Free PDF Download of CBSE Class 10 Maths Chapter 14 Statistics Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Statistics MCQs with Answers to know their preparation level.

## Class 10 Maths MCQs Chapter 14 Statistics

1. One of the methods for determining mode is

(a) Mode = 2 Median -3 Mean

(b) Mode = 3 Median – 2 Mean

(c) Mode = 2 Mean – 3 Median

(d) Mode = 3 Mean – 2 Median

**Answer**

Answer: b

2. Mode is the

(a) middle most frequent value

(b) least frequent value

(c) maximum frequent value

(d) none of these

**Answer**

Answer: c

3. The algebraic sum of the deviations of a frequency distribution from its mean is always,

(a) greater than zero

(b) less than zero

(c) zero

(d) a non-zero number

**Answer**

Answer: c

4. While computing mean of grouped data, we assume that the frequencies are

(a) centred at the upper limits of the classes

(b) centred at the lower limits of the classes

(c) centred at the classmarks of the classes

(d) evenly distributed over all the classes

5. Construction of a cumulative frequency table is useful in determining the

(a) mean

(b) median

(c) mode

(d) none of these

**Answer**

Answer: c

6. Which of the following can not be determined graphically?

(a) Mean

(b) Median

(c) Mode

(d) None of these

**Answer**

Answer: a

7. The absccissa of the point of intersection of the 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) None of these

**Answer**

Answer: b

8. For the following distribution

C.I. | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

f | 20 | 30 | 24 | 40 | 18 |

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

(a) 20

(b) 30

(c) 40

(d) 50

**Answer**

Answer: d

9. For the following distribution

C.I. | 0-5 | 6-11 | 12-17 | 18-23 | 24-29 |

f | 26 | 20 | 30 | 16 | 22 |

the upper limit of the median class is

(a) 18.5

(b) 18

(c) 17.5

(d) 17

**Answer**

Answer: c

10. For the following distribution

Marks | No. of students |

Less than 20 | 4 |

Less than 40 | 12 |

Less than 60 | 25 |

Less than 80 | 56 |

Less than 100 | 74 |

Less than 120 | 80 |

the modal class is

(a) 20 – 40

(b) 40 – 60

(c) 60 – 80

(d) 80 -100

**Answer**

Answer: c

11. For the following distribution

Monthly Expenditure (?) | No. of families |

Expenditure les than ? 10,000 | 15 |

Expenditure les than ? 13,000 | 31 |

Expenditure les than ? 16,000 | 50 |

Expenditure les than ? 19,000 | 67 |

Expenditure les than ?22,000 | 85 |

Expenditure les than ?25,000 | 100 |

The number of families having expenditure range (in ?) 16,000 – 19,000 is

(a) 15

(b) 16

(c) 17

(d) 19

**Answer**

Answer: c

12. In the given data:

C.I. | f |

65-85 | 4 |

85 – 105 | 5 |

105 – 125 | 13 |

125 – 145 | 20 |

145 – 165 | 14 |

165 – 185 | 7 |

185 – 205 | 4 |

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

(a) 38

(b) 20

(c) 19

(d) 0

**Answer**

Answer: b

13. For the following distribution

Cl | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 |

f | 10 | 15 | 12 | 20 | 9 |

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

(a) 0

(b) 5

(c) 10

(d) -5

**Answer**

Answer: a

14. For the following distribution

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

No. of students | 3 | 9 | 13 | 10 | 5 |

the number of students who got marks less than 30 is

(a) 13

(b) 25

(c) 10

(d) 12

**Answer**

Answer: b

15. For the following distribution

Marks obtained | No. of students |

More than or equal to 0 | 63 |

More than or equal to 10 | 58 |

More than or equal to 20 | 55 |

More than or equal to 30 | 51 |

More than or equal to 40 | 48 |

More than or equal to 50 | 42 |

the frequency of the class 20-30 is

(a) 35

(b) 4

(c) 48

(d) 51

**Answer**

Answer: b

16. The times, in seconds, taken by 150 atheletes to run a 100 m hurdle race are tabulated below:

C.I. | f |

13.8-14 | 3 |

14 – 14.2 | 4 |

14.2 – 14.4 | 6 |

14.4 – 14.6 | 69 |

14.6 – 14.8 | 48 |

14.8-15 | 20 |

The number of atheletes who completed the race in less than 14.6 seconds is

(a) 13

(b) 69

(c) 82

(d) 130

**Answer**

Answer: c

17. d_{i} is the deviation of x_{i} from assumed mean a. If mean = \(x+\frac{\sum f_{i} d_{i}}{\sum f_{i}}\), then x is

(a) class size ‘

(b) number of observations

(c) assumed mean

(d) none of these

**Answer/Explanation**

Answer: c

Explaination:

(c) ∵ Mean = assumed mean + \(\frac{\sum f_{i} d_{i}}{\sum f_{i}}\)

∴ x = assumed mean.

18. Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is

(a) 48

(b) 49

(c) 50

(d) 60

**Answer/Explanation**

Answer: c

Explaination:

(c) Sum of 100 observations

= 100 × 49 = 4900

Correct sum

= 4900 – [40 + 20 + 50 ] + [60 + 70 + 80] = 5000

∴ Correct mean = \(\frac{5000}{100}\) = 50.

19. Choose the correct answer from the given four options : In the formula

for finding the mean of grouped data d_{1} s are deviation from a of [NCERT Exemplar Problems]

(a) lower limits of the classes

(b) upper limits of the classes

(c) mid points of the classes

(d) frequencies of the class marks

**Answer/Explanation**

Answer: c

Explaination: (c) mid-points of the classes.

20. While computing mean of grouped data, we assume that the frequencies are [NCERT Exemplar Problems]

(a) evenly distributed over all the classes

(b) centred at the classmarks of the classes

(c) centred at the upper limits of the classes

(d) centred at the lower limits of the classes

**Answer/Explanation**

Answer: b

Explaination: (b) centred at the classmarks of the classes.

21. A car travels from city A to city B, 120 km apart at an average speed of 50 km/h. It then makes a return trip at an average speed of 60 km/h. It covers another 120 km distance at an average speed of 40 km/h. The average speed over the entire 360 km will be

**Answer/Explanation**

Answer: b

Explaination:

22. Mean of n numbers x_{1}, x_{2}, … x_{n} is m. If x_{n} is replaced by x, then new mean is

**Answer/Explanation**

Answer: b

Explaination:

23. In the formula \(\bar{x}=a+h\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right)\) finding the mean of grouped frequency distribution, u_{i} = [NCERT Exemplar Problems]

**Answer/Explanation**

Answer: c

Explaination: \(\frac{x_{i}-a}{h}\)

24. The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its [NCERT Exemplar Problems]

(a) mean

(b) median

(c) mode

(d) all the three above

**Answer/Explanation**

Answer: b

Explaination: median

25. For the following distribution:

Marks | Number of Students |

Below 10 | 3 |

Below 20 | 12 |

Below 30 | 27 |

Below 40 | 57 |

Below 50 | 75 |

Below 60 | 80 |

the modal class is

(a) 10 – 20

(b) 20 – 30

(c) 30 – 40

(d) 50 – 60

**Answer**

Answer: c

26. The times, in seconds, taken by 150 atheletes to run a 110 m hurdle race are tabulated below:

Class | Frequency |

13.8 – 14.0 | 2 |

14.0 – 14.2 | 4 |

14.2 – 14.4 | 5 |

14.4 – 14.6 | 71 |

14.6 – 14.8 | 48 |

14.8 – 15.0 | 20 |

The number of atheletes who completed the race in less then 14.6 seconds is: [NCERT Exemplar Problems]

(a) 11

(b) 71

(c) 82

(d) 130

**Answer/Explanation**

Answer: c

Explaination: (c) 82

27. Mode is the value of the variable which has:

(a) maximum frequency

(b) minimum frequency

(c) mean frequency

(d) middle most frequency [CBSE 2012]

**Answer/Explanation**

Answer: a

Explaination: (a) maximum frequency.

28. Mode and mean of a data are 12k and 15A. Median of the data is

(a) 12k

(b) 14k

(c) 15k

(d) 16k

**Answer/Explanation**

Answer: b

Explaination:

(b) ∵ Mode = 3 median – 2 mean

⇒ 12k = 3 median – 2 × 15k

⇒ 42k = 3 median

⇒ Median = 14k.

29. If mean = (3 median – mode) . k, then the value of k is

(a) 1

(b) 2

(C) \(\frac{1}{2}\)

(d) \(\frac{3}{2}\)

**Answer/Explanation**

Answer: c

Explaination:

(c) ∵ Mode = 3 median – 2 mean

⇒ 2 mean = 3 median – mode

⇒ mean = \(\frac{1}{2}\) (3 median – mode)

⇒ k= \(\frac{1}{2}\)

30. The median of set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set

(a) is increased by 2

(b) is decreased by 2

(c) is two times of the original number

(d) Remains the same as that of the original set.

**Answer/Explanation**

Answer: d

Explaination:

(d) No. of observations = 9

∴ median = 5th observation

∵ The largest four observations are increased

∴ 5th observation remains unchanged.

31. The median from the table is

Value | Frequency |

7 | 2 |

8 | 1 |

9 | 4 |

10 | 5 |

11 | 6 |

12 | 1 |

13 | 3 |

(a) 11

(b) 10

(c) 12

(d) 11.5

**Answer/Explanation**

Answer:

Explaination:

32. The relationship between mean, median and mode for a moderately skewed distribution is

(a) mode = median – 2 mean

(b) mode = 3 median – 2 mean

(c) mode = 2 median – 3 mean

(d) mode = median – mean

**Answer/Explanation**

Answer: b

Explaination: (b) Mode = 3 median – 2 mean

33. The abscissa of the point of intersection

of both types (less than & more than) of cumulative frequency curves help in finding

(a) mean

(b) median

(c) mode

(d) None of these

**Answer**

Answer: b

34. Cumulative frequency curve is also called

(a) histogram

(b) ogive

(c) bar graph

(d) median

**Answer**

Answer: b

35. The A.M. of a set of 50 numbers is 38. If two numbers of the set namely 55 and 45 are discarded, the A.M. of the remaining set of numbers is ______ .

**Answer/Explanation**

Answer:

Explaination: 37.5

36. If x is the mean of a distribution, then Σf_{i}(x_{i} – x) is eclual t0 _____ .

**Answer/Explanation**

Answer:

Explaination: (zero)

37. Find the class marks of classes 10 – 25 and 35 – 55.

**Answer/Explanation**

Answer:

Explaination:

38. If Σf_{i} = 11, Σf_{i}x_{i} = 2p + 52 and the mean of any distribution is 6, find the value of p.

**Answer/Explanation**

Answer:

Explaination:

39. A set of numbers consists of four 5’s, six 7’s, ten 9’s, eleven 12’s, three 13’s, two 14’s. The mode of this set of numbers is ____ .

**Answer/Explanation**

Answer:

Explaination: 12

40. The mode of the numbers 2, 3, 4, 4. 3, 5, 3, 6 is ______ .

**Answer/Explanation**

Answer:

Explaination: 3

41. Find the value of x, if the mode of the following data is 25.

15, 20, 25, 18, 14, 15, 25, 15, 18, 16, 20. 25, 20, x, 18

**Answer/Explanation**

Answer:

Explaination:

Observations | Tally Marks | f |

14 | 1 | 1 |

15 | III | 3 |

16 | 1 | 1 |

18 | III | 3 |

20 | III | 3 |

25 | III | 3 |

Observations 15,18,20, and 25 all have the same frequency i.e., 3. For 25 be the mode of the data it should have the maximum frequency.

∴ 25 should repeat itself at least once more i.e., x should be 25.

42. If the mode of a data is 18 and the mean is 24. then median is _____ .

**Answer/Explanation**

Answer:

Explaination: Median = \(\frac{66}{3}\) = 22.

43. Construction of cumulative frequency table is useful in determining the

**Answer/Explanation**

Answer:

Explaination: Median

44. In an arranged series of an even number of 2n terms the median is _______.

**Answer/Explanation**

Answer:

Explaination: mean of n^{th} term and (n + 1)th term

45.

x |
f |
c.f. |

1 | 8 | 8 |

2 | 10 | 18 |

3 | 11 | 29 |

4 | 16 | 45 |

5 | 20 | 65 |

6 | 25 | 90 |

7 | 15 | 105 |

8 | 9 | 114 |

9 | 6 | 120 |

Find the median distribution.

**Answer/Explanation**

Answer:

Explaination:

n = 120

⇒ \(\frac{n}{2}\) = 60

∴ Median is average of 60th and 61st observation

⇒ Median =\(\frac{5+5}{2}\) = 5

46.

Marks | Number of students |

0 and above | 80 |

10 and above | 77 |

20 and above | 72 |

30 and above | 65 |

40 and above | 55 |

50 and above | 43 |

60 and above | 28 |

70 and above | 16 |

80 and above | 10 |

90 and above | 8 |

100 and above | 0 |

Write the proper data (marks) and frequency (Number of students).

**Answer/Explanation**

Answer:

Explaination:

Marks | Frequency |

0 – 10 | 3 |

10 – 20 | 5 |

20 – 30 | 7 |

30 – 40 | 10 |

40 – 50 | 12 |

50 – 60 | 15 |

60 – 70 | 12 |

70 – 80 | 6 |

80 – 90 | 2 |

90 – 100 | 8 |

47. Find the median class of the following data:

Marks obtained | Frequency |

0-10 | 8 |

10-20 | 10 |

20-30 | 12 |

30-40 | 22 |

40-50 | 30 |

50-60 | 18 |

**Answer/Explanation**

Answer:

Explaination:

48. There are ____ types of cumulative frequency curve.

**Answer/Explanation**

Answer:

Explaination: two

49. Which measure of central tendency is given by the x-coordinate of the point of intersection of the “more than ogive” and “less than ogive”?

**Answer/Explanation**

Answer:

Explaination: Median

50. In drawing ogive, cumulative frequencies are marked on _____ axis.

**Answer/Explanation**

Answer:

Explaination: y

51. Which measure of central tendency is given by the x-coordinate of the point of intersection of “more than ogive”.and “less than ogive”?

**Answer/Explanation**

Answer:

Explaination: Median

We hope the given MCQ Questions for Class 10 Maths Statistics with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 14 Statistics Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.