Q: The average cost of 5 items in a shopping list is Rs 650. If another item costing Rs 1400 is added to the list, what will be the new average cost in rupees?

Introduction / Context: This is a straightforward question on averages where an additional value is added to an existing data set. It illustrates how the mean changes when a new item with a different cost is introduced. Given Data / Assumptions: - The original number of items is 5. - The average cost of these 5 items is Rs 650. - A sixth item costing Rs 1400 is added. - We need the new average cost of all 6 items. Concept / Approach: The average is equal to total sum divided by the number of items. So we first find the total cost of the original 5 items from their average. Then we add the cost of the sixth item to get the new total and finally divide by 6 to get the new average. Step-by-Step Solution: Step 1: Total cost of 5 items = average cost * number of items = 650 * 5. Step 2: Compute 650 * 5 = 3250 rupees. Step 3: Add the cost of the sixth item: new total = 3250 + 1400 = 4650 rupees. Step 4: The number of items is now 6, so the new average cost = 4650 / 6. Step 5: Compute 4650 / 6 = 775. Verification / Alternative check: Check by approximate reasoning: the new item is more expensive than the old average, so the new average must be higher than 650. The value 775 is higher than 650 but not extremely large, which fits the data well. Calculations also confirm this exactly. Why Other Options Are Wrong: - Option 875: This would imply a total of 875 * 6 = 5250, which is greater than 4650, so it is impossible. - Option 725: This corresponds to a total of 4350, which is less than the true total 4650 and therefore too small. - Option 825: This would imply a total of 4950, which again does not match the correct total. Common Pitfalls: A common mistake is to simply average 650 and 1400, which would ignore the fact that there are already 5 items at 650. Always return to the fundamental definition of average: total sum divided by count. From there, all such questions can be solved reliably. Final Answer: The new average cost of the items is 775 rupees.

Q: The average age of the district level hockey team of eleven players is 22 years. When the age of the coach is also included, the average age increases by 1 year. What is the age of the coach?

Introduction / Context: This problem uses the idea of average age before and after adding one more person. It shows how including an older person raises the group average and how you can work backward to find that person's age. Given Data / Assumptions: - There are 11 players with an average age of 22 years. - When the coach is added, there are 12 people and the new average age is 23 years (one year higher). - We need to find the age of the coach. Concept / Approach: Average age is total age divided by the number of people. We find the total age of the players alone from the first average. Then we find the total age of players plus coach from the second average. The difference between these two totals is the coach's age. Step-by-Step Solution: Step 1: Total age of the 11 players = 22 * 11. Step 2: Compute 22 * 11 = 242 years. Step 3: When the coach is added, there are 12 people with average 23 years, so the new total age is 23 * 12. Step 4: Compute 23 * 12 = 276 years. Step 5: The age of the coach = total with coach - total without coach = 276 - 242. Step 6: Calculate 276 - 242 = 34 years. Verification / Alternative check: Check by recomputing the final average. If you have 11 players totaling 242 years and a coach of 34 years, the total is 276. Dividing by 12 gives 23 years as required, so the age 34 is consistent. Why Other Options Are Wrong: - Option 33 yrs: This would give a total of 275 and an average slightly less than 23. - Option 35 yrs: This would give a total of 277 and an average greater than 23. - Option 36 yrs: This would yield an even higher total and average, inconsistent with the given new average of 23. Common Pitfalls: Some learners mistakenly average the two averages (22 and 23) instead of working with totals. Others forget to multiply the average by the number of people to get the sum. Always remember that changes in average correspond to changes in total sum, which you can track directly. Final Answer: The coach is 34 yrs old.

Q: The average marks of 40 students in an examination is 25. Later it is found that the marks of one student were wrongly entered as 73 instead of 37. What is the correct average marks for the class?

Introduction / Context: This is another example of correcting an average when one observation has been entered incorrectly. It reinforces the method of adjusting the total sum first and then recomputing the mean. Given Data / Assumptions: - There are 40 students in the class. - The initially reported average marks is 25. - One student's marks were wrongly recorded as 73 instead of the correct value 37. - We must find the corrected class average. Concept / Approach: As before, the average equals total marks divided by the number of students. We begin with the incorrect total implied by the wrong average, then adjust this total by removing the incorrect value and adding the correct one. Finally, we divide the corrected total by 40. Step-by-Step Solution: Step 1: Incorrect total marks = 25 * 40. Step 2: Compute 25 * 40 = 1000 marks. Step 3: The wrong entry is 73, the correct value is 37, so we must subtract 73 and add 37 to correct the total. Step 4: Correct total = 1000 - 73 + 37. Step 5: Compute 1000 - 73 = 927, then 927 + 37 = 964. Step 6: Correct average = 964 / 40. Step 7: 964 / 40 = 24.1 marks. Verification / Alternative check: The net error in the total is 73 - 37 = 36 marks too high. So we can also compute the corrected total as 1000 - 36 = 964 directly. Dividing 964 by 40 again gives 24.1, confirming the answer. Why Other Options Are Wrong: - Option 24.3: This would correspond to a total of 972 marks, which does not match the corrected sum 964. - Option 24.5: This implies a total of 980 marks, again inconsistent. - Option 24.7: This would imply 988 marks, which is too high given the nature of the correction. Common Pitfalls: Students sometimes try to average 25 and the corrected mark or directly adjust the average by the error divided by the number of students but with incorrect sign. The safe method is always to reconstruct the total, correct it, and then divide by the class strength. Final Answer: The correct average marks of the class is 24.1.

Question 1

Which one of the following fractions is greater than 1/2?

Accepted Answer

Introduction / Context: This question checks your ability to compare simple fractions with the benchmark value 1/2. Being comfortable with comparing fractions is very important in aptitude tests because it appears in ratio, proportion and data interpretation questions as well. Given Data / Assumptions: - The fractions to compare are 2/5, 4/7, 4/9 and 5/11. - We need to decide which single fraction is greater than 1/2. - All fractions are positive proper fractions. Concept / Approach: To compare a fraction a/b with 1/2, a fast method is to compare 2a with b. If 2a is greater than b, then a/b is greater than 1/2. If 2a is less than b, then the fraction is less than 1/2. This avoids converting to decimals and saves time in exams. Step-by-Step Solution: Step 1: For 2/5, compute 2 * 2 = 4 and compare with 5. Since 4 < 5, 2/5 is less than 1/2. Step 2: For 4/7, compute 2 * 4 = 8 and compare with 7. Since 8 > 7, 4/7 is greater than 1/2. Step 3: For 4/9, compute 2 * 4 = 8 and compare with 9. Since 8 < 9, 4/9 is less than 1/2. Step 4: For 5/11, compute 2 * 5 = 10 and compare with 11. Since 10 < 11, 5/11 is less than 1/2. Step 5: Among the given options, only 4/7 satisfies the condition of being greater than 1/2. Verification / Alternative check: You can also approximate the fractions as decimals. 2/5 = 0.4, 4/7 is about 0.571, 4/9 is around 0.444, and 5/11 is about 0.455. Only 0.571 is greater than 0.5, confirming that 4/7 is the correct choice. Why Other Options Are Wrong: - Option 2/5: Its decimal value is 0.4, clearly less than 0.5. - Option 4/9: Its value is approximately 0.444, still below 0.5. - Option 5/11: This is about 0.455, again less than 0.5, so it cannot be greater than 1/2. Common Pitfalls: A frequent mistake is to compare only numerators or only denominators, which does not work for fractions with different denominators. Another error is to assume that the fraction with the largest numerator or smallest denominator is the largest. Using the simple 2a versus b trick is a reliable and fast way to compare with 1/2. Final Answer: The only fraction greater than 1/2 is 4/7.

Question 2

The average of a non zero number and its square is 11 times the number. What is the value of the number?

Accepted Answer

Introduction / Context: This question combines the idea of average with a simple quadratic equation. You are given a relationship between a number, its square and a multiple of the number. Solving it correctly requires forming an equation and then using standard algebraic techniques to find the non zero solution. Given Data / Assumptions: - Let the number be x, where x is non zero. - The average of x and x squared is 11 times the number x. - Mathematically, this is expressed as (x + x^2) / 2 = 11x. - We must find the value of x from the options 22, 21, 11 and 17. Concept / Approach: The average of two numbers is their sum divided by 2. So the statement directly becomes an equation in x. After simplifying, we get a quadratic equation. Quadratic equations can be solved by factoring or by using the quadratic formula. Here the equation factors easily, and one of the roots is zero, which we discard because the number is stated to be non zero. Step-by-Step Solution: Step 1: Write the equation for the average: (x + x^2) / 2 = 11x. Step 2: Multiply both sides by 2 to remove the denominator: x + x^2 = 22x. Step 3: Rearrange all terms to one side: x^2 + x - 22x = 0. Step 4: Simplify: x^2 - 21x = 0. Step 5: Factor the equation: x(x - 21) = 0. Step 6: The solutions are x = 0 or x = 21. Since the number is non zero, we take x = 21. Verification / Alternative check: Check x = 21 in the original condition. The average of 21 and 21^2 is (21 + 441) / 2 = 462 / 2 = 231. On the other side, 11 times 21 is 231. Both sides match, so x = 21 satisfies the condition. Why Other Options Are Wrong: - Option 22: For x = 22, (22 + 484) / 2 = 253, but 11 * 22 = 242, so the condition fails. - Option 11: For x = 11, (11 + 121) / 2 = 66, but 11 * 11 = 121, not equal. - Option 17: For x = 17, (17 + 289) / 2 = 153, but 11 * 17 = 187, again not equal. Common Pitfalls: Some students forget to multiply both sides by 2 correctly, or they move terms incorrectly, leading to sign errors. Another common mistake is forgetting to discard the zero root even when the question states the number is non zero. Always check which solutions are allowed by the wording of the question. Final Answer: The required non zero number is 21.

Question 3

Amitabh spends on average Rs 4200 per month from January to June. His expenditure in January is Rs 1200 and in July it is Rs 1500. What is his average monthly expenditure from February to July?

Accepted Answer

Introduction / Context: This question is about recomputing an average when you change the time period. You are given the average for one block of months and specific monthly values for some months. Then you are asked to find the average over a different block of months that partly overlaps with the first. Given Data / Assumptions: - From January to June (6 months) the average monthly expenditure is Rs 4200. - Expenditure in January is Rs 1200. - Expenditure in July is Rs 1500. - We must find the average expenditure from February to July, which is 6 months again (February, March, April, May, June and July). Concept / Approach: First, we convert the average for January to June into the total expenditure for that period. Then we subtract the amount spent in January to get the total for February to June. After that we add the expenditure in July to obtain the total for February to July. Finally, we divide this total by 6, because this second period also has six months. Step-by-Step Solution: Step 1: Total expenditure from January to June = average * number of months = 4200 * 6. Step 2: Compute 4200 * 6 = 25200 rupees. Step 3: Expenditure from February to June = 25200 - 1200 = 24000 rupees. Step 4: Add July expenditure to get expenditure from February to July: 24000 + 1500 = 25500 rupees. Step 5: Number of months from February to July is 6, so the new average is 25500 / 6. Step 6: Compute 25500 / 6 = 4250 rupees per month. Verification / Alternative check: We can check that the total from February to July, 25500, divided by 6 indeed gives 4250. This is consistent and within a realistic range given that several months must have higher expenditure to counterbalance the very low January value of Rs 1200. Why Other Options Are Wrong: - Option Rs 4520: This would imply a total of 27120 for six months, which exceeds the possible total given the average for January to June. - Option Rs 4060: This total (24360) is too low because February to June plus July must exceed the January to June total due to replacing January 1200 with July 1500. - Option None of these: This is incorrect because Rs 4250 matches the correct calculation. Common Pitfalls: One common error is to add January and July directly or to mistakenly average January to July instead of February to July. Another is forgetting that the number of months in the new period is still six. Always write out which months you are including and carefully compute totals for the exact range specified. Final Answer: The average monthly expenditure from February to July is Rs 4250.

Question 4

At a cinema hall for a matinee show from 2.30 pm to 5.30 pm, the parking lot contains only three wheelers (auto rickshaws) and four wheelers (cars). The total number of wheels counted in the parking area is 240. How many three wheelers are parked th…

Accepted Answer

Introduction / Context: This is a classic puzzle involving a mixture of two types of vehicles with different numbers of wheels. You are given only the total number of wheels and must find how many three wheelers are present, using the fact that the rest are four wheelers. Given Data / Assumptions: - The parking lot has only three wheelers and four wheelers. - Each three wheeler has 3 wheels and each four wheeler has 4 wheels. - The total number of wheels is 240. - We need to find the number of three wheelers from among the options. Concept / Approach: Let the number of three wheelers be t and the number of four wheelers be f. The total wheels equation is 3t + 4f = 240. With only this equation there are infinitely many pairs (t, f) in theory, but the options restrict t to specific values. So we can test each option by substituting its value for t and checking whether f comes out as a non negative integer. Step-by-Step Solution: Step 1: Let t be the number of three wheelers. Then the total number of wheels is 3t + 4f = 240. Step 2: Rearrange to find f in terms of t: 4f = 240 - 3t, so f = (240 - 3t) / 4. Step 3: Try option t = 48. Then f = (240 - 3 * 48) / 4 = (240 - 144) / 4 = 96 / 4 = 24. Step 4: For t = 48, f is a whole number (24), so this is a valid solution. Step 5: We check other options to verify that they do not give integer values of f. Verification / Alternative check: With t = 48 three wheelers and f = 24 four wheelers, the total wheels are 48 * 3 + 24 * 4 = 144 + 96 = 240, which matches the given total. Therefore, this pair of values is consistent with the problem statement. Why Other Options Are Wrong: - Option 41: f = (240 - 3 * 41) / 4 = (240 - 123) / 4 = 117 / 4, which is not an integer. - Option 45: f = (240 - 3 * 45) / 4 = (240 - 135) / 4 = 105 / 4, again not an integer. - Option 43: f = (240 - 3 * 43) / 4 = (240 - 129) / 4 = 111 / 4, also not an integer. Common Pitfalls: Some learners try to guess by proportion or by approximate reasoning without using the equation. Others forget that the number of vehicles must be a whole number and accept a fractional value for f. Always set up the equation correctly and insist on integer solutions when counting real objects like vehicles. Final Answer: The number of three wheelers parked in the lot is 48.

Question 5

For the data set 3, 10, 10, 4, 7, 10 and 5, what is the mean deviation about the mean?

Accepted Answer

Introduction / Context: This question is about a measure of dispersion called mean deviation about the mean. It tells you on average how far the observations lie from the arithmetic mean of the data. Understanding this helps in describing how spread out a data set is. Given Data / Assumptions: - The data points are 3, 10, 10, 4, 7, 10 and 5. - We need to calculate mean deviation about the mean, not about the median. - There are 7 observations in total. Concept / Approach: Mean deviation about the mean is defined as the average of the absolute deviations from the arithmetic mean. The steps are: find the mean, compute each observation minus the mean in absolute value, sum these absolute deviations, and finally divide by the number of observations. Step-by-Step Solution: Step 1: Compute the sum of the data: 3 + 10 + 10 + 4 + 7 + 10 + 5. Step 2: The sum is 49, so the mean is 49 / 7 = 7. Step 3: Find absolute deviations from the mean 7: |3 - 7| = 4, |10 - 7| = 3, |10 - 7| = 3, |4 - 7| = 3, |7 - 7| = 0, |10 - 7| = 3, |5 - 7| = 2. Step 4: Sum of absolute deviations = 4 + 3 + 3 + 3 + 0 + 3 + 2 = 18. Step 5: Mean deviation about the mean = 18 / 7. Verification / Alternative check: Check that the mean 7 lies between the smallest and largest values and that deviations are correctly computed. Also confirm that the total of 49 is correct. Once those are confirmed, 18 / 7 is the only possible correct value. Why Other Options Are Wrong: - Option 49/7: This equals 7, which would mean each value differs from the mean on average by 7 units, far too large for this data set. - Option 19/7: This is approximately 2.714, which does not match the exact sum of absolute deviations, which is 18. - Option 50/7: This is based on a mistaken sum or deviation calculation and does not correspond to the true total absolute deviation. Common Pitfalls: Learners often forget to take absolute values and instead use signed deviations that add up to zero. Another common error is to divide by n - 1 instead of n, which is used in variance estimation, not in mean deviation. Careful arithmetic with each deviation is essential. Final Answer: The mean deviation about the mean for this data set is 18/7.

Question 6

The average cost of 5 items in a shopping list is Rs 650. If another item costing Rs 1400 is added to the list, what will be the new average cost in rupees?

Accepted Answer

Introduction / Context: This is a straightforward question on averages where an additional value is added to an existing data set. It illustrates how the mean changes when a new item with a different cost is introduced. Given Data / Assumptions: - The original number of items is 5. - The average cost of these 5 items is Rs 650. - A sixth item costing Rs 1400 is added. - We need the new average cost of all 6 items. Concept / Approach: The average is equal to total sum divided by the number of items. So we first find the total cost of the original 5 items from their average. Then we add the cost of the sixth item to get the new total and finally divide by 6 to get the new average. Step-by-Step Solution: Step 1: Total cost of 5 items = average cost * number of items = 650 * 5. Step 2: Compute 650 * 5 = 3250 rupees. Step 3: Add the cost of the sixth item: new total = 3250 + 1400 = 4650 rupees. Step 4: The number of items is now 6, so the new average cost = 4650 / 6. Step 5: Compute 4650 / 6 = 775. Verification / Alternative check: Check by approximate reasoning: the new item is more expensive than the old average, so the new average must be higher than 650. The value 775 is higher than 650 but not extremely large, which fits the data well. Calculations also confirm this exactly. Why Other Options Are Wrong: - Option 875: This would imply a total of 875 * 6 = 5250, which is greater than 4650, so it is impossible. - Option 725: This corresponds to a total of 4350, which is less than the true total 4650 and therefore too small. - Option 825: This would imply a total of 4950, which again does not match the correct total. Common Pitfalls: A common mistake is to simply average 650 and 1400, which would ignore the fact that there are already 5 items at 650. Always return to the fundamental definition of average: total sum divided by count. From there, all such questions can be solved reliably. Final Answer: The new average cost of the items is 775 rupees.

Question 7

Sripad scores an average of 65 marks in three subjects. In no subject does he score less than 58 marks, and he scores more marks in Mathematics than in the other two subjects. What is the maximum possible score he could obtain in Mathematics?

Accepted Answer

Introduction / Context: This question is about averages and constraints. You are given a fixed average over three subjects and lower bounds on each subject, and you must find how large one particular subject score can be while respecting all conditions. Given Data / Assumptions: - There are three subjects in total, including Mathematics. - The average of the three subject scores is 65 marks, so the total is fixed. - In every subject, the score is at least 58 marks. - The Mathematics score is higher than the scores in the other two subjects. - We need the maximum possible score in Mathematics. Concept / Approach: Since the average is fixed, the sum of the three scores is fixed. To maximise one subject, we minimise the other two while respecting their minimum allowed values. Once we fix the two smaller scores at their minimum values, the remaining marks go to Mathematics, which yields the maximum possible Mathematics score. Step-by-Step Solution: Step 1: Let the three subject scores be M (Maths), S1 and S2. Step 2: Average = 65, so M + S1 + S2 = 3 * 65 = 195. Step 3: Each subject must have at least 58 marks, and Mathematics must be greater than both S1 and S2. Step 4: To maximise M, choose S1 = 58 and S2 = 58, the smallest values allowed. Step 5: Then M = 195 - 58 - 58 = 195 - 116 = 79. Step 6: This value 79 is greater than 58, so the condition that Maths has the highest score is satisfied. Verification / Alternative check: If you try to increase either S1 or S2 above 58, the total 195 forces M to become smaller. Therefore, choosing both non Maths subjects at their minimum gives the largest possible M. No value greater than 79 is possible because that would make the sum exceed 195. Why Other Options Are Wrong: - Option 77: If M = 77 and S1, S2 are at least 58, then total is at least 77 + 58 + 58 = 193, which is less than 195, meaning the average would not be 65. - Option 76: Similar reasoning gives a total less than 195 with minimum other scores, or would require one subject under 58. - Option 73: This is a possible Maths score but not the maximum; we have already shown 79 is possible under the constraints. Common Pitfalls: Students sometimes forget to consider the minimum bound of 58 for each subject and instead assume the other subjects can go arbitrarily low. Another issue is trying to distribute marks evenly, which contradicts the condition that Maths has the highest score. Always use the idea that to maximise one quantity under a sum constraint, you minimise the others within their allowed ranges. Final Answer: The maximum possible score in Mathematics is 79 marks.

Question 8

The average age of the district level hockey team of eleven players is 22 years. When the age of the coach is also included, the average age increases by 1 year. What is the age of the coach?

Accepted Answer

Introduction / Context: This problem uses the idea of average age before and after adding one more person. It shows how including an older person raises the group average and how you can work backward to find that person's age. Given Data / Assumptions: - There are 11 players with an average age of 22 years. - When the coach is added, there are 12 people and the new average age is 23 years (one year higher). - We need to find the age of the coach. Concept / Approach: Average age is total age divided by the number of people. We find the total age of the players alone from the first average. Then we find the total age of players plus coach from the second average. The difference between these two totals is the coach's age. Step-by-Step Solution: Step 1: Total age of the 11 players = 22 * 11. Step 2: Compute 22 * 11 = 242 years. Step 3: When the coach is added, there are 12 people with average 23 years, so the new total age is 23 * 12. Step 4: Compute 23 * 12 = 276 years. Step 5: The age of the coach = total with coach - total without coach = 276 - 242. Step 6: Calculate 276 - 242 = 34 years. Verification / Alternative check: Check by recomputing the final average. If you have 11 players totaling 242 years and a coach of 34 years, the total is 276. Dividing by 12 gives 23 years as required, so the age 34 is consistent. Why Other Options Are Wrong: - Option 33 yrs: This would give a total of 275 and an average slightly less than 23. - Option 35 yrs: This would give a total of 277 and an average greater than 23. - Option 36 yrs: This would yield an even higher total and average, inconsistent with the given new average of 23. Common Pitfalls: Some learners mistakenly average the two averages (22 and 23) instead of working with totals. Others forget to multiply the average by the number of people to get the sum. Always remember that changes in average correspond to changes in total sum, which you can track directly. Final Answer: The coach is 34 yrs old.

Question 9

A cricketer has a certain batting average for 10 innings. In the eleventh innings he scores 108 runs, thereby increasing his average by 6 runs. What is his new batting average?

Accepted Answer

Introduction / Context: This question tests your understanding of how a single very high score can increase a batting average and how to use algebra to find both the old and the new averages. Given Data / Assumptions: - Before the eleventh innings, the cricketer has played 10 innings with some average A runs. - In the eleventh innings he scores 108 runs. - As a result, his average increases by 6 runs, becoming A + 6. - We are asked to find the new average A + 6. Concept / Approach: Let the original average be A. The total runs before the eleventh innings are 10A. After scoring 108 runs, his total runs become 10A + 108 over 11 innings, and his new average is A + 6. Equating these two expressions for the new total allows us to solve for A and then compute the new average. Step-by-Step Solution: Step 1: Let the initial average after 10 innings be A runs. Then total runs before the 11th innings are 10A. Step 2: In the 11th innings he scores 108 runs, so new total runs = 10A + 108. Step 3: The new average is A + 6 runs over 11 innings, so new total = 11 * (A + 6). Step 4: Set the two expressions equal: 10A + 108 = 11(A + 6). Step 5: Expand the right side: 10A + 108 = 11A + 66. Step 6: Rearrange: 108 - 66 = 11A - 10A, giving 42 = A. Step 7: The new average is A + 6 = 42 + 6 = 48 runs. Verification / Alternative check: Check totals: with average 42 over 10 innings, total runs are 420. Adding 108 gives 528 runs over 11 innings. Then 528 / 11 = 48 runs, confirming that the new average is indeed 48. Why Other Options Are Wrong: - Option 44 runs: This would mean an increase of only 2 runs from a hypothetical average of 42, not 6. - Option 46 runs: This corresponds to a smaller increase and does not fit the equation 10A + 108 = 11(A + 6). - Option 49 runs: This would imply an average increase of 7 runs, which contradicts the given information that the increase is 6 runs. Common Pitfalls: A common mistake is to directly average 108 with some guessed value or to forget that the average is over 11 innings, not 10. Others mis-handle the algebra when distributing 11 over (A + 6). Writing the equation slowly and carefully helps avoid these errors. Final Answer: The new batting average of the cricketer is 48 runs.

Question 10

The average weight of a group of boys is 30 kg. After a boy weighing 35 kg joins the group, the average weight of the group increases by 1 kg. How many boys were there in the group originally?

Accepted Answer

Introduction / Context: This problem shows how adding one new value above the existing average increases the group average. You are asked to infer the original group size based on the change in average caused by the new member. Given Data / Assumptions: - The original average weight of the boys is 30 kg. - A new boy with weight 35 kg joins the group. - After he joins, the average weight increases by 1 kg to 31 kg. - We must find the original number of boys in the group. Concept / Approach: Let the original number of boys be n. The original total weight is 30n. After adding a boy of 35 kg, the new total weight is 30n + 35 and the number of boys becomes n + 1. The new average is 31, so (30n + 35) / (n + 1) = 31. Solving this equation gives n. Step-by-Step Solution: Step 1: Let the original number of boys be n. Step 2: Original total weight = 30n kg. Step 3: After adding the new boy, total weight = 30n + 35 kg and number of boys = n + 1. Step 4: New average is 31, so (30n + 35) / (n + 1) = 31. Step 5: Multiply both sides by n + 1: 30n + 35 = 31(n + 1). Step 6: Expand the right side: 30n + 35 = 31n + 31. Step 7: Rearrange: 35 - 31 = 31n - 30n, so 4 = n. Step 8: Therefore, there were originally 4 boys in the group. Verification / Alternative check: If there were 4 boys with average weight 30 kg, total weight is 120 kg. Adding a boy of 35 kg gives 155 kg over 5 boys. The new average is 155 / 5 = 31 kg, which matches the condition. So n = 4 is correct. Why Other Options Are Wrong: - Option 6: If n = 6, original total is 180, new total 215, and new average 215 / 7, which is not 31. - Option 8: If n = 8, original total is 240, new total 275, and new average 275 / 9, again not equal to 31. - Option 5: If n = 5, original total is 150, new total 185, and new average 185 / 6, which is not 31. Common Pitfalls: Some students incorrectly assume that the change in average compares directly with the difference between 30 and 35 and try proportional reasoning instead of setting up an equation. Others forget to change both the numerator and the denominator when the new boy is added. Always reflect both the change in total weight and the change in group size. Final Answer: Originally there were 4 boys in the group.

Question 11

The average marks of 40 students in an examination is 34. Later it is found that the marks of one student were wrongly entered as 62 instead of 26. What is the correct average marks for the class?

Accepted Answer

Introduction / Context: This question involves correcting an average when one value has been mis-recorded. It is a very common type of aptitude problem related to data accuracy and averages. Given Data / Assumptions: - There are 40 students in the class. - The initially computed average is 34 marks. - One student's marks were incorrectly recorded as 62 instead of the correct value 26. - We must find the corrected average. Concept / Approach: First, we find the incorrect total marks that led to the wrong average by multiplying the average by the number of students. Then we adjust this total by subtracting the wrong value and adding the correct value. Finally, we divide the corrected total by 40 to obtain the correct average. Step-by-Step Solution: Step 1: Incorrect total marks = 34 * 40. Step 2: Compute 34 * 40 = 1360. Step 3: One student's mark was entered as 62 instead of 26. So we must subtract 62 and add 26 to correct the total. Step 4: Correct total marks = 1360 - 62 + 26. Step 5: Compute 1360 - 62 = 1298, then 1298 + 26 = 1324. Step 6: Correct average = 1324 / 40. Step 7: 1324 / 40 = 33.1. Verification / Alternative check: You can check that the correction made is a net reduction of 36 marks (because 62 minus 26 is 36). So the correct total should be 1360 - 36 = 1324, which agrees with the detailed calculation. Dividing by 40 gives 33.1, confirming the answer. Why Other Options Are Wrong: - Option 33.3: This corresponds to a total of 33.3 * 40 = 1332, which is 8 marks higher than the corrected total. - Option 33.5: This would imply a total of 1340, which does not match the proper correction. - Option 33.7: This corresponds to an even higher total, again inconsistent with the given error. Common Pitfalls: A common mistake is to directly average 34 and the difference between 62 and 26 or to treat the correction as affecting the number of students. Another frequent error is to forget that the difference between wrong and correct value must be applied to the total, not to the average directly. Always adjust the total first, then recompute the average. Final Answer: The corrected average marks of the class is 33.1.

Question 12

What is the average of the first 19 odd natural numbers?

Accepted Answer

Introduction / Context: This question tests your knowledge of a well known property of the sequence of odd natural numbers. Knowing the pattern of odd numbers helps you quickly compute their averages without listing all of them. Given Data / Assumptions: - We consider the first 19 odd natural numbers. - These are 1, 3, 5, ..., up to the 19th odd number. - We need to find their average. Concept / Approach: The nth odd natural number is given by 2n - 1. So the first 19 odd numbers are 1, 3, 5, ..., 37. A key property is that the average of the first n odd numbers is equal to n itself. This follows from the total sum formula for odd numbers and the definition of average. Step-by-Step Solution: Step 1: The nth odd number is 2n - 1, so the 19th odd number is 2 * 19 - 1 = 37. Step 2: The sum of the first n odd numbers is n^2. For n = 19, the sum is 19^2. Step 3: Compute 19^2 = 361. Step 4: There are 19 terms, so the average is sum divided by number of terms = 361 / 19. Step 5: Compute 361 / 19 = 19. Verification / Alternative check: You can quickly verify the pattern for smaller values. For n = 3, the first three odd numbers are 1, 3 and 5. Their sum is 9, and average is 9 / 3 = 3, which equals n. The same rule applies for n = 19, so the average must be 19. Why Other Options Are Wrong: - Option 9.5: This is less than the middle term of the sequence and cannot be the average of symmetric odd numbers. - Option 15.5: This is also too low relative to the spread from 1 to 37. - Option 38: This is larger than every term except the last one and cannot be an average of the whole set. Common Pitfalls: Some students try to list and add all 19 numbers, increasing the chance of calculation error. Others misunderstand the pattern and think the average is the middle term in every case, without verifying. In fact, for an arithmetic progression, the average equals the middle term, and the odd numbers form an arithmetic progression. Recognising this gives a quick route to the answer. Final Answer: The average of the first 19 odd natural numbers is 19.

Question 13

The average weight of P, Q and R is 47 kg. The average weight of P and Q together is 32.5 kg, and the average weight of Q and R together is 48.5 kg. What is the weight of Q in kilograms?

Accepted Answer

Introduction / Context: This problem illustrates how to use multiple average equations to find individual values. You are given averages for triples and pairs of people and must extract the weight of a single person. Given Data / Assumptions: - The average weight of P, Q and R is 47 kg. - The average weight of P and Q is 32.5 kg. - The average weight of Q and R is 48.5 kg. - We want the weight of Q alone. Concept / Approach: We convert each average into a total sum equation. From the first average we get P + Q + R. From the second we get P + Q, and from the third we get Q + R. Using these, we can find R and then Q by simple subtraction. Step-by-Step Solution: Step 1: From the average of P, Q and R: (P + Q + R) / 3 = 47, so P + Q + R = 47 * 3 = 141. Step 2: From the average of P and Q: (P + Q) / 2 = 32.5, so P + Q = 32.5 * 2 = 65. Step 3: From the average of Q and R: (Q + R) / 2 = 48.5, so Q + R = 48.5 * 2 = 97. Step 4: Use P + Q + R = 141 and P + Q = 65 to find R: R = 141 - 65 = 76. Step 5: Use Q + R = 97 and R = 76 to get Q: Q = 97 - 76. Step 6: Compute Q = 21 kg. Verification / Alternative check: Check all equations. With Q = 21 and R = 76, Q + R = 97, so their average is 48.5. To find P, use P + Q = 65, giving P = 44. Then P + Q + R = 44 + 21 + 76 = 141, whose average is 47. All conditions are satisfied. Why Other Options Are Wrong: - Option 25: If Q = 25, the equations for P + Q and Q + R would not match the given averages. - Option 29: This would force inconsistent values for P and R compared with the given combined totals. - Option 33: Again, this value would break at least one of the three average relationships provided in the statement. Common Pitfalls: Some learners try to solve directly with three variables without first using the simplest subtractions. Others mis-multiply the averages when converting to totals. Always check units, multiply correctly, and use subtraction carefully to isolate each variable step by step. Final Answer: The weight of Q is 21 kg.

Question 14

The average marks of 40 students in an examination is 25. Later it is found that the marks of one student were wrongly entered as 73 instead of 37. What is the correct average marks for the class?

Accepted Answer

Introduction / Context: This is another example of correcting an average when one observation has been entered incorrectly. It reinforces the method of adjusting the total sum first and then recomputing the mean. Given Data / Assumptions: - There are 40 students in the class. - The initially reported average marks is 25. - One student's marks were wrongly recorded as 73 instead of the correct value 37. - We must find the corrected class average. Concept / Approach: As before, the average equals total marks divided by the number of students. We begin with the incorrect total implied by the wrong average, then adjust this total by removing the incorrect value and adding the correct one. Finally, we divide the corrected total by 40. Step-by-Step Solution: Step 1: Incorrect total marks = 25 * 40. Step 2: Compute 25 * 40 = 1000 marks. Step 3: The wrong entry is 73, the correct value is 37, so we must subtract 73 and add 37 to correct the total. Step 4: Correct total = 1000 - 73 + 37. Step 5: Compute 1000 - 73 = 927, then 927 + 37 = 964. Step 6: Correct average = 964 / 40. Step 7: 964 / 40 = 24.1 marks. Verification / Alternative check: The net error in the total is 73 - 37 = 36 marks too high. So we can also compute the corrected total as 1000 - 36 = 964 directly. Dividing 964 by 40 again gives 24.1, confirming the answer. Why Other Options Are Wrong: - Option 24.3: This would correspond to a total of 972 marks, which does not match the corrected sum 964. - Option 24.5: This implies a total of 980 marks, again inconsistent. - Option 24.7: This would imply 988 marks, which is too high given the nature of the correction. Common Pitfalls: Students sometimes try to average 25 and the corrected mark or directly adjust the average by the error divided by the number of students but with incorrect sign. The safe method is always to reconstruct the total, correct it, and then divide by the class strength. Final Answer: The correct average marks of the class is 24.1.

Question 15

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

Accepted Answer

Introduction / Context: This question brings together three basic measures of central tendency: mean, median and mode. You must compute each of them for the same data set and then find their sum. Given Data / Assumptions: - The data points are 6, 3, 8, 4, 3, 11, 7, 4, 5 and 4. - We must calculate the mode (most frequent value), median (middle value when sorted) and mean (arithmetic average). - Finally, we must add these three measures together. Concept / Approach: To find the mode, we count how many times each value appears. To find the median, we sort the data and look at the middle value or the average of the two middle values. For the mean, we divide the total sum by the number of observations. Then we simply add the three results. Step-by-Step Solution: Step 1: Sort the data in ascending order: 3, 3, 4, 4, 4, 5, 6, 7, 8, 11. Step 2: Mode: 4 appears three times, more often than any other number, so the mode is 4. Step 3: Median: There are 10 values, so the median is the average of the 5th and 6th values in the sorted list. Step 4: The 5th value is 4 and the 6th value is 5, so median = (4 + 5) / 2 = 4.5. Step 5: Mean: First find the sum of all values: 6 + 3 + 8 + 4 + 3 + 11 + 7 + 4 + 5 + 4 = 55. Step 6: Mean = total sum / number of observations = 55 / 10 = 5.5. Step 7: Sum of mode, median and mean = 4 + 4.5 + 5.5 = 14. Verification / Alternative check: Double check each piece: the frequency count clearly shows 4 as the most common value, the sorted positions for median are correct for 10 elements, and the sum 55 is straightforward to verify by grouping (for example, (6 + 4) + (3 + 7) + (8 + 4) + (3 + 5) + (11 + 4) = 10 + 10 + 12 + 8 + 15 = 55). Why Other Options Are Wrong: - Option 13.5: This would be the result of a small arithmetic slip when summing 4, 4.5 and 5.5. - Option 13: This is even lower, inconsistent with the correctly computed values. - Option 14.5: This would require at least one of the mean, median or mode to be larger than we have calculated. Common Pitfalls: Students sometimes confuse the positions for the median when the number of terms is even, or misidentify the mode when several values repeat. Another frequent mistake is in the total sum calculation, which affects the mean. Carefully listing the sorted data and checking each computation reduces such errors. Final Answer: The sum of the mode, median and mean for the given data is 14.

Question 16

A certain amount of money is fully distributed among A, B and C in such a way that A receives 1/2 of the total amount, B receives 1/3 of the total amount and C receives Rs 1200. How much money does A receive?

Accepted Answer

Introduction / Context: This question involves sharing an amount of money according to given fractional parts. It tests your ability to work with fractions of a whole and to reconstruct the total from one person's share. Given Data / Assumptions: - The total amount of money is completely distributed among A, B and C. - A receives 1/2 of the total amount. - B receives 1/3 of the total amount. - C receives Rs 1200. - We must find A's share. Concept / Approach: Let the total amount be T. Then A receives T / 2, B receives T / 3 and C receives the remaining amount. Since the full amount is distributed, the sum of their shares equals T. This allows us to express C's share as a fraction of T, then equate it to Rs 1200 and solve for T. After that we compute A's share as T / 2. Step-by-Step Solution: Step 1: Let the total amount be T. Step 2: A's share = T / 2, B's share = T / 3. Step 3: Total distributed = T / 2 + T / 3 + C's share = T. Step 4: Combine A and B shares: T / 2 + T / 3 = (3T + 2T) / 6 = 5T / 6. Step 5: Therefore, C's share is T - 5T / 6 = T / 6. Step 6: We are told C receives Rs 1200, so T / 6 = 1200. Step 7: Multiply both sides by 6: T = 1200 * 6 = 7200. Step 8: A's share = T / 2 = 7200 / 2 = 3600. Verification / Alternative check: Check B's share as well: B receives 1/3 of 7200, which is 2400. A receives 3600, B receives 2400, and C receives 1200. Their sum is 3600 + 2400 + 1200 = 7200, which matches the total T, so the distribution is consistent. Why Other Options Are Wrong: - Option 4000: This would give a total greater than 7200 when combined with correct shares for B and C. - Option 1600: This is less than C's share, which contradicts the fact that A receives the largest fraction. - Option 1800: This would imply a much smaller total, inconsistent with C receiving 1200 as only one sixth of the total. Common Pitfalls: Some learners add the fractions 1/2 and 1/3 incorrectly or forget that the remaining share belongs entirely to C. Others do not recognise that the full distribution implies 1/2 + 1/3 + C's fraction = 1. Always convert the known shares to fractions of the total and ensure they sum to 1. Final Answer: A receives 3600 rupees.

Question 17

In an aptitude question on averages, the average weight of Somdev, Gurdeep and Ritu is 85 kg. If the average weight of Somdev and Gurdeep is 79 kg and the average weight of Gurdeep and Ritu is 73 kg, what is the weight of Gurdeep in kilograms?

Accepted Answer

Introduction / Context:This quantitative aptitude question tests the concept of averages and how to form and solve simple linear equations from average information. Three friends, Somdev, Gurdeep and Ritu, have a known overall average weight and you are also given the averages of two person groups. From this overlapping data, you must isolate and compute the individual weight of Gurdeep. Given Data / Assumptions:- Average weight of Somdev, Gurdeep and Ritu together is 85 kg.- Average weight of Somdev and Gurdeep is 79 kg.- Average weight of Gurdeep and Ritu is 73 kg.- All weights are measured in kilograms.- We need to find the weight of Gurdeep only. Concept / Approach:The key idea is that average multiplied by the number of persons gives the total weight. By converting each average into a total and introducing variables for the unknown weights, we can form a small system of linear equations. Solving this system step by step will give us the individual weight of Gurdeep. This method is standard in aptitude exams whenever overlapping group averages are given. Step-by-Step Solution:Step 1: Let the weights of Somdev, Gurdeep and Ritu be S, G and R kg respectively.Step 2: From the first statement, (S + G + R) / 3 = 85, so S + G + R = 3 * 85 = 255.Step 3: From the second statement, (S + G) / 2 = 79, so S + G = 2 * 79 = 158.Step 4: From the third statement, (G + R) / 2 = 73, so G + R = 2 * 73 = 146.Step 5: Use S + G + R = 255 and S + G = 158 to get R = 255 - 158 = 97.Step 6: Now substitute R into G + R = 146, so G + 97 = 146 which gives G = 146 - 97 = 49.Step 7: Therefore, Gurdeep has a weight of 49 kg. Verification / Alternative check:Check the averages using S + G = 158 and R = 97. Then S + G + R = 158 + 97 = 255 and 255 / 3 = 85, which matches the given average of 85 kg. Next, G + R = 49 + 97 = 146 and 146 / 2 = 73, which matches the given average for Gurdeep and Ritu. Finally, S + G = 158 so their average is 158 / 2 = 79, matching the second given condition. All statements are satisfied, confirming that Gurdeep must weigh 49 kg. Why Other Options Are Wrong:Weights such as 78 kg, 72 kg or 90 kg do not satisfy all three average relations when substituted into the equations. With those values, at least one of the calculated group averages differs from the values provided in the question. Therefore, they cannot represent Gurdeep weight in this scenario. Common Pitfalls:Some learners try to average the given averages directly, which does not work for overlapping groups. Others forget to convert averages into totals or miscalculate the multiplication by 2 or 3. Writing clean equations for S + G + R, S + G and G + R and carefully substituting is the safest way to reach the correct solution. Final Answer:The weight of Gurdeep is 49 kg.

Question 18

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

Accepted Answer

Introduction / Context:This is a classic average and total expenditure problem often asked in banking and management entrance tests. A hostel increases its number of students and sees that the total food expenditure rises, but the average expenditure per student falls. You must translate this situation into equations and determine the original total expenditure on food per day. Given Data / Assumptions:- Initial number of students in the hostel = 35.- When 7 more students join, the total number becomes 42.- Total food expenditure increases by Rs. 42 per day after the increase in strength.- Average expenditure per student per day is reduced by Rs. 1 after the increase.- We must find the initial total food expenditure per day in rupees. Concept / Approach:The average expenditure per student is defined as total expenditure divided by number of students. If we let the initial average be x rupees per student per day, then the initial total is 35x. After the increase, the new average is (x - 1) and the new total is 42(x - 1). The increase in total expenditure is equal to the difference between these totals, and it is given as Rs. 42. Setting up an equation from this information allows us to solve for x and then find the original total expenditure. Step-by-Step Solution:Step 1: Let the initial average expenditure per student per day be x rupees.Step 2: Initial total expenditure per day = 35 * x.Step 3: After 7 more students join, total students = 42 and the new average becomes x - 1.Step 4: New total expenditure per day = 42 * (x - 1) = 42x - 42.Step 5: Given that the total expenditure increases by Rs. 42, we have 42x - 42 - 35x = 42.Step 6: Simplify the left side: 7x - 42 = 42.Step 7: Add 42 to both sides: 7x = 84, so x = 84 / 7 = 12.Step 8: Initial total expenditure per day = 35 * 12 = Rs. 420. Verification / Alternative check:If the initial average is Rs. 12, initial total expenditure is 35 * 12 = Rs. 420. After 7 students join, total students = 42 and new average = 11 rupees per student per day. The new total expenditure is 42 * 11 = Rs. 462. The increase in total expenditure is 462 - 420 = Rs. 42, and the change in average is 12 - 11 = Rs. 1, both of which match the conditions in the problem. This confirms that the answer is correct. Why Other Options Are Wrong:If the initial total were Rs. 432, 442 or 400, the corresponding averages per student would not lead to a new total that is exactly Rs. 42 higher while also reducing the average by Rs. 1. Substituting these totals into the scenario leads to inconsistent averages or incorrect increases in total expenditure. Only Rs. 420 satisfies all given conditions. Common Pitfalls:Many students confuse the change in average with the change in total and try to subtract 1 directly from the total instead of from the per student value. Others forget that the number of students changes from 35 to 42 and therefore use the wrong multiplier when computing new expenditure. Always start with a variable for the average, compute both totals, and then use the given increase to construct a correct equation. Final Answer:The initial total expenditure on food per day was Rs. 420.

Question 19

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

Accepted Answer

Introduction / Context:This is a standard average age problem used in aptitude exams. A sports team has a known average age for its players. When the coach is added to the group, the average age increases by a fixed amount. From this new information you are asked to calculate the coach age. The question checks your comfort with converting averages to totals and then working backward to find an unknown value. Given Data / Assumptions:- Number of players in the team = 12.- Average age of the 12 players = 23 years.- When the coach is included, total members = 13.- New average age including the coach = 23 + 2 = 25 years.- We must find the age of the coach in years. Concept / Approach:The fundamental concept is that average age multiplied by the number of persons gives the total of all ages. First compute the total age of the 12 players. Then compute the total age of the 12 players plus the coach using the new average. The difference between these two totals must be the coach age. This simple method avoids unnecessary algebra and works very well in age based questions. Step-by-Step Solution:Step 1: Total age of the 12 players = average * number of players = 23 * 12.Step 2: Compute this value: 23 * 12 = 276 years.Step 3: After including the coach, there are 13 persons and the average age is 25 years.Step 4: Total age of players plus coach = 25 * 13.Step 5: Compute this value: 25 * 13 = 325 years.Step 6: Coach age = combined total age minus total age of players = 325 - 276.Step 7: Therefore, coach age = 49 years. Verification / Alternative check:To verify, consider the group with the coach. If the coach is 49 and the players together are 276 years, then the total age is 276 + 49 = 325 years. Dividing by 13 gives an average of 325 / 13 = 25 years, which matches the stated new average. The original average of 23 years for 12 players is already satisfied by the total 276, so all the given conditions are consistent with a coach age of 49 years. Why Other Options Are Wrong:If the coach were 41, 47 or 51 years old, then the total age for 13 persons would be different and the resulting average would not be exactly 25 years. You can quickly check by adding each option to 276 and dividing by 13. None of those values except 49 yields an average of 25, so they cannot be correct. Common Pitfalls:Some learners mistakenly increase the total age by 2 years instead of understanding that the average increased by 2 across all 13 members. Others forget to update the number of persons from 12 to 13 when computing the new total. Always remember that an increase in average across a larger group corresponds to a much larger increase in total, which must be handled by multiplication. Final Answer:The age of the coach is 49 years.

Question 20

The average of 50 consecutive natural numbers is x. If the next four consecutive natural numbers are also included, what will be the new average in terms of x?

Accepted Answer

Introduction / Context:This question tests understanding of averages for consecutive natural numbers and how the average changes when more consecutive terms are added. You are told that the average of 50 consecutive natural numbers is x and you must find the new average when the next four consecutive numbers are included. Because the numbers are consecutive, their pattern is very regular and can be handled with simple algebra. Given Data / Assumptions:- There are 50 consecutive natural numbers.- The average of these 50 numbers is x.- Four more consecutive natural numbers immediately after this block are included, making a total of 54 numbers.- All numbers are consecutive and belong to the same arithmetic progression with common difference 1.- We need the new average expressed in terms of x. Concept / Approach:The average of consecutive natural numbers is equal to the middle value of the sequence. For an even count of numbers, the average is the mean of the two central numbers. When you extend a block of consecutive numbers by adding more terms at the upper end, the new average will shift toward the newer, larger values. A direct algebraic method starts by expressing the original average and then recomputing the sum and average after adding four more numbers, and then comparing it with x. Step-by-Step Solution:Step 1: Let the first of the 50 consecutive natural numbers be n. Then the numbers are n, n + 1, n + 2, ..., n + 49.Step 2: The average of these 50 numbers is x, given by x = (first + last) / 2 = (n + (n + 49)) / 2.Step 3: Simplify this: x = (2n + 49) / 2.Step 4: The next four consecutive numbers are n + 50, n + 51, n + 52 and n + 53.Step 5: The sum of the original 50 numbers is 50 * x.Step 6: The sum of the four new numbers is (n + 50) + (n + 51) + (n + 52) + (n + 53) = 4n + 206.Step 7: New total sum for 54 numbers = 50x + 4n + 206.Step 8: Use x = (2n + 49) / 2 to replace 4n + 206. From x, we get 2x = 2n + 49, so 4n = 2 * (2n) = 4x - 98. Therefore 4n + 206 = 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 numerical example. Suppose the 50 consecutive numbers are 1 to 50. Their average is (1 + 50) / 2 = 25.5, so x = 25.5. If we include the next four numbers 51, 52, 53 and 54, we now have numbers 1 to 54. The new average is (1 + 54) / 2 = 27.5. The change in average is 27.5 - 25.5 = 2, which matches x + 2. This confirms that the algebraic result is correct. Why Other Options Are Wrong:The average cannot increase by only 1 because four larger numbers are added to the high end of the range, shifting the average more. It also cannot increase by 4 or by x / 54, which would not match either the algebraic deduction or the numerical example. Therefore only x + 2 correctly describes the new average. Common Pitfalls:Students sometimes assume that the average remains the same for any extended set of consecutive numbers, which is not true when you only extend at one end. Others forget that the average is closely tied to the first and last terms in an arithmetic progression. Remember that for consecutive numbers, the average moves toward the side where you add more terms and the change can be calculated exactly using sums and counts. Final Answer:The new average when the next four numbers are included is x + 2.

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