Table of contents  
Previous Year Questions 2024  
Previous Year Questions 2023  
Previous Year Questions 2022  
Previous Year Questions 2021  
Previous Year Questions 2020  
Previous Year Questions 2019 
Q1: Vocational training complements traditional education by providing practical skills and
From the above answer the following questions: (2024)
(A) What is the lower limit of the modal class of the above data?
(B) Find the median class of the above data.
OR
Find the number of participants of age less than 50 years who undergo vocational training.
(C) Give the empirical relationship between mean, median and mode.
Ans:
First convert the given table in exclusive form subtract 0.5 from lower limit and add 0.5 to upper limit, so the new table will be:
(i) Modal class is the class with highest frequency, so, it is 19.5 – 24.5. hence, lower limit will be ‘19.5’.
(ii) (a)
N/2 = 365/2 = 182.5
Medium class will be 19.5 – 24.5.
OR
(b) Approx 361 participants are there at Class Interval 44.549.5 showing 361 cummulative frequency.
Hence 361 participants are less than 50 year of age who undergo vocational training.
(iii) Empirical relationship between mean, median and mode.
Mode = 3 Median – 2 Mean
Ans: (b)
If each value of observation is increased by 3. then mean is also increased by 3.
Q3: For the following distribution. (2023)
The sum of the lower limits of the median and modal class is
(a) 15
(b) 25
(c) 30
(d) 35
Ans: (b)
Here, n/2 = 66/2 = 33
Cumulative frequency just greater than 33 is 37.
So. median class is 10  15.Lower limit of median class = 10
Highest frequency is 20 so modal class is 15  20.
Sum of the lower limits of the median and modaI class is 10 + 15 = 25
Q4: India meteorological department observes seasonal and annual rainfall every year in different subdivisions of our country.
It helps them to compare and analyse the results. The table given below shows subdivision wise seasonal (monsoon] rainfall [mm) in 2018:
Based on the above information, answer the following questions.
(I) Write the modal class.
(II) Find the median of the given data.
Find the mean rainfall in this season.
(Ill) If subdivision having at least 1000 mm rainfall during monsoon season, is considered good rainfall subdivision, then how many subdivisions had good rainfall? (2023)
Ans:
(i) Here, maximum class frequency is 7 and class corresponding to this frequency is 600800, so the modal class is 600800.
(ii) Here n/2 = 24/2 = 12
Class whose cumulative frequency just greater than and nearest to n/2 is called median class.
Here, cf. = 13 (>12) and corresponding class 600  800 is : median class.
1 = 600. c.f. = 6,f= 7, h = 200
So. the median of the given data is 771.429
Assumed mean a = 1100 and crass size, h = 400  200 = 200
So, mean rainfall in the season is 850 mm.
(iii) Number of subdivision having good rainfall
= 2 + 3 +1 + 1 = 7
Q5: 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. (2023)
Ans:
Since, 200 = 172 + x ⇒ x = 28
Let the assumed mean, a = 3250 and class size, h = 500
= 3250  587.5 = 2,662.5
Mean expenditure = Rs. 2,662.5
Also, we have n/2= 100, which lies in the class interval 2500  3000.
Median class is 2500  3000.
Here l = 2500, c.f. = 9, f = 28, h = 500
Ans: Table for the given data is as follows:
Now Mean =
Solving for p:
∴ p = 96/48= 2
Q7: Find the mean of the following frequency distribution: (2022)
View Answer
Ans:
Q8: The weights (in kg) of 50 wild animals of a National Park were recorded and the following data was obtained is
Find the mean weight (in kg) of animals, using assumed mean method. (2022)
View AnswerAns: Let the assumed mean, a = 125 We have the frequency distribution table for the given data as follows :
Hence, mean weight of animals = 123.B kg.
Q9: The mean of the following frequency distribution is 25. Find the value off. (2022)
View Answer
Ans: The frequency distribution table from the given data is as follows:
Hence, the value of f is 16.
Q10: Find the mean of the following data using assumed mean method. [2022, 2 Marks]
Ans: Let the assumed mean, a = 12.5 .. d
⇒ d = x_{i}  a = x_{i}  12.5
Now, we have the frequency distribution table as follows:
= 13.9
Q11: The mode of a grouped frequency distribution is 75 and the modal class is 6580. The frequency of the class preceding the modal class is 6 and the frequency of the class succeeding the modal class is 8. Find the frequency of the modaI class. (2022)
Ans: We know that
Here given l = 65,f_{0} = 6, f_{1}= f, h = 15, f_{2} = 8 and mode = 75
So, from equation (i), we get
Q12: Find the missing frequency 'x ' of the following data, if its mode is 240: (2022)
View Answer
Ans: Here the given mode = 240, which lies in interval 200300.
l = 200, f_{0} = 230, f_{1}= 270, f_{2} = x (missing frequency] and h = 100
Missing frequency, x = 210
Q13: If mode of the following frequency distribution is 55, then find the value of x. (2022)
View Answer
Ans: Here, mode of the frequency distribution = 55.
which lies in the class interval 4560.
∴ Modal class is 45  60
Lower limit (l) = 45
Class interval (h) = 15
Also, f _{0} = 15, f _{1} = x and f_{ 2} = 10
⇒ 10(30 – x – 10) = 225 – 15x
⇒ 300 –10x – 100 = 225 – 15x
⇒ 5x = 25
⇒ x = 5
Q14: Heights of 50 students in class X of a school are recorded and following data is obtained: (2022)
Find the median height of the students.
Ans: The cumulative frequency distribution table is as follows:
Now, we have N = 50
Since, the cumulative frequency just greater than 25 is 27.
∴ The median class is 140  145
and also, I = 140, c.f. = 15, f = 12 and h = 5
∴ Median height of the students = 144.16 cm.
Q15: Health insurance is an agreement whereby the insurance company agrees to undertake a guarantee of compensation for medical expenses in case the insured faffs ill or meets with an accident which leads to Hospitalisation of the insured. The government aIso promotes health insurance by providing a deduction from income tax.
An SB I health insurance agent found the following data for distribution of ages of 100 policyholders.
The health insurance policies art given to persons having age 15 years a nd onwards briefest han 60 years.
(i) Find the modal age of the policy holders.
(ii) Find the median age of the policy holders. (2022)
Ans: (i) It is clear from the given data, maximum frequency is 33. which lies in 3540
∴ Modal class is 3540.
So, modal age of policy holders is 37 years approx.
Q16: During the annual sports meet in a school, all the athletes were very enthusiastic. They all wanted to be the winner so that their house could stand first. The instructor noted down the time taken by a group of students to complete a certain race. The data recorded is given below:
Based on the above, answer the following questions: (2021)
We need to make the following frequency table as follows:
(i) What is the class mark of the modal class ?
(a) 60
(b) 70
(c) 80
(d) 140
Ans: (b)
Here the greatest frequency is 7, which lies in the interval 6080.
So, modal class is 6080.
Class mark of modal class = upper limit + lower limit / 2
= 60 + 80 / 2 = 70
So, class mark of modal class is 70.
(ii) The mode of the given data is
(a) 7033
(b) 7133
(c) 7233
(d) 7333
Ans: (d)
(iii) The median class of the given data is
(a) 2040
(b) 4060
(c) 80100
(d) 6080
Ans: (d)
Here n = 20 ⇒ n /2 = 10
Cumulative frequency just greater than 10 is 15 and corresponding interval is 6080.
So, median class is 6080.
(iv) The sum of the lower limits of median class and modal class is 1
(a) 80
(b) 140
(c) 120
(d) 100
Ans: (c)
Median class = 6080 .
∴ Lower limit of median = 60
Modal class  6080 = 120
∴ Lower limit of modal class = 60
So, the sum of lower limit of median and modal class = 60 + 60 = 120
(v) The median time (in seconds) of the given data is
(a) 657
(b) 857
(c) 457
(d) 257
Ans: (a)
From the above data, we have
l = 60, f = 7, c.f. = 8, h = 20
So, median time (in sec) of the given data = 65.7 sec.
Ans: Given, mean of first n natural numbers is 15.
Q18: In the formula (2020)
Ans: In the formula
where a is assumed mean and h = class size.
Q19: Find the mean of the following distribution: (2020)
View Answer
Ans: The frequency distribution table from the given data can be drawn as :
∴ Mean = 326/40 = 8.15
Q20: Find the mode of the following distribution: (2020)
View Answer
Ans: From the given data, we have maximum frequency 75. which lies in the interval 2025.
Modal class is 2025
So, l = 20, f_{0}= 30, f_{1} = 75, f_{2}= 20, h = 5
Mode = 20 + 2.25
= 22.25
Q21: Find the mode of the following distribution: (2020)
View Answer
Ans: From the given data, we have maximum frequency 12.
which lies in the interval 3040
Modal class is 3040
So, l = 30, f_{0}= 12, f_{1} = 7, f_{2}= 5, h = 10
Q22: Find the mode of the following distribution: (2020)
Ans: From the given data, we have maximum frequency 12.
which lies in the interval 6080.
Modal class is 6080
So, l = 60, f_{0}= 12, f_{1} = 10, f_{2}= 6, h = 20
Q23: The mean and median of a distribution are 14 and 15 respectively. The value of mode is (2020)
(a) 16
(b) 17
(c) 13
(d) 18
Ans: (b)
We know that Mode = 3 Median  2 Mean
So, Mode = 3 x 15  2 x 14
= 45  28 = 17
Q24: The distribution given below shows the number of wickets taken by bowlers in oneday cricket matches. Find the mean and the median of the number of wickets taken. (2020)
View Answer
Ans: The frequency distribution table for the given data can be drawn as:
= 100 + 26.25
= 126.25
Hence, mean number of wickets is 125.33 and median number of wickets is 126.25.
Q25: The arithmetic mean of the following frequency distribution is 53. Find the value of k. (2019)
View Answer
Ans: The frequency distribution table from the given data is as follows:
Now, [Given]
Q26: Find the mean of the following frequency distribution:
Ans: he frequency distribution table from the given data is as follows:
∴
= 50
Q27: If the mean of the following frequency distribution is 62.8, then find the missing frequency x: (2019)
Ans: Here h = 20
Let us construct the following table far the given data.
We know that Mean =
Missing frequency, x= 10
Q28: The weights of tea in 70 packets is given in the following table: (2019)
Find the modal weight.
Ans: From the given data, we observe that, highest frequency is 20,
which lies in the classinterval 4050.
∴ l = 40, f_{1} =_{ }20, f_{o}= 12, f_{2} = 11, h = 10
Q29: If the median of the following frequency distribution is 32.5, find the values of f_{1}and f_{2}. (2019)
View Answer
Ans: The frequency distribution table for the given data is as follows:
Here, N = 40 ⇒ 31 + f_{1} + f_{2} = 40
⇒ f_{1} + f_{2 }= 9 ...(i)
Given, median = 32.5, which lies in the class interval 3040.
So, median class is 3040.
I = 30, h = 10, f = 12, N = 40 and c.f. of preceding class = f_{1} + 14
Q30: Find the values a frequencies x and y in the following frequency distribution table, if N = 100 and median is 32. (2019)
Ans: The frequency distribution table for the given data is as follows:
Here. N = 100, median = 32, it lies in the Interval 30  40.
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