Looking for complete, step-by-step textbook solutions for SEBA Class 10 Mathematics Chapter 13 (Statistics - Exercise 13.3) for the 2026–27 academic session? Access verified, board-pattern solutions to all textual questions on finding the Median of grouped data using cumulative frequency distribution tables (cf). Master the high-yield median formula (Median = l + [((n/2) - cf) / f] * h), solving the famous 5-mark missing frequencies x and y problem with median 28.5, converting "less than" cumulative data into continuous class intervals (life insurance policies), calculating lengths of 40 leaves with continuous boundary adjustments, comparing mean, median, and mode for empirical verification (3 Median = Mode + 2 Mean), and exam-oriented model solutions designed strictly according to the Assam Board (SEBA) HSLC blueprint.
1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
| Monthly consumption (in units) | Number of consumers |
| 65 – 85 | 4 |
| 85 – 105 | 5 |
| 105 – 125 | 13 |
| 125 – 145 | 20 |
| 145 – 165 | 14 |
| 165 – 185 | 8 |
| 185 – 205 | 4 |
Soln:
| Monthly consumption (in units) | Number of consumers (fi) | Cumulative freq. (cf) | Class mark (xi) | ui=20xi−135 | fiui |
| 65 – 85 | 4 | 4 | 75 | -3 | -12 |
| 85 – 105 | 5 | 9 | 95 | -2 | -10 |
| 105 – 125 | 13 | 22 | 115 | -1 | -13 |
| 125 – 145 | 20 | 42 | 135 (a) | 0 | 0 |
| 145 – 165 | 14 | 56 | 155 | 1 | 14 |
| 165 – 185 | 8 | 64 | 175 | 2 | 16 |
| 185 – 205 | 4 | 68 | 195 | 3 | 12 |
| Total | sum f_i = 68 | — | — | — |
(i) Median:
n = 68 ⇒ n/2 = 68 / 2 = 34
The cumulative frequency just greater than 34 is 42, which corresponds to the class interval 125 – 145.
Median class = 125 – 145
Here:
l = 125, n/2 = 34, cf = 22, f = 20, h = 20
Formula:
Median = l + [((n/2) – cf) / f] × h
Substitution:
Median = 125 + [(34 – 22) / 20] × 20
Median = 125 + 12
Median = 137 units
(ii) Mean: Assumed mean (a) = 135 Class size (h) = 20
Formula:
Mean (x̄) = a + (Σfᵢuᵢ / Σfᵢ) × h
Substitution:
x̄ = 135 + (7 / 68) × 20
x̄ = 135 + 140 / 68
x̄ = 135 + 2.06
x̄ = 137.06 units
(iii) Mode: The maximum frequency is 20, corresponding to the class interval 125 – 145.
Modal class = 125 – 145
Here:
l = 125, f₁ = 20, f₀ = 13, f₂ = 14, h = 20
Formula:
Mode = l + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h
Substitution:
Mode = 125 + [(20 – 13) / (2(20) – 13 – 14)] × 20
Mode = 125 + [7 / (40 – 27)] × 20
Mode = 125 + (140 / 13)
Mode = 125 + 10.77
Mode = 135.77 units
Comparison:
Median = 137 units
Mean = 137.06 units
Mode = 135.77 units
All three measures of central tendency are approximately equal in this data.
2. If the median of the distribution given below is 28.5, find the values of x and y.
| Class interval | Frequency |
| 0 – 10 | 5 |
| 10 – 20 | x |
| 20 – 30 | 20 |
| 30 – 40 | 15 |
| 40 – 50 | y |
| 50 – 60 | 5 |
| Total | 60 |
Soln: Given:
Median = 28.5
Total frequency (n) = 60
| Class interval | Frequency (fi) | Cumulative frequency (cf) |
| 0 – 10 | 5 | 5 |
| 10 – 20 | x | 5 + x |
| 20 – 30 | 20 | 25 + x |
| 30 – 40 | 15 | 40 + x |
| 40 – 50 | y | 40 + x + y |
| 50 – 60 | 5 | 45 + x + y |
| Total | n = 60 | — |
From the table:
45 + x + y = 60
x + y = 60 – 45
x + y = 15 …(1)
Since Median = 28.5, it lies in the class interval 20 – 30. Median class = 20 – 30
Here:
l = 20, n/2 = 60 / 2 = 30, cf = 5 + x, f = 20, h = 10
Formula:
Median = l + [((n/2) – cf) / f] × h
Substitution:
28.5 = 20 + [{30 – (5 + x)} / 20] × 10
28.5 – 20 = (25 – x) / 2
8.5 = (25 – x) / 2
17 = 25 – x
x = 25 – 17
x = 8
Substituting x = 8 in equation (1):
8 + y = 15
y = 15 – 8
y = 7
Ans: x = 8, y = 7
3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
| Age (in years) | Number of policy holders |
| Below 20 | 2 |
| Below 25 | 6 |
| Below 30 | 24 |
| Below 35 | 45 |
| Below 40 | 78 |
| Below 45 | 89 |
| Below 50 | 92 |
| Below 55 | 98 |
| Below 60 | 100 |
Soln: Converting the given cumulative frequency distribution into class intervals:
| Age (in years) | Number of policy holders (cf) | Frequency (fi) |
| 18 – 20 | 2 | 2 |
| 20 – 25 | 6 | 6 – 2 = 4 |
| 25 – 30 | 24 | 24 – 6 = 18 |
| 30 – 35 | 45 | 45 – 24 = 21 |
| 35 – 40 | 78 | 78 – 45 = 33 |
| 40 – 45 | 89 | 89 – 78 = 11 |
| 45 – 50 | 92 | 92 – 89 = 3 |
| 50 – 55 | 98 | 98 – 92 = 6 |
| 55 – 60 | 100 | 100 – 98 = 2 |
| Total | — | n = 100 |
n = 100 ⇒ n/2 = 100 / 2 = 50
The cumulative frequency just greater than 50 is 78, which corresponds to the class interval 35 – 40.
Median class = 35 – 40
Here:
l = 35, n/2 = 50, cf = 45, f = 33, h = 5
Formula:
Median = l + [((n/2) – cf) / f] × h
Substitution:
Median = 35 + [(50 – 45) / 33] × 5
Median = 35 + (5 / 33) × 5
Median = 35 + 25 / 33
Median = 35 + 0.76
Median = 35.76 years
SEBA Class 10 Maths Exercise 13.3 Solutions
4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
| 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 |
Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, …, 171.5 – 180.5.)
Soln: Converting the given discontinuous classes into continuous classes by subtracting 0.5 from lower limits and adding 0.5 to upper limits:
| Length (in mm) | Number of leaves (fi) | Cumulative frequency (cf) |
| 117.5 – 126.5 | 3 | 3 |
| 126.5 – 135.5 | 5 | 8 |
| 135.5 – 144.5 | 9 | 17 |
| 144.5 – 153.5 | 12 | 29 |
| 153.5 – 162.5 | 5 | 34 |
| 162.5 – 171.5 | 4 | 38 |
| 171.5 – 180.5 | 2 | 40 |
| Total | n = 40 | — |
n = 40 ⇒ n/2 = 40 / 2 = 20
The cumulative frequency just greater than 20 is 29, which corresponds to the class interval 144.5 – 153.5.
Median class = 144.5 – 153.5
Here:
l = 144.5, n/2 = 20, cf = 17, f = 12, h = 9
Formula:
Median = l + [((n/2) – cf) / f] × h
Substitution:
Median = 144.5 + [(20 – 17) / 12] × 9
Median = 144.5 + (3 / 12) × 9
Median = 144.5 + (1 / 4) × 9
Median = 144.5 + 2.25
Median = 146.75 mm
Ans: Median length of the leaves = 146.75 mm
5. The following table gives the distribution of the life time of 400 neon lamps:
| Life time (in hours) | Number of lamps |
| 1500 – 2000 | 14 |
| 2000 – 2500 | 56 |
| 2500 – 3000 | 60 |
| 3000 – 3500 | 86 |
| 3500 – 4000 | 74 |
| 4000 – 4500 | 62 |
| 4500 – 5000 | 48 |
Find the median life time of a lamp.
Soln:
| Life time (in hours) | Number of lamps (fi) | Cumulative frequency (cf) |
| 1500 – 2000 | 14 | 14 |
| 2000 – 2500 | 56 | 70 |
| 2500 – 3000 | 60 | 130 |
| 3000 – 3500 | 86 | 216 |
| 3500 – 4000 | 74 | 290 |
| 4000 – 4500 | 62 | 352 |
| 4500 – 5000 | 48 | 400 |
| Total | n = 400 | — |
n = 400 ⇒ n/2 = 400 / 2 = 200
The cumulative frequency just greater than 200 is 216, which corresponds to the class interval 3000 – 3500.
Median class = 3000 – 3500
Here:
l = 3000, n/2 = 200, cf = 130, f = 86, h = 500
Formula:
Median = l + [((n/2) – cf) / f] × h
Substitution:
Median = 3000 + [(200 – 130) / 86] × 500
Median = 3000 + (70 / 86) × 500
Median = 3000 + 35000 / 86
Median = 3000 + 406.98
Median = 3406.98 hours
Ans: Median life time of a lamp = 3406.98 hours
6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
| Number of letters | Number of surnames |
| 1 – 4 | 6 |
| 4 – 7 | 30 |
| 7 – 10 | 40 |
| 10 – 13 | 16 |
| 13 – 16 | 4 |
| 16 – 19 | 4 |
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Soln: Assumed mean (a) = 8.5
Class size (h) = 3
uᵢ = (xᵢ – 8.5) / 3
| Number of letters | fi | cf | xi | ui=3xi−8.5 | fiui |
| 1 – 4 | 6 | 6 | 2.5 | -2 | -12 |
| 4 – 7 | 30 | 36 | 5.5 | -1 | -30 |
| 7 – 10 | 40 | 76 | 8.5 (a) | 0 | 0 |
| 10 – 13 | 16 | 92 | 11.5 | 1 | 16 |
| 13 – 16 | 4 | 96 | 14.5 | 2 | 8 |
| 16 – 19 | 4 | 100 | 17.5 | 3 | 12 |
| Total | sum f_i = 100 | — | — | — |
(i) Median:
n = 100 ⇒ n/2 = 50
Median class = 7 – 10 l = 7, cf = 36, f = 40, h = 3
Median = l + [((n/2) – cf) / f] × h
Median = 7 + [(50 – 36) / 40] × 3
Median = 7 + (14 / 40) × 3
Median = 7 + 42 / 40
Median = 7 + 1.05
Median = 8.05
(ii) Mean:
Mean (x̄) = a + (Σfᵢuᵢ / Σfᵢ) × h
x̄ = 8.5 + (-6 / 100) × 3
x̄ = 8.5 – 18 / 100
x̄ = 8.5 – 0.18
x̄ = 8.32
(iii) Mode:
Modal class = 7 – 10 (maximum frequency = 40) l = 7, f₁ = 40, f₀ = 30, f₂ = 16, h = 3
Mode = l + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h
Mode = 7 + [(40 – 30) / (2(40) – 30 – 16)] × 3
Mode = 7 + [10 / (80 – 46)] × 3
Mode = 7 + 30 / 34
Mode = 7 + 0.88
Mode = 7.88
Ans:
Median = 8.05
Mean = 8.32
Mode = 7.88
Class 10 Maths Chapter 13.3 Statistics SEBA 2026-27
7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
| Weight (in kg) | Number of students |
| 40 – 45 | 2 |
| 45 – 50 | 3 |
| 50 – 55 | 8 |
| 55 – 60 | 6 |
| 60 – 65 | 6 |
| 65 – 70 | 3 |
| 70 – 75 | 2 |
Soln:
| Weight (in kg) | Number of students (fi) | Cumulative frequency (cf) |
| 40 – 45 | 2 | 2 |
| 45 – 50 | 3 | 5 |
| 50 – 55 | 8 | 13 |
| 55 – 60 | 6 | 19 |
| 60 – 65 | 6 | 25 |
| 65 – 70 | 3 | 28 |
| 70 – 75 | 2 | 30 |
| Total | n = 30 | — |
n = 30 ⇒ n/2 = 30 / 2 = 15
The cumulative frequency just greater than 15 is 19, which corresponds to the class interval 55 – 60.
Median class = 55 – 60
Here:
l = 55, n/2 = 15, cf = 13, f = 6, h = 5
Formula:
Median = l + [((n/2) – cf) / f] × h
Substitution:
Median = 55 + [(15 – 13) / 6] × 5
Median = 55 + (2 / 6) × 5
Median = 55 + 5 / 3
Median = 55 + 1.67
Median = 56.67 kg
Ans: Median weight = 56.67 kg
8. Determine the value of median for the following data
| Marks | No. of Students |
| Less than 10 | 5 |
| Less than 20 | 15 |
| Less than 30 | 98 |
| Less than 40 | 242 |
| Less than 50 | 367 |
| Less than 60 | 405 |
| Less than 70 | 425 |
| Less than 80 | 438 |
| Less than 90 | 439 |
Soln:
| Marks | Cumulative frequency (cf) | Class Interval | Frequency (fi) |
| Less than 10 | 5 | 0 – 10 | 5 |
| Less than 20 | 15 | 10 – 20 | 15 – 5 = 10 |
| Less than 30 | 98 | 20 – 30 | 98 – 15 = 83 |
| Less than 40 | 242 | 30 – 40 | 242 – 98 = 144 |
| Less than 50 | 367 | 40 – 50 | 367 – 242 = 125 |
| Less than 60 | 405 | 50 – 60 | 405 – 367 = 38 |
| Less than 70 | 425 | 60 – 70 | 425 – 405 = 20 |
| Less than 80 | 438 | 70 – 80 | 438 – 425 = 13 |
| Less than 90 | 439 | 80 – 90 | 439 – 438 = 1 |
| Total | n = 439 | — | — |
n = 439 ⇒ n/2 = 439 / 2 = 219.5
The cumulative frequency just greater than 219.5 is 242, which corresponds to the class interval 30 – 40.
Median class = 30 – 40
Here:
l = 30, n/2 = 219.5, cf = 98, f = 144, h = 10
Formula:
Median = l + [((n/2) – cf) / f] × h
Substitution:
Median = 30 + [(219.5 – 98) / 144] × 10
Median = 30 + (121.5 / 144) × 10
Median = 30 + 1215 / 144
Median = 30 + 8.44
Median = 38.44
Ans: Median = 38.44
9. During the medical check-up of 35 students of a class, their weights were recorded as follows:
| Weight (in kg) | No. of Students |
| Less than 38 | 0 |
| Less than 40 | 3 |
| Less than 42 | 5 |
| Less than 44 | 9 |
| Less than 46 | 14 |
| Less than 48 | 28 |
| Less than 50 | 32 |
| Less than 52 | 35 |
Calculate the median weight from the data
Soln:
| Weight (in kg) | Cumulative frequency (cf) | Class Interval | Frequency (fi) |
| Less than 38 | 0 | Below 38 | 0 |
| Less than 40 | 3 | 38 – 40 | 3 – 0 = 3 |
| Less than 42 | 5 | 40 – 42 | 5 – 3 = 2 |
| Less than 44 | 9 | 42 – 44 | 9 – 5 = 4 |
| Less than 46 | 14 | 44 – 46 | 14 – 9 = 5 |
| Less than 48 | 28 | 46 – 48 | 28 – 14 = 14 |
| Less than 50 | 32 | 48 – 50 | 32 – 28 = 4 |
| Less than 52 | 35 | 50 – 52 | 35 – 32 = 3 |
| Total | n = 35 | — | — |
n = 35 ⇒ n/2 = 35 / 2 = 17.5
The cumulative frequency just greater than 17.5 is 28, which corresponds to the class interval 46 – 48.
Median class = 46 – 48
Here:
l = 46, n/2 = 17.5, cf = 14, f = 14, h = 2
Formula:
Median = l + [((n/2) – cf) / f] × h
Substitution:
Median = 46 + [(17.5 – 14) / 14] × 2
Median = 46 + (3.5 / 14) × 2
Median = 46 + 7 / 14
Median = 46 + 0.5
Median = 46.5 kg
Ans: Median weight = 46.5 kg
SEBA Class 10 Maths Ex 13.3 Question Answer
10. Which of the following statements about statistics are true?
(i) Statistics involves data collection
(ii) Statistics helps in drawing conclusions
(iii) All data must be quantitative
(iv) Statistics can be misleading if misused
Choose correct option
(a) (i) and (ii) are true
(b) (i), (ii) and (iv) are true
(c) (iii) and (iv) are true
(d) (i), (ii) and (iii) are true
Soln: Analysis of statements:
(i) Statistics involves data collection: True
(ii) Statistics helps in drawing conclusions: True
(iii) All data must be quantitative: False (data can also be qualitative)
(iv) Statistics can be misleading if misused: True
Hence, statements (i), (ii), and (iv) are true.
Correct Option: (b)
11. Below is the data for weekly earnings (in Rs.) of workers in a factory.
| Earnings (in Rs.) | No. of workers |
| 5000-6000 | 6 |
| 6000-7000 | 10 |
| 7000-8000 | 22 |
| 8000-9000 | 10 |
| 9000-10000 | 2 |
What is the cumulative frequency up to income Rs. 9000?
(a) 16
(b) 38
(c) 48
(d) 50
Soln:
| Earnings (in Rs.) | No. of workers (fi) | Cumulative frequency (cf) |
| 5000 – 6000 | 6 | 6 |
| 6000 – 7000 | 10 | 6 + 10 = 16 |
| 7000 – 8000 | 22 | 16 + 22 = 38 |
| 8000 – 9000 | 10 | 38 + 10 = 48 |
| 9000 – 10000 | 2 | 48 + 2 = 50 |
| Total | n = 50 | — |
Cumulative frequency up to income Rs. 9000 corresponds to the upper limit of the class 8000 – 9000, which is 48.
Correct Option: (c)
Ans: (c) 48
12. Match the formulas with the measure given in list (I) and list (II), then choose the correct option from the given alternatives
| List-(I) | List-(II) |
| P. Σfx / Σf | (i) Median |
| Q. l + [((n/2) – c.f.) / f] × h | (ii) Mode |
| R. l + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h | (iii) Mean |
| S. Highest value – Lowest value | (iv) Range |
(a) P → (iii), Q → (i), R → (ii), S → (iv)
(b) P → (i), Q → (ii), R → (iii), S → (iv)
(c) P → (iii), Q → (ii), R → (iv), S → (i)
(d) P → (iii), Q → (iv), R → (i), S → (ii)
Soln: Matching:
- P. Σfx / Σf = Mean ⇒ (iii)
- Q. l + [((n/2) – cf) / f] × h = Median ⇒ (i)
- R. l + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h = Mode ⇒ (ii)
- S. Highest value – Lowest value = Range ⇒ (iv)
Order: P → (iii), Q → (i), R → (ii), S → (iv)
Correct Option: (a)
SEBA HSLC Class 10 Maths Chapter 13 Exercise 13.3 Solved PDF
13. The following table shows the marks obtained by 60 students in a mathematics test
| Class interval (Marks) | Number of Students (frequency) |
| 0-10 | 2 |
| 10-20 | 4 |
| 20-30 | 10 |
| 30-40 | 11 |
| 40-50 | 15 |
| 50-60 | 10 |
| 60-70 | 3 |
| 70-80 | 2 |
| 80-90 | 2 |
| 90-100 | 1 |
Which is the median class of the above frequency table?
(a) 40-50
(b) 50-60
(c) 60-70
(d) 30-40
Soln:
| Class interval (Marks) | Frequency (fi) | Cumulative frequency (cf) |
| 0 – 10 | 2 | 2 |
| 10 – 20 | 4 | 6 |
| 20 – 30 | 10 | 16 |
| 30 – 40 | 11 | 27 |
| 40 – 50 | 15 | 42 |
| 50 – 60 | 10 | 52 |
| 60 – 70 | 3 | 55 |
| 70 – 80 | 2 | 57 |
| 80 – 90 | 2 | 59 |
| 90 – 100 | 1 | 60 |
| Total | n = 60 | — |
n = 60 ⇒ n/2 = 60 / 2 = 30
The cumulative frequency just greater than 30 is 42, which corresponds to the class interval 40 – 50.
Median class = 40 – 50
Correct Option: (a)
14. If the mean is 20 and the median is 18, then the mode is. (a) 18 (b) 22 (c) 14 (d) 26
Soln: Given:
Mean = 20
Median = 18
Empirical Formula:
Mode = 3(Median) – 2(Mean)
Substitution:
Mode = 3(18) – 2(20)
Mode = 54 – 40
Mode = 14
Correct Option: (c)
15. In annual sports meet of a school, a 100 m race was organized. A table shown below describes the time in which the race was completed by the number of students
| Time (in sec) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 PNG |
| No. of students | 6 | 13 | 10 | 3 | 8 |
Based on the above data answer the following questions:
(i) The lower limit of the modal class is
(a) 30
(b) 40
(c) 60
(d) 20
(ii) The average time taken by the student to finish the race is
(a) 40
(b) 47
(c) 50
(d) 30
(iii) The cumulative frequency table is constucted to determine
(a) Mean
(b) Mode
(c) Median
(d) All of above
(iv) How many students finished the race within 1 minute?
(a) 30
(b) 40
(c) 19
(d) 29
Soln:
| Time (in sec) | No. of students (fi) | Class mark (xi) | fixi | cf |
| 0 – 20 | 6 | 10 | 60 | 6 |
| 20 – 40 | 13 | 30 | 390 | 19 |
| 40 – 60 | 10 | 50 | 500 | 29 |
| 60 – 80 | 3 | 70 | 210 | 32 |
| 80 – 100 | 8 | 90 | 720 | 40 |
| Total | Sum of fi = 40 | — | Sum of (fi * xi) = 1880 | — |
(i) Lower limit of the modal class:
Highest frequency is 13, corresponding to the modal class 20 – 40.
Lower limit = 20
Correct Option: (d)
(ii) Average time taken to finish the race:
Mean (x̄) = Σfᵢxᵢ / Σfᵢ
x̄ = 1880 / 40
= 47 seconds
Correct Option: (b)
(iii) Purpose of cumulative frequency table:
Cumulative frequency table is constructed to determine the Median.
Correct Option: (c)
(iv) Number of students who finished within 1 minute (≤ 60 seconds):
Students with time 0 – 60 seconds = 6 + 13 + 10 = 29
Correct Option: (d)
📊 HSLC Class 10 Maths Exercise 13.3 Notice: This page features the complete, step-by-step SEBA Class 10 Mathematics Chapter 13 (Statistics) Exercise 13.3 Official Textual Solutions updated for the 2026–27 academic session. Formatted according to the strict HSLC board marking scheme, every numerical problem—from setting up cumulative frequency tables(cf), locating the median class using n/2, to working through the high-weightage missing frequencies x and y (where total frequency is 60 and median is 28.5)—is worked out with clear step-by-step algebraic simplification to ensure you secure full marks in your 4-mark and 5-mark board exam questions.
💡 Master SEBA Class 10 Median & Statistics:
- Having difficulty converting “below 20, below 25…” into continuous class intervals in Question 3, adjusting boundary limits by $0.5$ for the leaves length problem in Question 4, or solving the simultaneous equations for $x$ and $y$ in Question 2? Drop your query in the comments section below!
- Bookmark this page for quick cumulative frequency verification and formula reference before your unit tests, pre-board exams, and the upcoming HSLC examinations.
- Share this link with your classmates and WhatsApp study groups to make Class 10 Mathematics median calculations fast, methodical, and scoring!