Board Exam 2025
3 Marks
Q28 (a). One healthcare center working for
the
welfare of the patients suffering from 'Dengue', recorded the following information :
If the modal age of the patients is 54, then find the value of x.| Age of Patients | 0 - 15 | 15 - 30 | 30 - 45 | 45 - 60 | 60 - 75 | 75 - 90 |
|---|---|---|---|---|---|---|
| Number of Patients | 8 | 5 | x | 16 | 12 | 9 |
OR
(b) Weekly expenditure on Ayurvedic medicines of few households in a locality is
recorded below.
| Weekly Expenditure (in ?) | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 | 300 - 350 |
|---|---|---|---|---|---|
| Number of Households | 4 | 5 | y | 2 | 2 |
(a) Modal class is 45-60 (since mode 54 lies in it).
\(l = 45, h = 15, f_1 = 16, f_0 = x, f_2 = 12\).
Mode = \(l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\)
\(54 = 45 + \frac{16 - x}{32 - x - 12} \times 15\)
\(9 = \frac{16 - x}{20 - x} \times 15 \Rightarrow 3(20 - x) = 5(16 - x)\)
\(60 - 3x = 80 - 5x \Rightarrow 2x = 20 \Rightarrow x = 10\).
(b) Mean = 211.
Midpoints (\(x_i\)): 125, 175, 225, 275, 325.
Freq (\(f_i\)): 4, 5, y, 2, 2. \(\sum f_i = 13 + y\).
\(f_i x_i\): 500, 875, 225y, 550, 650. \(\sum f_i x_i = 2575 + 225y\).
\(\frac{2575 + 225y}{13 + y} = 211\)
\(2575 + 225y = 2743 + 211y\)
\(14y = 168 \Rightarrow y = 12\).
\(l = 45, h = 15, f_1 = 16, f_0 = x, f_2 = 12\).
Mode = \(l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\)
\(54 = 45 + \frac{16 - x}{32 - x - 12} \times 15\)
\(9 = \frac{16 - x}{20 - x} \times 15 \Rightarrow 3(20 - x) = 5(16 - x)\)
\(60 - 3x = 80 - 5x \Rightarrow 2x = 20 \Rightarrow x = 10\).
(b) Mean = 211.
Midpoints (\(x_i\)): 125, 175, 225, 275, 325.
Freq (\(f_i\)): 4, 5, y, 2, 2. \(\sum f_i = 13 + y\).
\(f_i x_i\): 500, 875, 225y, 550, 650. \(\sum f_i x_i = 2575 + 225y\).
\(\frac{2575 + 225y}{13 + y} = 211\)
\(2575 + 225y = 2743 + 211y\)
\(14y = 168 \Rightarrow y = 12\).
3 Marks
Q28. The median of the following distribution
is
50. Find the value of p.
| Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
|---|---|---|---|---|---|
| Frequency | 17 | p | 32 | 24 | 19 |
Median = 50. Median class is 40-60.
\(l=40, h=20, f=32\).
Cumulative Frequency before class: \(cf = 17+p\).
Total frequency \(N = 17+p+32+24+19 = 92+p\).
Formula: \(Median = l + (\frac{N/2 - cf}{f}) \times h\).
\(50 = 40 + (\frac{\frac{92+p}{2} - (17+p)}{32}) \times 20\).
\(10 = (\frac{92+p - 34 - 2p}{2 \times 32}) \times 20 = (\frac{58-p}{64}) \times 20\).
\(10/20 = \frac{58-p}{64} \Rightarrow \frac{1}{2} = \frac{58-p}{64}\).
\(32 = 58 - p \Rightarrow p = 58 - 32 = 26\).
Value of p is 26.
\(l=40, h=20, f=32\).
Cumulative Frequency before class: \(cf = 17+p\).
Total frequency \(N = 17+p+32+24+19 = 92+p\).
Formula: \(Median = l + (\frac{N/2 - cf}{f}) \times h\).
\(50 = 40 + (\frac{\frac{92+p}{2} - (17+p)}{32}) \times 20\).
\(10 = (\frac{92+p - 34 - 2p}{2 \times 32}) \times 20 = (\frac{58-p}{64}) \times 20\).
\(10/20 = \frac{58-p}{64} \Rightarrow \frac{1}{2} = \frac{58-p}{64}\).
\(32 = 58 - p \Rightarrow p = 58 - 32 = 26\).
Value of p is 26.
5 Marks
Q33. Find the values of the missing
frequencies p and q in the following distribution of 100 observations. The median of the
distribution is given as 47.
| Class | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 | 60 – 65 |
|---|---|---|---|---|---|---|---|
| Frequency | 12 | p | 17 | 20 | q | 12 | 8 |
Total = 100. So \(12 + p + 17 + 20 + q + 12 + 8 = 100\).
\(69 + p + q = 100 \Rightarrow p + q = 31\).
Median = 47. Median class is 45-50.
\(l = 45, h = 5, f = 20, N = 100, cf = 12 + p + 17 = 29 + p\).
\(Median = l + \frac{N/2 - cf}{f} \times h\)
\(47 = 45 + \frac{50 - (29 + p)}{20} \times 5\)
\(2 = \frac{21 - p}{4} \Rightarrow 8 = 21 - p \Rightarrow p = 13\).
Since \(p + q = 31\), \(q = 31 - 13 = 18\).
p = 13, q = 18
\(69 + p + q = 100 \Rightarrow p + q = 31\).
Median = 47. Median class is 45-50.
\(l = 45, h = 5, f = 20, N = 100, cf = 12 + p + 17 = 29 + p\).
\(Median = l + \frac{N/2 - cf}{f} \times h\)
\(47 = 45 + \frac{50 - (29 + p)}{20} \times 5\)
\(2 = \frac{21 - p}{4} \Rightarrow 8 = 21 - p \Rightarrow p = 13\).
Since \(p + q = 31\), \(q = 31 - 13 = 18\).
p = 13, q = 18
1 Mark
Q1. What is the mode of a data if median and
mean
of the same data are 9.6 and 10.5, respectively?
Mode = \(3(\text{Median}) - 2(\text{Mean}) = 3(9.6) - 2(10.5) = 28.8 - 21.0 = 7.8\).
Answer: (A)
Answer: (A)
5 Marks
Q34. The length of leaves of a plant are
measured
correct to nearest millimetre. Find the median length of leaves.
| Length (in mm) | Number of Leaves |
|---|---|
| 118 - 126 | 3 |
| 127 - 135 | 5 |
| 136 - 144 | 9 |
| 145 - 153 | 12 |
| 154 - 162 | 5 |
| 163 - 171 | 4 |
| 172 - 180 | 2 |
Classes are discontinuous. Convert to continuous: 117.5-126.5, 126.5-135.5, etc.
\(N=40, N/2=20\). CF: 3, 8, 17, 29, 34, 38, 40.
Median class (cf > 20) is 144.5-153.5. (Index 145-153).
\(l = 144.5, cf = 17, f = 12, h = 9\).
Median \( = 144.5 + \frac{20-17}{12} \times 9 \) \( = 144.5 + \frac{3}{12} \times 9 \) \( = 144.5 + 2.25 = 146.75 \) mm.
\(N=40, N/2=20\). CF: 3, 8, 17, 29, 34, 38, 40.
Median class (cf > 20) is 144.5-153.5. (Index 145-153).
\(l = 144.5, cf = 17, f = 12, h = 9\).
Median \( = 144.5 + \frac{20-17}{12} \times 9 \) \( = 144.5 + \frac{3}{12} \times 9 \) \( = 144.5 + 2.25 = 146.75 \) mm.
5 Marks
Q37. The India Meteorological Department observes
seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to
compare and analyse the results.
The table below shows sub-divisions wise seasonal (monsoon) rainfall (in mm) in 2023.
Based on the information given above, answer the following questions :
seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to
compare and analyse the results.
The table below shows sub-divisions wise seasonal (monsoon) rainfall (in mm) in 2023.
| Rainfall (mm) | No. of Sub-divisions |
|---|---|
| 200 - 400 | 3 |
| 400 - 600 | 4 |
| 600 - 800 | 7 |
| 800 - 1000 | 4 |
| 1000 - 1200 | 3 |
| 1200 - 1400 | 3 |
(i) Write the modal class.
(ii) (a) Find the median of the given data.
OR
(ii) (b) Find the mean rainfall in the season.(iii) If a sub-division having at least 800 mm rainfall during monsoon season is considered a good rainfall sub-division, then how many sub-divisions had good rainfall ?
(i) Max frequency 7. Class: 600-800.
(ii) (a) Total N = 24. N/2 = 12. Cumulative: 3, 7, 14... Median class 600-800.
\(l=600, h=200, f=7, CF=7\).
Median \( = 600 + \frac{12-7}{7} \times 200 \) \( = 600 + \frac{1000}{7} \) \( = 600 + 142.85 \approx 742.85 \).
(iii) \(4 + 3 + 3 = 10\).
(ii) (a) Total N = 24. N/2 = 12. Cumulative: 3, 7, 14... Median class 600-800.
\(l=600, h=200, f=7, CF=7\).
Median \( = 600 + \frac{12-7}{7} \times 200 \) \( = 600 + \frac{1000}{7} \) \( = 600 + 142.85 \approx 742.85 \).
(iii) \(4 + 3 + 3 = 10\).
1 Mark
Q3. The cumulative frequency for calculating
median is obtained by adding the frequencies of all the :
Preceding and including current. Answer states (a) classes up to median
class.
Answer: (a)
Answer: (a)
1 Mark
Q4. If mode and median of given set of
observations are 13 and 11 respectively, then the value of mean is :
\(3Median = Mode + 2Mean\). \(33 = 13 + 2M \Rightarrow 20 = 2M \Rightarrow
M=10\).
Answer: (c)
Answer: (c)
5 Marks
Q34. Following table shows the absentees record
of 40 students in an academic year :
Find the 'mean' and the 'mode' of the above data.
| Number of Days | Number of Students |
|---|---|
| 2-6 | 11 |
| 6-10 | 10 |
| 10-14 | 7 |
| 14-18 | 4 |
| 18-22 | 4 |
| 22-26 | 3 |
| 26-30 | 1 |
Mode: Max freq is 11 (Class 2-6).
\(l = 2, f_1 = 11, f_0 = 0, f_2 = 10, h = 4\).
\(\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\) \(= 2 + \frac{11-0}{22-0-10} \times 4\) \(= 2 + \frac{44}{12} = 5.67\).
Mean: \(x_i\): 4, 8, 12, 16, 20, 24, 28.
\(f_ix_i\): 44, 80, 84, 64, 80, 72, 28. Sum = 452.
\(\text{Mean} = 452/40 = 11.3\).
\(l = 2, f_1 = 11, f_0 = 0, f_2 = 10, h = 4\).
\(\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\) \(= 2 + \frac{11-0}{22-0-10} \times 4\) \(= 2 + \frac{44}{12} = 5.67\).
Mean: \(x_i\): 4, 8, 12, 16, 20, 24, 28.
\(f_ix_i\): 44, 80, 84, 64, 80, 72, 28. Sum = 452.
\(\text{Mean} = 452/40 = 11.3\).
1 Mark
Q7. If x median + y mean = z mode is the
empirical relationship between mean, median and mode, then \(x + y + z =\)
Empirical formula: 3Median = Mode + 2Mean. Rearranging: 3Median - 2Mean =
1Mode.
Here \(x = 3, y = -2, z = 1\). \(x + y + z = 3 + (-2) + 1 = 2\).
Correct Option: (C) 2
Here \(x = 3, y = -2, z = 1\). \(x + y + z = 3 + (-2) + 1 = 2\).
Correct Option: (C) 2
1 Mark
Q8. Following data shows marks obtained by
100 students. The median will be the average of which two observations?
| Marks | 20 | 29 | 28 | 33 | 42 | 38 | 43 | 25 |
|---|---|---|---|---|---|---|---|---|
| Students | 6 | 28 | 24 | 15 | 2 | 4 | 1 | 20 |
Arrange in ascending order: 20(6), 25(20), 28(24), 29(28), 33(15)...
Cumulative freq: 6, 26, 50, 78...
N = 100. Median = avg of 50th and 51st observations.
50th observation is 28, 51st observation is 29.
Correct Option: (C) 28 and 29
Cumulative freq: 6, 26, 50, 78...
N = 100. Median = avg of 50th and 51st observations.
50th observation is 28, 51st observation is 29.
Correct Option: (C) 28 and 29
5 Marks
Q34. The following table gives the daily
income of 50 cab drivers:
Find the mean income and the modal income.
| Income (?) | 500-600 | 600-700 | 700-800 | 800-900 | 900-1000 |
|---|---|---|---|---|---|
| No. of Drivers | 12 | 14 | 8 | 6 | 10 |
Mean: Class marks: 550, 650, 750, 850, 950.
\(\Sigma f_i x_i = 12(550) + 14(650) + 8(750) + 6(850) + 10(950) = 36300\).
Mean = \(36300/50 = 726\).
Mode: Max freq is 14 (Class 600-700). \(l = 600, h = 100, f_1 = 14, f_0 = 12, f_2 = 8\).
Mode \( = 600 + \frac{14-12}{28-12-8} \times 100 \) \( = 600 + \frac{2}{8} \times 100 = 625 \).
\(\Sigma f_i x_i = 12(550) + 14(650) + 8(750) + 6(850) + 10(950) = 36300\).
Mean = \(36300/50 = 726\).
Mode: Max freq is 14 (Class 600-700). \(l = 600, h = 100, f_1 = 14, f_0 = 12, f_2 = 8\).
Mode \( = 600 + \frac{14-12}{28-12-8} \times 100 \) \( = 600 + \frac{2}{8} \times 100 = 625 \).
5 Marks
Q34. The population of lions was noted in
different regions
across the world in the following table:
If the median of the given data is 525, find the values of \(x\) and \(y\).
| Number of lions | Number of regions |
|---|---|
| 0 - 100 | 2 |
| 100 - 200 | 5 |
| 200 - 300 | 9 |
| 300 - 400 | 12 |
| 400 - 500 | X |
| 500 - 600 | 20 |
| 600 - 700 | 15 |
| 700 - 800 | 9 |
| 800 - 900 | y |
| 900 - 1000 | 2 |
| Total | 100 |
Total \(N = 100\). Median = 525.
Median lies in class 500-600. So \(l = 500, h = 100, f = 20\).
Cumulative Frequency (CF) preceding class = \(2+5+9+12+x = 28+x\).
Formula: \(\text{Median} = l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h\)
\(525 = 500 + \left(\frac{50 - (28+x)}{20}\right) \times 100\)
\(25 = (22-x) \times 5\)
\(5 = 22 - x \Rightarrow x = 17\).
Sum of frequencies = 100.
\(74 + x + y = 100 \Rightarrow 74 + 17 + y = 100 \Rightarrow y = 9\).
Answer: \(x=17, y=9\).
Median lies in class 500-600. So \(l = 500, h = 100, f = 20\).
Cumulative Frequency (CF) preceding class = \(2+5+9+12+x = 28+x\).
Formula: \(\text{Median} = l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h\)
\(525 = 500 + \left(\frac{50 - (28+x)}{20}\right) \times 100\)
\(25 = (22-x) \times 5\)
\(5 = 22 - x \Rightarrow x = 17\).
Sum of frequencies = 100.
\(74 + x + y = 100 \Rightarrow 74 + 17 + y = 100 \Rightarrow y = 9\).
Answer: \(x=17, y=9\).
1 Mark
Q6. If the mode of some observations is 10 and
sum of mean and median is 25, then the mean and median respectively are
Mode \( = 10 \).
Mean \( + \) Median \( = 25 \Rightarrow \) Median \( = 25 - \) Mean.
Empirical Formula: Mode \( = 3(\text{Median}) - 2(\text{Mean}) \).
\( 10 = 3(25 - \text{Mean}) - 2(\text{Mean}) \)
\( 10 = 75 - 3(\text{Mean}) - 2(\text{Mean}) \)
\( 10 = 75 - 5(\text{Mean}) \)
\( 5(\text{Mean}) = 65 \Rightarrow \text{Mean} = 13 \).
Median \( = 25 - 13 = 12 \).
Answer: (B)
Mean \( + \) Median \( = 25 \Rightarrow \) Median \( = 25 - \) Mean.
Empirical Formula: Mode \( = 3(\text{Median}) - 2(\text{Mean}) \).
\( 10 = 3(25 - \text{Mean}) - 2(\text{Mean}) \)
\( 10 = 75 - 3(\text{Mean}) - 2(\text{Mean}) \)
\( 10 = 75 - 5(\text{Mean}) \)
\( 5(\text{Mean}) = 65 \Rightarrow \text{Mean} = 13 \).
Median \( = 25 - 13 = 12 \).
Answer: (B)
1 Mark
Q7. If the maximum number of students has
obtained 52 marks out of 80, then
The score obtained by the maximum number of students represents the value with the highest frequency,
which is the definition of **Mode**.
Answer: (C)
Answer: (C)