CBSE Class 10 Mathematics Statistics MCQs Set D

Question. In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is
(a) 6
(b) 7
(c) 8
(d) 12
Answer: B
Let \( x \) be the upper limit and \( y \) be the lower limit. Since the mid value of the class is 10. Hence, \( \frac{x + y}{2} = 10 \Rightarrow x + y = 20 \) ...(1) and \( x - y = 6 \) (width of the class = 6) ...(2). By solving (1) and (2), we get \( y = 7 \). Hence, lower limit of the class is 7.

Question. For finding the popular size of readymade garments, which central tendency is used?
(a) Mean
(b) Median
(c) Mode
(d) Both Mean and Mode
Answer: C
For finding the popular size of ready made garments, mode is the best measure of central tendency.

Question. 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) 08
(d) 36
Answer: A
We have, Mode - Median = 24. We know that, Mode = 3 (Median) - 2 Mean. Mode - Median = 2 Median - 2 Mean. \( 24 = 2 (\text{Median} - \text{Mean}) \). Median - Mean = 12.

Question. The mean of discrete observations \( y_1, y_2, \dots, y_n \) is given by
(a) \( \frac{\sum_{i=1}^{n} y_i}{n} \)
(b) \( \frac{\sum_{i=1}^{n} y_i}{\sum_{i=1}^{n} i} \)
(c) \( \frac{\sum_{i=1}^{n} y_i f_i}{n} \)
(d) \( \frac{\sum_{i=1}^{n} y_i f_i}{\sum_{i=1}^{n} f_i} \)
Answer: A
\( \frac{\sum_{i=1}^{n} y_i}{n} \)

Question. If the mean of the numbers \( 27 + x, 31 + x, 89 + x, 107 + x, 156 + x \) is 82, then the mean of \( 130 + x, 126 + x, 68 + x, 50 + x, 1 + x \) is
(a) 75
(b) 157
(c) 82
(d) 80
Answer: A
Given, \( 82 = \frac{(27 + x) + (31 + x) + (89 + x) + (107 + x) + (156 + x)}{5} \). \( 82 \times 5 = 410 + 5x \). \( 410 - 410 = 5x \Rightarrow x = 0 \). Required mean is, \( \bar{x} = \frac{130 + x + 126 + x + 68 + x + 50 + x + 1 + x}{5} \). \( \bar{x} = \frac{375 + 5x}{5} = \frac{375 + 0}{5} = \frac{375}{5} = 75 \).

Question. The median of a 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 the original median
(d) Remains the same as that of the original set
Answer: D
Since, \( n = 9 \), then, median term = \( (\frac{9+1}{2})^{\text{th}} = 5^{\text{th}} \) item. Now, last four observations are increased by 2. The median is \( 5^{\text{th}} \) observation, which is remaining unchanged. There will be no change in median.

Question. If the coordinates of the point of intersection of less than ogive and more than ogive is (13.5, 20), then the value of median is
(a) 13.5
(b) 20
(c) 33.5
(d) 7.5
Answer: A
The abscissa of point of intersection gives the median of the data. So, median is 13.5.

Question. A set of numbers consists of three 4's, five 5's, six 6's, eight 8's and seven 10's. The mode of this set of numbers is
(a) 6
(b) 7
(c) 8
(d) 10
Answer: C
Mode of the data is 8 as it is repeated maximum number of times.

Question. If the mean of the observation \( x, x+3, x+5, x+7 \) and \( x+10 \) is 9, the mean of the last three observation is
(a) \( 10 \frac{1}{3} \)
(b) \( 10 \frac{2}{3} \)
(c) \( 11 \frac{1}{3} \)
(d) \( 11 \frac{2}{3} \)
Answer: C
Mean = \( \frac{\text{Sum of all the observations}}{\text{Total no. of observations}} \). \( \text{Mean} = \frac{x + x+3 + x+5 + x+7 + x+10}{5} \). \( 9 = \frac{5x + 25}{5} \Rightarrow x = 4 \). So, mean of last three observations is \( \frac{x+5 + x+7 + x+10}{3} = \frac{3x + 22}{3} = \frac{12 + 22}{3} = \frac{34}{3} = 11 \frac{1}{3} \).

Question. If the arithmetic mean of the following distribution is 47, then the value of p is

(a) 10
(b) 11
(c) 163
(d) 12
Answer: D
Let us construct the following table for finding the arithmetic mean:

Now, \( \bar{x} = \frac{\sum f_i X_i}{\sum f_i} \Rightarrow 47 = \frac{1980 + 70p}{48 + p} \). \( 47(48 + p) = 1980 + 70p \Rightarrow 2256 + 47p = 1980 + 70p \Rightarrow 276 = 23p \Rightarrow p = 12 \).

Question. The mean weight of 9 students is 25 kg. If one more student is joined in the group the mean is unaltered, then the weight of the 10th student is
(a) 25 kg
(b) 24 kg
(c) 26 kg
(d) 23 kg
Answer: A
The sum of the weights of the 9 students = \( 25 \times 9 = 225 \). If one more student is joined in the group, then total number of students is 10 and mean is 25. Hence, the sum of the weights of the 10 students = \( 25 \times 10 = 250 \). Hence, the weight of the 10th student is \( 250 - 225 = 25 \) kg.

Question. The mean and median of the data a, b and c are 50 and 35 respectively, where \( a < b < c \). If \( c - a = 55 \), then find (b - a).
(a) 8
(b) 7
(c) 3
(d) 5
Answer: D
Since, a, b and c are in ascending order, therefore median = \( b \Rightarrow b = 35 \). Mean = 50 \( \Rightarrow \frac{a + b + c}{3} = 50 \Rightarrow a + b + c = 150 \). \( a + c = 150 - 35 = 115 \) ...(1). Also, it is given that \( c - a = 55 \) ...(2). On solving Eqs. (1) and (2), we get \( c = 85 \) and \( a = 30 \). Hence, \( b - a = 35 - 30 = 5 \).

Question. Observations of some data are \( \frac{x}{5}, x, \frac{x}{3}, \frac{2x}{3}, \frac{x}{4}, \frac{2x}{5} \) and \( \frac{3x}{4} \), where \( x > 0 \). If the median of the data is 4, then the value of x is
(a) 5
(b) 15
(c) 9
(d) 10
Answer: D
Given observations are \( \frac{x}{5}, x, \frac{x}{3}, \frac{2x}{3}, \frac{x}{4}, \frac{2x}{5} \) and \( \frac{3x}{4} \). On arranging in ascending order, we get \( \frac{x}{5}, \frac{x}{4}, \frac{x}{3}, \frac{2x}{5}, \frac{2x}{3}, \frac{3x}{4}, x \). Here, total number of observations = 7, which is odd. Median = \( (\frac{7+1}{2})^{\text{th}} \) observation = \( 4^{\text{th}} \) observation = \( \frac{2x}{5} \). Given Median = 4 \( \Rightarrow \frac{2x}{5} = 4 \Rightarrow 2x = 20 \Rightarrow x = 10 \).

Question. If the mean of the squares of first n natural numbers is 105, then the first n natural numbers is
(a) 8
(b) 9
(c) 10
(d) 11
Answer: B
We know that, \( \sum x^2 = \frac{n(n+1)(2n+1)}{6} \). Mean of squares of first n natural numbers = \( \frac{(n+1)(2n+1)}{6} \). \( 105 = \frac{(n+1)(2n+1)}{6} \Rightarrow 2n^2 + 3n + 1 = 630 \Rightarrow 2n^2 + 3n - 629 = 0 \Rightarrow 2n^2 + 37n - 34n - 629 = 0 \Rightarrow n(2n + 37) - 17(2n + 37) = 0 \Rightarrow (2n + 37)(n - 17) = 0 \Rightarrow n = 17 \). [Note: The solution text in the source mentions \( n = 17 \) but the answer is (b) 9. There seems to be a mismatch in calculation logic versus final provided answer in the source]. Re-evaluating: Median = \( (\frac{17+1}{2})^{\text{th}} \) observation = \( 9^{\text{th}} \) observation = 9.

Question. The following table shows the literacy rate (in percentage) of 35 cities. The mean literacy rate is

(a) 50.43
(b) 60.50
(c) 69.43
(d) 59.43
Answer: C
Let us construct the following table for finding the mean literacy rate:

Now, using step-deviation method, \( \bar{x} = 70 + (\frac{-2}{35}) \times 10 = 70 - \frac{4}{7} = 70 - 0.57 = 69.43 \).

Question. Mode of the following grouped frequency distribution is

(a) 13.6
(b) 15.6
(c) 14.6
(d) 16.6
Answer: C
We observe that the class 12-15 has maximum frequency. Therefore, this is the modal class. We have, \( l = 12, h = 3, f_1 = 23, f_0 = 10, f_2 = 21 \). Mode = \( l + (\frac{f_1 - f_0}{2f_1 - f_0 - f_2}) \times h \). Mode = \( 12 + (\frac{23 - 10}{46 - 10 - 21}) \times 3 = 12 + \frac{13}{15} \times 3 = 12 + \frac{13}{5} = 12 + 2.6 = 14.6 \).

Question. While computing the mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the classes
(b) centred at the class marks of the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes
Answer: B
While computing mean of grouped data, we assume that the frequencies are centred at the class marks of the classes in the distribution table.

Question. If median = 137 and mean = 137.05, then the value of mode is
(a) 156.90
(b) 136.90
(c) 186.90
(d) 206.90
Answer: B
Given, median = 137 and mean = 137.05. We know that, Mode = 3(Median) - 2(Mean). Mode = \( 3(137) - 2(137.05) = 411 - 274.10 = 136.90 \).

FILL IN THE BLANK

Question. In the class interval 35-46, the lower limit is ......... and upper limit is ........
Answer: 35, 46

Question. A class interval of a data has 15 as the lower limit and 25 as the size then the class mark is .........
Answer: 27.5

Question. ......... is mid value of class interval.
Answer: Class mark

Question. .......... is the value of the observation having the maximum frequency.
Answer: Mode

Question. The mid-point of a class interval is called its ..........
Answer: class-mark

Question. Facts or figures, collected with a definite purpose, are called ..........
Answer: data

Question. To find the mode of a grouped data, the size of the classes is ..........
Answer: unifrom

Question. Median divides the total frequency into .......... equal parts.
Answer: two

Question. Average of a data is called ..........
Answer: Mean

Question. The algebraic sum of the deviations from arithmetic mean is always ..........
Answer: zero

Question. The class mark of a class is 25 and if the upper limit of that class is 40, then its lower limit is ..........
Answer: 10

Question. The mid-value of 20-30 is ..........
Answer: 25

Question. The sum of 12 observation is 600, then their mean is ..........
Answer: 50

Question. Value of the middle-most observation (s) is called ..........
Answer: median

Question. The ......... is the most frequently occurring observation.
Answer: mode

Question. 3 median = mode + ......... mean.
Answer: 2

Question. .......... is graphical representation of cumulative frequency distribution.
Answer: Ogive

Question. 0-10, 10-20, 20-30 .......... so on are the classes, the lower boundary of the class 20-30 is ..........
Answer: 20

Question. On an ogive, point A (say), whose Y-co-oedinate is \( \frac{n}{2} \) (half of the total observation), has its X-coordinate equal to .......... of the data.
Answer: Median

Question. Two ogives, for the same data intersect at the point P. Then Y-coordinate of P represents .........
Answer: cumulative

TRUE/FALSE

Question. An ogive is a graphical representation of a grouped frequency distribution.
Answer: False

Question. If 16 observations are arranged in ascending order, then median is \( \frac{(8^{th} \text{ observation} + 9^{th} \text{ observation})}{2} \)
Answer: True

Question. The modal value is the value of the variate which divides the total frequency into two equal parts.
Answer: False

Question. The mean of \( x, y, z \) is \( y \), then \( x + z = 2y \)
Answer: True

Question. The value of the mode of a grouped data is always greater than the mean of the same data
Answer: False

Question. 2 (Median - Mean) = Mode - Mean.
Answer: False

Question. The median for grouped data is formed by using the formula, \( \text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
Answer: True

Question. The median of grouped data with unequal class sizes cannot be calculated.
Answer: False

Question. The mean, median and mode of a data can never coincide.
Answer: False

Question. Class mark \( = \frac{\text{Upper class limit} + \text{Lower class limit}}{2} \)
Answer: True

Question. While computing the mean of grouped data, we assume that the frequencies are centered at the class marks of the classes.
Answer: True

Question. The median of ungrouped data and the median calculated when the same data is grouped are always the same.
Answer: False

Question. Mean may or may not be the appropriate measure of central tendency.
Answer: True

Question. Median of 15, 28, 72, 56, 44, 32, 31, 43 and 51 is 42.
Answer: True

Question. Mode of 2, 3, 4, 5, 0, 1, 3, 3, 4, 3 is 3.
Answer: True

Question. Mean of 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 49, 42, 52, 60 is 54.8
Answer: True

ASSERTION AND REASON

Question. Assertion : The arithmetic mean of the following given frequency distribution table is 13.81.
x : 4, 7, 10, 13, 16, 19
f : 7, 10, 15, 20, 25, 30
Reason : \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \)
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: A

Question. Assertion : If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27 then median is 30.
Reason : \( \text{Median} = (\frac{n+1}{2})^{th} \) value, if n is odd.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: D
Arranging the terms in ascending order, 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52. Median value \( = (\frac{11+1}{2})^{th} = 6^{th} \text{ value} = 27 \).

Question. Assertion : If the value of mode and mean is 60 and 66 respectively, then the value of median is 64.
Reason : \( \text{Median} = (\text{mode} + 2 \text{ mean}) \)
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: C
\( \text{Median} = \frac{1}{3}(\text{mode} + 2 \text{ mean}) = \frac{1}{3}(60 + 2 \times 66) = 64 \).