Question 1

A bowler concedes runs in 5 matches at an average of 45 runs per match and in another 4 matches at an average of 15.75 runs per match. What is the average runs conceded by the bowler over all 9 matches?

Accepted Answer

Introduction / Context: This question tests the skill of combining averages from two different blocks of matches to obtain a single overall average. The bowler has different performance levels in two sets of matches, and we need to find the combined average runs conceded. This is a direct application of weighted averages where the weights are the number of matches in each block. Given Data / Assumptions: First block: 5 matches with average runs conceded = 45 runs per match. Second block: 4 matches with average runs conceded = 15.75 runs per match. Total matches = 5 + 4 = 9. Concept / Approach: To combine averages, convert each block's average into a total by multiplying by the number of matches. Then add these totals to get the overall runs conceded and divide by the total number of matches to get the combined average. Step-by-Step Solution: Step 1: Total runs in the first 5 matches = 5 * 45 = 225 runs.Step 2: Total runs in the next 4 matches = 4 * 15.75 = 63 runs.Step 3: Total runs conceded in all 9 matches = 225 + 63 = 288 runs.Step 4: Total matches = 9.Step 5: Overall average runs conceded = 288 / 9 = 32 runs per match. Verification / Alternative check: We can see that the second block has a much lower average, so the overall average must fall between 15.75 and 45. The value 32 lies between these bounds and is closer to 45 because more matches belong to the first block. This rough check supports our detailed calculation. Why Other Options Are Wrong: Option 15 is too low and close to the second block average, ignoring the heavier weight of the first set of matches. Option 35 is higher than the computed 32 and would require more runs conceded in the lower average block. Option 53.5 is even higher than the first block average, which is impossible. Option 28 is lower than the calculated value and inconsistent with the weighted contributions. Common Pitfalls: Taking a simple average of 45 and 15.75 without weighting by the different numbers of matches. Mistakes while multiplying 15.75 by 4 or when adding totals. Dividing by 2 instead of 9 after adding the two block totals. Final Answer: The average runs conceded by the bowler over all 9 matches is 32 runs per match.

Question 2

A canteen requires 56 kg of rice for 7 days. Assuming the same daily consumption, how many kilograms of rice are required together for the months of April and May?

Accepted Answer

Introduction / Context: This question is a direct application of unitary method and average daily consumption. You are given the total rice needed for 7 days and must scale this daily consumption to cover the combined length of April and May. It reinforces understanding of how to move between weekly and monthly requirements by working with a consistent per day rate. Given Data / Assumptions: Rice required for 7 days = 56 kg. Daily requirement is constant. Month of April has 30 days and month of May has 31 days. Total days considered = 30 + 31 = 61 days. Concept / Approach: We first compute the rice required per day by dividing the 7 day requirement by 7. Once we know the per day consumption, we multiply it by the total number of days in April and May combined. This is the standard unitary method: first find the value for 1 day, then scale up to many days. Step-by-Step Solution: Step 1: Daily rice requirement = 56 kg / 7 days = 8 kg per day.Step 2: Total days for April and May combined = 30 + 31 = 61 days.Step 3: Total rice required for 61 days = 8 kg per day * 61 days.Step 4: 8 * 61 = 488 kg.Step 5: Therefore, the canteen needs 488 kg of rice for April and May together. Verification / Alternative check: We can check that 8 kg per day for a week correctly gives 56 kg, so the daily rate is consistent. Then, extending the same rate to 61 days must simply be 8 times 61. The arithmetic 8 * 61 = 488 can be verified by 8 * 60 = 480 and adding 8 more to get 488. Why Other Options Are Wrong: Option 468 corresponds to using a slightly smaller daily value or fewer days. Option 498 and 508 overshoot the needed amount and would require either a higher daily rate or more days. Option 520 is even larger and clearly inconsistent with the given weekly requirement. Common Pitfalls: Forgetting that April and May together have 61 days, not 60. Dividing by 30 or multiplying incorrectly due to rushed arithmetic. Not reducing the 7 day information to a per day basis before scaling up. Final Answer: The canteen needs 488 kg of rice for the months of April and May together.

Question 3

Due to a price hike of 20 percent, 4 kg less sugar can be bought for Rs. 120 than before. What was the initial price per kilogram of sugar before the increase?

Accepted Answer

Introduction / Context: This problem combines unitary method with percentage increase. The same amount of money, Rs. 120, now buys 4 kg less sugar because the price has gone up by 20 percent. You must work backward to find the original price per kilogram. Such questions are very common in topics involving inflation, cost of living, and price rise calculations. Given Data / Assumptions: Money available = Rs. 120. Initial price of sugar = x rupees per kg. New price after 20 percent increase = 1.2x rupees per kg. 4 kg less sugar can be bought for the same Rs. 120 after the price hike. Concept / Approach: We express the quantity of sugar that can be bought before and after the price rise in terms of x. Before the rise, quantity = 120 / x kg. After the rise, quantity = 120 / (1.2x) kg. The difference between the two quantities is given as 4 kg. Setting up this equation and solving for x gives the initial price per kilogram. Step-by-Step Solution: Step 1: Quantity before hike = 120 / x kg.Step 2: New price = 1.2x, so quantity after hike = 120 / (1.2x) = 100 / x kg.Step 3: Given that the new quantity is 4 kg less, so (120 / x) − (100 / x) = 4.Step 4: Simplify the left side: (120 − 100) / x = 20 / x.Step 5: Therefore 20 / x = 4, giving x = 20 / 4 = 5.Step 6: Initial price per kg of sugar = Rs. 5 per kg. Verification / Alternative check: At Rs. 5 per kg, before the increase, quantity = 120 / 5 = 24 kg. New price after 20 percent rise = 1.2 * 5 = Rs. 6 per kg, so new quantity = 120 / 6 = 20 kg. The difference = 24 − 20 = 4 kg, which matches the problem statement. This confirms that the original price is correct. Why Other Options Are Wrong: Option 4 per kg would yield 30 kg before and 25 kg after a 20 percent increase to Rs. 4.8 per kg, a difference of 5 kg, not 4 kg. Option 6 per kg and option 5.5 per kg do not satisfy the difference condition when checked similarly. Option 3 per kg would produce an even larger error in the quantity difference. Common Pitfalls: Using 20 rupees rather than 20 percent as the increase in price. Forgetting to apply the new price of 1.2x correctly to find the new quantity. Incorrect algebra when forming or simplifying the equation involving x. Final Answer: The initial price of sugar was Rs. 5 per kg.

Question 4

The average temperature for Monday, Tuesday, Wednesday and Thursday was 48°. The average temperature for Tuesday, Wednesday, Thursday and Friday was 52°. If the temperature on Monday was 42°, what was the temperature on Friday, in degrees?

Accepted Answer

Introduction / Context: This question tests understanding of averages over overlapping groups of days and how to use them to recover an unknown value. We are given average temperatures over two overlapping four day periods in a week and the exact temperature for Monday. Using this information, we need to determine the temperature on Friday. The key idea is that the total temperature over each four day block can be written from the average and then compared. Given Data / Assumptions: Average temperature for Monday, Tuesday, Wednesday and Thursday = 48°. Average temperature for Tuesday, Wednesday, Thursday and Friday = 52°. Temperature on Monday = 42°. Temperatures are measured in degrees and can be treated as simple numbers. We must find the temperature on Friday. Concept / Approach: The average of a group of values is equal to the sum of those values divided by the number of values. For a group of four days, average = total temperature over those four days / 4. So we can convert each average into a total. Because the two four day blocks overlap on Tuesday, Wednesday and Thursday, subtracting one total from the other will isolate the difference between Monday and Friday, which leads directly to the required Friday temperature. Step-by-Step Solution: Step 1: Let the temperatures on Tuesday, Wednesday, Thursday and Friday be T, W, Th and F respectively. Step 2: Use the average for Monday to Thursday. (M + T + W + Th) / 4 = 48. Total for Monday to Thursday = 4 * 48 = 192. Step 3: Use the average for Tuesday to Friday. (T + W + Th + F) / 4 = 52. Total for Tuesday to Friday = 4 * 52 = 208. Step 4: Express Monday's value and relate totals. M = 42, so M + T + W + Th = 192. T + W + Th + F = 208. Subtract the first total from the second: (T + W + Th + F) − (M + T + W + Th) = 208 − 192. This simplifies to F − M = 16. Step 5: Substitute M = 42. F − 42 = 16, so F = 42 + 16 = 58. Verification / Alternative check: If Friday is 58°, the Tuesday to Friday total is Tuesday + Wednesday + Thursday + 58. The Monday to Thursday total is Monday + Tuesday + Wednesday + Thursday = 192. Replace Monday with 42 to check consistency: 42 + (T + W + Th) = 192 gives T + W + Th = 150. Then Tuesday to Friday total is 150 + 58 = 208, and the average is 208 / 4 = 52°, which matches the given data. Why Other Options Are Wrong: A Friday temperature of 56°, 52°, 50° or 54° would give a total for Tuesday to Friday that is not 208, so the average would no longer be 52°. Each of these alternatives contradicts at least one of the original average statements. Common Pitfalls: A common mistake is to try to average 48° and 52° directly, which has no meaning here because the sets of days differ. Another mistake is to forget that each average refers to four days and to multiply or divide by the wrong number when computing totals. Final Answer: The temperature on Friday was 58°.

Question 5

What is the average of the first 93 natural numbers (that is, the numbers from 1 to 93 inclusive)?

Accepted Answer

Introduction / Context: This question checks understanding of averages and arithmetic progressions. The first 93 natural numbers form a simple series starting at 1 and increasing by 1 each time. Instead of adding all 93 numbers one by one, we can use the properties of this sequence to find the average quickly and accurately. Given Data / Assumptions: We are dealing with the first 93 natural numbers. The numbers are 1, 2, 3, ..., 93. The sequence is an arithmetic progression with common difference 1. We are asked to find the average value of these 93 numbers. Concept / Approach: For an arithmetic progression with first term a and last term l, and with an odd or even number of terms, the average of all terms is equal to (a + l) / 2. This is because the sum of the terms is given by n * (a + l) / 2 and the average is that sum divided by n, which simplifies to (a + l) / 2. Here the first term is 1 and the last term is 93, so the average is simply the midpoint between these two numbers. Step-by-Step Solution: Step 1: Identify the first and last terms of the sequence. First term a = 1. Last term l = 93. Number of terms n = 93. Step 2: Use the average formula for an arithmetic progression. Average = (first term + last term) / 2. Average = (1 + 93) / 2. Step 3: Compute the sum in the numerator. 1 + 93 = 94. Step 4: Divide by 2. Average = 94 / 2 = 47. Verification / Alternative check: If we wanted, we could confirm by using the sum formula. Sum of first 93 natural numbers = n * (a + l) / 2 = 93 * 94 / 2. Compute 93 * 94 = 8742 and then divide by 2 to get 4371. Average = total sum / number of terms = 4371 / 93 = 47, which confirms the earlier result. Why Other Options Are Wrong: Values like 45, 46, 48 or 49 correspond to midpoints between different first and last numbers and do not match the exact midpoint of 1 and 93. Any other value would violate the property that, in a symmetric sequence from 1 to 93, the average must lie exactly in the middle. Common Pitfalls: Students sometimes think they must add all 93 numbers explicitly, which is unnecessary and error prone. Another error is to divide the last term 93 by 2 instead of taking the average of first and last terms, which gives 46.5 and is not correct for this question. Final Answer: The average of the first 93 natural numbers is 47.

Question 6

The average marks in Science for a class of 20 students was calculated as 68. Later it was found that the marks of two students had been misread as 48 and 65 instead of the correct values 72 and 61 respectively. What is the correct average (in marks…

Accepted Answer

Introduction / Context: This question is about correcting an average when some of the data values were recorded incorrectly. The original class average for Science was computed using two misread marks. Once we know both the wrong entries and their correct values, we can adjust the total and find the new, accurate average. This type of question regularly appears in aptitude tests on averages and data correction. Given Data / Assumptions: Number of students in the class = 20. Original average marks (with errors) = 68. Two marks were recorded as 48 and 65. The correct marks for these two students are 72 and 61 respectively. We need the corrected average after replacing the wrong values with the correct ones. Concept / Approach: Average is defined as total marks divided by number of students. When we discover that some marks are incorrect, we do not recompute everything from scratch. Instead, we adjust the incorrect total by subtracting the misread marks and adding the correct marks. The new average is the corrected total divided by the same number of students. This method is efficient and avoids reprocessing all individual marks. Step-by-Step Solution: Step 1: Compute the original total marks based on the incorrect average. Original total = number of students * original average. Original total = 20 * 68 = 1360. Step 2: Find how much the two incorrect entries differ from the correct values. Wrong marks were 48 and 65. Correct marks are 72 and 61. Total wrong contribution = 48 + 65 = 113. Total correct contribution = 72 + 61 = 133. Step 3: Adjust the total marks. Corrected total = original total − wrong contribution + correct contribution. Corrected total = 1360 − 113 + 133. Compute 1360 − 113 = 1247; then 1247 + 133 = 1380. Step 4: Compute the corrected average. Correct average = corrected total / number of students = 1380 / 20. 1380 / 20 = 69. Verification / Alternative check: Notice that the net increase to the total from the correction is 133 − 113 = 20. Since there are 20 students, adding 20 marks in total increases the class average by 20 / 20 = 1 mark. Original average was 68, so new average must be 68 + 1 = 69, confirming the calculation. Why Other Options Are Wrong: Options 68.5 and 69.5 suggest a half mark change which would require a total change of 10 marks, not 20, which does not fit the data. An average of 70 would require an even larger increase to the total marks. An average of 67.5 would imply a decrease, which contradicts the fact that the correct marks are higher overall than the misread marks. Common Pitfalls: A common error is to simply replace the wrong marks with the correct ones without adjusting the total used for the original average. Another mistake is miscomputing the difference between the wrong and correct marks, which directly affects the new average. Final Answer: The correct average marks of the class is 69.

Question 7

A group of boys has an average weight of 36 kg. One boy weighing 42 kg leaves the group, and another boy weighing 30 kg joins the group in his place. If the new average weight of the group becomes 35.7 kg, how many boys are there in the group?

Accepted Answer

Introduction / Context: This problem is a classic example of how the average of a group changes when one member leaves and another joins. We are told the original average weight of a group of boys, the weight of the boy who leaves, the weight of the boy who joins, and the new average. From this information we must determine the original number of boys in the group. The question tests understanding of average, total, and how small changes affect group statistics. Given Data / Assumptions: Original average weight of the group = 36 kg. One boy with weight 42 kg leaves the group. A new boy with weight 30 kg joins the group. New average weight becomes 35.7 kg. The number of boys remains the same, only membership changes. We must find the number of boys in the group, call it n. Concept / Approach: Average weight is equal to total weight divided by number of boys. Initially, total weight is 36 * n. When the boy of 42 kg leaves, the total decreases by 42. When the boy of 30 kg joins, the total increases by 30. The new total is therefore 36n − 42 + 30. Dividing this new total by n gives the new average 35.7. Setting up this equation and solving for n will give the group size. Step-by-Step Solution: Step 1: Let the original number of boys be n. Original total weight = 36n. Step 2: Account for the boy leaving and the boy joining. When a 42 kg boy leaves, new total = 36n − 42. When a 30 kg boy joins, updated total = 36n − 42 + 30 = 36n − 12. Step 3: Use the new average. New average = 35.7 = (36n − 12) / n. Step 4: Multiply both sides by n. 36n − 12 = 35.7n. Step 5: Bring like terms together. 36n − 35.7n = 12. 0.3n = 12. Step 6: Solve for n. n = 12 / 0.3 = 40. Verification / Alternative check: Original total weight = 36 * 40 = 1440 kg. After the 42 kg boy leaves, total = 1440 − 42 = 1398 kg. After the 30 kg boy joins, total = 1398 + 30 = 1428 kg. New average = 1428 / 40 = 35.7 kg, matching the given condition. Why Other Options Are Wrong: If n were 30, 32, 56 or 28, substituting into the equation would not give a new average of exactly 35.7 kg. Each incorrect n would yield a different total and hence an inconsistent average, so these options cannot be correct. Common Pitfalls: A common mistake is to treat the average change as if it directly adds or subtracts a fixed number of kilograms from each boy, without using the total-weight equation. Another typical error is mishandling decimal arithmetic with 35.7, which can lead to incorrect values for 0.3n. Final Answer: There are 40 boys in the group.

Question 8

Three real numbers a, b and c satisfy a + b = 5, b + c = 7.5 and c + a = 8.5. What is the average value of a, b and c?

Accepted Answer

Introduction / Context: This question deals with three unknown numbers a, b and c where only pairwise sums are given. Instead of individual values, we know a + b, b + c and c + a. The task is to find the average of a, b and c. This is a good example of how to use symmetry and simple algebra to work with partial information about several variables. Given Data / Assumptions: a + b = 5. b + c = 7.5. c + a = 8.5. We are asked to find the average of a, b and c, that is (a + b + c) / 3. Concept / Approach: The key idea is to add the three pairwise equations. When we add (a + b), (b + c) and (c + a), each variable a, b and c appears exactly twice. This means the sum of the three equations equals 2(a + b + c). From that, we can solve for a + b + c directly, without needing each variable separately. Finally, the average is that total divided by 3. This approach is efficient and avoids unnecessary algebra. Step-by-Step Solution: Step 1: Write out the three given equations clearly. a + b = 5. b + c = 7.5. c + a = 8.5. Step 2: Add all three equations. (a + b) + (b + c) + (c + a) = 5 + 7.5 + 8.5. Left side becomes 2a + 2b + 2c = 2(a + b + c). Right side is 5 + 7.5 + 8.5 = 21. So 2(a + b + c) = 21. Step 3: Solve for a + b + c. a + b + c = 21 / 2 = 10.5. Step 4: Compute the average. Average of a, b and c = (a + b + c) / 3 = 10.5 / 3. 10.5 / 3 = 3.5. Verification / Alternative check: We could find individual values if needed. From a + b = 5 and a + b + c = 10.5, we get c = 10.5 − 5 = 5.5. From b + c = 7.5, b = 7.5 − 5.5 = 2. From a + b = 5, a = 5 − 2 = 3. Check: c + a = 5.5 + 3 = 8.5, consistent with the data. Average of 3, 2 and 5.5 is (3 + 2 + 5.5) / 3 = 10.5 / 3 = 3.5. Why Other Options Are Wrong: Values 1.5, 3, 2.5 and 4.5 do not match the exact total 10.5 when multiplied by 3. Any other average would imply a different value for a + b + c and would break at least one of the pairwise equations. Common Pitfalls: Some learners attempt to solve for a, b and c individually before finding the average, which is longer and more error prone. Another typical mistake is adding the right hand sides incorrectly or forgetting that the left side becomes 2(a + b + c) rather than a + b + c. Final Answer: The average of a, b and c is 3.5.

Question 9

Seven friends each spent Rs 14 on a project, and an eighth friend spent Rs 21 more than the average expenditure of all eight friends together. What is the total amount of money (in rupees) spent by the eight friends on the project?

Accepted Answer

Introduction / Context: This question involves a slightly circular average situation. Seven friends spend a fixed known amount each. The eighth friend spends an amount defined in terms of the overall average of all eight, not as a fixed number. We are asked to find the total expenditure of all eight friends. To do this, we must set up an equation where the average itself is an unknown and the eighth friend's spending depends on that unknown average. Given Data / Assumptions: Seven friends each spend Rs 14. The eighth friend spends Rs 21 more than the average expenditure of all eight friends together. Let the average expenditure of all eight be A rupees. We need to find the total amount spent by all eight friends. Concept / Approach: If the average expenditure is A, then the total expenditure is 8A. We know seven of the friends each spend 14, so their combined spending is 7 * 14. The eighth friend spends A + 21. Therefore, total expenditure can also be written as 7 * 14 + (A + 21). Since both expressions represent the same total, we equate 8A to this sum and solve for A. Once A is known, we can compute the total expenditure as 8A directly. Step-by-Step Solution: Step 1: Let A be the average expenditure of all eight friends. Step 2: Total expenditure in terms of A is 8A. Step 3: Compute the spending of the seven friends. Seven friends spend 14 each, so total for seven friends = 7 * 14 = 98. Step 4: Express the eighth friend's spending. Eighth friend spends A + 21. Step 5: Equate total expenditure from both viewpoints. 8A = 98 + (A + 21). 8A = 98 + A + 21 = 119 + A. Step 6: Solve for A. 8A − A = 119. 7A = 119. A = 119 / 7 = 17. Step 7: Find the total expenditure. Total = 8A = 8 * 17 = 136. Verification / Alternative check: Seven friends together spend 7 * 14 = 98. Average expenditure is A = 17, so the eighth friend spends A + 21 = 17 + 21 = 38. Total = 98 + 38 = 136, and 136 / 8 = 17, confirming that A is indeed the correct average. Why Other Options Are Wrong: Totals of 133, 141, 155 or 147 would give averages that are not consistent with the condition that the eighth friend spends exactly Rs 21 more than the average. For instance, if total were 141, average would be 141 / 8, which is not an integer and would not give a clean value for A + 21 matching the eighth friend's spending. Common Pitfalls: A common mistake is to assume the eighth friend spends 21 more than the other friends, rather than 21 more than the overall average. Another error is to forget that the expression for total expenditure must be set equal to 8A, not just the sum of the known amounts. Final Answer: The total amount of money spent by the eight friends is Rs 136.

Question 10

In a class, the ratio of the number of boys to the number of girls is 2 : 3. The average weight of the boys is 18 kg and the average weight of the girls is 21 kg. What is the average weight (in kilograms) of all the students in the class together?

Accepted Answer

Introduction / Context: This problem tests the concept of weighted averages. The class is composed of boys and girls in a fixed ratio, and each group has its own average weight. The aim is to find the combined average weight of all students. Instead of assuming a specific number of students, we can use the ratio directly and treat it as a scaled model of the class. Given Data / Assumptions: Ratio of boys to girls = 2 : 3. Average weight of boys = 18 kg. Average weight of girls = 21 kg. All students are either boys or girls, no other categories. We must find the overall average weight of all students. Concept / Approach: When groups have different averages, the combined average is a weighted average. The weights are the sizes of the groups. Because the ratio of boys to girls is 2 : 3, we can assume there are 2k boys and 3k girls for some positive integer k. Then we compute total weight of boys and girls using their averages and divide the combined total by the total number of students 5k. The value of k cancels, leaving a simple expression that depends only on the given averages and ratio. Step-by-Step Solution: Step 1: Assume numbers based on the ratio. Let number of boys = 2k and number of girls = 3k. Total number of students = 2k + 3k = 5k. Step 2: Compute total weight of boys. Average weight of boys = 18 kg, so total weight of boys = 2k * 18 = 36k. Step 3: Compute total weight of girls. Average weight of girls = 21 kg, so total weight of girls = 3k * 21 = 63k. Step 4: Combine totals. Total weight of all students = 36k + 63k = 99k. Step 5: Find overall average. Overall average weight = total weight / total students = 99k / 5k. The k cancels, leaving 99 / 5 kg. Thus, average weight = 99/5 kg. Verification / Alternative check: We can choose a convenient value for k, for example k = 1. Then we have 2 boys and 3 girls, 5 students total. Total boy weight = 2 * 18 = 36. Total girl weight = 3 * 21 = 63. Combined total = 36 + 63 = 99. Average = 99 / 5 = 19.8, which matches 99/5. Why Other Options Are Wrong: 101/5, 109/6 and 96/5 correspond to different combinations of group sizes or different averages and do not match the computed weighted average. The value 20 is close but rounded and does not equal the exact value 99/5. Common Pitfalls: A common error is to take the simple average of 18 and 21 and compute (18 + 21) / 2 = 19.5, which ignores the 2 : 3 ratio. Another mistake is to pick arbitrary numbers of boys and girls that do not respect the given ratio, leading to inconsistent totals. Final Answer: The average weight of all the boys and girls together is 99/5 kilograms.

Question 11

The average of 8 consecutive natural numbers is 38.5. What is the largest of these 8 consecutive natural numbers?

Accepted Answer

Introduction / Context: This question focuses on consecutive natural numbers and their averages. When numbers are consecutive, they form an arithmetic progression with common difference 1. The average of such a sequence has a simple relationship to the first and last terms. We are given the average and asked to recover the largest of the 8 consecutive numbers. Given Data / Assumptions: There are 8 consecutive natural numbers. Their average is 38.5. We must find the largest of these 8 numbers. Concept / Approach: For an arithmetic progression, the average is equal to the midpoint between the first and last terms. For 8 consecutive numbers, if the first term is x, the numbers are x, x+1, x+2, ..., x+7, and the last term is x+7. The average is then (x + (x + 7)) / 2 = (2x + 7) / 2. We can set this equal to 38.5 and solve for x. Once the first term is found, the last term is x + 7. This is a direct linear equation and does not require listing all numbers explicitly. Step-by-Step Solution: Step 1: Let the first of the 8 consecutive numbers be x. Then the numbers are x, x + 1, x + 2, ..., x + 7. The last number is x + 7. Step 2: Use the average formula for an arithmetic progression. Average = (first term + last term) / 2. Given average = 38.5, so (x + (x + 7)) / 2 = 38.5. This simplifies to (2x + 7) / 2 = 38.5. Step 3: Multiply both sides by 2. 2x + 7 = 77. Step 4: Solve for x. 2x = 77 − 7 = 70. x = 70 / 2 = 35. Step 5: Find the largest number. Largest number = x + 7 = 35 + 7 = 42. Verification / Alternative check: The 8 numbers are 35, 36, 37, 38, 39, 40, 41, 42. Sum = 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42. Pairs from extremes sum to 77: (35 + 42), (36 + 41), (37 + 40), (38 + 39). Each pair sums to 77 and there are 4 pairs, so total sum = 4 * 77 = 308. Average = 308 / 8 = 38.5, matching the given average. Why Other Options Are Wrong: If the largest number were 41, then the first would be 34, and the average would be (34 + 41) / 2 = 37.5, not 38.5. For largest 39 or 40, the averages would be even smaller, while 45 would produce a much higher average than 38.5. Common Pitfalls: Some students try to compute the numbers starting from the average by adding and subtracting 1 without using a systematic equation, which can lead to mistakes. Others incorrectly use average = last term − first term, which is not a valid formula. Final Answer: The largest of the 8 consecutive natural numbers is 42.

Question 12

Consider the following grouped frequency distribution of a continuous variable x: 0–10 (frequency 8), 10–20 (frequency 12), 20–30 (frequency 10), 30–40 (frequency p), 40–50 (frequency 9). If the mean of this data is 25.2, what is the value of the mi…

Accepted Answer

Introduction / Context: This question involves a grouped frequency distribution with one missing frequency. We are given class intervals and frequencies for all but one class, along with the mean of the distribution. The task is to find the missing frequency. This tests understanding of how the mean of grouped data is calculated using class midpoints and frequencies. Given Data / Assumptions: Class intervals and frequencies are: 0–10: frequency 8 10–20: frequency 12 20–30: frequency 10 30–40: frequency p (unknown) 40–50: frequency 9 Mean of the distribution = 25.2. Class midpoints will be used for mean calculation. We must find the value of p. Concept / Approach: For grouped data, mean is calculated as sum(f * x) / sum(f), where x is the class midpoint and f is the frequency. We will compute the midpoints of each class, express the total frequency and the total of f * x in terms of p, then use the given mean to form an equation. Solving that equation will yield the missing frequency p. Careful arithmetic is essential, especially when handling decimals. Step-by-Step Solution: Step 1: Compute class midpoints. 0–10 midpoint = (0 + 10) / 2 = 5. 10–20 midpoint = 15. 20–30 midpoint = 25. 30–40 midpoint = 35. 40–50 midpoint = 45. Step 2: Write frequencies. f values: 8, 12, 10, p, 9. Total frequency = 8 + 12 + 10 + p + 9 = 39 + p. Step 3: Compute f * x products. For 0–10: 8 * 5 = 40. For 10–20: 12 * 15 = 180. For 20–30: 10 * 25 = 250. For 30–40: p * 35 = 35p. For 40–50: 9 * 45 = 405. Step 4: Sum f * x. Fixed part = 40 + 180 + 250 + 405 = 875. So sum(f * x) = 875 + 35p. Step 5: Use the mean formula. Mean = sum(f * x) / sum(f). 25.2 = (875 + 35p) / (39 + p). Step 6: Cross multiply to remove the fraction. 875 + 35p = 25.2 * (39 + p). Compute 25.2 * 39 = 982.8. Right side becomes 982.8 + 25.2p. Step 7: Solve for p. 875 + 35p = 982.8 + 25.2p. 35p − 25.2p = 982.8 − 875. 9.8p = 107.8. p = 107.8 / 9.8 = 11. Verification / Alternative check: With p = 11, total frequency = 39 + 11 = 50. Sum(f * x) = 875 + 35 * 11 = 875 + 385 = 1260. Mean = 1260 / 50 = 25.2, which matches the given value. Why Other Options Are Wrong: If p were 9, 10, 12 or 8, the sum(f * x) and total frequency would change, and the resulting mean would not equal 25.2. A quick check with any of these alternatives shows the mean deviating from 25.2. Common Pitfalls: A frequent error is to forget to include p in both the numerator and the denominator when writing the mean formula. Another common mistake is incorrect multiplication of 25.2 by 39 or incorrect handling of decimals when solving for p. Final Answer: The missing frequency p is 11.

Question 13

Consider the following discrete frequency distribution of a variable x: x = 8 with frequency 6, x = 5 with frequency 4, x = 6 with frequency 5, x = 10 with frequency 8, x = 9 with frequency 9, x = 4 with frequency 6 and x = 7 with frequency 4. What …

Accepted Answer

Introduction / Context: This question asks for the median of a discrete frequency distribution. The median is the value that lies in the middle of the data when all observations are arranged in ascending order. When a frequency table is given, we do not list every value individually. Instead, we use cumulative frequencies to identify the position of the median and then determine which value of x corresponds to that position. Given Data / Assumptions: Discrete values and their frequencies are: x = 8, frequency 6 x = 5, frequency 4 x = 6, frequency 5 x = 10, frequency 8 x = 9, frequency 9 x = 4, frequency 6 x = 7, frequency 4 We must find the median of the entire data set. Concept / Approach: To find the median from discrete frequency data, first arrange the values of x in ascending order along with their frequencies. Then compute cumulative frequencies and determine the position of the median. If total frequency N is even, the median is the average of the N/2th and (N/2 + 1)th observations; in many discrete cases, both positions fall in the same value of x, giving a clear median. We will follow this systematic procedure. Step-by-Step Solution: Step 1: Arrange x values in ascending order with their frequencies. x = 4, f = 6. x = 5, f = 4. x = 6, f = 5. x = 7, f = 4. x = 8, f = 6. x = 9, f = 9. x = 10, f = 8. Step 2: Compute total frequency N. N = 6 + 4 + 5 + 4 + 6 + 9 + 8 = 42. Step 3: Find median positions. For N = 42 (even), median lies between the 21st and 22nd observations. Step 4: Compute cumulative frequencies. For x = 4: cumulative = 6 (observations 1 to 6). For x = 5: cumulative = 6 + 4 = 10 (observations 7 to 10). For x = 6: cumulative = 10 + 5 = 15 (observations 11 to 15). For x = 7: cumulative = 15 + 4 = 19 (observations 16 to 19). For x = 8: cumulative = 19 + 6 = 25 (observations 20 to 25). Step 5: Locate the 21st and 22nd observations. The cumulative frequency jumps from 19 to 25 at x = 8, so observations 20 to 25 are all equal to 8. Thus, both the 21st and 22nd observations have x = 8. Verification / Alternative check: Because both central observations fall within the same value of x, the median is simply that value. There is no need to average different x values, because the 21st and 22nd data points are identical. Why Other Options Are Wrong: Values 6, 7, 9 or 5 correspond to different regions of the ordered data set and do not contain both the 21st and 22nd observations. Choosing any of them would misidentify the central part of the distribution. Common Pitfalls: One common error is to ignore ordering and use the raw listing order from the statement, which is not sorted. Another is miscomputing cumulative frequencies, which shifts the identified median position to a wrong x value. Final Answer: The median of the distribution is 8.

Question 14

In Class X of a school, the mean marks of 65 students in Division A is 54, of 30 students in Division B is 50, and of 55 students in Division C is 48. What is the mean of the marks of all the students of the three divisions of Class X taken together?

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Introduction / Context: This problem involves combining means from several groups with different sizes. We are given the mean marks and number of students in three divisions of Class X and asked to find the overall mean mark of all students combined. This is a direct application of the concept of weighted average, where each division's mean is weighted by the number of students in that division. Given Data / Assumptions: Division A: 65 students, mean marks = 54. Division B: 30 students, mean marks = 50. Division C: 55 students, mean marks = 48. All students belong to exactly one of these three divisions. We must find the mean marks for all students together. Concept / Approach: The combined mean of several groups is equal to the total of all marks divided by the total number of students. For each division, total marks = mean * number of students. We compute the total marks for each division, add them, and then divide by the combined strength of all three divisions. This gives a weighted average that correctly accounts for the different sizes of the divisions. Step-by-Step Solution: Step 1: Compute total marks for Division A. Total for A = 65 * 54. 65 * 54 = 65 * 50 + 65 * 4 = 3250 + 260 = 3510. Step 2: Compute total marks for Division B. Total for B = 30 * 50 = 1500. Step 3: Compute total marks for Division C. Total for C = 55 * 48. 55 * 48 = 55 * 40 + 55 * 8 = 2200 + 440 = 2640. Step 4: Add all totals to find the grand total of marks. Grand total = 3510 + 1500 + 2640 = 7650. Step 5: Compute the total number of students. Total students = 65 + 30 + 55 = 150. Step 6: Find the combined mean. Combined mean = grand total / total students = 7650 / 150. 7650 / 150 = 51. Verification / Alternative check: We can simplify by dividing numerator and denominator by 10, giving 765 / 15. 15 * 50 = 750 and 15 * 51 = 765, confirming that 765 / 15 = 51. Thus, the combined mean is consistent with all given group means and sizes. Why Other Options Are Wrong: Means such as 50.3, 49.6 or 52.4 correspond to different grand totals and would result from incorrect calculations of total marks or student counts. A value like 53 would imply a much higher grand total than 7650 when multiplied by 150 students. Common Pitfalls: A common mistake is to simple average the three means (54, 50 and 48) without considering the different numbers of students in each division. Another error is mismultiplying or adding the totals for each division, which leads to an incorrect numerator for the combined mean. Final Answer: The mean of the marks of all students in the three divisions is 51.

Question 15

Four students are sitting in a row. The average age of the last three students is 20 years, and the average age of the first three students is 21 years. If the age of the first student is 26 years, what is the age (in years) of the last student?

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Introduction / Context: This question examines reasoning with overlapping averages. Four students are arranged in a row, and we are given the average ages of the first three and of the last three. Additionally, the age of the first student is known. The aim is to determine the age of the last student. Solving this requires translating the average information into equations and then eliminating the middle students to isolate the last student's age. Given Data / Assumptions: There are four students in a row with ages a, b, c and d in order. Average age of the last three (b, c, d) = 20 years. Average age of the first three (a, b, c) = 21 years. Age of the first student a = 26 years. We must find the age of the last student d. Concept / Approach: From each average, we can find a total. The total age of the last three is 3 * 20 and of the first three is 3 * 21. By writing these totals as sums involving a, b, c and d, we can form two linear equations. Subtracting the equation for the first three from the equation for the last three eliminates b and c and leaves an expression involving only a and d. Substituting the known value of a then gives d directly. Step-by-Step Solution: Step 1: Represent the average of the last three students. (b + c + d) / 3 = 20. So b + c + d = 3 * 20 = 60. Step 2: Represent the average of the first three students. (a + b + c) / 3 = 21. So a + b + c = 3 * 21 = 63. Step 3: Use the known value of a. a = 26, so 26 + b + c = 63. Thus b + c = 63 − 26 = 37. Step 4: Use b + c in the equation for the last three. From b + c + d = 60 and b + c = 37, we get d = 60 − 37 = 23. Verification / Alternative check: We can test with a concrete example. Let b + c = 37; pick any b and c that add to 37, for example b = 17 and c = 20. Then a = 26, d = 23 from the calculation. Average of first three (26, 17, 20) = (26 + 17 + 20) / 3 = 63 / 3 = 21. Average of last three (17, 20, 23) = (17 + 20 + 23) / 3 = 60 / 3 = 20. Both conditions are satisfied, confirming that d = 23 is consistent. Why Other Options Are Wrong: If d were 24, the sum b + c + d would be 37 + 24 = 61, giving an average of 61 / 3, which is more than 20. If d were 29 or 21 or 37, similar checks would show averages that do not match 20 and 21 as required. Common Pitfalls: One common error is to average the two given averages or to treat them as independent rather than recognizing the shared students b and c. Another is to forget to multiply the averages by 3 when forming equations for total ages, which leads to incorrect sums and a wrong value of d. Final Answer: The age of the last student is 23 years.

Question 16

Three numbers have an average of 40. The first number is equal to one third of the sum of the other two numbers. What is the value of the first number?

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Introduction / Context: This question involves three numbers linked by an average and a special relationship. We are told that the average of the three numbers is 40 and that the first number is equal to one third of the sum of the remaining two. The task is to find the first number. This requires setting up a pair of equations using the average and the relationship between the numbers and then solving for the unknowns. Given Data / Assumptions: Let the three numbers be x, y and z. The average of x, y and z is 40. The first number x is one third of the sum of the other two: x = (y + z) / 3. We must find the value of x. Concept / Approach: From the average, we know that (x + y + z) / 3 = 40, so x + y + z = 120. From the second condition, y + z = 3x. Substituting y + z in the first equation gives x + 3x = 120, which directly yields x. This is a straightforward system of linear equations. The problem emphasizes recognizing how to combine the average equation with the relationship equation effectively. Step-by-Step Solution: Step 1: Convert the average into a sum equation. (x + y + z) / 3 = 40. So x + y + z = 3 * 40 = 120. Step 2: Use the relationship between the first number and the other two. x = (y + z) / 3. Multiply both sides by 3 to remove the fraction. 3x = y + z. Step 3: Substitute y + z into the sum equation. From x + y + z = 120 and y + z = 3x, we get x + 3x = 120. 4x = 120. Step 4: Solve for x. x = 120 / 4 = 30. Verification / Alternative check: If x = 30, then y + z = 3x = 90. Total sum x + y + z = 30 + 90 = 120, which gives an average of 120 / 3 = 40, matching the given average. Also, x is indeed one third of y + z because (y + z) / 3 = 90 / 3 = 30. Why Other Options Are Wrong: If x were 20, then y + z would be 60, giving a total of 80 and an average of 80 / 3, which is not 40. If x were 25, 35 or 50, similar checks show that the total and average would not satisfy both given conditions. Common Pitfalls: Some students mistakenly set x equal to one third of the total of all three numbers instead of one third of (y + z). Another frequent error is to forget to multiply the average by 3 when converting it into a sum equation, leading to a wrong total. Final Answer: The first number is 30.

Question 17

The average of 14 numbers is 32. If the average of the last 5 numbers is 26, what is the average of the remaining 9 numbers?

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Introduction / Context: Here we are given the average of a total set of numbers and the average of a subset of those numbers. Specifically, 14 numbers have a certain average, and the last 5 numbers within this group have a lower average. The aim is to find the average of the remaining 9 numbers. This is a typical mixture problem where we need to make use of sums derived from averages and then separate the contributions of subgroups. Given Data / Assumptions: Total number of numbers = 14. Average of all 14 numbers = 32. Average of the last 5 numbers = 26. We must find the average of the first 9 numbers (the remaining numbers). Concept / Approach: The sum of all 14 numbers can be obtained by multiplying the overall average by 14. Similarly, the sum of the last 5 numbers is the average of those 5 multiplied by 5. The sum of the first 9 numbers is simply the total sum minus the sum of the last 5. Once we know that sum, dividing by 9 gives the required average. This approach uses the idea that total = subgroup one + subgroup two, and is very common in average problems. Step-by-Step Solution: Step 1: Compute the total sum of all 14 numbers. Total sum = 14 * 32. 14 * 32 = 14 * (30 + 2) = 420 + 28 = 448. Step 2: Compute the sum of the last 5 numbers. Average of last 5 numbers = 26. Sum of last 5 numbers = 5 * 26 = 130. Step 3: Find the sum of the first 9 numbers. Sum of first 9 numbers = total sum − sum of last 5 numbers. Sum of first 9 = 448 − 130 = 318. Step 4: Compute the average of the first 9 numbers. Average of first 9 numbers = 318 / 9. 9 * 35 = 315, remainder 3, so 318 / 9 = 35 + 3/9 = 35.333..., which we write as 35.33 to two decimal places. Verification / Alternative check: We can verify that mixing averages works out. Weighted contribution of first 9 numbers = 9 * 35.33 ≈ 318. Weighted contribution of last 5 numbers = 5 * 26 = 130. Total ≈ 318 + 130 = 448, and 448 / 14 = 32, which matches the original average. Why Other Options Are Wrong: An average of 41.33 or 44.5 would give a total for the first 9 numbers that is too large and would raise the overall average above 32. An average of 27.5 or 32 for the first 9 numbers would produce totals that do not combine with the last 5 numbers to give an overall average of 32. Common Pitfalls: Some students mistakenly average 32 and 26 directly, without taking into account the fact that they apply to different group sizes. Another mistake is to mix up the number of terms, using 14 instead of 9 when computing the average of the remaining numbers. Final Answer: The average of the remaining 9 numbers is 35.33 (approximately).

Question 18

The average weight of Shubha, Govinda and Reshma is 65 kg. The average weight of Shubha and Govinda together is 63 kg, and the average weight of Govinda and Reshma together is 70 kg. What is the weight (in kg) of Govinda?

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Introduction / Context: This question is very similar to other average problems where individual values must be found from the averages of different groupings. We are given the average weight of three people together and the averages of two person groups within them. From this data, we must determine the weight of Govinda. The problem is solved by converting averages into total sums and then using simple algebra to isolate the unknown weight. Given Data / Assumptions: Let weights be S for Shubha, G for Govinda and R for Reshma. Average of S, G and R = 65 kg, so S + G + R = 3 * 65. Average of S and G = 63 kg, so S + G = 2 * 63. Average of G and R = 70 kg, so G + R = 2 * 70. We must find G. Concept / Approach: We translate each average into a sum by multiplying by the number of people in that group. This gives three equations in S, G and R. Adding the equations for S + G and G + R produces an expression involving S + 2G + R. We then subtract the equation for S + G + R from this to isolate G. This is a standard approach for such average problems and leads to a quick solution. Step-by-Step Solution: Step 1: Convert the given averages into equations. S + G + R = 3 * 65 = 195. S + G = 2 * 63 = 126. G + R = 2 * 70 = 140. Step 2: Add the equations S + G and G + R. (S + G) + (G + R) = 126 + 140. Left side becomes S + 2G + R. Right side is 266. So S + 2G + R = 266. Step 3: Subtract the equation for all three. (S + 2G + R) − (S + G + R) = 266 − 195. Left side simplifies to G; right side is 71. Therefore, G = 71 kg. Verification / Alternative check: Find S and R to verify. From S + G = 126, S = 126 − 71 = 55. From G + R = 140, R = 140 − 71 = 69. Check the overall average. S + G + R = 55 + 71 + 69 = 195. Average = 195 / 3 = 65 kg, which matches the given overall average. Why Other Options Are Wrong: Weights such as 41, 46, 60 or 88 do not satisfy all three average conditions when substituted into the equations. Each incorrect choice leads to inconsistencies in at least one of the three average relations. Common Pitfalls: A common mistake is to attempt solving for S and R first or to confuse the roles of the averages, which can lead to algebraic errors. Another error is miscomputing the products 3 * 65, 2 * 63 or 2 * 70, which directly affects the final value of G. Final Answer: Govinda's weight is 71 kilograms.

Question 19

The average of 100 items was found to be 30. Later it was noticed that two items had been wrongly recorded as 32 and 12 instead of the correct values 23 and 11. What is the correct average of the 100 items after rectifying these two errors?

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Introduction / Context: This is another example of correcting an average when mistakes in recorded values are discovered. The original average of 100 items was computed using two incorrect values. Once we know which values were wrong and what their correct values should be, we can adjust the total and recompute the average. The question reinforces how averages depend on the total sum of all items. Given Data / Assumptions: Number of items = 100. Original average (with errors) = 30. Two items were recorded as 32 and 12. Their correct values are 23 and 11. We must find the corrected average of the 100 items. Concept / Approach: Average is defined as total sum divided by number of items. The original total was computed using the wrong values 32 and 12. To correct the total, we subtract these wrong values and then add the correct ones 23 and 11. This gives the true total sum. Dividing this corrected total by 100 yields the correct average. It is important to handle the arithmetic carefully and interpret whether the total increases or decreases after correction. Step-by-Step Solution: Step 1: Compute the original total based on the incorrect average. Original total = 100 * 30 = 3000. Step 2: Determine the wrong and correct totals for the two items. Wrong contribution = 32 + 12 = 44. Correct contribution = 23 + 11 = 34. Step 3: Find the net change in the total when correcting the two items. The original total used 44, but it should have used 34. So we must reduce the total by 44 − 34 = 10. Step 4: Calculate the corrected total. Corrected total = original total − 10 = 3000 − 10 = 2990. Step 5: Compute the corrected average. Corrected average = 2990 / 100 = 29.9. Verification / Alternative check: We can directly adjust the original total. Corrected total = 3000 − (32 + 12) + (23 + 11). Corrected total = 3000 − 44 + 34 = 3000 − 10 = 2990. Dividing by 100 again gives 29.9, confirming our earlier result. Why Other Options Are Wrong: An average of 29.8 or 29.5 would imply a greater reduction in the total than the 10 units actually required. An average of 29 or 30.1 would correspond to total sums that do not match the corrected values (they would be 2900 and 3010 respectively). Common Pitfalls: One frequent error is to add instead of subtracting the net difference, mistakenly increasing the total when it should decrease. Another mistake is to replace only one of the wrong values or to miscalculate the difference between 44 and 34. Final Answer: The correct average of the 100 items is 29.9.

Question 20

The average annual revenue of a company over 11 consecutive years is Rs 69 lakhs. The average for the first 6 years is Rs 64 lakhs, and the average for the last 6 years is Rs 76 lakhs. What is the revenue (in Rs lakhs) for the 6th year?

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Introduction / Context: This question deals with averages over overlapping time periods. The company's average revenue over 11 years is known, as well as the averages for the first 6 and the last 6 years. Notice that the 6th year is included in both the first 6 and the last 6. We are asked to compute the revenue for this overlapping year. Solving this requires writing sums in terms of yearly revenues and using the fact that the 6th year is counted twice when adding the two 6 year totals. Given Data / Assumptions: Average revenue for 11 years = Rs 69 lakhs per year. Average revenue for the first 6 years = Rs 64 lakhs per year. Average revenue for the last 6 years = Rs 76 lakhs per year. Years are consecutive and the 6th year belongs to both first 6 and last 6 years. We must find revenue in the 6th year, call it X lakhs. Concept / Approach: First, convert each average into a total revenue. Let the revenues for the 11 years be R1, R2, ..., R11. The total for the first 6 years is R1 + R2 + R3 + R4 + R5 + R6, and the total for the last 6 years is R6 + R7 + R8 + R9 + R10 + R11. Adding these two totals counts R6 twice, while all other years are counted once. On the other hand, the total revenue for all 11 years is given by 11 * 69. Using these relationships, we can solve for R6. Step-by-Step Solution: Step 1: Let total revenue for all 11 years be T. T = 11 * 69 = 759 lakhs. Step 2: Compute total for the first 6 years. Sum of first 6 years = 6 * 64 = 384 lakhs. Denote this as S1_6 = R1 + R2 + R3 + R4 + R5 + R6. Step 3: Compute total for the last 6 years. Sum of last 6 years = 6 * 76 = 456 lakhs. Denote this as S6_11 = R6 + R7 + R8 + R9 + R10 + R11. Step 4: Add S1_6 and S6_11. S1_6 + S6_11 = (R1 + R2 + R3 + R4 + R5 + R6) + (R6 + R7 + R8 + R9 + R10 + R11). This equals (R1 + R2 + R3 + R4 + R5 + R7 + R8 + R9 + R10 + R11) + 2R6. But R1 + R2 + ... + R11 = T, so S1_6 + S6_11 = T + R6. Step 5: Use the numerical values. Left side S1_6 + S6_11 = 384 + 456 = 840. Right side is T + R6 = 759 + X. So 840 = 759 + X. Step 6: Solve for X. X = 840 − 759 = 81. Verification / Alternative check: If R6 = 81, we can confirm that all averages match. Total for first 6 years is 384; total for last 6 years is 456. Adding gives 840, which equals T + 81, so T = 759. Overall average = 759 / 11 = 69 lakhs per year, as given. Why Other Options Are Wrong: Values like 77, 79, 83 or 75 would not satisfy the equation 384 + 456 = 759 + X. Each of these alternatives would lead to a total T different from 759, and hence to a different overall average than 69 lakhs. Common Pitfalls: A common mistake is to think that the 6th year's revenue is simply the difference between the two averages 76 and 64, which is not correct. Another error is to forget that the 6th year appears in both 6 year averages, leading to miscounting in the equations. Final Answer: The revenue for the 6th year is Rs 81 lakhs.

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