Exercise 7.1 — Mean, Median and Mode

Problems based on mean, median and mode.

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Exercise 7.1 – Arithmetic Mean, Median, Mode & Deviation Method

Exercise 7.1 of Class 8 Mathematics (Chapter 7: Frequency Distribution Tables and Graphs) covers 20 problems testing all three measures of central tendency — Arithmetic Mean, Median, and Mode — as well as the powerful Deviation Method for finding the mean when data values are large or close together. This exercise is part of the CBSE, Telangana, and Andhra Pradesh Class 8 syllabi and is a direct preparation for similar questions in Class 9 and Class 10 Statistics chapters.

QuestionsTopicKey Skill
Q1 – Q6Arithmetic MeanDirect mean, reverse formula, error correction, age problems
Q7 – Q11Deviation MethodFinding mean using assumed mean; sum of deviations
Q12 – Q13MedianEven/odd n; finding unknown from given median
Q14 – Q17ModeIdentifying mode; effect of operations on mode
Q18 – Q20Mixed / Higher OrderMean of transformed data; maximising median; median unchanged

Questions 1–6: Arithmetic Mean

Arithmetic Mean

The first six questions test the core mean formula and its applications — finding the total from a known mean, correcting a miscalculated mean, and using the property that adding a constant to all observations shifts the mean by the same constant.

Mean (x̄) = Sum of all observations / Number of observations (n)
Sum of observations = Mean × n
Q1
Find the arithmetic mean of the daily sales of a fair price shop in a week: ₹10,000 | ₹10,250 | ₹10,790 | ₹9,865 | ₹15,350 | ₹10,110
DayMonTueWedThuFriSatTotal
Sales (₹) 10,00010,25010,7909,86515,35010,110 66,365
  1. Sum of observations = 10000 + 10250 + 10790 + 9865 + 15350 + 10110 = ₹66,365
  2. Number of observations (n) = 6
  3. Mean = 66365 ÷ 6 = ₹11,060.83 (approx.)
Mean = 66,365 / 6 = ₹11,060.83
∴ Mean daily sales = ₹11,060.83
📌 Notice that Thursday's unusually high sales (₹15,350) pulls the mean above most other days. This shows why mean is sensitive to extreme values — a key exam point.
Q2
Find the mean of: 10.25, 9, 4.75, 8, 2.65, 12, 2.35
Observations10.2594.7582.65122.35Sum
Values10.2594.7582.65122.3549
  1. Sum = 10.25 + 9 + 4.75 + 8 + 2.65 + 12 + 2.35 = 49
  2. n = 7
  3. Mean = 49 ÷ 7 = 7
Mean = 49 / 7 = 7
∴ Mean = 7
Q3
Mean of eight observations is 25. If one observation 11 is excluded, find the mean of the remaining observations.

This problem uses the reverse formula: Sum = Mean × n, then removes one value.

  1. Sum of 8 observations = Mean × n = 25 × 8 = 200
  2. After removing 11: New sum = 200 − 11 = 189
  3. Remaining observations = 8 − 1 = 7
  4. New mean = 189 ÷ 7 = 27
New Mean = (25×8 − 11) / 7 = 189 / 7 = 27
∴ Mean of remaining 7 observations = 27
Q4
Mean of 9 observations is 38. But observation 27 was mistakenly used instead of 72. Find the actual mean.

This is a error-correction problem. The wrong value inflated or deflated the sum — we fix the sum and recalculate.

  1. Incorrect sum = 38 × 9 = 342
  2. Correct sum = 342 − 27 (wrong) + 72 (correct) = 342 − 27 + 72 = 387
  3. Actual mean = 387 ÷ 9 = 43
Actual Mean = (38×9 − 27 + 72) / 9 = 387 / 9 = 43
∴ Actual mean = 43
💡 General formula for error correction:
Actual Mean = Incorrect Mean + (Correct value − Wrong value) / n
Q5
Five years ago, the mean age of a family was 25 years. What is the present mean age?

This question applies the addition property of mean: if a fixed number is added to every observation, the mean increases by the same number.

  1. Mean age 5 years ago = 25 years
  2. Each member is now 5 years older, so every observation increases by 5.
  3. Therefore, the mean also increases by 5.
  4. Present mean age = 25 + 5 = 30 years
Present Mean = Old Mean + Years passed = 25 + 5 = 30
∴ Present mean age = 30 years
Q6
Two years ago, mean age of 40 people was 11 years. Now a person left the group and mean age changed to 12 years. Find the age of the person who left.

This combines the age increase property and the reverse formula.

  1. Mean age 2 years ago = 11 years → Present mean of 40 people = 11 + 2 = 13 years
  2. Sum of ages of 40 people (now) = 13 × 40 = 520
  3. After 1 person leaves, 39 people remain with mean age 12.
  4. Sum of ages of 39 people = 12 × 39 = 468
  5. Age of person who left = 520 − 468 = 52 years
Age of person = (13×40) − (12×39) = 520 − 468 = 52
∴ Age of person who left = 52 years
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Deviation Method — Concept & Questions 7–11

Deviation Method

The Deviation Method (also called the Assumed Mean Method) is a shortcut for finding the mean when the observations are large numbers. Instead of summing all values directly, you choose any convenient number as an Assumed Mean (A), calculate how much each value deviates from A, and then use the formula below.

x̄ = A + (Σdᵢ / n)
where dᵢ = (xᵢ − A) is the deviation of each observation from the assumed mean A

Worked Illustration — Marks of 6 subjects

Marks: 18, 14, 13, 15, 17, 19. Let A = 17 (assumed mean).

Observation (xᵢ)Assumed Mean (A)Deviation dᵢ = xᵢ − A
1817+1
1417−3
1317−4
1517−2
17170
1917+2
n = 6—Σdᵢ = −6
x̄ = 17 + (−6 / 6) = 17 − 1 = 16
💡 Key insight: No matter which number you choose as the assumed mean A, the final answer is always the same. Q9 demonstrates this directly.
Q7
Find the sum of deviations of all observations 5, 8, 10, 15, 22 from their mean.
  1. Sum = 5 + 8 + 10 + 15 + 22 = 60; n = 5
  2. Mean = 60 ÷ 5 = 12
  3. Deviations from mean: (5−12), (8−12), (10−12), (15−12), (22−12) = −7, −4, −2, +3, +10
  4. Sum of deviations = −7 − 4 − 2 + 3 + 10 = 0
Sum of deviations from mean = −7 − 4 − 2 + 3 + 10 = 0
∴ Sum of deviations from the mean = 0
🔎 Universal property: The sum of deviations of all observations from their arithmetic mean is always zero. This holds for any dataset — it is a fundamental property of the arithmetic mean.
Q8
Sum of 20 deviations from the mean is 100. Find the mean deviation.
  1. Sum of deviations = 100
  2. Number of deviations = 20
  3. Mean deviation = 100 ÷ 20 = 5
Mean deviation = Sum of deviations / n = 100 / 20 = 5
∴ Mean deviation = 5
Q9
Marks of 12 students: 4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20, 14. Calculate mean using two different assumed means and verify the result is the same.

This problem beautifully proves that the arithmetic mean is independent of the assumed mean.

Using Assumed Mean A = 16

Observation421131759102019122014Σdᵢ
Deviation (xᵢ−16) −12+5−3+1−11−7−6+4+3−4+4−2 −28
x̄ = 16 + (−28 / 12) = 16 − 2.33 = 13.67

Using Assumed Mean A = 12

Observation421131759102019122014Σdᵢ
Deviation (xᵢ−12) −8+9+1+5−7−3−2+8+70+8+2 +20
x̄ = 12 + (20 / 12) = 12 + 1.67 = 13.67
∴ Mean = 13.67 in both cases ✓
📝 Conclusion: Regardless of the assumed mean chosen, the arithmetic mean of the data is always the same. This confirms the deviation method is a reliable shortcut.
Q10
Mean of marks scored by 10 students = 15. Karishma's deviations from the other 9 students' marks are −8, −6, −3, −1, 0, 2, 3, 4, 6. Find Karishma's marks.

Here, Karishma's own marks (x) are the assumed mean. The deviation of each of the other 9 students from x is given. Her own deviation from herself is 0.

  1. Let Karishma's marks = x (this is also the assumed mean A)
  2. All 10 deviations including Karishma's own (= 0): 0, −8, −6, −3, −1, 0, 2, 3, 4, 6
  3. Σdᵢ = 0 − 8 − 6 − 3 − 1 + 0 + 2 + 3 + 4 + 6 = −3
  4. Apply deviation formula: 15 = x + (−3 / 10) = x − 0.3
  5. x = 15 + 0.3 = 15.3
15 = x + (−3/10) → x = 15 + 0.3 = 15.3
∴ Karishma's marks = 15.3
Q11
Sum of deviations of n observations from 25 is 25, and from 35 is −25. Find the mean.

Set up two equations using the deviation formula, then solve simultaneously.

  1. Using A = 25:   x̄ = 25 + (25/n)   → equation (1)
  2. Using A = 35:   x̄ = 35 + (−25/n) = 35 − (25/n)   → equation (2)
  3. Equate (1) and (2):   25 + 25/n = 35 − 25/n
  4. 25/n + 25/n = 35 − 25 → 50/n = 10 → n = 5
  5. Substitute in (1): x̄ = 25 + (25/5) = 25 + 5 = 30
Eq.(1): x̄ = 25 + 25/n
Eq.(2): x̄ = 35 − 25/n
Adding: 2x̄ = 60 → x̄ = 30  |  n = 5
∴ Mean = 30, n = 5

Questions 12–13: Median

Median
If n is ODD: Median = value at position (n+1)/2
If n is EVEN: Median = average of values at positions n/2 and (n/2 + 1)
Q12
Find the median of: 3.3, 3.5, 3.1, 3.7, 3.2, 3.8
  1. Arrange in ascending order: 3.1, 3.2, 3.3, 3.5, 3.7, 3.8
  2. n = 6 (even)
  3. Median positions: n/2 = 3rd and n/2 + 1 = 4th
  4. 3rd value = 3.3, 4th value = 3.5
  5. Median = (3.3 + 3.5) / 2 = 6.8 / 2 = 3.4
Median = (3.3 + 3.5) / 2 = 3.4
∴ Median = 3.4
Q13
The data in ascending order is 10, 12, 14, x−3, x, x+2, 25 and its median is 15. Find x.
  1. n = 7 (odd)
  2. Median position = (7 + 1) / 2 = 4th observation
  3. 4th observation = x − 3
  4. Given median = 15, so: x − 3 = 15 → x = 18
4th term = x − 3 = 15 → x = 18
∴ x = 18
📝 Verify: sorted data becomes 10, 12, 14, 15, 18, 20, 25 — the 4th value is 15 ✓

Questions 14–17: Mode

Mode

The mode is the most frequently occurring value. These questions test your ability to identify mode, understand how operations on data affect the mode, and work with real-world counting contexts.

Q14
Find the mode of: 10, 12, 11, 10, 15, 20, 19, 21, 11, 9, 10
Value910111215192021
Frequency13211111

The value 10 occurs 3 times — more than any other value.

∴ Mode = 10
Q15
Mode of certain scores is x. If each score is decreased by 3, find the mode of the new series.

Just like the mean, the mode also shifts by the same constant when every observation is increased or decreased by a fixed number.

  1. Original mode = x
  2. Each score decreases by 3 → the most frequent value also decreases by 3
  3. New mode = x − 3
∴ New mode = x − 3
Q16
Find the mode of all digits used in writing the natural numbers from 1 to 100.

This is a counting problem. We need to count how many times each digit (0–9) appears when writing 1, 2, 3, … 100.

DigitWhere it appearsCount
010, 20, 30, …90, 100 (units); 100 (hundreds)11
11–9 (once each in 1,10–19 = 11 times for tens/units), 21,31,…91, 100 → total21
2–9Each appears in units + tens across 1–10020 each

Digit 1 appears most — once in 1, then in 10–19 (as the tens digit, 10 times, plus units digit giving 11, 21, 31, …), and once in 100. Total = 21 times. Digits 2–9 each appear exactly 20 times. Digit 0 appears 11 times.

∴ Mode of digits = 1 (appears 21 times)
Q17
Data: 5, 28, 15, 10, 15, 8, 24. Add 4 numbers so that mean and median remain the same, but mode increases by 1.

This is a creative higher-order thinking question combining all three measures.

  1. Current mode = 15 (appears twice). Target mode = 15 + 1 = 16.
  2. For 16 to be the new mode, it must appear more often than 15. Since 15 appears twice, add three 16s.
  3. Let the fourth number = x. New dataset has 11 observations.
  4. Current mean = (5 + 28 + 15 + 10 + 15 + 8 + 24) / 7 = 105 / 7 = 15
  5. Mean must stay 15: (105 + 16 + 16 + 16 + x) / 11 = 15
  6. 153 + x = 165 → x = 12
(105 + 48 + x) / 11 = 15 → 153 + x = 165 → x = 12
∴ Four numbers to add: 16, 16, 16, 12
📝 Verify median: New sorted data: 5, 8, 10, 12, 15, 15, 16, 16, 16, 24, 28 → 6th value = 15 = original median ✓

Questions 18–20: Higher Order Thinking

Mixed / HOTS
Q18
Mean of x₁, x₂, … x₁₀ is 20. Find the mean of (x₁+4), (x₂+8), (x₃+12), …, (x₁₀+40).

Each observation is increased by a different amount (4, 8, 12, …, 40), so we cannot simply add 4 to the mean.

  1. Sum of original data = 20 × 10 = 200
  2. Extra sum added = 4 + 8 + 12 + … + 40 = 4 × (1 + 2 + 3 + … + 10) = 4 × 55 = 220
  3. New sum = 200 + 220 = 420
  4. New mean = 420 / 10 = 42
New Mean = (200 + 4×55) / 10 = (200 + 220) / 10 = 420 / 10 = 42
∴ New mean = 42
💡 Recall: 1 + 2 + 3 + … + n = n(n+1)/2. Here n = 10, so sum = 10×11/2 = 55.
Q19
Six of nine integers are 7, 8, 3, 5, 9, 5. Find the largest possible median of all nine integers.
  1. Arrange the 6 known integers: 3, 5, 5, 7, 8, 9
  2. For n = 9 (odd), median = 5th observation.
  3. To maximise the median, we want the 5th position value to be as large as possible.
  4. Place the 3 unknown numbers after 9 (i.e., all greater than 9).
  5. Sorted full list: 3, 5, 5, 7, 8, 9, a, b, c (where a, b, c > 9)
  6. 5th observation = 8
Median = 5th observation = 8
∴ Largest possible median = 8
Q20
Median of 9 distinct observations is 20. If each of the 4 largest observations is increased by 2, find the new median.
  1. Let sorted data be: a, b, c, d, e, f, g, h, i (n = 9, odd)
  2. Median = 5th observation = e = 20
  3. The 4 largest observations are f, g, h, i — all to the right of the median.
  4. After increasing them by 2: a, b, c, d, e, f+2, g+2, h+2, i+2
  5. The 5th observation is still e = 20 — unchanged.
∴ Median of new set = 20 (unchanged)
🔎 Why? The median only depends on the middle value. Changing values that are above the median leaves the middle position untouched — the median stays the same.

Common Mistakes to Avoid in Exercise 7.1

  • Forgetting to sort before finding median: The median formula only works on data arranged in ascending or descending order. Always sort first.
  • Using wrong n in age problems (Q6): When a person leaves or joins, n changes. Recalculate the sum for the new n separately.
  • Not updating the sum in error-correction problems (Q4): Subtract the wrong value and add the correct value — do not just add the difference.
  • Mixing up deviation signs: In the deviation method, dᵢ = xᵢ − A. Negative deviations reduce the mean below A; positive ones raise it.
  • Assuming the mode is unique: A dataset can have two or more modes, or no mode at all. State this clearly in board exam answers.
  • In Q18: The additions are not all equal (4, 8, 12, …), so you cannot simply add 4 to the mean. You must sum the additional terms using the formula for sum of an AP.
📐 Telangana / AP / CBSE Board Exam Tip: Write the formula first, then substitute values, then state the final answer with a "∴" (therefore) symbol. For deviation method questions, always show the deviation table — examiners award step marks for it even if the final answer is wrong.

What Exercise 7.1 Prepares You For

Having mastered mean, median, and mode through these 20 problems, you are ready for the next exercises in Chapter 7, where you will learn to organise this kind of data into frequency distribution tables and represent it graphically using histograms and frequency polygons. The arithmetic mean of grouped data — calculated using the direct method and the assumed mean method — builds directly on Q7–Q11 here.

These same concepts reappear in more depth in Class 9 and Class 10. In Class 9 Statistics, you will calculate mean, median, and mode for grouped frequency distributions. In Class 10 Statistics, you will draw cumulative frequency curves (ogives) and find the median graphically — a skill that has carried marks in the AP and Telangana SSC board exams every year.

For more practice on the underlying number properties used in this exercise, revisit the Introduction lesson, which covers the core definitions and properties of all three central tendency measures.

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