Question 1

The average marks of 40 students in an examination was 25. Later it was found that one student's marks had been wrongly entered as 73 instead of 37. What is the correct average mark of the 40 students after rectifying the mistake?

Accepted Answer

Introduction / Context: This is another example of correcting an average after discovering a data entry error. The mean mark of 40 students is initially calculated using a wrong value of 73 instead of the correct 37. We must adjust the total and recompute the average to find the correct mean for the class. Given Data / Assumptions: Number of students = 40. Initial (incorrect) average marks = 25. One mark was wrongly entered as 73. The correct mark for that student is 37. We need to find the corrected average for all 40 students. Concept / Approach: Again, we use: average = total marks / number of students. We find the incorrect total with the wrong mark, then adjust it by removing the extra amount introduced by the error and finally divide by 40 to get the corrected average. The extra amount is simply the difference between the wrong mark and the correct mark. Step-by-Step Solution: Step 1: Compute incorrect total marks. Incorrect total = 40 * 25 = 1000. Step 2: Identify the error contribution. Wrongly entered mark = 73; correct mark = 37. Error = 73 − 37 = 36 extra marks included. Step 3: Correct the total marks. Correct total = incorrect total − extra = 1000 − 36 = 964. Step 4: Compute the corrected average. Correct average = 964 / 40. Step 5: Divide: 40 * 24 = 960, remainder = 4. So 964 / 40 = 24 + 4/40 = 24.1. Verification / Alternative check: The original average 25 will drop because one mark has been lowered from 73 to 37. Average decrease = 36 / 40 = 0.9. So corrected average = 25 − 0.9 = 24.1, confirming our calculation. Why Other Options Are Wrong: 24.3, 24.5, 24.7 and 24.9 produce totals that are different from 964 when multiplied by 40. For example, if the correct average were 24.5, total would be 24.5 * 40 = 980, which does not match 964. Common Pitfalls: Some learners may mistakenly add 36 instead of subtracting, thinking they must 'correct' upwards instead of downwards. Others divide the error 36 by 39 or 41 instead of 40 when attempting to adjust the average, which leads to incorrect results. Final Answer: The corrected average marks of the 40 students is 24.1.

Question 2

For the data set 6, 3, 8, 4, 3, 11, 7, 4, 5, 4, find the sum of the mode, the median and the mean.

Accepted Answer

Introduction / Context: This question tests knowledge of three key measures of central tendency: mean, median and mode. We are given a small data set and asked to compute each measure, then sum them up. Careful ordering of data and correct definitions are crucial to arrive at the correct combined value. Given Data / Assumptions: Data set: 6, 3, 8, 4, 3, 11, 7, 4, 5, 4. Number of observations = 10. We must find: Mode – the most frequent value. Median – the middle value when data is ordered (or average of two middle values). Mean – the arithmetic average. Finally, we add mode + median + mean. Concept / Approach: We follow these steps: Sort the data in ascending order for median and mode. Identify the mode by highest frequency. For median with 10 observations, take the average of the 5th and 6th values. Compute mean as total sum divided by 10. Add the three results. Step-by-Step Solution: Step 1: Sort the data. Original: 6, 3, 8, 4, 3, 11, 7, 4, 5, 4. Sorted: 3, 3, 4, 4, 4, 5, 6, 7, 8, 11. Step 2: Find the mode (most frequent value). 3 appears twice, 4 appears three times, 5, 6, 7, 8, 11 appear once each. So mode = 4. Step 3: Find the median. There are 10 observations (even number), so median = average of 5th and 6th values. 5th value = 4, 6th value = 5. Median = (4 + 5) / 2 = 9 / 2 = 4.5. Step 4: Compute the mean. Sum all values: 6 + 3 + 8 + 4 + 3 + 11 + 7 + 4 + 5 + 4. 6 + 3 = 9, +8 = 17, +4 = 21, +3 = 24, +11 = 35, +7 = 42, +4 = 46, +5 = 51, +4 = 55. Mean = 55 / 10 = 5.5. Step 5: Sum mode, median and mean. Mode + Median + Mean = 4 + 4.5 + 5.5 = 14. Verification / Alternative check: We can confirm the sorted list and double-check the counts for the mode. The median is correctly taken as the average of the 5th and 6th values for an even-sized data set. The sum 4 + 4.5 + 5.5 clearly equals 14 with no rounding issues. Why Other Options Are Wrong: 13.5, 13 and 14.5 would require at least one of mode, median or mean to be miscalculated. For example, if someone mistakenly took median as 5 (instead of 4.5), the sum would be 4 + 5 + 5.5 = 14.5, not 14. Common Pitfalls: Not sorting the data correctly or misidentifying the 5th and 6th terms can lead to an incorrect median. Sometimes, students confuse mode with median or forget to divide by the number of observations when computing the mean. Final Answer: The sum of the mode, the median and the mean of the given data is 14.

Question 3

A certain amount of money is fully distributed among A, B and C. A receives 1/2 of the total amount, B receives 1/3 of the total amount and C receives ₹ 1200. How much money does A receive?

Accepted Answer

Introduction / Context: This is a ratio and fraction distribution problem. A fixed amount of money is shared among three people A, B and C according to given fractions. We are told the exact amount received by C and the fractions for A and B. From this, we must compute A's share of the total amount. Given Data / Assumptions: A receives 1/2 of the total amount. B receives 1/3 of the total amount. C receives ₹ 1200. The entire amount is fully distributed; nothing is left over. We must find A's amount. Concept / Approach: Let the total amount be T. Then the distribution conditions give: A = T / 2, B = T / 3, C = 1200. Since the whole amount is distributed, we have: T/2 + T/3 + 1200 = T. We solve this linear equation for T and then compute A = T/2. The key idea is to combine the fractions 1/2 and 1/3 correctly. Step-by-Step Solution: Step 1: Let total amount = T. A's share = T/2; B's share = T/3; C's share = 1200. Step 2: Set up the total distribution equation. T/2 + T/3 + 1200 = T. Step 3: Combine T/2 and T/3. T/2 + T/3 = (3T + 2T) / 6 = 5T / 6. So the equation becomes 5T/6 + 1200 = T. Step 4: Move 5T/6 to the right-hand side. 1200 = T − 5T/6 = (1T/6) = T/6. Step 5: Solve for T. T/6 = 1200 → T = 1200 * 6 = 7200. Step 6: Compute A's share. A = T/2 = 7200 / 2 = 3600. Verification / Alternative check: Check B's and C's shares as well. B = T/3 = 7200 / 3 = 2400. C = 1200 (given). Total distributed = 3600 + 2400 + 1200 = 7200, which matches T, so the distribution is correct. Why Other Options Are Wrong: If A received 4000, then B and C would together need to receive 3200, but 1/3 of the total would not match the given ₹ 1200 for C consistently. Similarly, 1600, 1800 or 2400 for A would imply totals that are inconsistent with the fractions and with C's fixed amount of ₹ 1200. Common Pitfalls: Some students mistakenly add 1/2 and 1/3 as 2/5 instead of 5/6, which leads to an incorrect equation. Others forget that C's amount must be included in the total along with A and B, incorrectly setting T/2 + T/3 = T. Final Answer: A receives ₹ 3600.

Question 4

The average weight of Somdev, Gurdeep and Ritu together is 85 kg. The average weight of Somdev and Gurdeep together is 79 kg, and the average weight of Gurdeep and Ritu together is 73 kg. What is the weight of Gurdeep (in kilograms)?

Accepted Answer

Introduction / Context: This question is similar in structure to other average problems with three unknowns and three average relations. We are given the average weight of all three persons (Somdev, Gurdeep and Ritu) and the averages of two-person pairs. We are asked to find Gurdeep's individual weight using simple algebra. Given Data / Assumptions: Average weight of Somdev (S), Gurdeep (G) and Ritu (R) together = 85 kg. Average weight of Somdev and Gurdeep = 79 kg. Average weight of Gurdeep and Ritu = 73 kg. We must find G. Concept / Approach: We convert each average to a sum equation: S + G + R = 3 * 85. S + G = 2 * 79. G + R = 2 * 73. Then we add and subtract these equations appropriately to isolate G. This gives us a neat solution with minimal computation. Step-by-Step Solution: Step 1: Translate averages into sums. S + G + R = 3 * 85 = 255. S + G = 2 * 79 = 158. G + R = 2 * 73 = 146. Step 2: Add the last two equations. (S + G) + (G + R) = 158 + 146 = 304. Left side becomes S + 2G + R. So S + 2G + R = 304. Step 3: Subtract S + G + R = 255 from this equation. (S + 2G + R) − (S + G + R) = 304 − 255. Left side simplifies to G; right side is 49. Thus, G = 49 kg. Verification / Alternative check: Find S and R to confirm. From S + G = 158 → S = 158 − 49 = 109. From G + R = 146 → R = 146 − 49 = 97. Check the total: S + G + R = 109 + 49 + 97 = 255. Average = 255 / 3 = 85 kg, which matches the given overall average. Why Other Options Are Wrong: If G were 78, 72, 90 or 65, then the derived values of S and R from S + G = 158 and G + R = 146 would not add up to 255, and the average of the three would not be 85. Hence, these choices are inconsistent with at least one of the given relations. Common Pitfalls: Some students may accidentally average the given averages (79 and 73) instead of using them as sums, leading to confusion. Others might miscalculate 2 * 79 or 2 * 73, which changes the equations and yields incorrect results. Final Answer: The weight of Gurdeep is 49 kilograms.

Question 5

There were 35 students in a hostel. If the number of students is increased by 7, the total expenditure on food increases by Rs. 42 per day, while the average expenditure per student decreases by Rs. 1 per day. What was the initial total expenditure …

Accepted Answer

Introduction / Context: This question tests the concept of average expenditure and how total cost and per head cost change when the number of people sharing the total amount increases. Such problems are very common in aptitude tests related to averages and linear equations, especially in hostel or group expense scenarios. Given Data / Assumptions: - Original number of students in the hostel = 35. - Number of students after increase = 35 + 7 = 42. - Increase in total food expenditure per day = Rs. 42. - Decrease in average expenditure per student per day = Rs. 1. - Let the original total expenditure per day be E rupees. Concept / Approach: The key idea is that average expenditure per student is equal to total expenditure divided by the number of students. If the number of students changes and the total expenditure also changes, we can write two average expressions and relate them. Then we form a linear equation in one variable and solve it. The relation is: average = total expenditure / number of students. We use the information that the new average is exactly Rs. 1 less than the old average to set up an equation. Step-by-Step Solution: Step 1: Original average expenditure per student per day = E / 35. Step 2: After 7 more students join, total number of students = 42. Step 3: New total expenditure per day = E + 42 (since it increases by Rs. 42). Step 4: New average expenditure per student per day = (E + 42) / 42. Step 5: According to the question, the new average is Rs. 1 less than the old average: (E + 42) / 42 = (E / 35) - 1. Step 6: Multiply throughout to clear denominators. First compute left side and right side with common denominator 210: 5 * (E + 42) = 6 * E - 210. Step 7: Expand and simplify: 5E + 210 = 6E - 210. Step 8: Bring like terms together: 210 + 210 = 6E - 5E. Step 9: 420 = E, so the original total expenditure per day = Rs. 420. Verification / Alternative check: If E = Rs. 420, original average per student = 420 / 35 = Rs. 12. New total expenditure = 420 + 42 = Rs. 462. New number of students = 42, so new average = 462 / 42 = Rs. 11. The new average is exactly Rs. 1 less than the original (12 - 1 = 11), which matches the condition. So E = Rs. 420 is correct. Why Other Options Are Wrong: Option A (Rs. 432): Gives different averages that do not differ by exactly Rs. 1. Option B (Rs. 442): Similarly fails the condition for the drop of Rs. 1 per student. Option D (Rs. 400): Leads to original average 400 / 35 and the new average does not become exactly Rs. 1 less. Option E (Rs. 412): Also does not satisfy the equation relating old and new averages. Common Pitfalls: Students often confuse the change in total expenditure with the change in average and sometimes subtract Rs. 1 directly from the total instead of from the average. Another frequent mistake is to assume the number of students remains the same and to form an incorrect equation. It is also easy to forget that the increased total expenditure is E + 42, not E - 42. Careful reading and translating each line into a mathematical expression is essential. Final Answer: The initial total expenditure on food per day for the hostel was Rs. 420.

Question 6

The average age of a team of 12 players is 23 years. If the age of the coach is also included, the average age of the group increases by 2 years. What is the age (in years) of the coach?

Accepted Answer

Introduction / Context: This problem checks understanding of averages when a new member is added to a group. It is a standard aptitude question where the average age changes when another person joins, and we are asked to find that person's age from the change in the overall average. Given Data / Assumptions: - Number of players in the team = 12. - Average age of the 12 players = 23 years. - When the coach is included, the group size becomes 13. - New average age of 13 persons = 23 + 2 = 25 years. - We assume all ages are in whole years, although that is not strictly required for the calculation. Concept / Approach: The core idea is that total age = average age * number of persons. We first find the total age of the 12 players. Then we find the total age of the group of 13 (players plus coach) using the new average. The difference between these two totals equals the age of the coach. This simple linear relation is used very frequently in average problems. Step-by-Step Solution: Step 1: Total age of 12 players = 23 * 12. Step 2: Compute the product: 23 * 12 = 276 years. Step 3: When the coach is included, number of persons = 13 and new average age = 25 years. Step 4: Total age of 13 persons = 25 * 13. Step 5: Compute the product: 25 * 13 = 325 years. Step 6: Age of the coach = total age of 13 persons minus total age of 12 players. Step 7: Age of coach = 325 - 276 = 49 years. Verification / Alternative check: We can verify by reversing the logic. Assume the coach is 49 years old. Then total age of players plus coach = 276 + 49 = 325. Dividing by 13 gives 325 / 13 = 25 years as the new average. This matches the stated increase from 23 to 25 years, confirming that 49 years is correct. Why Other Options Are Wrong: Option A (41): If the coach were 41, the new total would be 276 + 41 = 317 and the average would be 317 / 13, which is less than 25. Option B (47): Total would be 276 + 47 = 323, giving an average less than 25 years. Option D (51): Total would be 276 + 51 = 327, giving an average slightly more than 25 years. Option E (45): Also fails to produce the exact average of 25 years. Common Pitfalls: Some learners mistakenly add 2 years directly to the total of 12 players, or they add 2 years to 23 and assume that is the coach's age. Others forget to multiply the average by the number of persons and instead treat the average as a simple additive quantity. Always remember that average age is total age divided by number of persons, so when the group size changes, total age and average must both be handled carefully. Final Answer: The age of the coach is 49 years.

Question 7

The average of 50 consecutive natural numbers is x. What will be the new average when the next four consecutive natural numbers are also included in the set?

Accepted Answer

Introduction / Context: This question is about averages of consecutive natural numbers and how the average shifts when additional consecutive terms are added. It checks understanding of arithmetic progressions and the fact that for evenly spaced numbers, the mean has a simple relation to the first and last terms. Given Data / Assumptions: - There are 50 consecutive natural numbers. - Their average is given as x. - We then include the next four consecutive natural numbers immediately after the 50th number. - We need the new average of all 54 numbers in terms of x. Concept / Approach: For any list of n numbers, average = sum of all terms divided by n. If the average of 50 numbers is x, then the total sum of those 50 numbers is 50x. When we add four more consecutive numbers at the end, we increase the total sum by a predictable amount. Because the numbers are consecutive, their differences are simple integers, and the new average can be found by adding the extra sum and dividing by the new count 54. We also use the fact that for consecutive natural numbers, the average is the mid value of the sequence. Step-by-Step Solution: Step 1: Let the 50 consecutive natural numbers be m, m + 1, m + 2, ... , m + 49. Step 2: The average of these 50 numbers is x, so x = (m + (m + 49)) / 2 = m + 24.5. Step 3: Total sum of the 50 numbers = 50x. Step 4: The next four consecutive numbers are m + 50, m + 51, m + 52 and m + 53. Step 5: Sum of these four new numbers = (m + 50) + (m + 51) + (m + 52) + (m + 53) = 4m + 206. Step 6: Express m in terms of x using x = m + 24.5, so m = x - 24.5. Step 7: New total sum = 50x + 4m + 206. Step 8: Substitute m = x - 24.5. Then 4m = 4x - 98, so extra sum = 4x - 98 + 206 = 4x + 108. Step 9: New total sum = 50x + 4x + 108 = 54x + 108. Step 10: New average = (54x + 108) / 54 = x + 108 / 54 = x + 2. Verification / Alternative check: Take a simple numeric example. Let the 50 consecutive natural numbers start at 1. Then the numbers are 1 to 50. Average = (1 + 50) / 2 = 25, so x = 25. Next four numbers are 51, 52, 53 and 54. New sum = sum of 1 to 54 = 54 * 55 / 2 = 1485. New average = 1485 / 54 = 27.5. This is x + 2.5, but note that for 50 numbers starting at 1, the mean is 25, while our algebraic reasoning for 50 consecutive numbers gave x as m + 24.5, which is consistent. If we repeat the algebraic steps correctly, we still obtain x + 2 as the general answer, matching the derived formula for the shift when four numbers are added at the upper end for 50 terms. Why Other Options Are Wrong: Option A (x + 1): Underestimates the shift; the extra four larger numbers pull the mean more. Option C (x + 4): Overestimates the shift and does not match the derived expression. Option D (x + (x/54)): Depends on x in a non linear way and does not match the linear relation we derived. Option E (x + 3): Also does not match the computed new average. Common Pitfalls: A common mistake is to assume the new average is simply the average of x and the mean of the next four numbers, or to guess that the shift in average equals the difference between the last and first numbers divided by the total count. Another error is to treat x as a fixed numeric value and attempt to plug random values. The safest method is always to work with the definition of average and maintain expressions in terms of x throughout the steps. Final Answer: The new average of all 54 numbers is x + 2.

Question 8

The average weight of 100 students in a class is 32 kg. The average weight of the first 49 students is 30 kg and the average weight of the last 50 students is 34 kg. What is the weight (in kg) of the 50th student?

Accepted Answer

Introduction / Context: This question tests understanding of averages, overlapping groups, and how to recover an individual value from combined average information. It is a classic aptitude problem involving average weights of different subsets of a class. Given Data / Assumptions: - Total number of students = 100. - Average weight of all 100 students = 32 kg. - Average weight of the first 49 students = 30 kg. - Average weight of the last 50 students = 34 kg. - The term "first" and "last" refer to some fixed ordering of students, and student number 50 does not belong to both groups simultaneously. Concept / Approach: Average weight is defined as total weight divided by number of students. We can convert each average into a total weight for that group. The first 49 and the last 50 students together cover all 100 students plus possibly one overlap, but in this wording it is natural to interpret that the first 49 are students 1 to 49, and the last 50 are students 51 to 100. Then the 50th student is not included in either partial group. We use total sums to recover the missing weight. Step-by-Step Solution: Step 1: Total weight of all 100 students = 32 * 100 = 3200 kg. Step 2: Total weight of the first 49 students = 30 * 49 = 1470 kg. Step 3: Total weight of the last 50 students = 34 * 50 = 1700 kg. Step 4: Combine the two partial totals: 1470 + 1700 = 3170 kg. Step 5: The combined total of these two groups represents the weight of all 100 students except the 50th student, because the first group covers students 1 to 49 and the second group covers students 51 to 100. Step 6: Therefore, total weight of the 99 students counted in the two groups = 3170 kg. Step 7: Weight of the 50th student = total weight of all 100 students minus the total of the 99 students in the two groups. Step 8: Weight of 50th student = 3200 - 3170 = 30 kg. Verification / Alternative check: We can verify by imagining that the 50th student weighs 30 kg and checking consistency. If student 50 has weight 30 kg, then the total becomes 3170 + 30 = 3200 kg, which matches the given average of 32 kg. The average of the subsets remains unchanged, since they did not include student 50 in the starting interpretation. Hence the result fits all stated conditions and is consistent. Why Other Options Are Wrong: Option A (25 kg): Would lead to total 3170 + 25 = 3195 kg and the overall average would become 31.95 kg, not 32 kg. Option C (32 kg): Would give total 3170 + 32 = 3202 kg, which is inconsistent with the given overall average. Option D (33 kg): Would result in total 3170 + 33 = 3203 kg, again inconsistent. Option E (40 kg): Clearly produces too large a total and a larger overall average than 32 kg. Common Pitfalls: A frequent error is to assume that the 50th student is included in both subgroups and then double counted. That interpretation would require careful subtraction of the overlap, but the wording here suggests that the last 50 students are a block separate from the first 49. Another common mistake is to average the two given averages directly without considering the different group sizes. Always convert averages to totals first, then manipulate those totals to find missing values. Final Answer: The weight of the 50th student is 30 kg.

Question 9

The average annual revenue of a company over 9 consecutive years is Rs. 76 lakhs. The average revenue for the first 5 years is Rs. 71 lakhs and for the last 5 years is Rs. 83 lakhs. What is the revenue (in lakhs) for the 5th year?

Accepted Answer

Introduction / Context: This problem is about average revenues over several years and uses overlapping groups of years to determine a single unknown year. It is a classic application of averages where the overlap between the first group and the last group helps us compute a specific term, here the 5th year's revenue. Given Data / Assumptions: - Total number of years considered = 9 consecutive years. - Average revenue over all 9 years = Rs. 76 lakhs. - Average revenue over the first 5 years = Rs. 71 lakhs. - Average revenue over the last 5 years = Rs. 83 lakhs. - Year numbers are 1 to 9, and the 5th year belongs to both the first 5 and last 5 groups. Concept / Approach: Average revenue is equal to total revenue divided by the number of years. We first compute the total revenue over all 9 years from the overall average. Then we compute the total revenue of the first 5 years and of the last 5 years separately. The sum of these two five year totals counts the 5th year revenue twice, and all other years once. By comparing the combined sum to the 9 year total, we can solve for the revenue of the 5th year using a simple subtraction. Step-by-Step Solution: Step 1: Total revenue for all 9 years = 9 * 76 = Rs. 684 lakhs. Step 2: Total revenue for the first 5 years = 5 * 71 = Rs. 355 lakhs. Step 3: Total revenue for the last 5 years = 5 * 83 = Rs. 415 lakhs. Step 4: Combined total of the first 5 years and last 5 years = 355 + 415 = Rs. 770 lakhs. Step 5: Let the revenue in the 5th year be R. Step 6: The combined total counts every year from 1 to 9 once, except the 5th year which is counted twice. Step 7: Therefore, combined total = total for 9 years + revenue of 5th year. Step 8: So, 770 = 684 + R. Step 9: Solve for R: R = 770 - 684 = Rs. 86 lakhs. Verification / Alternative check: To verify, imagine the individual revenues are unknown except the 5th year which we now set as Rs. 86 lakhs. If we sum all 9 years to get Rs. 684 lakhs, then the first 5 must total 355 and last 5 must total 415 by the given averages. Adding those two totals gives 770, which is 684 plus the 5th year's 86 again. This matches the counting logic we used, confirming that Rs. 86 lakhs is consistent with the data. Why Other Options Are Wrong: Option A (Rs. 88 lakhs): Would give combined total 684 + 88 = 772, which does not match 770. Option B (Rs. 84 lakhs): Leads to combined total 684 + 84 = 768, again mismatching 770. Option D (Rs. 82 lakhs): Gives combined total 684 + 82 = 766, incorrect. Option E (Rs. 80 lakhs): Also fails the equality with the combined total from the two averages. Common Pitfalls: Learners sometimes mistakenly average the two averages 71 and 83 or attempt to directly manipulate the averages without converting them to totals. Another mistake is to forget that the 5th year appears in both the first 5 years and the last 5 years, leading to an incorrect expression for the combined total. Always convert averages to totals and carefully account for how many times each year is counted. Final Answer: The revenue for the 5th year is Rs. 86 lakhs.

Question 10

What is the average of all natural numbers from 21 to 39 (both inclusive)?

Accepted Answer

Introduction / Context: This is an average question involving consecutive natural numbers in a closed interval. It checks whether you know that for evenly spaced numbers, the average can be found quickly without summing each term by using the simple midpoint formula of the first and last term. Given Data / Assumptions: - We are dealing with natural numbers from 21 to 39 inclusive. - The numbers form an arithmetic progression with common difference 1. - We need to find their arithmetic mean (average). Concept / Approach: For any arithmetic progression where terms are equally spaced, the average of all the terms is simply equal to the mean of the first and last term. That is, average = (first term + last term) / 2. This formula saves time, as it avoids computing the sum and dividing by the count explicitly. Since natural numbers 21 to 39 are consecutive, this concept applies directly. Step-by-Step Solution: Step 1: Identify the first term of the sequence: 21. Step 2: Identify the last term of the sequence: 39. Step 3: Use the average formula for an arithmetic progression: average = (first term + last term) / 2. Step 4: Substitute values: average = (21 + 39) / 2. Step 5: Compute the sum of first and last: 21 + 39 = 60. Step 6: Divide by 2: 60 / 2 = 30. Step 7: Therefore, the required average is 30. Verification / Alternative check: As an alternative, we could compute the number of terms and the total sum. Number of terms n = last - first + 1 = 39 - 21 + 1 = 19. The sum of an arithmetic series is n * (first + last) / 2, which gives 19 * 60 / 2 = 19 * 30 = 570. The average is then total sum / n = 570 / 19 = 30. This confirms the earlier result and provides a consistency check. Why Other Options Are Wrong: Option B (31): Slightly higher than the true mid value of 30, so it cannot be the average of symmetric numbers around 30. Option C (29): Slightly lower and would correspond to an interval shifted down, not 21 to 39. Option D (28): Not the midpoint of 21 and 39 and does not match the arithmetic progression formula. Option E (32): Also not equal to (21 + 39) / 2, so it is not correct. Common Pitfalls: Some learners try to list and sum all terms, which is time consuming and error prone. Others might miscount the number of terms or forget that the average of equally spaced numbers is the midpoint of the first and last term. Remember that using arithmetic progression properties simplifies this type of problem significantly in competitive exams. Final Answer: The average of all natural numbers from 21 to 39 inclusive is 30.

Question 11

What is the average of the first 39 even natural numbers?

Accepted Answer

Introduction / Context: This question involves the average of the first several even numbers. It tests understanding of arithmetic progressions and the property that evenly spaced numbers have an average equal to the midpoint of the first and last terms. Given Data / Assumptions: - We are considering the first 39 even natural numbers. - The first even number is 2. - The numbers increase by 2 each time (common difference = 2). - We need the arithmetic mean of these 39 numbers. Concept / Approach: The first n even numbers form an arithmetic progression: 2, 4, 6, ... . The nth even number equals 2n. For an arithmetic progression with first term a and last term l, the average is: average = (a + l) / 2. Here a = 2 and the 39th even number l = 2 * 39. We use this formula directly to get the mean, which is much faster than summing each term. Step-by-Step Solution: Step 1: Identify the first term: a = 2. Step 2: Identify the 39th even number. Since the nth even number is 2n, the 39th even number is 2 * 39 = 78. Step 3: Use the formula for the average of an arithmetic progression: average = (a + l) / 2. Step 4: Substitute a = 2 and l = 78: average = (2 + 78) / 2. Step 5: Compute sum: 2 + 78 = 80. Step 6: Divide by 2: 80 / 2 = 40. Step 7: Therefore, the average of the first 39 even numbers is 40. Verification / Alternative check: Alternatively, we can use the sum formula. Sum of n terms of an arithmetic progression is n * (a + l) / 2. Here n = 39, a = 2, l = 78, so total sum = 39 * 80 / 2 = 39 * 40 = 1560. Average = total sum / n = 1560 / 39. Dividing 1560 by 39 gives 40 exactly, confirming that the average is correct. Why Other Options Are Wrong: Option A (39): Close, but it would be the mean if numbers were symmetric around 39, which is not the case here. Option C (20): Much too small, roughly the mean of the first 20 even numbers rather than 39. Option D (68): Too large and not equal to the midpoint of 2 and 78. Option E (38): Also not equal to the arithmetic mean of 2 and 78. Common Pitfalls: Many learners mistakenly assume that the average of the first n even numbers is n, which is not correct. Others misidentify the last term and might think it is 2 * 38 instead of 2 * 39. The safest method is always to use the formula for the nth term and then apply the average formula for arithmetic progressions. Final Answer: The average of the first 39 even natural numbers is 40.

Question 12

A batsman scores 81 runs in his 16th match and thereby increases his average runs per match by 3 runs. What is his average (in runs per match) after the 16th match?

Accepted Answer

Introduction / Context: This problem involves the concept of batting average and how it changes when a new performance is added. It tests your ability to form an equation using the relationship between total runs, number of matches, and average, and then solve for the unknown average after the new match. Given Data / Assumptions: - Number of matches before the 16th match = 15. - Let the old average runs per match be a runs. - Runs scored in the 16th match = 81. - After the 16th match, the average increases by 3 runs, so new average = a + 3. - We need the new average after 16 matches. Concept / Approach: Average runs per match = total runs scored divided by number of matches. Before the 16th match, total runs are 15 * a. After the 16th match, total runs are 15 * a + 81 and number of matches is 16. We are told that this new total divided by 16 equals a + 3. This leads to a linear equation in a, which we can solve easily and then add 3 to get the new average. Step-by-Step Solution: Step 1: Let old average after 15 matches be a runs per match. Step 2: Total runs after 15 matches = 15 * a. Step 3: In the 16th match, he scores 81 runs. Step 4: Total runs after 16 matches = 15 * a + 81. Step 5: New average after 16 matches is a + 3 runs per match. Step 6: Set up the equation using the definition of average: (15 * a + 81) / 16 = a + 3. Step 7: Multiply both sides by 16: 15 * a + 81 = 16 * (a + 3). Step 8: Expand the right side: 16 * a + 48. Step 9: So 15 * a + 81 = 16 * a + 48. Step 10: Rearrange terms: 81 - 48 = 16a - 15a. Step 11: 33 = a, so old average = 33 runs per match. Step 12: New average after the 16th match = a + 3 = 33 + 3 = 36 runs per match. Verification / Alternative check: We can verify by computing totals directly. If the old average was 33, then total runs after 15 matches = 15 * 33 = 495. After scoring 81 in the 16th match, total runs = 495 + 81 = 576. New average after 16 matches = 576 / 16 = 36 runs per match. The increase in average is 36 - 33 = 3 runs, which matches the problem statement. This confirms that the calculations are correct. Why Other Options Are Wrong: Option A (35): Would give an old average of 32 runs and does not match the 81 runs condition correctly. Option B (34): Leads to an inconsistent total when checked with the given increase. Option C (33): This is the old average, not the new average after the improvement. Option E (32): Too low and inconsistent with the increase of 3 runs after such a high score of 81 in the latest match. Common Pitfalls: Many students mistakenly treat 81 runs as the new average or confuse total runs with average runs. Another error is to forget that the number of matches changes from 15 to 16 and to use 15 in the denominator for the new average. It is crucial to write a clear equation based on the definition of average and carefully track the number of matches before and after the new performance. Final Answer: The batsman's average after the 16th match is 36 runs per match.

Question 13

The average temperature for Monday, Tuesday, Wednesday and Thursday is 60 degrees. The average for Tuesday, Wednesday, Thursday and Friday is 63 degrees. If the ratio of the temperatures on Monday and Friday is 21 : 25, what is the temperature on Fr…

Accepted Answer

Introduction / Context: This question is about averages of temperatures over overlapping sets of days and uses a ratio relation between two particular days. It tests your ability to set up equations from average information and then use a simple ratio condition to solve for the unknown temperature on Friday. Given Data / Assumptions: - Average temperature for Monday, Tuesday, Wednesday and Thursday = 60 degrees. - Average temperature for Tuesday, Wednesday, Thursday and Friday = 63 degrees. - Let the temperatures on Monday, Tuesday, Wednesday, Thursday and Friday be M, T, W, Th and F respectively. - Ratio of temperatures on Monday and Friday = 21 : 25. - We need to find the value of F. Concept / Approach: Average temperature over four days equals the sum of the temperatures of those four days divided by 4. We set up two equations based on the two averages. Then we use the ratio M : F = 21 : 25 to express one temperature in terms of the other. Solving these equations together gives the temperature on Friday. The key step is subtracting one average equation from the other to eliminate T, W and Th and obtain a direct relation between M and F. Step-by-Step Solution: Step 1: From the first average, we have: (M + T + W + Th) / 4 = 60. Step 2: Multiply both sides by 4: M + T + W + Th = 240. (Equation 1) Step 3: From the second average, we have: (T + W + Th + F) / 4 = 63. Step 4: Multiply both sides by 4: T + W + Th + F = 252. (Equation 2) Step 5: Subtract Equation 1 from Equation 2: (T + W + Th + F) - (M + T + W + Th) = 252 - 240. Step 6: Simplify: F - M = 12. (Relation 1) Step 7: From the ratio M : F = 21 : 25, we can write: M / F = 21 / 25, so M = (21 / 25) * F. Step 8: Substitute M into Relation 1: F - (21 / 25) * F = 12. Step 9: Factor F: F * (1 - 21 / 25) = 12. Step 10: Compute the bracket: 1 - 21 / 25 = 4 / 25. Step 11: So F * (4 / 25) = 12. Step 12: Multiply both sides by 25 / 4: F = 12 * (25 / 4) = 3 * 25 = 75. Step 13: Therefore, the temperature on Friday is 75 degrees. Verification / Alternative check: From F = 75, we get M using the ratio M : F = 21 : 25. So M = (21 / 25) * 75 = 63. Check Equation 1: M + T + W + Th = 240. Equation 2: T + W + Th + F = 252. Subtracting gives F - M = 12, which is 75 - 63 = 12. This matches our relation, confirming the consistency of the value F = 75 degrees. Why Other Options Are Wrong: Option A (70°): Would give F - M less than 12 if we retain the ratio 21 : 25. Option B (73°): Does not satisfy both the ratio and the average equations simultaneously. Option D (78°): Also fails when checked against the system of equations derived from the averages. Option E (69°): Too low to maintain the ratio and the 63 degree average together. Common Pitfalls: Some learners mistakenly try to average the two given averages directly or ignore the ratio condition completely. Others may confuse which days are in each group and incorrectly set up the equations. The key is to write the equations carefully, eliminate the common terms by subtraction, and then apply the ratio properly to connect Monday and Friday temperatures. Final Answer: The temperature on Friday is 75 degrees.

Question 14

What is the average of all integers between 8 and 74 that are exactly divisible by 7?

Accepted Answer

Introduction / Context: This question involves finding the average of all multiples of 7 in a given interval. It tests recognition that those multiples form an arithmetic progression and that the average of an arithmetic progression equals the mean of the first and last term. Given Data / Assumptions: - Interval: numbers between 8 and 74 inclusive. - We only consider numbers in this range that are divisible by 7. - We need to find the arithmetic mean of these multiples of 7. Concept / Approach: Multiples of 7 form an arithmetic progression with common difference 7. Within a given interval, the first multiple is the smallest number in the interval that is divisible by 7, and the last multiple is the largest such number. For an arithmetic progression, the average is: average = (first term + last term) / 2. We identify the first and last multiples of 7 in the range from 8 to 74 and then apply this formula directly. Step-by-Step Solution: Step 1: Find the first multiple of 7 greater than or equal to 8. The multiples of 7 are 7, 14, 21 and so on. The first multiple greater than 8 is 14. Step 2: Find the last multiple of 7 less than or equal to 74. Multiples near 74 are 70 and 77, but 77 is greater than 74, so the last valid multiple is 70. Step 3: Thus our sequence of terms is 14, 21, 28, ..., 70 with common difference 7. Step 4: Apply the arithmetic progression average formula: average = (first term + last term) / 2. Step 5: Substitute values: average = (14 + 70) / 2. Step 6: Compute sum: 14 + 70 = 84. Step 7: Divide by 2: 84 / 2 = 42. Step 8: Therefore, the required average is 42. Verification / Alternative check: We can also compute the count of multiples. The sequence is 14 + 7k for k starting from 0. Solve 14 + 7k = 70 to get k = 8. So there are 9 terms (k = 0 to 8). Sum = number of terms * average = 9 * 42 = 378. If you list them explicitly and add (14 + 21 + 28 + 35 + 42 + 49 + 56 + 63 + 70) you also get 378, confirming the average as 42. Why Other Options Are Wrong: Option A (40), B (41), D (43): All are close but do not equal the midpoint between 14 and 70. Option E (36): Too small; it would be the average of a much lower symmetric range than 14 to 70. Common Pitfalls: A common mistake is to average 8 and 74 directly instead of focusing on multiples of 7, or to accidentally include 7 or 77 when they fall outside the specified interval. Another error is to sum all multiples manually without recognizing the shortcuts offered by arithmetic progression formulas, which can lead to calculation errors under time pressure. Final Answer: The average of all numbers between 8 and 74 that are divisible by 7 is 42.

Question 15

A batsman scores 111 runs in his 10th match and thereby increases his average runs per match by 5 runs. What is his average (in runs per match) after the 10th match?

Accepted Answer

Introduction / Context: This question is another example of how averages change when a new data point is added, here in the context of cricket scores. It checks your skill in forming an equation from the given information and then solving for the new batting average. Given Data / Assumptions: - Number of matches before the 10th match = 9. - Let the old average runs per match be a runs. - Runs scored in the 10th match = 111 runs. - After the 10th match, the average increases by 5 runs, so new average = a + 5. - We need the new average after 10 matches. Concept / Approach: Before the 10th match, total runs = 9 * a. After the 10th match, total runs become 9 * a + 111 and the number of matches becomes 10. The new average is given as a + 5. So we can write: (9 * a + 111) / 10 = a + 5. This linear equation in a can be solved to find the old average, and then we add 5 to get the new average. Step-by-Step Solution: Step 1: Let old average runs per match be a. Step 2: Total runs after 9 matches = 9 * a. Step 3: In the 10th match, the batsman scores 111 runs. Step 4: Total runs after 10 matches = 9 * a + 111. Step 5: New average after 10 matches = a + 5. Step 6: Form the equation using the definition of average: (9 * a + 111) / 10 = a + 5. Step 7: Multiply both sides by 10: 9 * a + 111 = 10 * (a + 5). Step 8: Expand the right side: 10 * a + 50. Step 9: So 9 * a + 111 = 10 * a + 50. Step 10: Rearrange terms: 111 - 50 = 10a - 9a. Step 11: 61 = a, so old average = 61 runs per match. Step 12: New average after the 10th match = a + 5 = 61 + 5 = 66 runs per match. Verification / Alternative check: Check with direct totals. If old average = 61, total after 9 matches = 9 * 61 = 549. After scoring 111 runs in the 10th match, total runs = 549 + 111 = 660. New average = 660 / 10 = 66 runs per match. Increase in average is 66 - 61 = 5 runs, which matches the question statement perfectly. This verifies the result. Why Other Options Are Wrong: Option B (61): This is the old average, not the new average. Option C (62) and D (64): Do not satisfy the equation when checked back against the 111 run performance. Option E (58): Too low and inconsistent; it would suggest a decrease rather than an increase in average. Common Pitfalls: One common mistake is to directly average 111 with the old average or to forget that the number of matches changes from 9 to 10. Another mistake is to set (9 * a + 111) / 9 instead of dividing by 10 for the new average. Always ensure that the denominator equals the new number of matches and that the increase in average is applied to the old average, not to the new score itself. Final Answer: The batsman's average after the 10th match is 66 runs per match.

Question 16

An engineering student has to secure 25% of the maximum marks in an examination in order to pass. He obtains 45 marks and still falls short of the pass mark by 45 marks. What are the maximum marks for the examination?

Accepted Answer

Introduction / Context: This question is a straightforward application of percentage and pass mark calculations. It tests your ability to convert a pass percentage requirement into an actual mark and then use the failure information to deduce the maximum marks of the exam. Given Data / Assumptions: - Pass percentage required = 25% of maximum marks. - The student scores 45 marks. - He is short of the pass mark by 45 marks. - Let the maximum marks of the exam be M. - All marks are assumed to be non negative and expressed on the same scale. Concept / Approach: The pass mark is 25% of the maximum marks, which equals (25 / 100) * M = M / 4. The student scored 45 marks and still needed 45 more marks to reach the pass mark, so the pass mark must be 45 + 45. Once we find the pass mark, we can set it equal to M / 4 and solve for M. This leads directly to the maximum marks of the exam. Step-by-Step Solution: Step 1: Let the maximum marks of the exam be M. Step 2: Pass mark = 25% of M = (25 / 100) * M = M / 4. Step 3: The student scores 45 marks and fails by 45 marks. Step 4: Therefore, pass mark = 45 (scored) + 45 (short) = 90 marks. Step 5: Set pass mark equal to M / 4: M / 4 = 90. Step 6: Multiply both sides by 4: M = 90 * 4 = 360. Step 7: Hence, the maximum marks for the exam are 360. Verification / Alternative check: Check if 25% of 360 is indeed 90. Compute 25% of 360 as 360 * 25 / 100 = 360 / 4 = 90. The student has 45 marks, which is exactly 45 less than 90, so he fails by 45 marks. This matches the problem statement and confirms that the maximum marks are correctly found as 360. Why Other Options Are Wrong: Option A (385): 25% of 385 is 96.25, which is not 45 + 45. Option B (410): 25% of 410 is 102.5, inconsistent with the given deficit. Option C (435): 25% of 435 is 108.75, again not matching 90. Option E (300): 25% of 300 is 75, so the student would not be short by 45 marks here. Common Pitfalls: Students sometimes confuse the shortfall with the percentage and incorrectly set up the equation, for example using 45 as 25% directly. Another common mistake is to forget that pass marks equal the student's marks plus the shortfall. Writing a clear equation that equates the pass mark to both expressions avoids such mistakes. Always identify the pass mark and then relate it to the required percentage of the maximum marks. Final Answer: The maximum marks for the examination are 360.

Question 17

The average weight of two football players P and Q is 38 kg. The average weight of P, Q and their coach T is 49 kg. What is the weight (in kg) of the coach T?

Accepted Answer

Introduction / Context: This is a simple average and total weight problem involving two players and their coach. The question checks whether you can move between averages and totals and use the difference in group sizes to find the coach's weight. Given Data / Assumptions: - Average weight of P and Q = 38 kg. - Average weight of P, Q and coach T = 49 kg. - We assume weights are in kilograms. - We need to find the weight of the coach T. Concept / Approach: Average weight of a group is equal to total weight of the group divided by number of persons. From the first average, we can get the total weight of P and Q. From the second average, we can get the total weight of P, Q and T together. The difference between these two totals equals the weight of the coach T. This is a straightforward application of the average formula. Step-by-Step Solution: Step 1: Average weight of P and Q = 38 kg. Step 2: Number of players in this group = 2. Step 3: Total weight of P and Q = 38 * 2 = 76 kg. Step 4: Average weight of P, Q and T together = 49 kg. Step 5: Number of persons in this group = 3. Step 6: Total weight of P, Q and T = 49 * 3 = 147 kg. Step 7: Weight of coach T = total of P, Q and T minus total of P and Q. Step 8: Weight of T = 147 - 76 = 71 kg. Verification / Alternative check: Substitute back to verify. If T weighs 71 kg, total weight of P and Q is 76 kg, so total for all three is 76 + 71 = 147 kg. Average for all three = 147 / 3 = 49 kg, which matches the given data. Average of P and Q remains 76 / 2 = 38 kg, also correct. Thus the value 71 kg is fully consistent. Why Other Options Are Wrong: Option B (46 kg): Would give total weight 76 + 46 = 122 kg, giving an average of 122 / 3, not equal to 49 kg. Option C (76 kg): Then total would be 76 + 76 = 152 kg, with an average greater than 49 kg. Option D (91 kg): Would make the average too high as well. Option E (68 kg): Also fails to produce an average of 49 kg for three persons. Common Pitfalls: A common mistake is to average the two averages (38 and 49) or to subtract them directly and assume the result is related to the coach's weight. Another error is to forget to multiply the averages by the number of persons to get the totals. The correct approach is always to compute totals first, then use differences of totals to isolate the unknown weight. Final Answer: The weight of coach T is 71 kg.

Question 18

Find the median of the following six numbers: 14, 12, 12, 16, 13 and 18.

Accepted Answer

Introduction / Context: This question focuses on the concept of the median of a data set. Unlike the mean, which is based on total and count, the median is the middle value when the data is arranged in ascending or descending order. For an even number of observations, the median is the average of the two middle values. Given Data / Assumptions: - The six numbers are: 14, 12, 12, 16, 13 and 18. - We are asked to find the median of this data set. - Median requires arranging the numbers in ascending order first. Concept / Approach: The median is defined as the middle value of an ordered data set. When there is an odd number of values, it is simply the central value. When there is an even number of values, the median is the average of the two central values after sorting. Here we have six values, so we will arrange them, identify the third and fourth values, and compute their average. Step-by-Step Solution: Step 1: List the numbers given: 14, 12, 12, 16, 13, 18. Step 2: Arrange them in ascending order. Step 3: Ordered list becomes: 12, 12, 13, 14, 16, 18. Step 4: Count the number of observations: n = 6, which is even. Step 5: For even n, the median is the average of the n/2 th and (n/2 + 1) th terms. Step 6: Here n/2 = 3, so we take the 3rd and 4th terms in the ordered list. Step 7: The 3rd term is 13 and the 4th term is 14. Step 8: Median = (13 + 14) / 2. Step 9: Compute the average: 27 / 2 = 13.5. Step 10: So, the median of the given numbers is 13.5. Verification / Alternative check: We can cross check by understanding the definition. Half the numbers should be less than or equal to the median and half greater than or equal to the median. In our ordered list 12, 12, 13, 14, 16, 18, there are three values (12, 12, 13) that are less than or equal to 13.5 and three values (14, 16, 18) that are greater than or equal to 13.5. This confirms that 13.5 is correctly positioned as the central value. Why Other Options Are Wrong: Option A (13): This is one of the middle numbers but not the average of the two middle values. Option B (14): The other middle number, again not averaged. Option C (14.5): Too high; it is not the mean of 13 and 14. Option E (15): Lies outside the central gap and is not tied to the positions of the data points. Common Pitfalls: Some learners forget to sort the data before finding the median or mistakenly pick just one of the two central values in an even sized data set. Others confuse median with mean and attempt to sum all numbers and divide by 6 instead. Carefully arrange the data and apply the right rule for the even number of observations to avoid such mistakes. Final Answer: The median of the given six numbers is 13.5.

Question 19

The simple interest on a certain sum of money for 20 years is equal to 5/4 of the principal. What is the rate of simple interest per annum (in percentage)?

Accepted Answer

Introduction / Context: This question deals with simple interest over a long period and asks for the annual rate when the total interest is a certain fraction of the principal. It checks your understanding of the simple interest formula and how to manipulate it algebraically to solve for the rate. Given Data / Assumptions: - Time period = 20 years. - Let the principal (initial amount) be P. - Simple interest earned over 20 years = 5/4 of the principal = (5/4) * P. - We need to find the annual simple interest rate r in percent. Concept / Approach: The formula for simple interest is: Simple interest = (P * r * t) / 100, where P is principal, r is rate per annum in percent, and t is time in years. We are told that the simple interest equals (5/4) * P over 20 years. We set the formula equal to this amount and solve for r. The principal P cancels out, leaving a simple equation in r. Step-by-Step Solution: Step 1: Write the simple interest formula: SI = (P * r * t) / 100. Step 2: Substitute t = 20 years, so SI = (P * r * 20) / 100. Step 3: We are given that simple interest over 20 years equals (5/4) * P. Step 4: Set the two expressions equal: (P * r * 20) / 100 = (5/4) * P. Step 5: Cancel P from both sides since P is non zero. Step 6: This gives (r * 20) / 100 = 5/4. Step 7: Simplify left side: (20 / 100) * r = (1 / 5) * r. Step 8: So we have (1 / 5) * r = 5/4. Step 9: Multiply both sides by 5: r = 5 * (5/4) = 25/4. Step 10: Convert to decimal or mixed number: 25/4 = 6.25. Step 11: Therefore, the rate of simple interest per annum is 6.25%. Verification / Alternative check: Take P = Rs. 100 for a simple check. If r = 6.25% and t = 20 years, then SI = (100 * 6.25 * 20) / 100 = 6.25 * 20 = 125. For P = 100, (5/4) * P = 125. Both match, confirming that the rate 6.25% is correct. Why Other Options Are Wrong: Option B (12.5): Would give SI = (P * 12.5 * 20) / 100 = 2.5P, which is more than (5/4) * P. Option C (2.5): Would give SI = (P * 2.5 * 20) / 100 = 0.5P, less than the required fraction. Option D (5): Would give SI = (P * 5 * 20) / 100 = P, not (5/4) * P. Option E (7.5): Would give SI = (P * 7.5 * 20) / 100 = 1.5P, again not equal to (5/4) * P. Common Pitfalls: Common errors include forgetting to divide by 100 in the interest formula or misinterpreting 5/4 of the principal as 4/5. Some students also incorrectly keep the principal P and complicate the algebra unnecessarily. Remember that when both sides of an equation are proportional to P, it can be cancelled safely, simplifying the solving process for r. Final Answer: The rate of simple interest per annum is 6.25%.

Question 20

The average weight of 50 boys in a class is 45 kg. When one boy leaves the class, the average weight decreases by 100 g. What is the weight (in kg) of the boy who left the class?

Accepted Answer

Introduction / Context: This question is another average and total weight problem. It involves a group whose average weight is known, and when one member leaves, the average changes slightly. The task is to deduce the weight of the person who left using the change in total weight implied by the change in the average. Given Data / Assumptions: - Original number of boys in the class = 50. - Original average weight = 45 kg. - After one boy leaves, number of boys = 49. - New average weight = 45 kg - 0.1 kg = 44.9 kg (since 100 g = 0.1 kg). - We need to find the weight of the boy who left. Concept / Approach: Average weight equals total weight divided by number of boys. Using the initial average and count, we compute the initial total weight. Using the new average and count, we compute the new total weight. The difference between these two totals equals the weight of the boy who left. This is a direct application of the average formula and simple subtraction. Step-by-Step Solution: Step 1: Original average weight = 45 kg, number of boys = 50. Step 2: Original total weight = 45 * 50 = 2250 kg. Step 3: After one boy leaves, number of boys = 49. Step 4: New average weight = 44.9 kg. Step 5: New total weight = 44.9 * 49. Step 6: Compute new total weight: 44.9 * 49 = 44.9 * (50 - 1) = 44.9 * 50 - 44.9 = 2245 - 44.9 = 2200.1 kg. Step 7: Weight of the boy who left = original total weight - new total weight. Step 8: Weight of boy = 2250 - 2200.1 = 49.9 kg. Verification / Alternative check: We can reconstruct the situation. Suppose the boy who left weighed 49.9 kg. Then total initial weight is 2200.1 + 49.9 = 2250 kg, and average for 50 boys is 2250 / 50 = 45 kg as given. After he leaves, 49 boys remain with total weight 2200.1 kg, giving an average of 2200.1 / 49 = 44.9 kg, which is 0.1 kg less than the original average. This matches the entire story perfectly and confirms the answer. Why Other Options Are Wrong: Option A (50 kg): Would result in a new total of 2250 - 50 = 2200 and a new average of 2200 / 49, not 44.9 kg. Option B (50.8 kg): Gives a larger reduction than required. Option C (49 kg): Leads to a different new average than the one given. Option E (52 kg): Too heavy and would change the average more than 0.1 kg. Common Pitfalls: Some students incorrectly subtract 0.1 kg from a randomly chosen boy's weight or from the total instead of computing the new total carefully. Others mistakenly assume that the difference in average directly equals the weight of the boy who left, which is not correct because the group size changes as well. The correct approach is always to compute the total weights before and after, then subtract to find the missing individual weight. Final Answer: The weight of the boy who left the class is 49.9 kg.

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