CBSE Class 10 · Maths 
Total frequency $N = 60$, so $\dfrac{N}{2} = 30$. The cumulative frequency just greater than or equal to 30 is in class $160-165$.
Upper limit of median class = 165 
Using the Mode formula: $$\text{Mode} = l + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times h = 30 + \frac{12-7}{24-7-5} \times 10 = 30 + \frac{5}{12} \times 10 = 30 + 4.17$$ Modal marks ≈ 34.17 marks 
From total frequency: $55 + a + b = 70 \Rightarrow a + b = 15$ …(i)
Using Median formula (median class $15-20$): $$16 = 15 + \frac{35-24-a}{15} \times 5 \Rightarrow 1 = \frac{11-a}{3} \Rightarrow a = 8$$ From (i): $8 + b = 15 \Rightarrow b = 7$
Therefore $a = 8$ and $b = 7$. 
$\Sigma f_i = 40$, $\Sigma f_i d_i = -81$, Assumed mean $A = 149$
$$\text{Mean} = A + \frac{\Sigma f_i d_i}{\Sigma f_i} = 149 + \frac{-81}{40} = 149 – 2.025 = \mathbf{146.975 \text{ mm}}$$ 
$N = 50$, $\dfrac{N}{2} = 25$. Median class = $140-145$. Here $l=140$, $h=5$, $cf=15$, $f=12$.
$$\text{Median} = 140 + 5 \times \frac{25-15}{12} = 140 + \frac{50}{12} = 140 + 4.167 = \mathbf{144.167 \text{ cm}}$$ 
Let $x$ = number of days in $20-40$ mm range.
$$35 = \frac{850+30x}{21+x} \Rightarrow 735+35x = 850+30x \Rightarrow 5x = 115 \Rightarrow \boxed{x = 23}$$ Number of days = 23 
$N = 100$, so $\dfrac{N}{2} = 50$. Cumulative frequency just greater than 50 is 75, corresponding to class $155-160$.
Median class = $155-160$ 
$N = 60$, $\dfrac{N}{2} = 30$. Median class = $20-30$. $l=20$, $cf=13$, $f=20$, $h=10$.
$$\text{Median} = 20 + \frac{30-13}{20} \times 10 = 20 + \frac{17}{2} = 20 + 8.5 = \mathbf{28.5}$$ 
$\Sigma f_i = 90$, $\Sigma f_i x_i = 664$
(i) $\text{Mean} = \dfrac{664}{90} \approx 7.38$ cars/day. If demand doubles: $2 \times 7.38 \approx \mathbf{15 \text{ cars/day}}$
(ii) Mean lies in range $4-8$, so at least on 33 days fewer than average cars were assembled. 
$N=80$, $\dfrac{N}{2}=40$. Median class = $50-60$.
Median: $l=50$, $cf=37$, $f=15$, $h=10$: $$\text{Median} = 50 + \frac{40-37}{15} \times 10 = 50 + 2 = \mathbf{52}$$ Mode: Modal class = $50-60$. $l=50$, $f_1=15$, $f_0=12$, $f_2=12$, $h=10$: $$\text{Mode} = 50 + \frac{15-12}{30-12-12} \times 10 = 50 + \frac{3}{6} \times 10 = 50 + 5 = \mathbf{55}$$ 
From total frequency: $96 + x + y = 150 \Rightarrow x + y = 54$ …(1)
From Mean $= 91$, setting up the second equation and solving simultaneously:
$x = 34$ and $y = 20$ 
From total: $p + q + 78 = 90 \Rightarrow p + q = 12$ …(i)
Using median formula (median class $50-60$): $$50 = 50 + \frac{45-(p+40)}{20} \times 10 \Rightarrow 0 = \frac{45-p-40}{2} \Rightarrow p = 5$$ From (i): $q = 7$
Mode: $l=40$, $f_1=25$, $f_0=15$, $f_2=20$, $h=10$: $$\text{Mode} = 40 + \frac{25-15}{50-15-20} \times 10 = 40 + \frac{100}{15} = 40 + 6.67 = \mathbf{46.67}$$ $p = 5$, $q = 7$, Mode $= 46.67$
Table 1
Table 2
Refer to Table 1:
1. The average age for which maximum cases occurred is: (A) 32.24 (B) 34.36 (C) 36.82 (D) 42.24
2. The upper limit of modal class is: (A) 15 (B) 25 (C) 35 (D) 45
3. The mean of the given data is: (A) 26.2 (B) 32.4 (C) 33.5 (D) 35.4
4. The mode of the given data is: (A) 41.4 (B) 48.2 (C) 55.3 (D) 64.6
5. The median of the given data is: (A) 32.7 (B) 40.2 (C) 42.3 (D) 48.6
CBSE Class 10 Maths Statistics Previous Year Questions
Help your child master CBSE Class 10 Maths Statistics with this carefully curated collection of previous year questions drawn from real board papers spanning 2020–2024. Every question comes with a detailed step-by-step solution covering mean, median, mode, and the empirical relationship — topics that consistently appear across all question types in the board exam.
CBSE Class 10 Maths Statistics — Questions with Solutions
The empirical relation between the mode, median and mean of a distribution is:
Solution
Answer: Option (A) is correct.
Explanation: The standard empirical formula relating the three measures of central tendency is: Mode $= 3$ Median $- 2$ Mean.
Explanation: The standard empirical formula relating the three measures of central tendency is: Mode $= 3$ Median $- 2$ Mean.
For the following distribution:
The modal class is:
The modal class is:
Solution
Answer: Option (C) is correct.
Maximum frequency is 30, which belongs to class $30-40$. Therefore, the modal class is $\mathbf{30-40}$.
Maximum frequency is 30, which belongs to class $30-40$. Therefore, the modal class is $\mathbf{30-40}$.
The upper limit of the modal class of the given distribution is:
Solution
Answer: Option (D) is correct.
Maximum frequency is 18, corresponding to class $145-150$. So the modal class is $145-150$ and its upper limit is 150.
Maximum frequency is 18, corresponding to class $145-150$. So the modal class is $145-150$ and its upper limit is 150.
Sweety, Nitesh, and Ashraf visited a hospital for their annual body checkup. Their systolic blood pressure readings: Sweety: 121 mmHg, Nitesh: 147 mmHg, Ashraf: 160 mmHg. The table below depicts the systolic blood pressure ranges of all patients who visited the hospital on the same day.
Who among the three friends have a blood pressure reading that falls in the modal class?
Who among the three friends have a blood pressure reading that falls in the modal class?
Solution
Answer: Option (B) is correct.
Explanation: Maximum frequency is 19, corresponding to class $145-155$. So the modal class is $145-155$. Nitesh’s blood pressure is 147 mmHg, which falls in this modal class. Therefore Nitesh is the answer.
Explanation: Maximum frequency is 19, corresponding to class $145-155$. So the modal class is $145-155$. Nitesh’s blood pressure is 147 mmHg, which falls in this modal class. Therefore Nitesh is the answer.
The table below depicts the weight of students of class 6 of Red Bricks Public School. There are 18 students who weigh above the median weight.
If there are no students with the same weight as the median, how many students weigh between the range of $37-40$ kgs?
If there are no students with the same weight as the median, how many students weigh between the range of $37-40$ kgs?
Solution
Answer: Option (A) is correct.
Let $x$ be students in $37-40$ kg. Since 18 students weigh above median: $$18 \times 2 > 31 + x \Rightarrow 36 – 31 > x \Rightarrow x = 5$$ Thus 5 students weigh between $37-40$ kgs.
Let $x$ be students in $37-40$ kg. Since 18 students weigh above median: $$18 \times 2 > 31 + x \Rightarrow 36 – 31 > x \Rightarrow x = 5$$ Thus 5 students weigh between $37-40$ kgs.
For the following distribution:
The sum of lower limits of median class and modal class is:
The sum of lower limits of median class and modal class is:
Solution
Answer: Option (B) is correct.
Maximum frequency 20 belongs to class $15-20$ → modal class lower limit = 15.
$\Sigma f_i = 66$, $\dfrac{N}{2} = 33$ falls in class $10-15$ → median class lower limit = 10.
Sum $= 15 + 10 = \mathbf{25}$
Maximum frequency 20 belongs to class $15-20$ → modal class lower limit = 15.
$\Sigma f_i = 66$, $\dfrac{N}{2} = 33$ falls in class $10-15$ → median class lower limit = 10.
Sum $= 15 + 10 = \mathbf{25}$
An outlier is a data point that differs significantly from other observations. If an outlier is included in the following data set, which measure(s) of central tendency would change?
12, 15, 22, 44, 44, 48, 50, 51
12, 15, 22, 44, 44, 48, 50, 51
Solution
Answer: Option (A) is correct.
Explanation: Including an outlier changes the sum and count of observations, so the mean changes. The median and mode are positional/frequency-based and are less sensitive to extreme values in this dataset.
Explanation: Including an outlier changes the sum and count of observations, so the mean changes. The median and mode are positional/frequency-based and are less sensitive to extreme values in this dataset.
If mode of some data is 7 and their mean is also 7, then their median is:
Solution
Answer: Option (D) is correct.
Using the empirical formula: $$\text{Median} = \frac{1}{3}\text{Mode} + \frac{2}{3}\text{Mean} = \frac{1}{3}(7) + \frac{2}{3}(7) = \frac{7}{3} + \frac{14}{3} = \frac{21}{3} = \mathbf{7}$$
Using the empirical formula: $$\text{Median} = \frac{1}{3}\text{Mode} + \frac{2}{3}\text{Mean} = \frac{1}{3}(7) + \frac{2}{3}(7) = \frac{7}{3} + \frac{14}{3} = \frac{21}{3} = \mathbf{7}$$
The class mark of the median class of the following data is:
Solution
Answer: Option (C) is correct.
Total = 30, so $\dfrac{N}{2} = 15$. Median class = $40-55$. $$\text{Class Mark} = \frac{40+55}{2} = \mathbf{47.5}$$
Total = 30, so $\dfrac{N}{2} = 15$. Median class = $40-55$. $$\text{Class Mark} = \frac{40+55}{2} = \mathbf{47.5}$$
The following distribution shows the number of runs scored by some batsmen in test matches:
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The lower limit of the modal class is:
The lower limit of the modal class is:
Solution
Answer: Option (B) is correct.
Explanation: Frequencies — 3000–4000: 5, 4000–5000: 10, 5000–6000: 9, 6000–7000: 8. Highest frequency (10) is in class $4000-5000$. Lower limit = 4000.
Explanation: Frequencies — 3000–4000: 5, 4000–5000: 10, 5000–6000: 9, 6000–7000: 8. Highest frequency (10) is in class $4000-5000$. Lower limit = 4000.
The Mode and Mean of a data are $15x$ and $18x$ respectively. Then the median of the data is:
Solution
Answer: Option (C) is correct.
Using Mode $= 3$ Median $- 2$ Mean: $$15x = 3 \times \text{Median} – 2 \times 18x \Rightarrow 15x = 3\,\text{Median} – 36x \Rightarrow 3\,\text{Median} = 51x \Rightarrow \text{Median} = \mathbf{17x}$$
Using Mode $= 3$ Median $- 2$ Mean: $$15x = 3 \times \text{Median} – 2 \times 18x \Rightarrow 15x = 3\,\text{Median} – 36x \Rightarrow 3\,\text{Median} = 51x \Rightarrow \text{Median} = \mathbf{17x}$$
The mean and median of a statistical data are 21 and 23 respectively. The mode of the data is:
Solution
Answer: Option (A) is correct.
$$\text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} = 3 \times 23 – 2 \times 21 = 69 – 42 = \mathbf{27}$$
$$\text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean} = 3 \times 23 – 2 \times 21 = 69 – 42 = \mathbf{27}$$
If a certain variable $x$ divides a statistical data arranged in order into two equal parts, then the value of $x$ is called the:
Solution
Answer: Option (B) is correct.
The median is the value that divides data arranged in order into two equal halves — with half the data points below it and half above it.
The median is the value that divides data arranged in order into two equal halves — with half the data points below it and half above it.
For some data $x_1, x_2, \ldots, x_n$ with respective frequencies $f_1, f_2, \ldots, f_n$, the value of $$\sum_{i=1}^{n} f_i(x_i – \bar{x})$$ is equal to:
Solution
Answer: Option (D) is correct.
The sum of deviations of all observations from the mean, weighted by their frequencies, is always zero. This is a fundamental property of the arithmetic mean.
The sum of deviations of all observations from the mean, weighted by their frequencies, is always zero. This is a fundamental property of the arithmetic mean.
The middle most observation of every data arranged in order is called:
Solution
Answer: Option (B) is correct.
The median is the middlemost value when data is arranged in order, dividing the dataset into two equal halves.
The median is the middlemost value when data is arranged in order, dividing the dataset into two equal halves.
Find the class-marks of the classes $10-25$ and $35-55$.
Answer
Class mark of $10-25 = \dfrac{10+25}{2} = \dfrac{35}{2} = \mathbf{17.5}$
Class mark of $35-55 = \dfrac{35+55}{2} = \dfrac{90}{2} = \mathbf{45}$
Class mark of $35-55 = \dfrac{35+55}{2} = \dfrac{90}{2} = \mathbf{45}$
Consider the following frequency distribution of the heights of 60 students of a class.
Find the upper limit of the median class in the given data.
Find the upper limit of the median class in the given data.
Answer
Total frequency $N = 60$, so $\dfrac{N}{2} = 30$. The cumulative frequency just greater than or equal to 30 is in class $160-165$.
Upper limit of median class = 165
Find the mode of the following distribution:
Answer
Using the Mode formula: $$\text{Mode} = l + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times h = 30 + \frac{12-7}{24-7-5} \times 10 = 30 + \frac{5}{12} \times 10 = 30 + 4.17$$ Modal marks ≈ 34.17 marks
If mode of the following frequency distribution is 55, then find the value of $x$.
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Answer
Mode $= 55$, so modal class is $45-60$. Here $l=45$, $h=15$, $f_1=15$, $f_0=x$, $f_2=10$.
$$55 = 45 + \frac{15-x}{30-x-10} \times 15 \Rightarrow 10(20-x) = 15(15-x) \Rightarrow 200-10x = 225-15x \Rightarrow 5x = 25$$ $$\boxed{x = 5}$$
$$55 = 45 + \frac{15-x}{30-x-10} \times 15 \Rightarrow 10(20-x) = 15(15-x) \Rightarrow 200-10x = 225-15x \Rightarrow 5x = 25$$ $$\boxed{x = 5}$$
The median of the following data is 16. Find the missing frequencies $a$ and $b$, if the total of the frequencies is 70.
Answer
From total frequency: $55 + a + b = 70 \Rightarrow a + b = 15$ …(i)
Using Median formula (median class $15-20$): $$16 = 15 + \frac{35-24-a}{15} \times 5 \Rightarrow 1 = \frac{11-a}{3} \Rightarrow a = 8$$ From (i): $8 + b = 15 \Rightarrow b = 7$
Therefore $a = 8$ and $b = 7$.
The length of 40 leaves of a plant are measured correct to nearest millimeter. Find the mean length of the leaves.
Answer
$\Sigma f_i = 40$, $\Sigma f_i d_i = -81$, Assumed mean $A = 149$
$$\text{Mean} = A + \frac{\Sigma f_i d_i}{\Sigma f_i} = 149 + \frac{-81}{40} = 149 – 2.025 = \mathbf{146.975 \text{ mm}}$$
Heights of 50 students of class X of a school are recorded. Find the median height.
Answer
$N = 50$, $\dfrac{N}{2} = 25$. Median class = $140-145$. Here $l=140$, $h=5$, $cf=15$, $f=12$.
$$\text{Median} = 140 + 5 \times \frac{25-15}{12} = 140 + \frac{50}{12} = 140 + 4.167 = \mathbf{144.167 \text{ cm}}$$
In a class test, the mean score of the class is 60. Half the students scored 80 marks or above. Dipti said, “Each of the remaining half would have definitely got 40 marks or below for the mean to be 60.” Prove or disprove Dipti’s statement with a valid example.
Answer
Dipti’s claim is NOT correct.
Consider 4 students with scores: 80, 90, 20, 50.
Mean $= \dfrac{80+90+20+50}{4} = \dfrac{240}{4} = 60$ ✓
Half the students (80 and 90) scored 80 or above, but one student scored 50 — which is above 40. This disproves Dipti’s statement.
Consider 4 students with scores: 80, 90, 20, 50.
Mean $= \dfrac{80+90+20+50}{4} = \dfrac{240}{4} = 60$ ✓
Half the students (80 and 90) scored 80 or above, but one student scored 50 — which is above 40. This disproves Dipti’s statement.
The mean temperature of a city for 31 consecutive days was $35.7°C$. Mean of first 8 days was $28.4°C$ and mean of next 12 days was $36.4°C$. Find the mean temperature of the remaining days.
Answer
Remaining days $= 31 – (8+12) = 11$
Sum of 31 days $= 35.7 \times 31 = 1106.7°C$
Sum of 8 days $= 28.4 \times 8 = 227.2°C$
Sum of 12 days $= 36.4 \times 12 = 436.8°C$
Sum of 11 days $= 1106.7 – (227.2 + 436.8) = 442.7°C$
$$\text{Mean} = \frac{442.7}{11} = \mathbf{40.2°C}$$
Sum of 31 days $= 35.7 \times 31 = 1106.7°C$
Sum of 8 days $= 28.4 \times 8 = 227.2°C$
Sum of 12 days $= 36.4 \times 12 = 436.8°C$
Sum of 11 days $= 1106.7 – (227.2 + 436.8) = 442.7°C$
$$\text{Mean} = \frac{442.7}{11} = \mathbf{40.2°C}$$
A bowler has taken 148 wickets with a strike rate of 27. In his next match, he bowls 48 balls and takes 2 wickets. What is his new strike rate?
Answer
Total balls so far $= 148 \times 27 = 3996$
Total balls after match $= 3996 + 48 = 4044$
Total wickets after match $= 148 + 2 = 150$
$$\text{New Strike Rate} = \frac{4044}{150} = \mathbf{26.96}$$
Total balls after match $= 3996 + 48 = 4044$
Total wickets after match $= 148 + 2 = 150$
$$\text{New Strike Rate} = \frac{4044}{150} = \mathbf{26.96}$$
The frequency distribution of daily rainfall in a town is shown below. The information on the $20-40$ mm range was deleted. If the mean daily rainfall was 35 mm, find the number of days when rainfall ranged between $20-40$ mm.
Answer
Let $x$ = number of days in $20-40$ mm range.
$$35 = \frac{850+30x}{21+x} \Rightarrow 735+35x = 850+30x \Rightarrow 5x = 115 \Rightarrow \boxed{x = 23}$$ Number of days = 23
Following distribution gives cumulative frequencies of ‘more than type’:
Change the above data to a continuous grouped frequency distribution.
Change the above data to a continuous grouped frequency distribution.
Answer
The required continuous grouped frequency distribution:
In the following frequency distribution, find the median class.
Answer
$N = 100$, so $\dfrac{N}{2} = 50$. Cumulative frequency just greater than 50 is 75, corresponding to class $155-160$.
Median class = $155-160$
Find the median of the following distribution:
Answer
$N = 60$, $\dfrac{N}{2} = 30$. Median class = $20-30$. $l=20$, $cf=13$, $f=20$, $h=10$.
$$\text{Median} = 20 + \frac{30-13}{20} \times 10 = 20 + \frac{17}{2} = 20 + 8.5 = \mathbf{28.5}$$
Find the mean of the following frequency distribution:
Answer
Using $\text{Mean} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$
Midpoints × frequencies: 0–10: $5×12=60$; 10–20: $15×18=270$; 20–30: $25×27=675$; 30–40: $35×20=700$; 40–50: $45×17=765$; 50–60: $55×6=330$
$\Sigma f_i x_i = 2800$, $\Sigma f_i = 100$
$$\text{Mean} = \frac{2800}{100} = \mathbf{28}$$
Midpoints × frequencies: 0–10: $5×12=60$; 10–20: $15×18=270$; 20–30: $25×27=675$; 30–40: $35×20=700$; 40–50: $45×17=765$; 50–60: $55×6=330$
$\Sigma f_i x_i = 2800$, $\Sigma f_i = 100$
$$\text{Mean} = \frac{2800}{100} = \mathbf{28}$$
The monthly expenditure on milk in 200 families of a Housing Society is given below. Find the value of $x$ and also find the median and mean expenditure on milk.
Answer
Finding $x$: $24+40+33+x+30+22+16+7=200 \Rightarrow \boxed{x=28}$
Mean: $A=2750$, $\Sigma f_i d_i = -17500$, $\Sigma f_i = 200$ $$\text{Mean} = 2750 + \frac{-17500}{200} = 2750 – 87.5 = \mathbf{₹2662.50}$$ Median: $\dfrac{N}{2}=100$ lies in class $2500-3000$. $l=2500$, $f=28$, $cf=97$, $h=500$ $$\text{Median} = 2500 + \frac{100-97}{28} \times 500 = 2500 + 53.57 = \mathbf{₹2553.57}$$
Mean: $A=2750$, $\Sigma f_i d_i = -17500$, $\Sigma f_i = 200$ $$\text{Mean} = 2750 + \frac{-17500}{200} = 2750 – 87.5 = \mathbf{₹2662.50}$$ Median: $\dfrac{N}{2}=100$ lies in class $2500-3000$. $l=2500$, $f=28$, $cf=97$, $h=500$ $$\text{Median} = 2500 + \frac{100-97}{28} \times 500 = 2500 + 53.57 = \mathbf{₹2553.57}$$
A car assembly unit assembles cars daily based on prevailing demand. The table shows cars assembled over three consecutive months:
(i) If demand doubles, estimate the average cars to be assembled per day to meet demand.
(ii) On at least how many days were fewer than average cars assembled? Show your work.
(i) If demand doubles, estimate the average cars to be assembled per day to meet demand.
(ii) On at least how many days were fewer than average cars assembled? Show your work.
Answer
$\Sigma f_i = 90$, $\Sigma f_i x_i = 664$
(i) $\text{Mean} = \dfrac{664}{90} \approx 7.38$ cars/day. If demand doubles: $2 \times 7.38 \approx \mathbf{15 \text{ cars/day}}$
(ii) Mean lies in range $4-8$, so at least on 33 days fewer than average cars were assembled.
The marks obtained by 80 students of class X in a mock test are given below. Find the median and the mode of the data.
Answer
$N=80$, $\dfrac{N}{2}=40$. Median class = $50-60$.
Median: $l=50$, $cf=37$, $f=15$, $h=10$: $$\text{Median} = 50 + \frac{40-37}{15} \times 10 = 50 + 2 = \mathbf{52}$$ Mode: Modal class = $50-60$. $l=50$, $f_1=15$, $f_0=12$, $f_2=12$, $h=10$: $$\text{Mode} = 50 + \frac{15-12}{30-12-12} \times 10 = 50 + \frac{3}{6} \times 10 = 50 + 5 = \mathbf{55}$$
If the mean of the following frequency distribution is 91, find the missing frequencies $x$ and $y$.
Answer
From total frequency: $96 + x + y = 150 \Rightarrow x + y = 54$ …(1)
From Mean $= 91$, setting up the second equation and solving simultaneously:
$x = 34$ and $y = 20$
The median of the following data is 50. Find the values of ‘$p$’ and ‘$q$’, if the sum of all frequencies is 90. Also find the mode.
Answer
From total: $p + q + 78 = 90 \Rightarrow p + q = 12$ …(i)
Using median formula (median class $50-60$): $$50 = 50 + \frac{45-(p+40)}{20} \times 10 \Rightarrow 0 = \frac{45-p-40}{2} \Rightarrow p = 5$$ From (i): $q = 7$
Mode: $l=40$, $f_1=25$, $f_0=15$, $f_2=20$, $h=10$: $$\text{Mode} = 40 + \frac{25-15}{50-15-20} \times 10 = 40 + \frac{100}{15} = 40 + 6.67 = \mathbf{46.67}$$ $p = 5$, $q = 7$, Mode $= 46.67$
Case Study — COVID-19 Pandemic
The COVID-19 pandemic is an ongoing pandemic of coronavirus disease caused by SARS-CoV-2. The following tables show the age distribution of cases admitted during a day in two different hospitals.Table 1
Table 2
Refer to Table 1:
1. The average age for which maximum cases occurred is: (A) 32.24 (B) 34.36 (C) 36.82 (D) 42.24
2. The upper limit of modal class is: (A) 15 (B) 25 (C) 35 (D) 45
3. The mean of the given data is: (A) 26.2 (B) 32.4 (C) 33.5 (D) 35.4
4. The mode of the given data is: (A) 41.4 (B) 48.2 (C) 55.3 (D) 64.6
5. The median of the given data is: (A) 32.7 (B) 40.2 (C) 42.3 (D) 48.6
Answer
1. Option (C) — 36.82 Modal class = $35-45$ (highest freq = 23). $$\text{Mode} = 35 + \frac{23-21}{46-21-14} \times 10 = 35 + \frac{20}{11} \approx 36.82$$
2. Option (D) — 45 Modal class is $35-45$, upper limit = 45.
3. Option (D) — 35.4
$\Sigma f_i=80$, $\Sigma f_i d_i=430$, $a=30$: $$\text{Mean} = 30 + \frac{430}{80} = 35.375 \approx \mathbf{35.4}$$ 4. Option (A) — 41.4 Modal class (Table 2) = $35-45$ (freq = 42). $$\text{Mode} = 35 + \frac{42-10}{84-10-24} \times 10 = 35 + 6.4 = \mathbf{41.4}$$ 5. Option (B) — 40.2
$\dfrac{N}{2}=56$. Median class $=35-45$. $l=35$, $cf=34$, $f=42$, $h=10$: $$\text{Median} = 35 + \frac{56-34}{42} \times 10 = 35 + 5.24 \approx \mathbf{40.2}$$
2. Option (D) — 45 Modal class is $35-45$, upper limit = 45.
3. Option (D) — 35.4
$\Sigma f_i=80$, $\Sigma f_i d_i=430$, $a=30$: $$\text{Mean} = 30 + \frac{430}{80} = 35.375 \approx \mathbf{35.4}$$ 4. Option (A) — 41.4 Modal class (Table 2) = $35-45$ (freq = 42). $$\text{Mode} = 35 + \frac{42-10}{84-10-24} \times 10 = 35 + 6.4 = \mathbf{41.4}$$ 5. Option (B) — 40.2
$\dfrac{N}{2}=56$. Median class $=35-45$. $l=35$, $cf=34$, $f=42$, $h=10$: $$\text{Median} = 35 + \frac{56-34}{42} \times 10 = 35 + 5.24 \approx \mathbf{40.2}$$
The weights (in kg) of 50 wrestlers are recorded in the following table:
1. What is the upper limit of modal class? Also find the modal class frequency?
2. How many wrestlers weigh 120 kg or more? Also find the class mark for class $130-140$.
OR Find the mode of the given data.
3. Which method is more suitable to find the mean?
1. What is the upper limit of modal class? Also find the modal class frequency?
2. How many wrestlers weigh 120 kg or more? Also find the class mark for class $130-140$.
OR Find the mode of the given data.
3. Which method is more suitable to find the mean?
Answer
1. Modal class $= 120-130$. Upper limit $= \mathbf{130}$. Modal class frequency $= \mathbf{21}$.
2. Wrestlers $\geq 120$ kg $= 21+8+3 = \mathbf{32}$. Class mark of $130-140 = \dfrac{130+140}{2} = \mathbf{135}$.
OR (Mode): $l=120$, $f_1=21$, $f_0=14$, $f_2=8$, $h=10$: $$\text{Mode} = 120 + \frac{21-14}{42-14-8} \times 10 = 120 + \frac{70}{20} = 120 + 3.5 = \mathbf{123.5 \text{ kg}}$$ 3. The Assumed Mean Method is more suitable for finding the mean of this data.
2. Wrestlers $\geq 120$ kg $= 21+8+3 = \mathbf{32}$. Class mark of $130-140 = \dfrac{130+140}{2} = \mathbf{135}$.
OR (Mode): $l=120$, $f_1=21$, $f_0=14$, $f_2=8$, $h=10$: $$\text{Mode} = 120 + \frac{21-14}{42-14-8} \times 10 = 120 + \frac{70}{20} = 120 + 3.5 = \mathbf{123.5 \text{ kg}}$$ 3. The Assumed Mean Method is more suitable for finding the mean of this data.
Frequently Asked Questions
What does the Statistics chapter cover in CBSE Class 10 Maths?⌄
The Statistics chapter in CBSE Class 10 covers three measures of central tendency — mean, median, and mode — for grouped data. Students learn to calculate these using the direct method, assumed mean method, and step deviation method. The empirical relationship between mean, median, and mode is also a key concept regularly tested in board exams.
How many marks does Statistics carry in the CBSE Class 10 board exam?⌄
Statistics is part of the Statistics and Probability unit, which together carry approximately 11 marks in the CBSE Class 10 board exam. Questions appear as 1-mark MCQs, 2-mark and 3-mark short answers, and 5-mark long answer or case study questions — making it one of the most mark-intensive topics your child can prepare thoroughly.
What are the most important topics in Class 10 Statistics for board exams?⌄
The most frequently tested topics are: finding the modal class and computing mode using the formula, finding the median class and applying the median formula, calculating mean using the assumed mean or step deviation method, and applying the empirical relationship Mode = 3 Median − 2 Mean. These four areas account for the majority of Statistics marks in previous year papers.
What common mistakes do students make when solving Statistics questions?⌄
A very common mistake is incorrectly identifying the median class — students sometimes pick the class where cumulative frequency equals N/2, instead of where it first exceeds N/2. Another frequent error is substituting wrong values of f₀ and f₂ in the mode formula. Practising with real board paper questions, as provided here, helps your child avoid these mistakes under exam pressure.
How does Angle Belearn help students score well in Statistics?⌄
Angle Belearn’s CBSE specialists curate chapter-wise question banks drawn directly from real board papers, with every question paired with a clear, step-by-step verified solution. Students who practise on Angle Belearn develop the habit of showing structured working — a key factor in earning full marks in CBSE board exams. Regular practice builds both speed and accuracy.
