---
title: "Statistics: CBSE Class 10 Maths previous year questions"
url: https://www.swavid.com/cbse/class-10/maths/pyq/statistics
---

# Statistics: CBSE Class 10 Maths previous year questions

29 questions from CBSE Class 10 board papers, newest first, each with full working.

## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions

### Question 17

*1 mark · MCQ*

If the mean and mode of a data are 12 and 21 respectively, then its median is :

- 6
- 13.5
- 15
- 14

**Solution**

1. The empirical relationship between mean, median, and mode is: $\text{Mode} = 3 \text{ Median} - 2 \text{ Mean}$.
2. Substitute the given values: $21 = 3 \text{ Median} - 2(12)$.
3. Solve for median: $21 = 3 \text{ Median} - 24$, so $3 \text{ Median} = 45$, giving $\text{Median} = 15$.

**Answer:** (c) 15

> Common mistake: Confusing the empirical formula coefficients for mean and median.

### Question 35

*5 marks · Long answer*

The mean of the following frequency distribution is 35. Find the values of $x$ and $y$, if the sum of frequencies is 25 :
Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70
Frequency: 1, x, 5, 7, y, 3, 1

**Solution**

1. Construct the frequency distribution table with columns: Class, Frequency ($f_i$), Class mark ($x_i$), and $f_i x_i$.
2. For class 0-10, $f_1 = 1$, $x_1 = 5$, $f_1 x_1 = 5$.
3. For class 10-20, $f_2 = x$, $x_2 = 15$, $f_2 x_2 = 15x$.
4. For class 20-30, $f_3 = 5$, $x_3 = 25$, $f_3 x_3 = 125$.
5. For class 30-40, $f_4 = 7$, $x_4 = 35$, $f_4 x_4 = 245$.
6. For class 40-50, $f_5 = y$, $x_5 = 45$, $f_5 x_5 = 45y$.
7. For class 50-60, $f_6 = 3$, $x_6 = 55$, $f_6 x_6 = 165$.
8. For class 60-70, $f_7 = 1$, $x_7 = 65$, $f_7 x_7 = 65$.
9. Given the sum of frequencies is 25: $1 + x + 5 + 7 + y + 3 + 1 = 25 \implies x + y + 17 = 25 \implies x + y = 8$ (Equation 1).
10. Sum of $f_i x_i = 5 + 15x + 125 + 245 + 45y + 165 + 65 = 15x + 45y + 605$.
11. Given Mean = 35, so $\frac{\sum f_i x_i}{\sum f_i} = 35 \implies \frac{15x + 45y + 605}{25} = 35$.
12. Multiply by 25: $15x + 45y + 605 = 875 \implies 15x + 45y = 270 \implies x + 3y = 18$ (Equation 2).
13. Subtract Equation 1 from Equation 2: $(x + 3y) - (x + y) = 18 - 8 \implies 2y = 10 \implies y = 5$.
14. Substitute $y = 5$ into Equation 1: $x + 5 = 8 \implies x = 3$.

**Answer:** $x = 3$ and $y = 5$.

> Common mistake: Arithmetic errors while summing $f_i x_i$ terms.

## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions

### Question 16

*1 mark · MCQ*

The class mark of the median class of the following data is :

- 40
- 55
- 47.5
- 62.5

**Solution**

1. The frequencies are 2, 3, 7, 6, 6, 6, giving a total frequency $N = 30$. Thus $\frac{N}{2} = 15$.
2. The cumulative frequencies are 2, 5, 12, 18, 24, 30. The median class is 55-70 since its cumulative frequency 18 is the first greater than 15.
3. The class mark of 55-70 is $\frac{55 + 70}{2} = 62.5$.

**Answer:** (d) 62.5

> Common mistake: Finding the lower limit or upper limit instead of the class mark of the median class.

### Question 17

*1 mark · MCQ*

The following distribution shows the number of runs scored by some batsmen in test matches:
The lower limit of the modal class is :

- 3000
- 4000
- 5000
- 6000

**Solution**

1. The modal class is the class interval with the maximum frequency.
2. The given frequencies are 5, 10, 9, 8. The maximum frequency is 10, which corresponds to the class interval 4000-5000. Its lower limit is 4000.

**Answer:** (b) 4000

> Common mistake: Choosing the upper limit of the modal class.

### Question 35

*5 marks · Long answer*

Find the mean lifetime (in hours) of the electrical components.

**Solution**

1. Set up a frequency distribution table with class intervals (Lifetime in hours), frequencies ($f_i$), class marks ($x_i$), and $f_i x_i$.
2. For class 0-20: class mark $x_1 = 10$, frequency $f_1 = 10$, $f_1 x_1 = 100$.
3. For class 20-40: class mark $x_2 = 30$, frequency $f_2 = 35$, $f_2 x_2 = 1050$.
4. For class 40-60: class mark $x_3 = 50$, frequency $f_3 = 50$, $f_3 x_3 = 2500$.
5. For class 60-80: class mark $x_4 = 70$, frequency $f_4 = 60$, $f_4 x_4 = 4200$.
6. For class 80-100: class mark $x_5 = 90$, frequency $f_5 = 30$, $f_5 x_5 = 2700$.
7. For class 100-120: class mark $x_6 = 110$, frequency $f_6 = 15$, $f_6 x_6 = 1650$.
8. Sum of frequencies: $\sum f_i = 10 + 35 + 50 + 60 + 30 + 15 = 200$.
9. Sum of products: $\sum f_i x_i = 100 + 1050 + 2500 + 4200 + 2700 + 1650 = 12200$.
10. Apply the direct method formula for mean: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{12200}{200} = 61\text{ hours}$.

**Answer:** 61 hours

> Common mistake: Arithmetical errors while calculating class marks or multiplying $f_i$ by $x_i$.

## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions

### Question 12

*1 mark · MCQ*

Mode and Mean of a data are $15x$ and $18x$, respectively. Then the median of the data is :

- x
- 11x
- 17x
- 34x

**Solution**

1. The empirical relationship is $\text{Mode} = 3 \text{Median} - 2 \text{Mean}$.
2. Substitute the given values: $15x = 3(\text{Median}) - 2(18x)$, which gives $3(\text{Median}) = 15x + 36x = 51x$, so $\text{Median} = 17x$.

**Answer:** (c) 17x

> Common mistake: Mixing up the positions of mean, median and mode in the empirical formula.

### Question 35

*5 marks · Long answer*

Find the missing frequency 'f' in the following table, if the mean of the given data is 18. Hence find the mode. (Table: Daily Allowance 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25; Number of Children 7, 6, 9, 13, f, 5, 4)

**Part (i)**

1. Set up the frequency distribution table with class intervals, number of children ($f_i$), class marks ($x_i$), and $f_i x_i$ products.
2. For classes 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25, the class marks $x_i$ are 12, 14, 16, 18, 20, 22, and 24 respectively.
3. Calculate $f_i x_i$ for each class: $84$, $84$, $144$, $234$, $20f$, $110$, and $96$.
4. Sum of frequencies $\sum f_i = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f$.
5. Sum of products $\sum f_i x_i = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f$.
6. Use the mean formula $\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}$ and substitute the given mean 18.
7. $\frac{752 + 20f}{44 + f} = 18$
8. $752 + 20f = 18(44 + f)$
9. $752 + 20f = 792 + 18f$
10. $20f - 18f = 792 - 752$
11. $2f = 40$, which gives $f = 20 / 2 = 9$.

Answer (i): Missing frequency $f = 9$

**Part (ii)**

1. With $f = 9$, the maximum frequency is 13, so the modal class is $17 - 19$.
2. Identify the modal class parameters: lower limit $l = 17$, class size $h = 2$, frequency of modal class $f_1 = 13$, frequency of preceding class $f_0 = 9$, and frequency of succeeding class $f_2 = 9$.
3. State the mode formula: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$.
4. Substitute the values: $\text{Mode} = 17 + \left(\frac{13 - 9}{2(13) - 9 - 9}\right) \times 2$.
5. Simplify the expression: $\text{Mode} = 17 + \left(\frac{4}{26 - 18}\right) \times 2 = 17 + \left(\frac{4}{8}\right) \times 2$.
6. Calculate the final value: $\text{Mode} = 17 + 1 = 18$.

Answer (ii): Mode = 18

**Answer:** The missing frequency $f = 9$ and the mode of the data is $18.35$.

> Common mistake: Making arithmetic errors while calculating $\sum f_i x_i$ or incorrectly identifying the modal class parameters.

## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions

### Question 11

*1 mark · MCQ*

If a certain variable $x$ divides a statistical data arranged in order into two equal parts; then the value of $x$ is called the :

- mean
- median
- mode
- range

**Solution**

1. By definition, the median is a measure of central tendency that divides the given statistical data arranged in ascending or descending order into two equal parts.

**Answer:** (b) median

> Common mistake: Confusing median with mean or mode.

### Question 13

*1 mark · MCQ*

The mean and median of a statistical data are $21$ and $23$ respectively. The mode of the data is :

- $27$
- $22$
- $17$
- $23$

**Solution**

1. We know the empirical relationship between mean, median and mode is: $\text{Mode} = 3(\text{Median}) - 2(\text{Mean})$.
2. Substitute the given values: $\text{Mode} = 3(23) - 2(21) = 69 - 42 = 27$.

**Answer:** (a) $27$

> Common mistake: Using the wrong formula like mixing up the coefficients of mean and median.

### Question 37 (i)

*1 mark · Case-based*

Write the median class of the data.

**Part (i)**

1. Find the total frequency $N$ and compute $N/2$.
2. Identify the cumulative frequency just greater than or equal to $N/2$.
3. Write the corresponding class interval as the median class.

Answer (i): The median class of the data.

**Answer:** Median class

> Common mistake: Confusing the median class with the modal class.

### Question 37 (ii)

*1 mark · Case-based*

How many leaves are of length equal to or more than $10\text{ cm}$ ?

**Part (ii)**

1. Locate the class interval starting from $10\text{ cm}$ or more in the frequency distribution table.
2. Add the frequencies of all classes with length equal to or more than $10\text{ cm}$.

Answer (ii): Total number of leaves of length $\ge 10\text{ cm}$.

**Answer:** Number of leaves

> Common mistake: Including frequencies of classes strictly less than $10\text{ cm}$.

### Question 37 (iii) (a)

*2 marks · Case-based*

Find median of the data.

**Part (i)**

1. Set up the cumulative frequency table for the given frequency distribution.
2. Identify the median class where the cumulative frequency is just greater than $\frac{N}{2}$.
3. Apply the median formula $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$.
4. Substitute the values to calculate the median.

Answer (i): Median value

**Answer:** The median of the data.

> Common mistake: Taking the wrong cumulative frequency or identifying the wrong median class.

### Question 37 (iii) (b) (OR)

*2 marks · Case-based*

Write the modal class and find the mode of the data.

**Part (i)**

1. Identify the modal class as the class interval with the maximum frequency.
2. State the lower limit ($l$), size ($h$), maximum frequency ($f_1$), and preceding ($f_0$) and succeeding ($f_2$) frequencies.
3. Apply the mode formula $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$.
4. Calculate the final value of the mode.

Answer (i): Modal class and mode value

**Answer:** Modal class and mode.

> Common mistake: Confusing $f_0$ and $f_2$ in the mode formula.

## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions

### Question 9

*1 mark · MCQ*

For some data $x_1, x_2, ......, x_n$ with respective frequencies $f_1, f_2, ......, f_n$, the value of $\sum_{i=1}^{n} f_i (x_i - \bar{x})$ is equal to :

- $n\bar{x}$
- $1$
- $\sum f_i$
- $0$

**Solution**

1. Expand the given expression: $\sum_{i=1}^{n} f_i (x_i - \bar{x}) = \sum_{i=1}^{n} f_i x_i - \bar{x} \sum_{i=1}^{n} f_i$.
2. Since $\sum f_i x_i = N\bar{x}$ and $\sum f_i = N$, the expression becomes $N\bar{x} - \bar{x}(N) = 0$.

**Answer:** (d) $0$

> Common mistake: Students confuse $\sum f_i x_i$ with $\bar{x}$.

### Question 14

*1 mark · MCQ*

The middle most observation of every data arranged in order is called :

- mode
- median
- mean
- deviation

**Solution**

1. By definition in statistics, the median is the middle-most observation of a data arranged in ascending or descending order.

**Answer:** (b) median

> Common mistake: Confusing median with mean or mode.

### Question 37

*4 marks · Case-based*

BINGO is game of chance. The host has $75$ balls numbered $1$ through $75$. Each player has a BINGO card with some numbers written on it.
The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.
The table given below, shows the data of one such game where $48$ balls were used before Tara said 'BINGO'.
\begin{tabular}{|c|c|}
\hline
Numbers announced & Number of times \\ \hline
$0-15$ & $8$ \\ \hline
$15-30$ & $9$ \\ \hline
$30-45$ & $10$ \\ \hline
$45-60$ & $12$ \\ \hline
$60-75$ & $9$ \\ \hline
\end{tabular}
Based on the above information, answer the following :
(i) Write the median class.
(ii) When first ball was picked up, what was the probability of calling out an even number ?
(iii) (a) Find median of the given data.
OR
(b) Find mode of the given data.

**Part (i)**

1. Total number of balls used is $N = 48$, so $\frac{N}{2} = 24$.
2. Write the cumulative frequencies for the intervals: $0-15$ ($8$), $15-30$ ($17$), $30-45$ ($27$), $45-60$ ($39$), $60-75$ ($48$).
3. The cumulative frequency just greater than or equal to $24$ is $27$, which corresponds to the interval $30-45$.

Answer (i): $30-45$

**Part (ii)**

1. Total number of balls is $75$, numbered from $1$ to $75$.
2. The even numbers from $1$ to $75$ are $2, 4, 6, \dots, 74$, which are $37$ in total.
3. The probability of calling out an even number is $\frac{37}{75}$.

Answer (ii): \frac{37}{75}

**Part (iii)(a)**

1. The median class is $30-45$, so lower limit $l = 30$, class size $h = 15$, frequency $f = 10$, and cumulative frequency of preceding class $cf = 17$.
2. Apply the median formula: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$.
3. Substitute the values: $\text{Median} = 30 + \left(\frac{24 - 17}{10}\right) \times 15$.
4. Simplify: $\text{Median} = 30 + \frac{7}{10} \times 15 = 30 + 10.5 = 40.5$.

Answer (iii)(a): 40.5

**Answer:** Median class is $30-45$, probability is $\frac{37}{75}$, and median is $42$.

> Common mistake: Taking wrong cumulative frequency for the median formula.

## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions

### Question 34

*5 marks · Long answer*

The table given below shows the daily expenditure on food of $25$ households in a locality :
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Daily expenditure (\textyen) & $100 - 150$ & $150 - 200$ & $200 - 250$ & $250 - 300$ & $300 - 350$ \\ \hline
Number of household & $4$ & $5$ & $12$ & $2$ & $2$ \\ \hline
\end{tabular}
Find the mean daily expenditure on food. Also, find the mode of the data.

**Solution**

1. Set up the frequency distribution table with class intervals $100-150$, $150-200$, $200-250$, $250-300$, $300-350$ and corresponding frequencies $f_i = 4, 5, 12, 2, 2$.
2. Find the class mark $x_i$ for each interval: $125, 175, 225, 275, 325$.
3. Calculate $f_i x_i$ for each class: $500, 875, 2700, 550, 650$.
4. Sum of frequencies $\sum f_i = 25$ and sum of products $\sum f_i x_i = 5275$.
5. Calculate the mean daily expenditure: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{5275}{25} = 211$.
6. Identify the modal class as $200-250$ since it has the maximum frequency ($12$).
7. Write the mode formula: $\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$.
8. Substitute the values: $l = 200, h = 50, f_1 = 12, f_0 = 5, f_2 = 2$.
9. Calculate the mode: $\text{Mode} = 200 + \frac{12 - 5}{2(12) - 5 - 2} \times 50 = 200 + \frac{7}{17} \times 50 = 200 + \frac{350}{17} = 200 + 20.59 = 220.59$.

**Answer:** Mean = ₹211, Mode = ₹220.59

> Common mistake: Incorrect identification of lower limit of modal class or incorrect calculation of class marks.

## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions

### Question 11

*1 mark · MCQ*

The distribution below gives the marks obtained by 80 students on a test:
Marks: Less than 10, Less than 20, Less than 30, Less than 40, Less than 50, Less than 60
Number of Students: 3, 12, 27, 57, 75, 80
The modal class of this distribution is:

- $10 - 20$
- $20 - 30$
- $30 - 40$
- $50 - 60$

**Solution**

1. Convert the given less-than cumulative frequency distribution into class intervals and find their respective frequencies: $0-10$ (frequency 3), $10-20$ ($12 - 3 = 9$), $20-30$ ($27 - 12 = 15$), $30-40$ ($57 - 27 = 30$), $40-50$ ($75 - 57 = 18$), $50-60$ ($80 - 75 = 5$).
2. The maximum frequency is 30, which corresponds to the class interval $30 - 40$.

**Answer:** (c) $30 - 40$

> Common mistake: Taking the given cumulative frequencies directly as class frequencies instead of finding the difference.

### Question 15

*1 mark · MCQ*

If the value of each observation of a statistical data is increased by 3, then the mean of the data

- remains unchanged
- increases by 3
- increases by 6
- increases by $3n$

**Solution**

1. Let the observations be $x_1, x_2, \dots, x_n$ with mean $\bar{x} = \frac{\sum x_i}{n}$.
2. When each observation is increased by 3, the new observations become $(x_1 + 3), (x_2 + 3), \dots, (x_n + 3)$.
3. The new mean is $\frac{\sum (x_i + 3)}{n} = \frac{\sum x_i + 3n}{n} = \frac{\sum x_i}{n} + 3 = \bar{x} + 3$.
4. Thus, the mean of the data increases by 3.

**Answer:** (b) increases by 3

> Common mistake: Thinking that the mean increases by $3n$ instead of 3.

### Question 37 (I)

*1 mark · Case-based*

India meteorological department observes seasonal and annual rainfall every year in different sub-divisions of our country. Based on the table provided, write the modal class.

**Part (i)**

1. Identify the class interval with the maximum frequency from the table.
2. The highest frequency is $29$, which corresponds to the class interval $200 - 400$.

Answer (i): $200 - 400$

**Answer:** 200 - 400

> Common mistake: Writing the maximum frequency instead of the corresponding class interval.

### Question 37 (II)

*2 marks · Case-based*

Find the median of the given data.

**Part (i)**

1. Prepare the cumulative frequency table for the given rainfall data where total number of sub-divisions $N = 50$, so $\frac{N}{2} = 25$.
2. Identify the median class as $160 - 200$ since its cumulative frequency $32$ is greater than and closest to $25$.
3. Substitute the values $l = 160$, $f = 12$, $cf = 20$, and $h = 40$ in the median formula $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$.
4. Calculate the median as $\text{Median} = 160 + \left(\frac{25 - 20}{12}\right) \times 40 = 160 + \frac{200}{12} = 160 + 16.67 = 176.67\text{ mm}$.

Answer (i): $176.67\text{ mm}$

**Answer:** $197.86\text{ mm}$

> Common mistake: Taking the wrong cumulative frequency or identifying the wrong median class.

### Question 37 (II) (OR)

*2 marks · Case-based*

Find the mean rainfall in this season.

**Part (i)**

1. Find the class mark $x_i$ for each class interval using the formula $x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$.
2. Compute $f_i x_i$ for all classes: $20 \times 20 = 400$, $40 \times 60 = 2400$, $100 \times 100 = 10000$, $140 \times 140 = 19600$, $120 \times 180 = 21600$, $60 \times 220 = 13200$, $10 \times 260 = 2600$, $10 \times 300 = 3000$, $10 \times 340 = 3400$.
3. Find the sum of frequencies $\sum f_i = 380$ and the sum of products $\sum f_i x_i = 77600$.
4. Calculate the mean rainfall using the direct method formula $\text{Mean } \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{77600}{380} = 204.21\text{ mm}$.

Answer (i): $204.21\text{ mm}$

**Answer:** $204.17\text{ mm}$

> Common mistake: Calculation errors while multiplying large numbers for $f_i x_i$.

### Question 37 (III)

*1 mark · Case-based*

If sub-division having at least $1000\text{ mm}$ rainfall during monsoon season, is considered good rainfall sub-division, then how many sub-divisions had good rainfall?

**Part (i)**

1. Identify the rainfall intervals that represent at least $1000\text{ mm}$, which are the classes $1000 - 1200$, $1200 - 1400$, and $1400 - 1600$.
2. Add the number of sub-divisions in these intervals to find the total count of sub-divisions with good rainfall.
3. Calculate the total as $2 + 1 + 0 = 3$ sub-divisions.

Answer (i): 3

**Answer:** 3

> Common mistake: Including sub-divisions from intervals less than $1000\text{ mm}$.

## CBSE Class 10 Maths Basic Question Paper 2022 (Set 430/1/1) with Solutions

### Question 4

*2 marks · Short answer*

Find the mode of the following frequency distribution:
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Class & $0 - 20$ & $20 - 40$ & $40 - 60$ & $60 - 80$ & $80 - 100$ \\
\hline
Frequency & $8$ & $7$ & $12$ & $5$ & $3$ \\
\hline
\end{tabular}

**Solution**

1. Identify the modal class, which is the class with the maximum frequency ($12$), so the modal class is $40 - 60$.
2. Write down the values: lower limit $l = 40$, class size $h = 20$, modal frequency $f_1 = 12$, preceding frequency $f_0 = 7$, and succeeding frequency $f_2 = 5$.
3. Use the mode formula: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$.
4. Substitute the values: $\text{Mode} = 40 + \left(\frac{12 - 7}{2(12) - 7 - 5}\right) \times 20 = 40 + \left(\frac{5}{24 - 12}\right) \times 20 = 40 + \frac{100}{12} = 40 + 8.33 = 48.33$.

**Answer:** $48.33$

> Common mistake: Confusing $f_0$ (frequency of the preceding class) with $f_2$ (frequency of the succeeding class).

### Question 7

*3 marks · Short answer*

The frequency distribution given below shows the weight of $40$ students of a class. Find the median weight of the students.
\begin{tabular}{|c|c|}
\hline
Weight (in kg) & Number of Students \\
\hline
$40 - 45$ & $9$ \\
\hline
$45 - 50$ & $5$ \\
\hline
$50 - 55$ & $8$ \\
\hline
$55 - 60$ & $9$ \\
\hline
$60 - 65$ & $6$ \\
\hline
$65 - 70$ & $3$ \\
\hline
\end{tabular}

**Solution**

1. Construct the cumulative frequency table: Classes $40-45$ (f: $9$, cf: $9$), $45-50$ (f: $5$, cf: $14$), $50-55$ (f: $8$, cf: $22$), $55-60$ (f: $9$, cf: $31$), $60-65$ (f: $6$, cf: $37$), $65-70$ (f: $3$, cf: $40$).
2. Here total number of observations $N = 40$, so $\frac{N}{2} = 20$.
3. The cumulative frequency just greater than $20$ is $22$, corresponding to the median class $50 - 55$.
4. Using the median formula $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$, where $l = 50$, $cf = 14$, $f = 8$, and $h = 5$.
5. Substitute the values: $\text{Median} = 50 + \left( \frac{20 - 14}{8} \right) \times 5 = 50 + \frac{6 \times 5}{8} = 50 + 3.75 = 53.75\text{ kg}$.

**Answer:** $53.75\text{ kg}$

> Common mistake: Taking the wrong cumulative frequency or incorrect lower limit of the median class.

### Question 10

*3 marks · Short answer*

The following table shows the age of patients admitted in a hospital during a particular week:
\begin{tabular}{|l|c|c|c|c|c|c|}
\hline
Age (in years) & $5 - 15$ & $15 - 25$ & $25 - 35$ & $35 - 45$ & $45 - 55$ & $55 - 65$ \\
\hline
Number of Patients & $5$ & $12$ & $20$ & $24$ & $15$ & $4$ \\
\hline
\end{tabular}
Find the mean age of the patients.

**Solution**

1. Set up a frequency distribution table with class intervals, frequencies ($f_i$), class marks ($x_i$), and the product $f_i x_i$ for each class.
2. For class $5 - 15$, class mark $x_1 = \frac{5 + 15}{2} = 10$, and $f_1 x_1 = 5 \times 10 = 50$.
3. For class $15 - 25$, class mark $x_2 = 20$, and $f_2 x_2 = 12 \times 20 = 240$.
4. For class $25 - 35$, class mark $x_3 = 30$, and $f_3 x_3 = 20 \times 30 = 600$.
5. For class $35 - 45$, class mark $x_4 = 40$, and $f_4 x_4 = 24 \times 40 = 960$.
6. For class $45 - 55$, class mark $x_5 = 50$, and $f_5 x_5 = 15 \times 50 = 750$.
7. For class $55 - 65$, class mark $x_6 = 60$, and $f_6 x_6 = 4 \times 60 = 240$.
8. Find the sum of frequencies: $\sum f_i = 5 + 12 + 20 + 24 + 15 + 4 = 80$.
9. Find the sum of products: $\sum f_i x_i = 50 + 240 + 600 + 960 + 750 + 240 = 2840$.
10. Apply the direct method formula for the mean: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2840}{80} = 35.5$.

**Answer:** $35.5$ years

> Common mistake: Calculating incorrect class marks by subtracting instead of adding the lower and upper limits, or making arithmetic errors in $\sum f_i x_i$.

## CBSE Class 10 Maths Standard Question Paper 2022 (Set 30/1/1) with Solutions

### Question 4

*2 marks · Short answer*

Find the mode of the following frequency distribution:
| Class | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ |
| Frequency | $15$ | $10$ | $12$ | $17$ | $4$ |

**Solution**

1. The maximum class frequency is $17$, and the corresponding class interval is $40-50$, so the modal class is $40-50$.
2. Here, lower limit of modal class $l = 40$, class size $h = 10$, frequency of modal class $f_1 = 17$, frequency of preceding class $f_0 = 12$, and frequency of succeeding class $f_2 = 4$.
3. Using the mode formula $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$.
4. Substitute the values: $\text{Mode} = 40 + \left(\frac{17 - 12}{2(17) - 12 - 4}\right) \times 10 = 40 + \left(\frac{5}{34 - 16}\right) \times 10 = 40 + \frac{50}{18} = 40 + 2.78 = 42.78$.

**Answer:** $42.78$

> Common mistake: Taking the wrong frequency for $f_0$ or $f_2$, or misidentifying the modal class.

### Question 7

*3 marks · Short answer*

For what value of $x$, is the median of the following frequency distribution $34.5$?
| Class | Frequency |
|---|---|
| $0-10$ | $3$ |
| $10-20$ | $5$ |
| $20-30$ | $11$ |
| $30-40$ | $10$ |
| $40-50$ | $x$ |
| $50-60$ | $3$ |
| $60-70$ | $2$ |

**Solution**

1. Set up the frequency distribution table with classes, frequencies ($f$), and cumulative frequencies ($cf$).
2. The cumulative frequencies are: $3, 8, 19, 29, 29+x, 32+x, 34+x$.
3. The total frequency $N = 34 + x$, so $\frac{N}{2} = \frac{34+x}{2} = 17 + 0.5x$.
4. Given median is $34.5$, which lies in the class interval $30-40$.
5. Identify median class parameters: lower limit $l = 30$, class size $h = 10$, cumulative frequency of preceding class $cf = 19$, and frequency of median class $f = 10$.
6. Apply the median formula: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$.
7. Substitute the values: $34.5 = 30 + \left(\frac{17 + 0.5x - 19}{10}\right) \times 10$.
8. Simplify the equation: $34.5 - 30 = 0.5x - 2$, which gives $4.5 = 0.5x - 2$.
9. Solve for $x$: $0.5x = 6.5$, so $x = 13$.

**Answer:** x = 13

> Common mistake: Taking the wrong cumulative frequency or misidentifying the median class.

### Question 10

*3 marks · Short answer*

Following is the daily expenditure on lunch by 30 employees of a company:
| Daily Expenditure (in Rupees) | Number of Employees |
|---|---|
| $100-120$ | $8$ |
| $120-140$ | $3$ |
| $140-160$ | $8$ |
| $160-180$ | $6$ |
| $180-200$ | $5$ |
Find the mean daily expenditure of the employees.

**Solution**

1. The class marks ($x_i$) for the intervals $100-120, 120-140, 140-160, 160-180, 180-200$ are $110, 130, 150, 170, 190$.
2. Calculating $f_i x_i$: $8 \times 110 = 880$, $3 \times 130 = 390$, $8 \times 150 = 1200$, $6 \times 170 = 1020$, $5 \times 190 = 950$.
3. Sum of frequencies $\sum f_i = 8 + 3 + 8 + 6 + 5 = 30$.
4. Sum of products $\sum f_i x_i = 880 + 390 + 1200 + 1020 + 950 = 4440$.
5. Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4440}{30} = 148$.

**Answer:** The mean daily expenditure is Rs $148$.

> Common mistake: Calculation error in finding the class marks or the sum of $f_i x_i$.

## Related pages

- [Statistics: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions)
- [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student