CBSE Class 10 · Maths 
(i) (d): The possible values of assumed mean A are class marks: 35, 45, 55, 65, 75. Among the options, the possible value is 55.
(ii) (c): When A is chosen as a class mark, $d_i = x_i – A = 0$ for that class. Minimum value of $|d_i|$ is 0.
(iii) (b): $$ \text{Mean} = A + \frac{\sum f_i d_i}{\sum f_i} = 55 – \frac{1200}{420} \approx 55 – 2.86 = \mathbf{₹\,52.14} $$
(iv) (a): Mean by direct method and assumed mean method are always equal.
(v) (d): Average toll per vehicle = ₹52.14; Total vehicles = 420.
Total toll = $₹(52.14 \times 420) = ₹\,21,898.80 \approx ₹\,21,899$. None of the given options (₹21000, ₹21900, ₹30000) is exactly correct, so the answer is none of these.
CBSE Class 10 Maths Statistics Competency Based Questions
Help your child build strong exam-ready skills with these CBSE Class 10 Maths Statistics Competency Based Questions. Curated by Angle Belearn’s CBSE specialists, every question covers mean, median, mode, and data interpretation — with detailed step-by-step solutions your child can learn from independently and score confidently in board exams.
CBSE Class 10 Maths Statistics — Questions with Solutions
Which of the following is a measure of central tendency?
Solution
(d) Median is a measure of central tendency. The other three — class limit, lower limit, and cumulative frequency — are terms related to the organisation of grouped data, not central tendency.
While computing the mean of grouped data, we assume that the frequencies are:
Solution
(a) Centered at the class marks of the class. When computing the mean of grouped data, we assume all observations in a class are concentrated at the midpoint (class mark) of that class interval.
If $$ x_i $$’s are the mid-points of the class intervals of grouped data, $$ f_i $$’s are the corresponding frequencies and $$ \overline{x} $$ is the mean, then $$ \sum f_i (x_i – \overline{x}) $$ is equal to:
Solution
(a) Given, $$ \overline{x} $$ is the mean. Therefore,
$$ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \Rightarrow \overline{x} \sum f_i = \sum f_i x_i $$
Now,
$$ \sum f_i (x_i – \overline{x}) = \sum f_i x_i – \overline{x} \sum f_i = \sum f_i x_i – \sum f_i x_i = \mathbf{0} $$
$$ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \Rightarrow \overline{x} \sum f_i = \sum f_i x_i $$
Now,
$$ \sum f_i (x_i – \overline{x}) = \sum f_i x_i – \overline{x} \sum f_i = \sum f_i x_i – \sum f_i x_i = \mathbf{0} $$
In the formula $$ \overline{x} = A + \frac{\sum f_i d_i}{\sum f_i} $$ for finding the mean of grouped data, $$ d_i $$’s are deviations from A of:
Solution
(c) Mid-points of the classes. In the assumed mean method, $$ d_i = x_i – A $$, where $$ x_i $$ is the class mark (mid-point) of each class and A is the assumed mean.
Construction of a cumulative frequency table is useful in determining the:
Solution
(c) Median. A cumulative frequency table is specifically used to identify the median class — the class interval in which $$ \frac{N}{2} $$ falls — which is the first step in computing the median of grouped data.
The range of the following data is:
Solution
(c) Here, upper class boundary of the highest interval $$ u = 80 $$ and lower class boundary of the lowest interval $$ l = 30 $$.
$$ \text{Range of grouped data} = u – l = 80 – 30 = \mathbf{50} $$
$$ \text{Range of grouped data} = u – l = 80 – 30 = \mathbf{50} $$
In the following distribution table, the preceding frequency value of the modal class is:
Solution
(b) In the given distribution table, the highest frequency is 18. Therefore its modal class is 10–15. The class preceding 10–15 is 5–10, whose frequency is 5.
For the following distribution:
The lower limit of modal class is:
The lower limit of modal class is:
Solution
(a) From the given distribution, the highest frequency is 20, which lies in the class interval 15–20. This class is the modal class. So, the lower limit of the modal class is 15.
Consider the following distribution:
The frequency of the class 30–40 is:
The frequency of the class 30–40 is:
Solution
(a) The frequency of the class 30–40 = (Number of students with marks ≥ 30) − (Number of students with marks ≥ 40)
$$ = 51 – 48 = \mathbf{3} $$
$$ = 51 – 48 = \mathbf{3} $$
Consider the following frequency distribution:
The upper limit of the median class is:
The upper limit of the median class is:
Solution
The given frequency table is not continuous, so convert it by subtracting $$ 0.5 $$ from lower limits and adding $$ 0.5 $$ to upper limits. After conversion, the cumulative frequency table is as below:
Here, $$ N = 57 \Rightarrow \frac{N}{2} = 28.5 $$, which lies in cumulative frequency 38. The corresponding median class is 11.5–17.5. Therefore, the upper limit of the median class is 17.5.
Here, $$ N = 57 \Rightarrow \frac{N}{2} = 28.5 $$, which lies in cumulative frequency 38. The corresponding median class is 11.5–17.5. Therefore, the upper limit of the median class is 17.5.
An agency has decided to install customised playground equipments at various colony parks. For that they decided to study the age-group of children playing in a park of the particular colony. The classification of children according to their ages, playing in a park is shown in the following table:
Based on the above information, solve the following questions:
Q1. The maximum number of children are of the age-group:
a. 12–14 b. 10–12 c. 14–16 d. 8–10
Q2. The lower limit of the modal class is:
a. 10 b. 12 c. 14 d. 8
Q3. Frequency of the class succeeding the modal class is:
a. 58 b. 70 c. 42 d. 27
Q4. The mode of the ages of children playing in the park is:
a. 9 yr b. 8 yr c. 11.5 yr d. 10.6 yr
Q5. If mean and mode of the ages of children playing in the park are same, then median will be equal to:
a. Mean b. Mode c. Both a. and b. d. Neither a. nor b.
Based on the above information, solve the following questions:
Q1. The maximum number of children are of the age-group:
a. 12–14 b. 10–12 c. 14–16 d. 8–10
Q2. The lower limit of the modal class is:
a. 10 b. 12 c. 14 d. 8
Q3. Frequency of the class succeeding the modal class is:
a. 58 b. 70 c. 42 d. 27
Q4. The mode of the ages of children playing in the park is:
a. 9 yr b. 8 yr c. 11.5 yr d. 10.6 yr
Q5. If mean and mode of the ages of children playing in the park are same, then median will be equal to:
a. Mean b. Mode c. Both a. and b. d. Neither a. nor b.
Answer
1. (b) Since the highest frequency is 70, the maximum number of children are of the age-group 10–12, which is also the modal class.
2. (a) Since the modal class is 10–12, the lower limit of the modal class $$ (l) = \mathbf{10} $$.
3. (c) From the table, the frequency of the modal class is 70. The frequency of the class succeeding the modal class (12–14) is 42.
4. (d) Here, $$ l = 10,\; f_1 = 70,\; f_0 = 58,\; f_2 = 42,\; h = 2 $$
$$ \text{Mode} = l + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times h $$
$$ = 10 + \frac{70 – 58}{2(70) – 58 – 42} \times 2 = 10 + \frac{12}{40} \times 2 = 10 + 0.6 = \mathbf{10.6 \text{ yr}} $$
5. (c) Given Mean = Mode. By the empirical relation:
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} $$
$$ \Rightarrow \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mode} $$
$$ \Rightarrow 3 \times \text{Mode} = 3 \times \text{Median} \Rightarrow \text{Median} = \text{Mode} = \text{Mean} $$
So the answer is Both a. and b.
2. (a) Since the modal class is 10–12, the lower limit of the modal class $$ (l) = \mathbf{10} $$.
3. (c) From the table, the frequency of the modal class is 70. The frequency of the class succeeding the modal class (12–14) is 42.
4. (d) Here, $$ l = 10,\; f_1 = 70,\; f_0 = 58,\; f_2 = 42,\; h = 2 $$
$$ \text{Mode} = l + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times h $$
$$ = 10 + \frac{70 – 58}{2(70) – 58 – 42} \times 2 = 10 + \frac{12}{40} \times 2 = 10 + 0.6 = \mathbf{10.6 \text{ yr}} $$
5. (c) Given Mean = Mode. By the empirical relation:
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} $$
$$ \Rightarrow \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mode} $$
$$ \Rightarrow 3 \times \text{Mode} = 3 \times \text{Median} \Rightarrow \text{Median} = \text{Mode} = \text{Mean} $$
So the answer is Both a. and b.
A petrol pump owner wants to analyse the daily need of diesel at the pump. For this he collected the data of vehicles visited in 1 hour. The following frequency distribution table shows the classification of the number of vehicles and quantity of diesel filled in them:
Q1. Which of the following is correct?
a. If $$ x_i $$ and $$ f_i $$ are sufficiently small, then direct method is appropriate choice for calculating mean.
b. If $$ x_i $$ and $$ f_i $$ are sufficiently large, then direct method is appropriate choice for calculating mean.
c. If $$ x_i $$ and $$ f_i $$ are sufficiently small, then assumed mean method is appropriate choice for calculating mean.
d. None of the above
Q2. Average diesel required for a vehicle is:
a. 8.15 L b. 6 L c. 7 L d. 5.5 L
Q3. If approximately 2000 vehicles come daily at the petrol pump, then how much litres of diesel should the pump have?
a. 16200 L b. 16300 L c. 10600 L d. 15000 L
Q4. The sum of upper and lower limit of median class is:
a. 22 b. 10 c. 16 d. None of these
Q5. If the median of the given data is 8 L, then mode will be equal to:
a. 7.5 L b. 7.7 L c. 5.7 L d. 8 L
Q1. Which of the following is correct?
a. If $$ x_i $$ and $$ f_i $$ are sufficiently small, then direct method is appropriate choice for calculating mean.
b. If $$ x_i $$ and $$ f_i $$ are sufficiently large, then direct method is appropriate choice for calculating mean.
c. If $$ x_i $$ and $$ f_i $$ are sufficiently small, then assumed mean method is appropriate choice for calculating mean.
d. None of the above
Q2. Average diesel required for a vehicle is:
a. 8.15 L b. 6 L c. 7 L d. 5.5 L
Q3. If approximately 2000 vehicles come daily at the petrol pump, then how much litres of diesel should the pump have?
a. 16200 L b. 16300 L c. 10600 L d. 15000 L
Q4. The sum of upper and lower limit of median class is:
a. 22 b. 10 c. 16 d. None of these
Q5. If the median of the given data is 8 L, then mode will be equal to:
a. 7.5 L b. 7.7 L c. 5.7 L d. 8 L
Answer
1. (a) If $$ f_i $$ and $$ x_i $$ are very small, the direct method is the appropriate method for calculating the mean.
2. (a) $$ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{326}{40} = \mathbf{8.15 \text{ L}} $$
3. (b) If 2000 vehicles come daily and average diesel per vehicle is 8.15 L:
$$ \text{Total diesel} = 2000 \times 8.15 = \mathbf{16300 \text{ L}} $$
4. (c) Here $$ N = 40 $$, so $$ \frac{N}{2} = 20 $$. The cumulative frequencies are 5, 15, 25, 32, 40. The cf just greater than 20 is 25, corresponding to class interval 7–9.
$$ \text{Sum of upper and lower limits} = 7 + 9 = \mathbf{16} $$
5. (b) Using the empirical relation:
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} = 3(8) – 2(8.15) = 24 – 16.3 = \mathbf{7.7 \text{ L}} $$
2. (a) $$ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{326}{40} = \mathbf{8.15 \text{ L}} $$
3. (b) If 2000 vehicles come daily and average diesel per vehicle is 8.15 L:
$$ \text{Total diesel} = 2000 \times 8.15 = \mathbf{16300 \text{ L}} $$
4. (c) Here $$ N = 40 $$, so $$ \frac{N}{2} = 20 $$. The cumulative frequencies are 5, 15, 25, 32, 40. The cf just greater than 20 is 25, corresponding to class interval 7–9.
$$ \text{Sum of upper and lower limits} = 7 + 9 = \mathbf{16} $$
5. (b) Using the empirical relation:
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} = 3(8) – 2(8.15) = 24 – 16.3 = \mathbf{7.7 \text{ L}} $$
A group of 71 people visited a museum on a certain day. The following table shows their ages:
Q1. If true class limits have been decided by making the classes of interval 10, then find the first class interval.
Q2. Find the cumulative frequency table.
Q3. Find the frequency of class preceding the median class.
Or If the price of a ticket for the age group 30–40 is ₹30, then find the total amount spent by this age group.
Q1. If true class limits have been decided by making the classes of interval 10, then find the first class interval.
Q2. Find the cumulative frequency table.
Q3. Find the frequency of class preceding the median class.
Or If the price of a ticket for the age group 30–40 is ₹30, then find the total amount spent by this age group.
Answer
1. The age of any person is a positive number, so the first class interval must be 0–10.
2. Cumulative frequency table:
3. From the table, $$ N = 71 $$, therefore $$ \frac{N}{2} = 35.5 $$. The class interval whose cumulative frequency is just greater than 35.5 is 30–40. So the median class = 30–40.
The frequency of the class preceding the median class (20–30) is 2.
Or
Number of persons in the age group 30–40 = 18.
$$ \text{Total amount} = ₹(30 \times 18) = \mathbf{₹\,540} $$
2. Cumulative frequency table:
3. From the table, $$ N = 71 $$, therefore $$ \frac{N}{2} = 35.5 $$. The class interval whose cumulative frequency is just greater than 35.5 is 30–40. So the median class = 30–40.
The frequency of the class preceding the median class (20–30) is 2.
Or
Number of persons in the age group 30–40 = 18.
$$ \text{Total amount} = ₹(30 \times 18) = \mathbf{₹\,540} $$
Electric buses are becoming popular nowadays. These buses have the electricity stored in a battery. Electric buses could have a range of approximately 280 km with just one charge. Electric buses are superior to diesel buses as they reduce brake wear and also reduce pollution. Transport department of a city wants to buy some electric buses for the city. So, the department wants to know the distance travelled by existing public transport buses in a day.
The following data shows the distance travelled by 50 existing public transport buses in a day:
Q1. Write the relation between mean, median and mode.
Q2. Find the modal class of the given distribution.
Q3. Find the ‘median’ distance travelled by a bus.
OR Find the ‘mean (average)’ distance travelled by a bus.
The following data shows the distance travelled by 50 existing public transport buses in a day:
Q1. Write the relation between mean, median and mode.
Q2. Find the modal class of the given distribution.
Q3. Find the ‘median’ distance travelled by a bus.
OR Find the ‘mean (average)’ distance travelled by a bus.
Answer
1. The relation between mean, median and mode (empirical formula):
$$ \text{Mode} = 3(\text{Median}) – 2(\text{Mean}) $$
2. From the given distribution, the highest frequency is 14, which lies in the class interval 120–140. This is the modal class.
3.
Here $$ N = 50,\; \frac{N}{2} = 25 $$, which lies in cumulative frequency 26. So median class = 120–140.
Here $$ l = 120,\; cf = 12,\; f = 14,\; h = 20 $$
$$ \text{Median} = l + \left(\frac{\frac{N}{2} – cf}{f}\right) \times h = 120 + \left(\frac{25 – 12}{14}\right) \times 20 = 120 + \frac{260}{14} = 120 + 18.57 = \mathbf{138.57 \text{ km}} $$
OR (Mean by step-deviation method)
Assumed mean $$ A = 150 $$:
$$ \text{Mean} = A + \frac{\sum f u_i}{\sum f} \times h = 150 + \frac{-12}{50} \times 20 = 150 – 4.8 = \mathbf{145.2 \text{ km}} $$
$$ \text{Mode} = 3(\text{Median}) – 2(\text{Mean}) $$
2. From the given distribution, the highest frequency is 14, which lies in the class interval 120–140. This is the modal class.
3.
Here $$ N = 50,\; \frac{N}{2} = 25 $$, which lies in cumulative frequency 26. So median class = 120–140.
Here $$ l = 120,\; cf = 12,\; f = 14,\; h = 20 $$
$$ \text{Median} = l + \left(\frac{\frac{N}{2} – cf}{f}\right) \times h = 120 + \left(\frac{25 – 12}{14}\right) \times 20 = 120 + \frac{260}{14} = 120 + 18.57 = \mathbf{138.57 \text{ km}} $$
OR (Mean by step-deviation method)
Assumed mean $$ A = 150 $$:
$$ \text{Mean} = A + \frac{\sum f u_i}{\sum f} \times h = 150 + \frac{-12}{50} \times 20 = 150 – 4.8 = \mathbf{145.2 \text{ km}} $$
An agency has decided to install customised playground equipments at various colony parks. For that they decided to study the age-group of children playing in a park of the particular colony. The classification of children according to their ages, playing in a park is shown in the following table:
Based on the above information, answer the following questions:
(i) The maximum number of children are of the age-group:
(a) 12–14 (b) 10–12 (c) 14–16 (d) 8–10
(ii) The lower limit of the modal class is:
(a) 10 (b) 12 (c) 14 (d) 8
(iii) Frequency of the class succeeding the modal class is:
(a) 58 (b) 70 (c) 42 (d) 27
(iv) The mode of the ages of children playing in the park is:
(a) 9 years (b) 8 years (c) 11.5 years (d) 10.6 years
(v) If mean and mode of the ages of children playing in the park are same, then median will be equal to:
(a) Mean (b) Mode (c) Both (a) and (b) (d) Neither (a) nor (b)
Based on the above information, answer the following questions:
(i) The maximum number of children are of the age-group:
(a) 12–14 (b) 10–12 (c) 14–16 (d) 8–10
(ii) The lower limit of the modal class is:
(a) 10 (b) 12 (c) 14 (d) 8
(iii) Frequency of the class succeeding the modal class is:
(a) 58 (b) 70 (c) 42 (d) 27
(iv) The mode of the ages of children playing in the park is:
(a) 9 years (b) 8 years (c) 11.5 years (d) 10.6 years
(v) If mean and mode of the ages of children playing in the park are same, then median will be equal to:
(a) Mean (b) Mode (c) Both (a) and (b) (d) Neither (a) nor (b)
Answer
(i) (b): The highest frequency is 70, so the maximum number of children are of the age-group 10–12.
(ii) (a): Modal class is 10–12, so lower limit $$ = \mathbf{10} $$.
(iii) (c): Here $f_0 = 58,\; f_1 = 70$ and $f_2 = 42$. The frequency of the class succeeding the modal class is 42.
(iv) (d):
$$ \text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h = 10 + \left(\frac{70 – 58}{140 – 58 – 42}\right) \times 2 = 10 + \frac{12}{40} \times 2 = 10 + 0.6 = \mathbf{10.6 \text{ years}} $$
(v) (c): Given Mean = Mode. Using empirical relation:
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} \Rightarrow \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mode} $$
$$ \Rightarrow 3 \times \text{Mode} = 3 \times \text{Median} \Rightarrow \text{Median} = \text{Mode} = \text{Mean} $$
Answer: Both (a) and (b).
(ii) (a): Modal class is 10–12, so lower limit $$ = \mathbf{10} $$.
(iii) (c): Here $f_0 = 58,\; f_1 = 70$ and $f_2 = 42$. The frequency of the class succeeding the modal class is 42.
(iv) (d):
$$ \text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h = 10 + \left(\frac{70 – 58}{140 – 58 – 42}\right) \times 2 = 10 + \frac{12}{40} \times 2 = 10 + 0.6 = \mathbf{10.6 \text{ years}} $$
(v) (c): Given Mean = Mode. Using empirical relation:
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} \Rightarrow \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mode} $$
$$ \Rightarrow 3 \times \text{Mode} = 3 \times \text{Median} \Rightarrow \text{Median} = \text{Mode} = \text{Mean} $$
Answer: Both (a) and (b).
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If $\sum f_i = 20$, $\sum f_i x_i = 3\lambda + 20$ and mean of the distribution is 4, then the value of $\lambda$ is 20.
Reason (R): If there are $x_1, x_2, \dots, x_n$ observations where corresponding frequencies are $f_1, f_2, \dots, f_m$, then mean is determined by the formula $$ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} $$
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If $\sum f_i = 20$, $\sum f_i x_i = 3\lambda + 20$ and mean of the distribution is 4, then the value of $\lambda$ is 20.
Reason (R): If there are $x_1, x_2, \dots, x_n$ observations where corresponding frequencies are $f_1, f_2, \dots, f_m$, then mean is determined by the formula $$ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} $$
Answer
(a) — Both A and R are true and R is the correct explanation of A.
Verification of Assertion (A):
Given $\sum f_i = 20$, $\sum f_i x_i = 3\lambda + 20$, $\bar{x} = 4$
$$ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \Rightarrow 4 = \frac{3\lambda + 20}{20} \Rightarrow 3\lambda + 20 = 80 \Rightarrow 3\lambda = 60 \Rightarrow \lambda = 20 $$
Assertion is true. Reason states the correct mean formula, so it is also true and correctly explains the assertion.
Verification of Assertion (A):
Given $\sum f_i = 20$, $\sum f_i x_i = 3\lambda + 20$, $\bar{x} = 4$
$$ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \Rightarrow 4 = \frac{3\lambda + 20}{20} \Rightarrow 3\lambda + 20 = 80 \Rightarrow 3\lambda = 60 \Rightarrow \lambda = 20 $$
Assertion is true. Reason states the correct mean formula, so it is also true and correctly explains the assertion.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The mode of the following frequency distribution is 52.25 kg.
Reason (R): A modal class is a class which has the highest frequency.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The mode of the following frequency distribution is 52.25 kg.
Reason (R): A modal class is a class which has the highest frequency.
Answer
(d) — A is false but R is true.
Checking Assertion (A):
The highest frequency is 35, which lies in class interval 50–55. So modal class = 50–55.
Here $l = 50,\; f_m = 35,\; f_p = 30,\; f_s = 20,\; h = 5$
$$ \text{Mode} = l + \left(\frac{f_m – f_p}{2f_m – f_p – f_s}\right) \times h = 50 + \left(\frac{35 – 30}{70 – 30 – 20}\right) \times 5 = 50 + \frac{25}{20} = 50 + 1.25 = \mathbf{51.25 \text{ kg}} $$
Assertion states 52.25 kg — this is false. Reason is true.
Checking Assertion (A):
The highest frequency is 35, which lies in class interval 50–55. So modal class = 50–55.
Here $l = 50,\; f_m = 35,\; f_p = 30,\; f_s = 20,\; h = 5$
$$ \text{Mode} = l + \left(\frac{f_m – f_p}{2f_m – f_p – f_s}\right) \times h = 50 + \left(\frac{35 – 30}{70 – 30 – 20}\right) \times 5 = 50 + \frac{25}{20} = 50 + 1.25 = \mathbf{51.25 \text{ kg}} $$
Assertion states 52.25 kg — this is false. Reason is true.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The median of the frequency distribution is 68.75.
Reason (R): The cumulative frequency of the median class is just next to the median class.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The median of the frequency distribution is 68.75.
Reason (R): The cumulative frequency of the median class is just next to the median class.
Answer
(c) — A is true but R is false.
Verification of Assertion (A):
$N = 50$, so $\frac{N}{2} = 25$, which lies in cumulative frequency 26. Median class = 60–70.
$l = 60,\; f = 8,\; cf = 18,\; h = 10$
$$ \text{Median} = 60 + \left(\frac{25 – 18}{8}\right) \times 10 = 60 + \frac{70}{8} = 60 + 8.75 = \mathbf{68.75} $$
Assertion is true. However, Reason is false — the cumulative frequency used is the one just before (preceding) the median class, not the one just after it.
Verification of Assertion (A):
$N = 50$, so $\frac{N}{2} = 25$, which lies in cumulative frequency 26. Median class = 60–70.
$l = 60,\; f = 8,\; cf = 18,\; h = 10$
$$ \text{Median} = 60 + \left(\frac{25 – 18}{8}\right) \times 10 = 60 + \frac{70}{8} = 60 + 8.75 = \mathbf{68.75} $$
Assertion is true. However, Reason is false — the cumulative frequency used is the one just before (preceding) the median class, not the one just after it.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The following table gives the marks scored by students in an examination:
The succeeding frequency of the modal class is 16.
Reason (R): The sum of frequency of modal class and its preceding frequency is 40.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The following table gives the marks scored by students in an examination:
The succeeding frequency of the modal class is 16.
Reason (R): The sum of frequency of modal class and its preceding frequency is 40.
Answer
(c) — A is true but R is false.
Assertion (A): In the given table, the highest frequency is 24, whose modal class is 15–20. The succeeding frequency is 16. Assertion is true.
Reason (R): Frequency of modal class = 24, preceding frequency = 15. Sum = $24 + 15 = 39 \neq 40$. Reason is false.
Assertion (A): In the given table, the highest frequency is 24, whose modal class is 15–20. The succeeding frequency is 16. Assertion is true.
Reason (R): Frequency of modal class = 24, preceding frequency = 15. Sum = $24 + 15 = 39 \neq 40$. Reason is false.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the median and mode of a frequency distribution are 150 and 154 respectively, then its mean is 148.
Reason (R): The relation between mean, mode, and median is: $$ \text{Mode} = 3 \times (\text{Median}) – 2 \times (\text{Mean}) $$
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the median and mode of a frequency distribution are 150 and 154 respectively, then its mean is 148.
Reason (R): The relation between mean, mode, and median is: $$ \text{Mode} = 3 \times (\text{Median}) – 2 \times (\text{Mean}) $$
Answer
(a) — Both A and R are true and R is the correct explanation of A.
Given: Median = 150, Mode = 154
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} $$
$$ \Rightarrow 154 = 3 \times 150 – 2 \times \text{Mean} \Rightarrow 2 \times \text{Mean} = 450 – 154 = 296 \Rightarrow \text{Mean} = \mathbf{148} $$
Both Assertion and Reason are true, and R correctly explains A.
Given: Median = 150, Mode = 154
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} $$
$$ \Rightarrow 154 = 3 \times 150 – 2 \times \text{Mean} \Rightarrow 2 \times \text{Mean} = 450 – 154 = 296 \Rightarrow \text{Mean} = \mathbf{148} $$
Both Assertion and Reason are true, and R correctly explains A.
If median of the following data is 345,
Then find the median class.
Then find the median class.
Solution
(c) Given median is 345, which lies in the interval 340–350. Therefore, the median class is 340–350.
Consider the following frequency distribution:
The median class is:
The median class is:
Solution
(b) Cumulative frequency table:
Here $\frac{56}{2} = 28$. Since 28 lies in cumulative frequency 37, the median class is 12–18.
| Class | Frequency (f) | Cumulative Frequency (cf) |
|---|---|---|
| 0–6 | 12 | 12 |
| 6–12 | 10 | 22 |
| 12–18 | 15 | 37 |
| 18–24 | 8 | 45 |
| 24–30 | 11 | 56 |
For the following distribution:
The sum of lower limits of the median class and modal class is:
The sum of lower limits of the median class and modal class is:
Solution
(b) The cumulative frequency table is shown below:
Here $N = 66 \Rightarrow \frac{66}{2} = 33$, which lies in cumulative frequency 37. So median class = 10–15 (lower limit = 10).
The highest frequency is 20, so modal class = 15–20 (lower limit = 15).
$$ \text{Sum} = 10 + 15 = \mathbf{25} $$
Here $N = 66 \Rightarrow \frac{66}{2} = 33$, which lies in cumulative frequency 37. So median class = 10–15 (lower limit = 10).
The highest frequency is 20, so modal class = 15–20 (lower limit = 15).
$$ \text{Sum} = 10 + 15 = \mathbf{25} $$
Consider the data:
The difference of the upper limit of the median class and the lower limit of the modal class is:
The difference of the upper limit of the median class and the lower limit of the modal class is:
Solution
(c) The cumulative frequency table is shown below:
$N = 67 \Rightarrow \frac{67}{2} = 33.5$, which lies in cumulative frequency 42. So median class = 125–145 (upper limit = 145).
The highest frequency is 20, so modal class = 125–145 (lower limit = 125).
$$ \text{Difference} = 145 – 125 = \mathbf{20} $$
$N = 67 \Rightarrow \frac{67}{2} = 33.5$, which lies in cumulative frequency 42. So median class = 125–145 (upper limit = 145).
The highest frequency is 20, so modal class = 125–145 (lower limit = 125).
$$ \text{Difference} = 145 – 125 = \mathbf{20} $$
Median and Mode of a distribution are 25 and 21 respectively. Mean of the data using the empirical relationship is:
Solution
(a) Given: Median = 25, Mode = 21
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} $$
$$ 21 = 3(25) – 2 \times \text{Mean} \Rightarrow 2 \times \text{Mean} = 75 – 21 = 54 \Rightarrow \text{Mean} = \mathbf{27} $$
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} $$
$$ 21 = 3(25) – 2 \times \text{Mean} \Rightarrow 2 \times \text{Mean} = 75 – 21 = 54 \Rightarrow \text{Mean} = \mathbf{27} $$
The mean of the following frequency distribution is 25. Find the missing frequency $f$.
Solution
(b) $$ \sum f = 3 + 7 + f + 2 = 12 + f $$
$$ \sum fx = (10 \times 3) + (20 \times 7) + (30 \times f) + (40 \times 2) = 30 + 140 + 30f + 80 = 250 + 30f $$
Given mean = 25:
$$ 25 = \frac{250 + 30f}{12 + f} \Rightarrow 300 + 25f = 250 + 30f \Rightarrow 50 = 5f \Rightarrow f = \mathbf{10} $$
$$ \sum fx = (10 \times 3) + (20 \times 7) + (30 \times f) + (40 \times 2) = 30 + 140 + 30f + 80 = 250 + 30f $$
Given mean = 25:
$$ 25 = \frac{250 + 30f}{12 + f} \Rightarrow 300 + 25f = 250 + 30f \Rightarrow 50 = 5f \Rightarrow f = \mathbf{10} $$
The following table shows the distribution of marks obtained by 100 students:
The median class is:
The median class is:
Solution
(b) $$ \frac{N}{2} = \frac{100}{2} = 50 $$
The cumulative frequency first exceeding 50 is 60, which corresponds to the class 20–30. Therefore the median class is 20–30.
The cumulative frequency first exceeding 50 is 60, which corresponds to the class 20–30. Therefore the median class is 20–30.
If the mean of the following distribution is 18, find the missing frequency $f$.
Solution
(b) Class marks: 0–10 → 5; 10–20 → 15; 20–30 → 25; 30–40 → 35
$$ \sum f = 5 + f + 15 + 10 = 30 + f $$
$$ \sum fx = (5)(5) + (f)(15) + (15)(25) + (10)(35) = 25 + 15f + 375 + 350 = 750 + 15f $$
Given mean = 18:
$$ 18 = \frac{750 + 15f}{30 + f} \Rightarrow 540 + 18f = 750 + 15f \Rightarrow 3f = 210 \Rightarrow f = \mathbf{10} $$
$$ \sum f = 5 + f + 15 + 10 = 30 + f $$
$$ \sum fx = (5)(5) + (f)(15) + (15)(25) + (10)(35) = 25 + 15f + 375 + 350 = 750 + 15f $$
Given mean = 18:
$$ 18 = \frac{750 + 15f}{30 + f} \Rightarrow 540 + 18f = 750 + 15f \Rightarrow 3f = 210 \Rightarrow f = \mathbf{10} $$
In a grouped frequency distribution, the mode is always:
Solution
(a) In grouped frequency distribution, the modal class is the class with the highest frequency. (Note: Option D describes mode for ungrouped/raw data, not grouped data.)
If the cumulative frequency just greater than $$ \frac{N}{2} $$ is 48, then:
Solution
(c) By definition, the median class is identified as the class whose cumulative frequency is just greater than $\frac{N}{2}$. Here that cumulative frequency is 48, so the median class is the class corresponding to CF = 48.
As the demand for products grew, a manufacturing company decided to hire more employees. For this they want to know the mean time required to complete the work for a worker. The following table shows the frequency distribution of the time required for each worker to complete a work:
Based on the above information, answer the following questions:
(i) The class mark of the class 25–29 is:
(a) 17 (b) 22 (c) 27 (d) 32
(ii) If $x_i$’s denotes the class marks and $f_i$’s denotes the corresponding frequencies, then the value of $\sum x_i f_i$ equals:
(a) 1200 (b) 1205 (c) 1260 (d) 1265
(iii) The mean time required to complete the work for a worker is:
(a) 22 hrs (b) 23 hrs (c) 24 hrs (d) none of these
(iv) If a worker works for 8 hrs in a day, then approximate time required to complete the work is:
(a) 3 days (b) 4 days (c) 5 days (d) 6 days
(v) The measures of central tendency are:
(a) Mean (b) Median (c) Mode (d) All of these
Based on the above information, answer the following questions:
(i) The class mark of the class 25–29 is:
(a) 17 (b) 22 (c) 27 (d) 32
(ii) If $x_i$’s denotes the class marks and $f_i$’s denotes the corresponding frequencies, then the value of $\sum x_i f_i$ equals:
(a) 1200 (b) 1205 (c) 1260 (d) 1265
(iii) The mean time required to complete the work for a worker is:
(a) 22 hrs (b) 23 hrs (c) 24 hrs (d) none of these
(iv) If a worker works for 8 hrs in a day, then approximate time required to complete the work is:
(a) 3 days (b) 4 days (c) 5 days (d) 6 days
(v) The measures of central tendency are:
(a) Mean (b) Median (c) Mode (d) All of these
Answer
(i) (c): Class mark of 25–29 $= \frac{25 + 29}{2} = \frac{54}{2} = \mathbf{27}$
(ii) (d): From the calculated table:
$\sum x_i f_i = \mathbf{1265}$
(iii) (d): $$ \text{Mean} = \frac{\sum x_i f_i}{\sum f_i} = \frac{1265}{50} = 25.3 \text{ hrs} $$ (None of the options a–c matches, so the answer is none of these.)
(iv) (a): Mean time = 25.3 hrs ≈ 25 hrs. At 8 hrs/day: $\frac{25.3}{8} \approx 3.16 \approx \mathbf{3 \text{ days}}$
(v) (d): The measures of central tendency are Mean, Median, and Mode — all of these.
(ii) (d): From the calculated table:
$\sum x_i f_i = \mathbf{1265}$
(iii) (d): $$ \text{Mean} = \frac{\sum x_i f_i}{\sum f_i} = \frac{1265}{50} = 25.3 \text{ hrs} $$ (None of the options a–c matches, so the answer is none of these.)
(iv) (a): Mean time = 25.3 hrs ≈ 25 hrs. At 8 hrs/day: $\frac{25.3}{8} \approx 3.16 \approx \mathbf{3 \text{ days}}$
(v) (d): The measures of central tendency are Mean, Median, and Mode — all of these.
On a particular day, the National Highway Authority of India (NHAI) checked the toll tax collection of a toll plaza in Rajasthan.
The following table shows the toll tax paid by drivers and the number of vehicles on that particular day:
Based on the above information, answer the following questions:
(i) If A is taken as assumed mean, then the possible value of A is:
(a) 32 (b) 42 (c) 85 (d) 55
(ii) If $x_i$’s denotes class marks and $d_i$’s denotes deviation of assumed mean (A) from $x_i$’s, then the minimum value of $|d_i|$ is:
(a) −200 (b) −100 (c) 0 (d) 100
(iii) The mean of toll tax received by NHAI by assumed mean method is:
(a) ₹52 (b) ₹52.14 (c) ₹52.50 (d) ₹53.50
(iv) The mean of toll tax received by NHAI by direct method is:
(a) equal to the mean by assumed mean method (b) greater (c) less (d) none of these
(v) The average toll tax received by NHAI in a day from that particular toll plaza is:
(a) ₹21000 (b) ₹21900 (c) ₹30000 (d) none of these
The following table shows the toll tax paid by drivers and the number of vehicles on that particular day:
Based on the above information, answer the following questions:
(i) If A is taken as assumed mean, then the possible value of A is:
(a) 32 (b) 42 (c) 85 (d) 55
(ii) If $x_i$’s denotes class marks and $d_i$’s denotes deviation of assumed mean (A) from $x_i$’s, then the minimum value of $|d_i|$ is:
(a) −200 (b) −100 (c) 0 (d) 100
(iii) The mean of toll tax received by NHAI by assumed mean method is:
(a) ₹52 (b) ₹52.14 (c) ₹52.50 (d) ₹53.50
(iv) The mean of toll tax received by NHAI by direct method is:
(a) equal to the mean by assumed mean method (b) greater (c) less (d) none of these
(v) The average toll tax received by NHAI in a day from that particular toll plaza is:
(a) ₹21000 (b) ₹21900 (c) ₹30000 (d) none of these
Answer
(i) (d): The possible values of assumed mean A are class marks: 35, 45, 55, 65, 75. Among the options, the possible value is 55.
(ii) (c): When A is chosen as a class mark, $d_i = x_i – A = 0$ for that class. Minimum value of $|d_i|$ is 0.
(iii) (b): $$ \text{Mean} = A + \frac{\sum f_i d_i}{\sum f_i} = 55 – \frac{1200}{420} \approx 55 – 2.86 = \mathbf{₹\,52.14} $$
(iv) (a): Mean by direct method and assumed mean method are always equal.
(v) (d): Average toll per vehicle = ₹52.14; Total vehicles = 420.
Total toll = $₹(52.14 \times 420) = ₹\,21,898.80 \approx ₹\,21,899$. None of the given options (₹21000, ₹21900, ₹30000) is exactly correct, so the answer is none of these.
Transport department of a city wants to buy some Electric buses for the city. For which they want to analyse the distance travelled by existing public transport buses in a day.
The following data shows the distance travelled by 60 existing public transport buses in a day:
Based on the above information, answer the following questions:
(i) The upper limit of a class and the lower limit of its succeeding class differ by:
(a) 9 (b) 1 (c) 10 (d) none of these
(ii) The median class is:
(a) 229.5–239.5 (b) 230–239 (c) 220–229 (d) 219.5–229.5
(iii) The cumulative frequency of the class preceding the median class is:
(a) 14 (b) 18 (c) 26 (d) 10
(iv) The median of the distance travelled is:
(a) 222 km (b) 225 km (c) 223 km (d) none of these
(v) If the mode of the distance travelled is 223.78 km, then mean of the distance travelled by the bus is:
(a) 225 km (b) 220 km (c) 230.29 km (d) none of these
The following data shows the distance travelled by 60 existing public transport buses in a day:
Based on the above information, answer the following questions:
(i) The upper limit of a class and the lower limit of its succeeding class differ by:
(a) 9 (b) 1 (c) 10 (d) none of these
(ii) The median class is:
(a) 229.5–239.5 (b) 230–239 (c) 220–229 (d) 219.5–229.5
(iii) The cumulative frequency of the class preceding the median class is:
(a) 14 (b) 18 (c) 26 (d) 10
(iv) The median of the distance travelled is:
(a) 222 km (b) 225 km (c) 223 km (d) none of these
(v) If the mode of the distance travelled is 223.78 km, then mean of the distance travelled by the bus is:
(a) 225 km (b) 220 km (c) 230.29 km (d) none of these
Answer
(i) (b): The upper limit of a class and the lower limit of its succeeding class (in this inclusive distribution) differ by 1.
(ii) (d): Convert class intervals to exclusive form by subtracting 0.5 from lower limits and adding 0.5 to upper limits:
$N = 60 \Rightarrow \frac{N}{2} = 30$. The cf just greater than 30 is in class 219.5–229.5.
(iii) (b): The cumulative frequency of the class preceding the median class is 18.
(iv) (d):
$l = 219.5,\; cf = 18,\; f = 26,\; h = 10$
$$ \text{Median} = 219.5 + \left(\frac{30 – 18}{26}\right) \times 10 = 219.5 + \frac{120}{26} = 219.5 + 4.62 = \mathbf{224.12 \text{ km}} $$
None of options a, b, c matches — answer is none of these.
(v) (d):
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} $$
$$ \Rightarrow \text{Mean} = \frac{3 \times 224.12 – 223.78}{2} = \frac{672.36 – 223.78}{2} = \frac{448.58}{2} = \mathbf{224.29 \text{ km}} $$
Answer is none of these.
(ii) (d): Convert class intervals to exclusive form by subtracting 0.5 from lower limits and adding 0.5 to upper limits:
$N = 60 \Rightarrow \frac{N}{2} = 30$. The cf just greater than 30 is in class 219.5–229.5.
(iii) (b): The cumulative frequency of the class preceding the median class is 18.
(iv) (d):
$l = 219.5,\; cf = 18,\; f = 26,\; h = 10$
$$ \text{Median} = 219.5 + \left(\frac{30 – 18}{26}\right) \times 10 = 219.5 + \frac{120}{26} = 219.5 + 4.62 = \mathbf{224.12 \text{ km}} $$
None of options a, b, c matches — answer is none of these.
(v) (d):
$$ \text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} $$
$$ \Rightarrow \text{Mean} = \frac{3 \times 224.12 – 223.78}{2} = \frac{672.36 – 223.78}{2} = \frac{448.58}{2} = \mathbf{224.29 \text{ km}} $$
Answer is none of these.
A group of 71 people visited a museum on a certain day. The following table shows their ages:
Based on the above information, answer the following questions:
(i) If true class limits have been decided by making the classes of interval 10, then the first class must be:
(a) 5–15 (b) 0–10 (c) 10–20 (d) none of these
(ii) The median class for the given data will be:
(a) 20–30 (b) 10–20 (c) 30–40 (d) 40–50
(iii) The cumulative frequency of class preceding the median class is:
(a) 22 (b) 13 (c) 25 (d) 35
(iv) The median age of the persons who visited the museum is:
(a) 30 years (b) 32.5 years (c) 34 years (d) 37.5 years
(v) If the price of a ticket for the age group 30–40 is ₹30, then the total amount spent by this age group is:
(a) ₹360 (b) ₹420 (c) ₹540 (d) ₹340
Based on the above information, answer the following questions:
(i) If true class limits have been decided by making the classes of interval 10, then the first class must be:
(a) 5–15 (b) 0–10 (c) 10–20 (d) none of these
(ii) The median class for the given data will be:
(a) 20–30 (b) 10–20 (c) 30–40 (d) 40–50
(iii) The cumulative frequency of class preceding the median class is:
(a) 22 (b) 13 (c) 25 (d) 35
(iv) The median age of the persons who visited the museum is:
(a) 30 years (b) 32.5 years (c) 34 years (d) 37.5 years
(v) If the price of a ticket for the age group 30–40 is ₹30, then the total amount spent by this age group is:
(a) ₹360 (b) ₹420 (c) ₹540 (d) ₹340
Answer
(i) (b): Age is a positive number, so the first class must be 0–10.
(ii) (c): From the cumulative frequency table:
$N = 71 \Rightarrow \frac{N}{2} = 35.5$. The cf just greater than 35.5 is in class 30–40.
(iii) (a): The cumulative frequency of the class preceding the median class (20–30) is 22.
(iv) (d): $l = 30,\; cf = 22,\; f = 18,\; h = 10$
$$ \text{Median} = 30 + \left(\frac{35.5 – 22}{18}\right) \times 10 = 30 + \frac{13.5 \times 10}{18} = 30 + 7.5 = \mathbf{37.5 \text{ years}} $$
(v) (c): Number of persons in age group 30–40 = 18.
$$ \text{Total amount} = 30 \times 18 = \mathbf{₹\,540} $$
(ii) (c): From the cumulative frequency table:
$N = 71 \Rightarrow \frac{N}{2} = 35.5$. The cf just greater than 35.5 is in class 30–40.
(iii) (a): The cumulative frequency of the class preceding the median class (20–30) is 22.
(iv) (d): $l = 30,\; cf = 22,\; f = 18,\; h = 10$
$$ \text{Median} = 30 + \left(\frac{35.5 – 22}{18}\right) \times 10 = 30 + \frac{13.5 \times 10}{18} = 30 + 7.5 = \mathbf{37.5 \text{ years}} $$
(v) (c): Number of persons in age group 30–40 = 18.
$$ \text{Total amount} = 30 \times 18 = \mathbf{₹\,540} $$
An electric scooter manufacturing company wants to declare the mileage of their electric scooters. For this, they recorded the mileage (km/charge) of 50 scooters of the same model. Details are given in the following table:
Based on the above information, answer the following questions:
(i) The average mileage is:
(a) 140 km/charge (b) 150 km/charge (c) 130 km/charge (d) 144.8 km/charge
(ii) The modal value of the given data is:
(a) 150 (b) 150.91 (c) 145.6 (d) 140.9
(iii) The median value of the given data is:
(a) 140 (b) 146.67 (c) 130 (d) 136.6
(iv) Assumed mean method is useful in determining the:
(a) Mean (b) Median (c) Mode (d) All of these
(v) The manufacturer can claim that the mileage for his scooter is:
(a) 144 km/charge (b) 155 km/charge (c) 165 km/charge (d) 175 km/charge
Based on the above information, answer the following questions:
(i) The average mileage is:
(a) 140 km/charge (b) 150 km/charge (c) 130 km/charge (d) 144.8 km/charge
(ii) The modal value of the given data is:
(a) 150 (b) 150.91 (c) 145.6 (d) 140.9
(iii) The median value of the given data is:
(a) 140 (b) 146.67 (c) 130 (d) 136.6
(iv) Assumed mean method is useful in determining the:
(a) Mean (b) Median (c) Mode (d) All of these
(v) The manufacturer can claim that the mileage for his scooter is:
(a) 144 km/charge (b) 155 km/charge (c) 165 km/charge (d) 175 km/charge
Answer
Frequency distribution table:
(i) (d): $$ \text{Mean} = \frac{7240}{50} = \mathbf{144.8 \text{ km/charge}} $$
(ii) (b): Highest frequency = 18 → modal class = 140–160. Here $l = 140,\; f_1 = 18,\; f_0 = 12,\; f_2 = 13,\; h = 20$
$$ \text{Mode} = 140 + \frac{18 – 12}{36 – 12 – 13} \times 20 = 140 + \frac{6}{11} \times 20 = 140 + 10.91 = \mathbf{150.91} $$
(iii) (b): $N = 50,\; \frac{N}{2} = 25$. Median class = 140–160 (cf just > 25 is 37).
$l = 140,\; cf = 19,\; f = 18,\; h = 20$
$$ \text{Median} = 140 + \frac{25 – 19}{18} \times 20 = 140 + \frac{120}{18} = 140 + 6.67 = \mathbf{146.67} $$
(iv) (a): Assumed mean method is useful in determining the Mean.
(v) (a): Mean = 144.8, Mode = 150.91, Median = 146.67. The minimum is approximately 144, so the manufacturer can claim 144 km/charge.
(i) (d): $$ \text{Mean} = \frac{7240}{50} = \mathbf{144.8 \text{ km/charge}} $$
(ii) (b): Highest frequency = 18 → modal class = 140–160. Here $l = 140,\; f_1 = 18,\; f_0 = 12,\; f_2 = 13,\; h = 20$
$$ \text{Mode} = 140 + \frac{18 – 12}{36 – 12 – 13} \times 20 = 140 + \frac{6}{11} \times 20 = 140 + 10.91 = \mathbf{150.91} $$
(iii) (b): $N = 50,\; \frac{N}{2} = 25$. Median class = 140–160 (cf just > 25 is 37).
$l = 140,\; cf = 19,\; f = 18,\; h = 20$
$$ \text{Median} = 140 + \frac{25 – 19}{18} \times 20 = 140 + \frac{120}{18} = 140 + 6.67 = \mathbf{146.67} $$
(iv) (a): Assumed mean method is useful in determining the Mean.
(v) (a): Mean = 144.8, Mode = 150.91, Median = 146.67. The minimum is approximately 144, so the manufacturer can claim 144 km/charge.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The arithmetic mean of the following given frequency distribution table is 13.81.
Reason (R): $$ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} $$
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The arithmetic mean of the following given frequency distribution table is 13.81.
Reason (R): $$ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} $$
Answer
(a) — Both A and R are true and R is the correct explanation of A.
The arithmetic mean formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ (Reason) correctly explains how to arrive at the mean value of 13.81 (Assertion).
The arithmetic mean formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ (Reason) correctly explains how to arrive at the mean value of 13.81 (Assertion).
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): 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 the median is 30.
Reason (R): $$ \text{Median} = \left(\frac{n + 1}{2}\right)^{\text{th}} \text{ value, if } n \text{ is odd.} $$
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): 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 the median is 30.
Reason (R): $$ \text{Median} = \left(\frac{n + 1}{2}\right)^{\text{th}} \text{ value, if } n \text{ is odd.} $$
Answer
(d) — A is false but R is true.
Arranging in ascending order: 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52 (n = 11, odd)
$$ \text{Median} = \left(\frac{11 + 1}{2}\right)^{\text{th}} = \text{6th value} = \mathbf{27} $$
Assertion states median = 30, which is false. The correct median is 27. Reason is true.
Arranging in ascending order: 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52 (n = 11, odd)
$$ \text{Median} = \left(\frac{11 + 1}{2}\right)^{\text{th}} = \text{6th value} = \mathbf{27} $$
Assertion states median = 30, which is false. The correct median is 27. Reason is true.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the number of runs scored by 11 players of a cricket team are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27, then the median is 27.
Reason (R): When the number of observations $n$ is odd, the median is the $\left(\frac{n+1}{2}\right)^{\text{th}}$ observation after arranging the data in ascending order.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the number of runs scored by 11 players of a cricket team are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27, then the median is 27.
Reason (R): When the number of observations $n$ is odd, the median is the $\left(\frac{n+1}{2}\right)^{\text{th}}$ observation after arranging the data in ascending order.
Answer
(a) — Both A and R are true and R is the correct explanation of A.
Ascending order: 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52
$n = 11$ (odd). Median position = $\frac{11+1}{2} = 6^{\text{th}}$ value = 27. Assertion is true, Reason is true and correctly explains it.
Ascending order: 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52
$n = 11$ (odd). Median position = $\frac{11+1}{2} = 6^{\text{th}}$ value = 27. Assertion is true, Reason is true and correctly explains it.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The mean of the following data can be calculated using the assumed mean method.
Reason (R): Assumed mean method is used to simplify calculations when values of $x$ are large or not convenient.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The mean of the following data can be calculated using the assumed mean method.
Reason (R): Assumed mean method is used to simplify calculations when values of $x$ are large or not convenient.
Answer
(a) — Both A and R are true and R is the correct explanation of A.
When class marks are large numbers, the assumed mean method simplifies calculations significantly. Assertion is true (the mean can be calculated using this method) and Reason correctly explains why this method is used.
When class marks are large numbers, the assumed mean method simplifies calculations significantly. Assertion is true (the mean can be calculated using this method) and Reason correctly explains why this method is used.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): For the following distribution, the median class is 20–30.
Reason (R): The median class is the class whose cumulative frequency is just greater than $\frac{N}{2}$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): For the following distribution, the median class is 20–30.
Reason (R): The median class is the class whose cumulative frequency is just greater than $\frac{N}{2}$.
Answer
(a) — Both A and R are true and R is the correct explanation of A.
The cumulative frequency just greater than $\frac{N}{2}$ falls in the class 20–30, making it the median class. Reason correctly states the rule for identifying the median class.
The cumulative frequency just greater than $\frac{N}{2}$ falls in the class 20–30, making it the median class. Reason correctly states the rule for identifying the median class.
Explore More CBSE Class 10 Maths Chapters
Real Numbers → Polynomials → Pair of Linear Equations in Two Variables → Quadratic Equations → Arithmetic Progressions → Triangles → Coordinate Geometry → Introduction to Trigonometry → Some Applications of Trigonometry → Circles → Areas Related to Circles → Surface Areas and Volumes → Probability →
Frequently Asked Questions
What is Statistics about in CBSE Class 10 Maths? ⌄
Statistics in CBSE Class 10 Maths deals with collecting, organising, and analysing numerical data. Your child will learn how to calculate the three measures of central tendency — mean, median, and mode — for grouped data, and will also study their interrelationship through the empirical formula. This chapter is practical, data-driven, and directly useful in everyday problem-solving.
How many marks does Statistics carry in the CBSE Class 10 board exam? ⌄
Statistics is part of the “Statistics and Probability” unit in CBSE Class 10 Maths, which typically carries around 11 marks in the board exam. Within this unit, Statistics alone can contribute 6–8 marks through MCQs, short answer, and case study questions. It is one of the more scoring chapters when practised consistently with structured questions.
What are the most important topics in Class 10 Statistics for board exams? ⌄
The most important topics are: (1) Mean of grouped data using direct method, assumed mean method, and step-deviation method; (2) Median of grouped data using the cumulative frequency formula; (3) Mode of grouped data using the modal class formula; and (4) the empirical relationship between mean, median, and mode — Mode = 3 × Median − 2 × Mean. Case study questions based on real-life contexts (transport, parks, hospitals) are especially important for the CBQ section.
What common mistakes do students make in Statistics questions? ⌄
The most common mistakes include: confusing the modal class with the median class, using the wrong cumulative frequency (cf should be the one just before the median class, not the median class itself), forgetting to convert discontinuous class intervals to continuous ones before applying the median formula, and incorrectly identifying the preceding or succeeding frequencies when calculating mode. Regular practice with varied data tables helps your child avoid these errors automatically.
How does Angle Belearn help students master Statistics for Class 10? ⌄
Angle Belearn’s CBSE specialists prepare competency-based questions that mirror the exact pattern of board exams, covering MCQs, case study questions, and assertion-reason problems all in one place. Each question comes with a detailed, step-by-step solution so your child can learn independently and understand the logic behind every answer. Our structured question banks are designed to build both conceptual clarity and exam speed for Statistics and every other Class 10 chapter.
