Statistics Class 10 MCQ Important Maths

Statistics Class 10 MCQ with Solution

1. The weight (in kg) of 5 men are 62, 65, 69, 66 and 61. The median is :

(A) 45 kg

(B) 66 kg

(D) 55 kg

Ans. (C)

Sol. Arrange in ascending order

Hence no. of item = 5

Hence median is ${\left( {\frac{{5 + 1}}{2}} \right)^{{\bf{th}}}}$ term

= 3^rd term

∴ 65 is the median

2. The mean of x, x +3, x + 6, x + 9 and x + 12 is :

(A) x + 6

(B) x + 3

(D) x + 12

Ans. (A)

Sol. Mean = $\frac{{x + (x + 3) + (x + 6) + (x + 9) + (x + 12)}}{5}$

= $\frac{{5x + 30}}{5} = \frac{{5(x + 6)}}{5}$ = x + 6

Hence (A)

3. The mean of a data is ‘p’. If each observation is multiplied by 3 and then 1 is added to each result, then the mean of the new observations so obtained is :

(A) p

(B) 3p

(D) 3p+1

Ans. (D)

Sol. If all the observation are multiplied by same number and added with same number. Mean gets changed by same.

i.e. new mean is 3p + 1

4. The mean of 20 observations is 12.5. By error, one observation was noted as – 15 instead of 15. Then the correct mean is:

(A) 11.75

(B) 11

(D) None of these

Ans. (C)

Sol. New mean = $\frac{{20 \times 12.5 - ( - 15) + 15}}{{20}} = \frac{{280}}{{20}} = 14$

5. The average age of 5 teachers is 28 years. If one teacher is excluded the mean gets reduced by 2 years. The age of the excluded teacher is :

(A) 26 years

(B) 33 years

(D) None of these

Ans. (C)

Sol. 26 = $\frac{{140 - x}}{4}$ ⇒ x = 36 year

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

(A) Mean

(B) Median

(D) Standard deviation

Ans. (D)

Sol. Standard deviation

7. The true statement for the data : 1, –1, 0, 2, 3, 5, 5, 6, 8, 10 and 11 is :

(A) Mean = Mode = Median

(B) Mode = Median=5

(D) Mean = Mode = 5

Ans. (B)

Sol. –1, 0, 1, 2, 3, 5, 5, 6, 8, 10, 11

Mode is 5

Median = 5

Mean = $\frac{{{\bf{Sum}}}}{{{\bf{11}}}} = \frac{{50}}{{11}} < 5$

8. The mean weight of a class of 34 students is 46.5 kg. If the weight of the teacher is included, the mean rises by 500g. Then the weight of the teacher is :

(A) 175 kg

(B) 62 kg

(D) 72 kg

Ans. (C)

Sol. Here weight of teacher = 46.5 + 500 g × 35

= 46.5 + 17.5 = 64.0 kg

i.e. 500 g is distributed over 35 people now (including teacher).

Hence (C)

9. If in a data, 10 numbers arranged in increasing order. If the 7^th entry is increased by 4, then the median increases by

(A) zero

(B) 4

(D) 5

Ans. (A)

Sol. Since median is here 5^th& 6^th data / observation.

Hence it is not affected by 7^th entry.

10. 20 years ago, when my parents got married, their average age was 23 years, now the average age of my family, consisting of myself and my parents only is 35 years. My present age is

(A) 34 years

(B) 42 years

(D) 19 years

Ans. (D)

Sol. Now total age of parents = 23 × 2 + 20 × 2 = 46 + 40 = 86 years

⇒ 35 = $\frac{{86 + x}}{3}$ ⇒ x = 19

11. The median of the series –5, 11, 10, –3, 5, 5, 8, –8, 3, –10 is :

(A) 2

(B) 3

(D) 5

Ans. (C)

Sol. Median of –10, –8, –5, –3, 3, 5, 5, 8, 10, 11

Median = $\left( {\frac{{{5^{th}} + {6^{th}}}}{2}} \right)$ = $\frac{{3 + 5}}{2} = 4$

12. Histogram is useful to determine graphically the value of :

(A) Arithmetic Mean

(B) Median

(D) None of these

Ans. (C)

Sol. Histrogram is used to find mode.

13. A histogram is a n-dimensional diagram where n is equal to :

(A) 0

(B) 1

(D) 3

Ans. (C)

Sol. A histogram is a 2-dimensional diagram.

14. A student got marks in 5 subjects in a monthly test is given as 2, 3, 4, 5, 6. In these obtained marks, 4 is the:

(A) Mean and median

(B) Mean but no median

(D) Mode

Ans. (A)

Sol. 2, 3, 4, 5, 6

Mean = $\frac{{20}}{5} = 4$

Median = 4

Mean = Median = 4.

15. In an examination, 10 students scored the following marks in Mathematics 35, 19, 28, 32, 63, 02, 47, 31, 13, 98 Its range is :

(A) 2

(B) 96

(D) 50

Ans. (B)

Sol. 02, 13, 19, 28, 31, 32, 35, 47, 63, 98

Range = Upper limit – lower limit

= 98 – 2 = 96

16. In a frequency distribution, the mid value of a class is 15 and the class interval is 4. The lower limit of the class is :

(A) 10

(B) 12

(D) 14

Ans. (C)

Sol. Mid value of a class is 15 and class interval is 4

So Range is 13 – 17

So, 13 will be lower limit and 17 is upper limit.

17. The mid value of a class interval is 42. If the class size is 10, then the upper and lower limits of the class are :

(A) 47 & 37

(B) 37 & 47

(D) 47.5 & 37.5

Ans. (A)

Sol. Mid vlaue is 42 and class size is 10.

So, class range should be 37 – 47

So lower and upper limit is 37 and 47.

18. The following marks were obtained by the students in a test :

81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62,

The range of the marks is :

(A) 9

(B) 17

(D) 33

Ans. (D)

Sol. Range = upperlimit – lower limit

= 95 – 62 = 33.

19. Find the mode of the following data :

8, 12, 9, 18, 11, 15, 12, 10, 16, 12, 15

(A) 12

(B) 15

(D) 8

Ans. (A)

Sol. Mode of the 8, 12, 9, 18, 11, 15, 12, 10, 16, 15, 12

Mode = 12 which occur 3 times.

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

(A) Histogram

(B) Frequency curve

(D) Ogive

Ans. (D)

Sol. Median can be find by ogive.

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

(A) Histogram

(B) Frequency polygon

(D) Frequency curve

Ans. (A)

Sol. Histogram is used to find mode.

22. The mean of a data is ‘p’. If each observation is multiplied by 4 and then 3 is added to each result, then the mean of the new observations so obtained is :

(A) 4p

(B) 4p + 1

(D) 3p + 1

Ans. (C)

Sol. Mean is P

It each observation is multipilled by 4 & then 3 is added to each result then the new mean is (4p + 3).

23. Which of the following is true :

(A) Mode = 2 Median – 3 Mean

(B) Mean = Mode + $\frac{2}{3}\,(Median\, - Mode)$

(D) Median = Mode + $\frac{2}{3}\,(Mean\, - Mode)$

Ans. (D)

Sol. Mode = 3 median – 2 mean (it is an emprical relation).

24. A set of numbers consists of three 4s, five 5s, six 6s, eight 8s nine 9s and seven 10s. The mode of this set of numbers is :

(A) 6

(B) 9

(D) 10

Ans. (B)

Sol. Mode = 9

as 9 occurs 9 times which is maximum.

25. The mean of x₁, x₂, ………, x₅₀ is M if each x_i where i = 1, 2, ………….., 50 is replaced by x_i/50 then new mean is :

(A) M

(B) M + $\frac{1}{{50}}$

(D) $\frac{M}{{50}}$

Ans. (D)

Sol. Mean of x₁, x₂ ……. x₅₀is M.

If each no.is divided by 50 then the new mean will be $\frac{M}{{50}}$

26. The average income of Sambhu and Ganesh is Rs. 3000 and that of Arun and Vinay is Rs. 500. What is the average income of Sambhu, Ganesh, Arun and Vinay?

(A) Rs. 1750

(B) Rs. 1850

(D) Rs. 2500

Ans. (A)

Sol. Sum of Income of Sambhu & Ganesh = 2 × 300 = Rs. 6000

Sum of Income of Arun & Vinay = 2 × 500 = Rs. 1000

Average Income of all = $\frac{{7000}}{4}$ = Rs. 1750

27. If the arithmetic mean of n numbers of a series is $\bar x$ and the sum of first (n – 1) numbers is K, then the n^th number is :

(A) n + k

(B) n $\bar x$ + k

(D) n – k

Ans. (C)

Sol. Sum of the n number = n $\bar x$

Sum of the first (n – 1) number = k

then the n^thnumber is = (n $\bar x$ – k)

28. The numbers 3, 5, 6, and 4 have frequencies of x, x + 2, x – 8 and x + 6 respectively. If their mean is 4 then find x :

(A) 5

(B) 6

(D) 8

Ans. (C)

Sol. The no.s are 3, 5, 6 & 4 frequencies of x, x + 2, x – 8 & x + 6

Mean = $\frac{{3x + 5(x + 2) + 6(x - 8) + 4(x + 6)}}{{x + x + 2 + x - 8 + x + 6}}$

= $\frac{{3x + 5x + 10 + 6x - 48 + 4x + 24}}{{4x}}$

Mean = 4 = $\frac{{18x-14}}{{4x}}$ ⇒ 2x = 14

x = 7

29. The mean of weight of 100 persons is 46 kg. The mean of weight of males being 50 kg and of females being 40 kg then the number of males is :

(A) 50

(B) 60

(D) 65

Ans. (B)

Sol. Let number of Boys = x and girls = (100 – x)

Total sum of weight = 100 × 46 = 4600 kg

Sum of weight of Boys = 50x

Sum of weight of girls = 40(100 – x)

4600 = 50x + (4000 – 40x) ⇒ 600 = 10x ⇒ x = 60

No. of Males = 60

30. The mean weight of a class of 34 students is 47 kg. If the weight of the teacher is included, the mean rises by 1 kg. Then the weight of the teacher is :

(A) 82 kg

(B) 62 kg

(D) 72 kg

Ans. (A)

Sol. Total weight of a class = 34 × 47 = 1598 kg

Total weight of a class including teacher = 35 × 48 = 1680 kg

Weight of teacher = (1680 – 1598) = 82 kg

31. The average age of 5 teachers is 30 years. If one teacher is excluded the mean gets reduced by 4 years. The age of the excluded teacher is :

(A) 26 years

(B) 33 years

(D) 46 years

Ans. (D)

Sol. 5 × 30 = 150 = sum of ages of five teachers

4 × 26 = 104 = sum of ages of four teachers

Age of the excluded teacher

= (150 – 104) = 46 years.

32. If the mean and median of a set of numbers are 8.7 and 8.8 respectively, then the mode will be :

(A) 8.2

(B) 9

(D) Can’t be determined

Ans. (B)

Sol. Mode = 3 median – 2 mean

= 3 × 8.8 – 2 × 8.7

= 26.4 – 17.4

Mode = 9

33. The median of a set of 11 distinct observations is 21.5. If each of the largest 5 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

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

Ans. (D)

Sol. In odd number like 11 number median is $\left( {\frac{{11 + 1}}{2}} \right)$ = 6^th term so median will not get change in changing the largest observation.

34. The median of the following incomplete frequency distribution is 4 :

The frequency of 8 is :

(A) 1

(B) 2

(D) 4

Ans. (A)

Sol.

as media is 4 so its CF = 10

as so $\frac{{18 + F + 1}}{2}$ = 10

19 + F = 20

F = 20 – 19

Value of F = 1

35. The average value of the median of 2, 8, 3, 7, 4, 6, 7 and the mode of 2, 9, 3, 4, 9, 6, 9 is :

(A) 9

(B) 8

(D) 6

Ans. (C)

Sol. Median of 2 3 4 6 7 7 8 is 6 and mode of 2, 9, 3, 4, 9, 6, 9 is 9.

So ⇒ $\frac{{9 + 6}}{2}$ = $\frac{{15}}{2}$ = 7.5

36. A, B, C are three sets of value 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

Select the correct statement from among the following :

(A) Mean of A is equal to mode of C.

(B) Mean of C is equal to median of B.

(D) Mean, median and mode of A are same.

Ans. (D)

Sol. A → 1 2 2 3 3 3 7

B → 3 5 5 7 8 9 12

C → 2 3 4 4 4 7 11

Mean of A is 3, Mode is 3 and Median is 3

So, mean = mode = median = 3.

37. The mean of 20 observations is 15. By error, one observation was noted as – 15 instead of 15. Then the correct mean is :

(A) 11.75

(B) 16.5

(D) 13.5

Ans.. (B)

Sol. No. of observations = 20

Mean = 15

⇒ Sum = 15 × 20

New mean = $\frac{{15 \times 20 + 15 + 15}}{{20}}$ = $\frac{{300 + 30}}{{20}} = \frac{{330}}{{20}} = 16.5$

38. Find the range of the following frequency distribution :

(A) 20

(B) 21

(D) 19.5

Ans. (A)

Sol. The range is the difference between the mid value of the least class-interval and the greatest class interval.

Mid value of least class interval = $\frac{{0 + 5}}{2} = 2.5$

Mid value of greatest class interval = $\frac{{20 + 25}}{2} = 22.5$

∴ Range = 22.5 – 2.5 = 20

39. If the mean of 6, 4, 7, P and 10 is 8 find P.

(A) 12

(B) 13

(D) 9

Ans. (B)

Sol. $\frac{{6 + 4 + 7 + P + 10}}{5}$ = 8

⇒ P = 13

40. Find the median of the following values :

37, 31, 42, 43, 46, 25, 39, 45, 32.

(A) 32

(B) 46

(D) 43

Ans. (C)

Sol. Arranging the data in ascending order, we have

25, 31, 32, 37, 39, 42, 43, 45, 46

Here the number of observation n = 9 (odd)

∴ Median = Value of ${\left( {\frac{{9 + 1}}{2}} \right)^{th}}$ observation

= Value of 5th observation

= 39

41. The weight of 6 mens are 30, 75, 28, 85, 23, 21. The median is :

(A) 56.5

(B) 30

(D) 75

Ans. (C)

Sol. Arrange in ascending order 21, 23, 28, 30, 75, 85

Median = $\frac{{28 + 30}}{2} = \frac{{58}}{2} = 29.$

42. The mean of x + 2, x, x + 3, x + 4, x + 1 is 20 then Find x :

(A) 3

(B) 12

(D) 1

Ans. (C)

Sol. Mean = $\frac{{5x + 10}}{5}$

20 = $\frac{{5x + 10}}{5}$

90 = 5x

18 = x

43. The average of 5 men is 25 if a new men of age 19 join then find new average.

(A) 25

(B) 24

(D) 20

Ans. (B)

Sol. New avg. = $\frac{{25 \times 5 + 19}}{6}$

= $\frac{{144}}{6} = 24$

44. Find the mode of the distribution 4, –1, 1, 1, 2, 3, 1, –1, 4, 5, 4, 5, 1

(A) 1

(B) –1

(D) 5

Ans. (A)

Sol. Mode = Highest number of frequency = 1

45. The mean of 10 observation is 50, By error the observation was noted as –30 instead of 50, Then correct mean is :

(A) 48

(B) 58

(D) 52

Ans. (B)

Sol. New mean = $\frac{{50 \times 10-(-30) + 50}}{{10}}$

= $\frac{{500 + 80}}{{10}} = 58$

46. The true statement for observation is 1, 5, 4, 3, 2, 7, 10, 9, 5

(A) Mean = 5

(B) Median = 5.5

(D) Mode = 1

Ans. (C)

Sol. 1, 2, 3, 4, 5, 5, 7, 9, 10

i.e. Mode = 5, Median = 5, and Mean = $\frac{{46}}{9} > 5$

47. The mean weight of class of 20 student is 50 kg if the weight of teacher is included the mean rised by 2 kg. The weight of teacher is :

(A) 82 kg

(B) 72 kg

(D) 91 kg

Ans. (C)

Sol. $\frac{{20 \times 50 + x}}{{21}}$ = 52

x = 52 × 21 – 1000

= 1092 – 1000 = 92 kg

48. If 12 data arranged in increasing order. If the 8^th entry is increased by 4, then median increased by :

(A) 4

(B) 1

(D) 2

Ans. (C)

Sol. Since median is 6^th & 7^th observation. Hence no change.

49. The average income of Ajay & Vijay is Rs. 500 and average of Ajay & Manish is Rs. 800 and their total income is Rs.2200 then find Ajay income.

(A) Rs. 300

(B) Rs. 400

(D) Rs. 600

Ans. (B)

Sol. Ajay + Vijay = 1000

Ajay + Manish = 1600

Total income = 2200

Ajay income = 2600 – 2200 = Rs.400

50. The average age of 6 teacher is 30 years. If one teacher is excluded the mean gets reduced by 4 years. The age of excluded teacher is :

(A) 40 yrs.

(B) 45 yrs.

(D) 52 yrs.

Ans. (C)

Sol. $\frac{{6 \times 30-x}}{5}$ = 26

180 – x = 130

x = 50 yrs.

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