Question 1

A shopkeeper recorded daily sales (in ₹100s) for 6 days: 12, 15, 18, 12, 20, 15. What is the mean daily sales?

Accepted Answer

Step 1: Sum all values: 12 + 15 + 18 + 12 + 20 + 15 = 92. Step 2: Count the number of days: 6. Step 3: Mean = 92 / 6 = 15.33 (in ₹100s). Step 4: In actual rupees: 15.33 × 100 = ₹1533.33 ≈ ₹1533.

Question 2

The mean of 8 numbers is 32. If one number is removed, the mean of the remaining 7 numbers becomes 30. What is the number that was removed?

Accepted Answer

Step 1: Sum of 8 numbers = Mean × Count = 32 × 8 = 256. Step 2: Sum of 7 numbers = 30 × 7 = 210. Step 3: The removed number = 256 - 210 = 46.

Question 3

The median of the dataset {7, 12, 5, 18, 9, 14, 11} is:

Accepted Answer

Step 1: Arrange the data in ascending order: {5, 7, 9, 11, 12, 14, 18}. Step 2: Count the number of values: 7 (odd number). Step 3: The median is the middle value at position (7+1)/2 = 4. Step 4: The 4th value is 11.

Question 4

The mode of the dataset {15, 22, 15, 18, 22, 15, 25, 22} is:

Accepted Answer

Step 1: Count the frequency of each value. 15 appears 3 times, 22 appears 3 times, 18 appears 1 time, 25 appears 1 time. Step 2: Both 15 and 22 have the highest frequency of 3. Step 3: When two values share the highest frequency, the dataset is bimodal. Step 4: The modes are 15 and 22.

Question 5

A dataset consists of 6 values: 10, 14, 14, 18, 22, 26. If the value 14 is replaced by 20, what is the change in the mean?

Accepted Answer

Step 1: Original sum = 10 + 14 + 14 + 18 + 22 + 26 = 104. Original mean = 104 ÷ 6 ≈ 17.33. Step 2: New sum = 10 + 20 + 14 + 18 + 22 + 26 = 110. New mean = 110 ÷ 6 ≈ 18.33. Step 3: Change in mean = 18.33 − 17.33 = 1 (or exactly 6 ÷ 6 = 1).

Question 6

In a class of 30 students, the mean score in a test is 72. If 5 students with a mean score of 60 are excluded, what is the mean score of the remaining 25 students?

Accepted Answer

Step 1: Total score of 30 students = 30 × 72 = 2160. Step 2: Total score of 5 excluded students = 5 × 60 = 300. Step 3: Total score of remaining 25 students = 2160 − 300 = 1860. Step 4: Mean of remaining 25 = 1860 ÷ 25 = 74.4.

Question 7

A dataset has 5 observations: 12, 15, 18, 15, 20. What is the difference between the median and mode of this dataset?

Accepted Answer

Step 1: Arrange in order: 12, 15, 15, 18, 20. Step 2: Median (middle value of 5 observations) = 15. Step 3: Mode (most frequent value) = 15 (appears twice). Step 4: Difference = 15 − 15 = 0.

Question 8

The mean of a set of 10 numbers is 35. If the mean of the first 6 numbers is 32, what is the mean of the remaining 4 numbers?

Accepted Answer

Step 1: Sum of all 10 numbers = 10 × 35 = 350. Step 2: Sum of first 6 numbers = 6 × 32 = 192. Step 3: Sum of remaining 4 numbers = 350 − 192 = 158. Step 4: Mean of remaining 4 numbers = 158 ÷ 4 = 39.5.

Question 9

The mean of 8 numbers is 24. If one number is removed, the mean of the remaining 7 numbers becomes 22. What is the value of the removed number?

Accepted Answer

Step 1: Sum of 8 numbers = 8 × 24 = 192. Step 2: Sum of remaining 7 numbers = 7 × 22 = 154. Step 3: Removed number = 192 − 154 = 38.

Question 10

A dataset contains 7 numbers. When the smallest number is removed, the mean of the remaining 6 numbers is 18. When the largest number is removed instead, the mean of the remaining 6 numbers is 16. If the median of all 7 numbers is 17, find the sum of the smallest and largest numbers.

Accepted Answer

Step 1: Let the sum of all 7 numbers = S. When smallest is removed, sum of 6 numbers = S - smallest = 6 × 18 = 108, so S = 108 + smallest. Step 2: When largest is removed, sum of 6 numbers = S - largest = 6 × 16 = 96, so S = 96 + largest. Step 3: From both equations: 108 + smallest = 96 + largest, therefore largest - smallest = 12. Step 4: Total sum S = 108 + smallest. Also S = 96 + largest = 96 + (smallest + 12) = 108 + smallest ✓. Step 5: The median of 7 numbers is the 4th value = 17. Step 6: Sum of all 7 = 108 + smallest. We need another constraint. From S - smallest = 108 and S - largest = 96, we get S = 108 + smallest and largest = smallest + 12. Step 7: The sum of smallest and largest = smallest + (smallest + 12) = 2·smallest + 12. Step 8: Using median property and the constraint that mean of all 7 = (108 + smallest)/7, and knowing largest = smallest + 12, we can deduce: if we denote smallest = a, then largest = a + 12, and S = 108 + a. The 7 numbers arranged have median 17 at position 4. Testing: if smallest = 10, largest = 22, sum = 118, mean = 118/7 ≈ 16.86. Verification: 118 - 10 = 108 ✓, 118 - 22 = 96 ✓. Sum of smallest and largest = 10 + 22 = 32.

Question 11

In a class of 40 students, the mean score is 72. A new student joins and the mean becomes 71.5. Later, one of the original 40 students (who scored 85) leaves. What is the new mean score of the remaining students?

Accepted Answer

Step 1: Sum of 40 students' scores = 40 × 72 = 2880. Step 2: After new student joins (41 students), sum = 41 × 71.5 = 2931.5. Step 3: New student's score = 2931.5 - 2880 = 51.5. Step 4: Now we have 41 students with total 2931.5. Step 5: One original student (score 85) leaves, so we remove 85 from the total. Step 6: New sum = 2931.5 - 85 = 2846.5. Step 7: Number of remaining students = 41 - 1 = 40. Step 8: New mean = 2846.5 ÷ 40 = 71.1625 ≈ 71.16 or exactly 2846.5/40 = 5693/80 = 71.1625. Rounding to one decimal: 71.2 or checking if answer expects 71.16 or 71.1625. Most likely answer is 71.2 or 71.16. Let me recalculate: 2846.5/40 = 71.1625. If options are given as decimals, this is approximately 71.16.

Question 12

A frequency distribution has 5 classes with frequencies 8, 12, 15, 10, and 5 respectively. The class midpoints are 10, 20, 30, 40, and 50. If the mode class is the class with midpoint 30, and the mode is calculated as Mode = L + (f₁ - f₀) / (2f₁ - f₀ - f₂) × h, where L is the lower boundary of the mode class, f₁ is the frequency of the mode class, f₀ and f₂ are frequencies of adjacent classes, and h is the class width, find the mode.

Accepted Answer

Step 1: Identify the mode class — it has the highest frequency. Frequencies are 8, 12, 15, 10, 5. The highest is 15, which corresponds to midpoint 30. Step 2: The problem states the mode class is the class with midpoint 30, confirming this. Step 3: Class width h = 20 - 10 = 10 (or any consecutive midpoint difference). Step 4: Lower boundary L of the class with midpoint 30: if midpoint is 30 and width is 10, then L = 30 - 10/2 = 25. Step 5: f₁ = 15 (frequency of mode class). Step 6: f₀ = 12 (frequency of previous class, midpoint 20). f₂ = 10 (frequency of next class, midpoint 40). Step 7: Mode = 25 + (15 - 12) / (2×15 - 12 - 10) × 10 = 25 + 3 / (30 - 22) × 10 = 25 + 3/8 × 10 = 25 + 30/8 = 25 + 3.75 = 28.75.

Question 13

A dataset of 9 numbers has a median of 24. When a 10th number is added, the new median becomes 25. Which of the following MUST be true about the 10th number?

Accepted Answer

Step 1: With 9 numbers, the median is the 5th value when sorted. So the 5th value = 24. Step 2: With 10 numbers, the median is the average of the 5th and 6th values. New median = (5th + 6th)/2 = 25, so 5th + 6th = 50. Step 3: We know the 5th value = 24 (from the original dataset). Step 4: Therefore, 24 + 6th = 50, so 6th value = 26. Step 5: For the 10th number to cause the 6th position to become 26, the 10th number must be ≥ 26 (so it either becomes the 6th value or pushes the original 6th value to position 6). Step 6: Actually, let's reconsider. The original 5th value is 24. After adding the 10th number, we have 10 values. The new median is average of 5th and 6th. Step 7: If the 10th number is ≤ 24, it would be inserted at or before position 5, shifting the original 5th (24) to position 6 or later. If it's between 24 and 26, it could be at position 5 or 6. If it's ≥ 26, it's at position 6 or later. Step 8: For the new 6th value to be 26, and the new 5th value to be 24, the 10th number must be ≥ 26. But we need to verify: if 10th ≥ 26, then the original 5th (24) stays at position 5, and the original 6th becomes position 6 or later. Actually, the original 6th value is unknown. Step 9: Let's denote original sorted values as a₁, a₂, a₃, a₄, 24, a₆, a₇, a₈, a₉. After adding x, the new median = (5th + 6th)/2 = 25. Step 10: If x ≤ 24, the new 5th is still 24, and new 6th is a₆. So (24 + a₆)/2 = 25 → a₆ = 26. If x > 24, then new 5th could be 24 or something else. If 24 < x < a₆, then new 5th = 24, new 6th = x. So (24 + x)/2 = 25 → x = 26. If x ≥ a₆, then new 5th = 24, new 6th = a₆. So (24 + a₆)/2 = 25 → a₆ = 26. Step 11: In all cases, either x = 26 or a₆ = 26. The 10th number must be ≥ 24 (otherwise it would be before position 5 and shift everything). More precisely, the 10th number must be ≥ 24 and could be exactly 26 or greater. But the safest statement: 10th number ≥ 24.

Question 14

In a survey of 100 employees, the mean salary is ₹50,000. The median salary is ₹48,000. If the mode salary is ₹45,000 and represents the salary of 15 employees, and we know that exactly 20 employees earn ₹48,000, how many employees earn more than ₹48,000?

Accepted Answer

Step 1: Total employees = 100. Mean salary = ₹50,000, so total salary = 100 × 50,000 = ₹50,00,000. Step 2: Median = ₹48,000 for 100 employees means the average of the 50th and 51st values (when sorted) = ₹48,000. Step 3: Mode = ₹45,000 with 15 employees. Median = ₹48,000 with 20 employees. Step 4: For the median to be ₹48,000, the 50th and 51st values must average to ₹48,000. Since 20 employees earn exactly ₹48,000, these could include the 50th and 51st positions. Step 5: If the 50th and 51st values are both ₹48,000, then the median = ₹48,000 ✓. Step 6: For this to happen, at least 50 employees must earn ≤ ₹48,000. Step 7: We have 15 employees at ₹45,000 (mode). We need to account for employees earning < ₹45,000 and those earning between ₹45,000 and ₹48,000. Step 8: If exactly 50 employees earn ≤ ₹48,000, then 50 employees earn > ₹48,000. Step 9: But we need to verify this is consistent with the mean. Let's denote: employees earning ≤ ₹45,000 = a, earning ₹45,000 = 15, earning between ₹45,000 and ₹48,000 = b, earning ₹48,000 = 20, earning > ₹48,000 = c. Step 10: a + 15 + b + 20 + c = 100 → a + b + c = 65. Step 11: For median to be ₹48,000 with 50th and 51st values = ₹48,000: a + 15 + b ≤ 49 (so 50th is at ₹48,000) and a + 15 + b + 20 ≥ 51 (so 51st is at ₹48,000). This gives: a + b ≤ 34 and a + b ≥ 31, so 31 ≤ a + b ≤ 34. Step 12: If a + b = 31, then c = 65 - 31 = 34. If a + b = 34, then c = 31. Step 13: We need to check which is consistent with the mean. Total salary = ₹50,00,000. Salary from mode group = 15 × 45,000 = ₹6,75,000. Salary from median group = 20 × 48,000 = ₹9,60,000. Remaining salary = 50,00,000 - 6,75,000 - 9,60,000 = ₹33,65,000. Step 14: This remaining salary is for (a + b + c) = 65 employees. Average for these 65 = 33,65,000 / 65 = ₹51,615.38. Step 15: Since employees earning > ₹48,000 must have average > ₹48,000, and the overall average for the 65 is ₹51,615.38, we need to determine c. Step 16: Using a + b + c = 65 and 31 ≤ a + b ≤ 34, we get 31 ≤ c ≤ 34. But we need more constraints. Let's assume a + b = 31, so c = 34. Then employees earning > ₹48,000 = 34. But let's verify: if a + b = 34, then c = 31. Step 17: The problem likely expects us to use the constraint that the 50th and 51st positions are exactly at ₹48,000. With 15 + 20 = 35 employees at ₹45,000 or ₹48,000, and needing the 50th position to be ₹48,000, we need a + b ≤ 30 (so that positions 31-50 include the ₹48,000 group). Actually, if a + 15 + b = 30, then positions 31-50 are the ₹48,000 group (20 employees), and position 51 is the first > ₹48,000. So median = (48,000 + salary of 51st)/2 = 48,000 → 51st salary = 48,000. This means at least 51 employees earn ≤ ₹48,000. Step 18: So employees earning > ₹48,000 = 100 - 51 = 49. Hmm, let me reconsider. If the 50th and 51st are both ₹48,000, then 50 employees earn < ₹48,000 and 50 employees earn ≥ ₹48,000. But 20 earn exactly ₹48,000, so 30 earn > ₹48,000. Wait, that's not right either. Step 19: Let me restart: if 50th value = 48,000 and 51st value = 48,000, then exactly 50 employees earn ≤ 48,000 (including the 50th). So 50 employees earn ≤ 48,000 and 50 earn ≥ 48,000. But the 50th and 51st are both 48,000, so 50 earn < 48,000 or = 48,000 (with the 50th being 48,000), and 50 earn = 48,000 or > 48,000 (with the 51st being 48,000). Step 20: If exactly 20 earn 48,000, and the 50th and 51st are both 48,000, then the 20 employees earning 48,000 span from position 31 to position 50 (or some range that includes 50 and 51). Actually, if 20 earn exactly 48,000 and they occupy positions 31-50, then the 50th is 48,000 ✓. But then the 51st is the first > 48,000, so median = (48,000 + salary of 51st)/2 ≠ 48,000 unless the 51st is also 48,000. Step 21: For both 50th and 51st to be 48,000, the 20 employees earning 48,000 must include both positions 50 and 51. So they could occupy positions 31-50 or 32-51 or any range including 50-51. If they occupy positions 31-50, then the 51st is > 48,000. If they occupy positions 32-51, then the 50th is < 48,000. So they must occupy positions 30-49 or 31-50 or 32-51 or 33-52, etc. For both 50 and 51 to be in this range: 30-49 (no), 31-50 (no, 51 is outside), 32-51 (yes, 50 and 51 are inside). So the 20 employees earning 48,000 occupy positions 32-51. This means 31 employees earn < 48,000 and 49 earn > 48,000. Step 22: Employees earning > 48,000 = 49.

Question 15

A teacher records test scores for 50 students. The mean score is 75. After reviewing, the teacher realizes that 5 students' scores were recorded incorrectly — each was recorded 10 points higher than the actual score. After correction, what is the new mean score?

Accepted Answer

Step 1: Original mean = 75 for 50 students. Original total = 50 × 75 = 3750. Step 2: Five students' scores were recorded 10 points too high. Step 3: Total error = 5 × 10 = 50 points (overcount). Step 4: Corrected total = 3750 - 50 = 3700. Step 5: New mean = 3700 / 50 = 74. Step 6: Alternatively, the mean decreases by 50/50 = 1 point. New mean = 75 - 1 = 74.

SBI PO Mean, Median, Mode

SBI PO Mean, Median, Mode — Past Exam Questions

Mean, Median, Mode— Rules & Concept

Key Points to Remember

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60-Second Revision — Mean, Median, Mode