+x+8+y+14 / 6 = 9

⇒ 35 + x + y = 54

⇒ x + y = 54 – 35 = 19

Hence , the correct option is b .

23.

**Solution**

24. If the mean of first n natural numbers is 5n/9 then n =

(a) 5

(b) 4

(c) 9

(d) 10

**Solution**

25. The arithmetic mean and mode of a data are 24 and 12 respectively , then its median is

(a) 25

(b) 18

(c) 20

(d) 22

**Solution**

Given :

Mean = 24 and Mode = 12

We know that

Mode = 3 Median – 2 Mean

⇒ 12 = 3 median – 2 × 24

⇒ 3 median = 12 + 48 = 60

⇒ Median = 20

26. The mean of first n odd natural number is

(a) n+1/2

(b) n/2

(c) n

(d) n

^{2}

**Solution**

27. The mean of first n odd natural numbers is n

^{2}/81, then n =

(a) 9

(b) 81

(c) 27

(d) 18

**Solution**

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

(a) 12

(b) 24

(c) 8

(d) 36

**Solution**

Given :

Mode – Median = 24

We know that

Mode = 3 median – 2 mean

Now,

Mode – median = 2(Median – Mean)

⇒ 24 = 2(Median – Mean)

⇒ Median – Mean = 12

29. If the arithmetic mean , 7,8,x,11,14 is x, then x =

(a) 9

(b) 9.5

(c) 10

(d) 10.5

**Solution**

The given observations are 7,8,x,11,14 .

Mean = x (Given)

Now,

Mean = 7+8+x+11+14 / 5

⇒ x = 40+x / 5

⇒ 5x = 40 + x

⇒ 4x = 40

⇒ x = 10

30. If the mode of a series exceeds its mean by 12, then mode exceeds the median by

(a) 4

(b) 8

(c) 6

(d) 10

**Solution**

Given: Mode – Mean = 12

We know that

Mode = 3 median – 2 Mean

∴ Mode – Mean = 3(Median – Mean)

⇒ 12 = 3(Median – Mean)

⇒ Median – Mean = 4 …(1)

Again ,

Mode = 3 Median – 2 Mean

⇒ 2 Mode = 6 Median – 4 Mean

⇒ Mode – Mean + Mode = 6 Median – 5 Mean

⇒ 12 + (Mode – Median) = 5(Median – Mean)

⇒ 12 + (Mode – Median) = 20 [Using (1)]

⇒ Mode – Median = 20 – 12 = 8

(a) 15

(b) 30

(c) 14

(d) 29

**Solution**

32.

**Solution**

**Solution**

34.

(a) 23

(b) 24(c) 27

(d) 25

**Solution**

(a) 2

(b) 1.5

(c) 1

(d) 0.5

**Solution**

(a) Evenly distributed over all the classes.

(b) Centred at the class marks of the classes.

(c) Centred at the upper limit of the classes .

(d) Centred at the lower limit of the classes .

**Solution**

We know that while computing the mean of a grouped data, the frequencies are centered at the class marks of the classes .

37.

**Solution**

38. For the following distributions :

Class : 0-5 5-10 10-15 15-20 20-25

Frequency : 10 15 12 20 9

The sum of the lower limits of the median and modal class is

(a) 15

(b) 25

(c) 30

(d) 35

**Solution**

Here, N = 66.

∴ N/2 = 33, which lies in the interval 10-15.

So, the lower limit of the median class is 10.The highest frequency is 20, which lies in the interval 15-20.

Therefore, the lower limit of modal class is 15.

So, the required sum is 10 +15 = 25 .

Below: 10 20 30 40 50 60

Number of students : 3 12 27 57 75 80

the modal class is

(a) 10-20

(b) 20-30

(c) 30-40

(d) 50-60

**Solution**

Here, N = 80

∴ N/2 = 40, which lies in the interval 30-40.Therefore, the modal class is 30-40.

40. Consider the following frequency distribution :

Class: 65-85 85-105 105-125 125-145 145-165 165-185 185-205Frequency : 4 5 13 20 14 7 4

The difference of the upper limit of the median class and the lower limit of the modal class is :

**Solution**

∴ N/2 = 33.5, which lies in the interval 125-145.

Therefore, the lower limit of the median class is 125.

The highest frequency is 20, which lies in the interval 125-145.

Therefore, the upper limit of modal class is 145.

So, the required difference is 145-125 = 20.

41.

**Solution**
The given formula represents the formula to find the mean by assumed mean method .

Here, d_{i}= x

_{i}– a where xi is the ith observation and a is assumed mean .

So, d

_{i}’s are the deviation from a of mid – points of the classes .

42. The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data gives its

(a) Mean

(b) Median

(c) Mode

(d) All the three above

**Solution**

The less than ogive and more than ogive when drawn on the same graph intersect at a point . From this point, if we draw a perpendicular on the x-axis, the point at which it cuts the x-axis gives us the median.

Thus, the abscissa of the point if intersection of less than type and of the more than type cumulative curves of a grouped data gives its median.

43. Consider the following frequency distribution :

Class: 0-5 6-11 12-17 18-23 24-29

Frequency: 13 10 15 8 11

The upper limit of the median class is

(a) 17

(b) 17.5

(c) 18

(d) 18.5

Class: 0-5 6-11 12-17 18-23 24-29

Frequency: 13 10 15 8 11

The upper limit of the median class is

(a) 17

(b) 17.5

(c) 18

(d) 18.5

**Solution**
The given classes in the table are non-continuous. So , we first make the classes continuous by adding 0.5 to the upper limit and substracting 0.5 from the lower limit in each class .

Now, from the table we see that N = 57.

So, N/2 = 57/2 = 28.5

28.5 lies in the class 11.5 – 17.5

The upper limit of the interval 11.5-17.5 is 17.5