Question 1

The mean of marks secured by 30 students in Division A of Class X is 67, by 55 students in Division B is 63, and by 40 students in Division C is 61. What is the mean of the marks of all the students of the three divisions of Class X together?

Accepted Answer

Introduction / Context: This is another weighted average problem involving three divisions of a class. Each division has a different number of students and a different mean mark. We are asked to find the overall mean for all students in Class X across these three divisions. The combined mean must account for the size of each division, and not just average the three given means directly. Given Data / Assumptions: Division A: 30 students, mean marks = 67. Division B: 55 students, mean marks = 63. Division C: 40 students, mean marks = 61. All students in Class X belong to one of these divisions. We need the combined mean marks of all students. Concept / Approach: For several groups with different sizes and means, the overall mean is found by computing total marks for each group (mean multiplied by group size), adding these totals, and dividing by the total number of students. This approach automatically weights each division by its size. It is crucial not to simply average 67, 63 and 61, as that would ignore the different numbers of students in each division. Step-by-Step Solution: Step 1: Compute the total marks for Division A. Total for A = 30 * 67 = 2010. Step 2: Compute the total marks for Division B. Total for B = 55 * 63. 55 * 63 = 55 * (60 + 3) = 3300 + 165 = 3465. Step 3: Compute the total marks for Division C. Total for C = 40 * 61 = 2440. Step 4: Find the grand total of marks. Grand total = 2010 + 3465 + 2440 = 7915. Step 5: Compute the total number of students. Total students = 30 + 55 + 40 = 125. Step 6: Calculate the combined mean. Combined mean = grand total / total students = 7915 / 125. 125 * 60 = 7500, remainder 415. 125 * 3 = 375, remainder 40, so 7915 / 125 = 63 + 40 / 125. 40 / 125 = 0.32, hence combined mean = 63.32. Verification / Alternative check: We can check approximate consistency. Division B has the largest number of students and a mean of 63, so the overall mean should be near 63, slightly pulled upward by Division A and downward by Division C. The value 63.32 fits this expectation. Why Other Options Are Wrong: Values such as 62.62, 61.92, 64.72 or 60.12 correspond to different total marks and would arise only if the group sizes or means were misused. For example, 62.62 would produce a grand total of 62.62 * 125, which is not equal to 7915. Common Pitfalls: One common mistake is to compute the simple average (67 + 63 + 61) / 3, ignoring the different numbers of students in each division. Another is to miscalculate one of the products, especially 55 * 63, which affects the final mean directly. Final Answer: The mean of marks of all the students of the three divisions together is 63.32.

Question 2

The mean of marks secured by 55 students in Division A of Class X is 58, by 45 students in Division B is 54, and by 75 students in Division C is 52. Find the mean of the marks of all the students of the three divisions of Class X together.

Accepted Answer

Introduction / Context: Once again, this problem asks for the combined mean of several groups with different sizes and means. We know the average marks in three divisions of a class and the number of students in each division. The objective is to compute the overall mean marks for all students from all three divisions. This reinforces the idea of weighted averages across multiple groups. Given Data / Assumptions: Division A: 55 students, mean marks = 58. Division B: 45 students, mean marks = 54. Division C: 75 students, mean marks = 52. We must find the combined mean for all students of Class X in these divisions. Concept / Approach: The overall mean is the total of all marks divided by the total number of students. For each division, the total marks are the product of the division mean and the division size. Once we find the total for each division, we add them together and divide by the combined number of students. This approach gives a properly weighted overall average that reflects the different group sizes. Step-by-Step Solution: Step 1: Compute total marks for Division A. Total for A = 55 * 58. 55 * 58 = 55 * (60 − 2) = 3300 − 110 = 3190. Step 2: Compute total marks for Division B. Total for B = 45 * 54. 45 * 54 = 45 * (50 + 4) = 2250 + 180 = 2430. Step 3: Compute total marks for Division C. Total for C = 75 * 52. 75 * 52 = 75 * (50 + 2) = 3750 + 150 = 3900. Step 4: Find the grand total of marks. Grand total = 3190 + 2430 + 3900 = 9520. Step 5: Compute total number of students. Total students = 55 + 45 + 75 = 175. Step 6: Calculate the combined mean. Combined mean = 9520 / 175. 175 * 50 = 8750, remainder 770. 175 * 4 = 700, remainder 70. 175 * 0.4 = 70, so 770 / 175 = 4.4. Therefore, 9520 / 175 = 54.4. Verification / Alternative check: Since the largest group is Division C with mean 52, but Division A has a higher mean of 58 and a substantial size, the overall mean should be between 52 and 58, most likely a little above the middle of the three means. The value 54.4 fits logically in this range and is consistent with the computed grand total. Why Other Options Are Wrong: A mean of 53 or 53.7 would correspond to lower total marks than 9520, while 55.8 or 56.1 would correspond to higher totals. These values do not match the exact ratio 9520 / 175 and hence are incorrect. Common Pitfalls: Some may incorrectly average 58, 54 and 52 directly, ignoring the different sizes of the divisions, which does not produce the correct result. Others might miscalculate one of the products, especially 55 * 58 or 75 * 52, leading to a wrong total and mean. Final Answer: The combined mean of marks of all students of the three divisions is 54.4.

Question 3

The mean of marks secured by 60 students in Division A of Class X is 64, by 40 students in Division B is 60, and by 60 students in Division C is 58. Find the mean of the marks of all the students of the three divisions of Class X together.

Accepted Answer

Introduction / Context: This final problem in the set is another weighted average question involving three divisions of a class. Each division has a known number of students and a known mean mark. The goal is to calculate the overall mean of marks when all three divisions are combined. Solving it reinforces the systematic method for combining multiple group means with different group sizes. Given Data / Assumptions: Division A: 60 students, mean marks = 64. Division B: 40 students, mean marks = 60. Division C: 60 students, mean marks = 58. We must determine the mean of the marks of all students across these three divisions. Concept / Approach: The combined mean is found by summing the total marks from each division and dividing by the total number of students. For each division, total marks are equal to the division mean multiplied by the number of students. Once we compute these totals for all three divisions, we add them together and divide by the total student count. This correctly weights each division in proportion to its size. Step-by-Step Solution: Step 1: Compute total marks for Division A. Total for A = 60 * 64. 60 * 64 = 3840. Step 2: Compute total marks for Division B. Total for B = 40 * 60 = 2400. Step 3: Compute total marks for Division C. Total for C = 60 * 58. 60 * 58 = 60 * (60 − 2) = 3600 − 120 = 3480. Step 4: Add these to find the grand total of marks. Grand total = 3840 + 2400 + 3480 = 9720. Step 5: Compute the total number of students. Total students = 60 + 40 + 60 = 160. Step 6: Find the combined mean. Combined mean = grand total / total students = 9720 / 160. Divide numerator and denominator by 10: 972 / 16. 16 * 60 = 960, remainder 12. 12 / 16 = 0.75, so 972 / 16 = 60.75. Verification / Alternative check: Division A and C each have 60 students, while Division B has 40. Their means 64, 60 and 58 suggest an overall mean between 58 and 64, leaning a bit toward the larger groups A and C. The value 60.75 fits naturally in this range and matches the calculation. Why Other Options Are Wrong: 60.05 and 59.35 are too low given that one large division has mean 64. 62.15 and 61.4 are higher than the correctly computed 60.75 and do not correspond to the ratio 9720 / 160. Common Pitfalls: One pitfall is to average 64, 60 and 58 directly without weighting, which does not account for the different group sizes. Another is to miscalculate one of the products, especially 60 * 58, resulting in an incorrect grand total and combined mean. Final Answer: The mean of marks of all the students of the three divisions together is 60.75.

Question 4

The average revenue of a company over 7 consecutive years is Rs 75 lakhs per year. If the average revenue for the first 4 years is Rs 70 lakhs and the average revenue for the last 4 years is Rs 82 lakhs, what is the revenue (in Rs lakhs) for the 4th…

Accepted Answer

Introduction / Context: This problem tests understanding of averages over consecutive years and how overlapping groups can be used to determine an unknown yearly revenue. Instead of handling each year separately, we use the relationship between the overall average, the average of the first subset of years, and the average of the last subset of years to find the revenue in the 4th year. Given Data / Assumptions: • There are 7 consecutive years of revenue.• Average revenue of all 7 years = Rs 75 lakhs per year.• Average revenue of the first 4 years = Rs 70 lakhs per year.• Average revenue of the last 4 years = Rs 82 lakhs per year.• Revenues are measured in lakhs of rupees. Concept / Approach: The total revenue over a period equals average revenue multiplied by the number of years. When we know averages for overlapping groups (first 4 years and last 4 years) and the overall group (7 years), we can add the sums of the two overlapping groups. This double counts the common year, which in this case is the 4th year. Using this fact, we can isolate and compute the revenue for that overlapping year. Step-by-Step Solution: Step 1: Let the revenues for the 7 consecutive years be R1, R2, R3, R4, R5, R6, R7.Step 2: Total revenue for all 7 years = 7 * 75 = 525 lakhs.Step 3: Total revenue for the first 4 years = (R1 + R2 + R3 + R4) = 4 * 70 = 280 lakhs.Step 4: Total revenue for the last 4 years = (R4 + R5 + R6 + R7) = 4 * 82 = 328 lakhs.Step 5: Add the two partial sums: (R1 + R2 + R3 + R4) + (R4 + R5 + R6 + R7) = 280 + 328 = 608 lakhs.Step 6: The left side equals (R1 + R2 + R3 + R4 + R5 + R6 + R7) + R4, which is the total of all 7 years plus R4 once more.Step 7: Total of all 7 years is 525 lakhs, so 525 + R4 = 608.Step 8: Therefore, R4 = 608 - 525 = 83 lakhs. Verification / Alternative check: We can verify by constructing a simple hypothetical distribution that respects the sums. Once we fix R4 = 83 lakhs, we can choose values for other years so that the first 4 years sum to 280 lakhs and the last 4 years sum to 328 lakhs, and the overall sum is 525 lakhs. The consistency of these totals confirms that R4 = 83 lakhs is correct. Why Other Options Are Wrong: Rs 85 lakhs: This would make the combined sum of first and last 4 years too large compared with the seven year total.Rs 81 lakhs: This is obtained if arithmetic steps are reversed or if the double counting logic is misapplied.Rs 79 lakhs: This is too low and does not satisfy both partial averages simultaneously. Common Pitfalls: Students often forget that the 4th year is included in both the first 4 years and the last 4 years, leading to double counting mistakes. Another common mistake is to average the two partial averages directly without considering the number of years in each subset. Care must be taken to convert averages to totals and then use algebra to isolate the unknown year. Final Answer: The revenue for the 4th year is Rs 83 lakhs.

Question 5

In a class of 66 students there are 33 girls and 33 boys. The average weight of the girls is 61 kg and the average weight of the entire class is 66 kg. What is the average weight (in kg) of the boys in the class?

Accepted Answer

Introduction / Context: This question checks understanding of how to work with averages when a group is split into two equal parts. You are given the average weight of the whole class and the average weight of the girls. Using this information, you have to determine the average weight of the boys in the class, which involves careful manipulation of totals derived from averages. Given Data / Assumptions: • Total number of students = 66.• Number of girls = 33, so number of boys = 33.• Average weight of girls = 61 kg.• Average weight of the full class = 66 kg.• We assume all given averages are correct and based on the same unit of kilograms. Concept / Approach: The key concept is that average = total sum of weights divided by number of students. If we know the average of the whole class and of a subgroup (girls), then we can find the sum of weights of the whole class and the girls, and by subtraction we get the total weight of the boys. Finally, dividing by the number of boys gives the average weight of the boys. Step-by-Step Solution: Step 1: Total weight of all 66 students = 66 * 66 = 4356 kg.Step 2: Total weight of the 33 girls = 33 * 61 = 2013 kg.Step 3: Total weight of the 33 boys = total weight of class minus total weight of girls = 4356 - 2013 = 2343 kg.Step 4: Average weight of the boys = total weight of boys divided by number of boys = 2343 / 33.Step 5: Compute 2343 / 33. Since 33 * 70 = 2310 and 2343 - 2310 = 33, we have 2343 = 33 * 71.Step 6: Therefore, average weight of boys = 71 kg. Verification / Alternative check: If the average of boys is 71 kg and average of girls is 61 kg, then the combined average with equal numbers of boys and girls is (61 + 71) / 2 = 132 / 2 = 66 kg, which matches the given class average. This consistency confirms that our answer is correct. Why Other Options Are Wrong: 72: This would make the combined average greater than 66 kg, which contradicts the given overall average.73: This is even larger and increases the overall average far beyond 66 kg.69: This is too low and would pull the combined average below 66 kg when mixed with 61 kg for girls. Common Pitfalls: One common mistake is to simply take the difference of averages or average the given averages directly, without considering group sizes. Another error is to divide by the wrong number of boys. Always convert averages into totals and then back to averages after performing the required subtractions. Final Answer: The average weight of the boys is 71 kg.

Question 6

The average of four consecutive odd natural numbers is 64. These four odd numbers follow one after another in increasing order. What is the value of the largest of these four odd numbers?

Accepted Answer

Introduction / Context: This problem tests understanding of averages and sequences of consecutive odd numbers. Consecutive odd numbers form an arithmetic progression with a common difference of 2. Knowing the average allows us to find the middle of this progression and then determine all the numbers in the set, including the largest one. Given Data / Assumptions: • There are 4 consecutive odd natural numbers.• The average of these 4 numbers is 64.• The numbers are increasing and each differs from the next by 2.• We need to find the largest of these four odd numbers. Concept / Approach: For any set of equally spaced numbers (an arithmetic progression), the average is equal to the middle value. For 4 consecutive odd numbers, the average equals the mean of the two central numbers and also equal to the first number plus 3. Alternatively, we can represent the numbers algebraically and use the average formula to solve for the first number, then compute the largest number. Step-by-Step Solution: Step 1: Let the four consecutive odd numbers be x, x + 2, x + 4, and x + 6.Step 2: Their average is given as 64.Step 3: Average of these numbers = (x + (x + 2) + (x + 4) + (x + 6)) / 4.Step 4: Simplify the numerator: x + x + 2 + x + 4 + x + 6 = 4x + 12.Step 5: So average = (4x + 12) / 4 = x + 3.Step 6: Set this equal to 64: x + 3 = 64.Step 7: Solve for x: x = 64 - 3 = 61.Step 8: The largest number is x + 6 = 61 + 6 = 67. Verification / Alternative check: The four odd numbers are 61, 63, 65, and 67. Their sum is 61 + 63 + 65 + 67 = 256. The average is 256 / 4 = 64, which matches the given value. This confirms that 67 is indeed the largest number. Why Other Options Are Wrong: 65: This would correspond to a lower average than 64 when taken with three smaller odd numbers.69: This would give an average greater than 64 if used as the largest in the sequence.71: This is even larger and cannot match the average of 64 for four consecutive odd numbers. Common Pitfalls: Students sometimes confuse even and odd sequences or assume that the average equals one of the given numbers without checking the arithmetic progression. Others may forget that the step size is 2 for odd numbers, not 1. Using algebra or the property that the average of equally spaced numbers is the central value helps avoid these errors. Final Answer: The largest number is 67.

Question 7

In a school, 391 boys and 323 girls have been divided into the largest possible number of equal classes, such that each class has the same number of boys and the same number of girls. How many such equal classes are formed in the school?

Accepted Answer

Introduction / Context: This question combines the idea of grouping with the concept of highest common factor. The school wants to divide 391 boys and 323 girls into the largest possible number of equal classes where every class has the same number of boys and the same number of girls. The phrase largest possible number of equal classes hints strongly at finding a divisor common to both groups. Given Data / Assumptions: • Number of boys = 391.• Number of girls = 323.• All classes must be equal in composition.• Each class must have the same number of boys as every other class.• Each class must have the same number of girls as every other class. Concept / Approach: If we form k equal classes, then the number of boys per class and the number of girls per class must both be integers. Therefore, k must be a common divisor of both 391 and 323. To maximize the number of classes, we must find the greatest common divisor (GCD) of 391 and 323. Once we know the GCD, that value gives the largest possible number of such equal classes. Step-by-Step Solution: Step 1: Let the number of classes be k.Step 2: Then boys per class = 391 / k and girls per class = 323 / k, both must be whole numbers.Step 3: Hence, k must divide both 391 and 323.Step 4: Compute the GCD of 391 and 323.Step 5: Find 391 - 323 = 68.Step 6: Now compute GCD(323, 68). Divide 323 by 68: 68 * 4 = 272, remainder 323 - 272 = 51.Step 7: Now find GCD(68, 51). We have 68 - 51 = 17.Step 8: Next, 51 divided by 17 is exact, so GCD(68, 51) = 17.Step 9: Therefore, GCD(391, 323) = 17, which is the maximum possible number of equal classes. Verification / Alternative check: If there are 17 classes, then boys per class = 391 / 17 = 23 boys and girls per class = 323 / 17 = 19 girls. Both are whole numbers, and any larger number of classes, such as 19 or 23 or 44, will not divide both numbers exactly. Hence 17 is indeed the largest possible number of classes. Why Other Options Are Wrong: 23: This divides 391 but does not divide 323 evenly.19: This divides 323 but not 391 without a remainder.44: This divides neither 391 nor 323 exactly, so equal classes cannot be formed. Common Pitfalls: Students sometimes mistakenly divide the total number of students instead of considering boys and girls separately, or they look for the least common multiple instead of the greatest common divisor. Remember that to divide each group evenly into the same number of classes, that number must be a common divisor of both group sizes, and to maximize classes we need the greatest such divisor. Final Answer: The number of equal classes formed is 17.

Question 8

The average of 9 observations was initially calculated as 35. Later, it was discovered that one of the observations was 81 but had been misread and recorded as 18. What is the correct average of the 9 observations after this error is corrected?

Accepted Answer

Introduction / Context: This problem illustrates how a single incorrect value can affect the average of a data set. You are given the original average, then told that one observation was misread. By adjusting the total sum to correct the mistake, you can compute the new, accurate average. Given Data / Assumptions: • Number of observations = 9.• Initially calculated average = 35.• One observation was actually 81.• It was misread and used in the calculation as 18.• All other observations were recorded correctly. Concept / Approach: The average is defined as total sum divided by the number of observations. When a value has been misread, the originally used total is incorrect. To correct the average, we first compute the initial total based on the wrong average, then adjust the total by removing the misread value and adding the correct one. Finally, we divide the corrected total by the same number of observations. Step-by-Step Solution: Step 1: Let S be the incorrect total sum used originally.Step 2: Since average = 35 for 9 observations, S = 9 * 35 = 315.Step 3: One value 81 was misread as 18, so 18 contributed to S instead of 81.Step 4: Correct total sum = S - misread value + correct value = 315 - 18 + 81.Step 5: Compute 315 - 18 = 297, then 297 + 81 = 378.Step 6: Correct average = corrected total sum / number of observations = 378 / 9.Step 7: Compute 378 / 9 = 42. Verification / Alternative check: We can reason that the difference between the correct observation and the misread observation is 81 - 18 = 63. So the correct total must be 63 more than the incorrect one. Hence the average must increase by 63 / 9 = 7. The original average was 35, so the corrected average is 35 + 7 = 42, which agrees with the detailed calculation. Why Other Options Are Wrong: 28: This would mean the average decreased, which is impossible because the corrected value is greater than the misread one.35: This is the original incorrect average before adjustment.45: This would require an increase larger than 7 per observation, which is not supported by the difference of 63 in the total. Common Pitfalls: Some learners forget to adjust the total by both subtracting the misread value and adding the correct one. Others mistakenly divide the difference directly by the new average or number of observations without adjusting the original total. Always recompute the total carefully and then divide by the number of data points. Final Answer: The correct average of the observations is 42.

Question 9

The mean of 100 observations was calculated as 40. Later, it was found that one observation had been misread as 83 instead of the correct value 53. What is the correct mean of the 100 observations after this error is corrected?

Accepted Answer

Introduction / Context: This question tests the ability to correct an average when one data value was recorded incorrectly. Such mistakes are common when working with large data sets, and the correct method is to adjust the total sum and then recompute the mean. Given Data / Assumptions: • Number of observations = 100.• Initially calculated mean = 40.• One observation was misread as 83.• The correct value for that observation is 53.• All other values are correctly recorded. Concept / Approach: The mean or average is total sum divided by the number of observations. If one observation is misread, the total sum is incorrect. To fix this, we subtract the incorrect value from the total and then add the correct value. The number of observations remains the same, so the new mean is computed using the corrected total divided by 100. Step-by-Step Solution: Step 1: Let S be the incorrect total based on the wrong mean.Step 2: Since the original mean is 40 for 100 observations, S = 100 * 40 = 4000.Step 3: The misread value is 83 but should have been 53.Step 4: Correct total = S - incorrect value + correct value = 4000 - 83 + 53.Step 5: Compute 4000 - 83 = 3917, then 3917 + 53 = 3970.Step 6: Correct mean = corrected total / number of observations = 3970 / 100.Step 7: 3970 / 100 = 39.7. Verification / Alternative check: The difference between the correct and incorrect values is 53 - 83 = -30. This means the correct total must be 30 less than the initially used total. Dividing this difference by 100 gives a change in mean of -30 / 100 = -0.3. The original mean was 40, so the corrected mean should be 40 - 0.3 = 39.7, which matches the computed result. Why Other Options Are Wrong: 39: This would imply a change of -1 in the mean, which is too large for a difference of 30 spread over 100 observations.40.3: This would require adding to the total, but the correct value is smaller than the misread value.42.7: This is far from the original mean and does not reflect the direction of change caused by the correction. Common Pitfalls: Students may wrongly add the difference instead of subtracting it or may forget that only one observation changed. Another common error is to divide the difference by the wrong number of observations. Always recompute the total sum carefully and then divide by the exact count of data points. Final Answer: The correct mean of the observations is 39.7.

Question 10

A student, by mistake, wrote 64 instead of 46 as one of the numbers while finding the average of 10 given numbers, and obtained the average as 50. What is the correct average of the 10 numbers after adjusting this mistake?

Accepted Answer

Introduction / Context: This question involves correcting an average when one of the numbers used in the calculation was wrong. Understanding how a single error affects the overall average is important in data interpretation and basic statistics. Given Data / Assumptions: • Number of values = 10.• Incorrect number used in the calculation = 64.• Correct number should have been = 46.• Average obtained using the wrong number = 50.• All other values are correctly recorded. Concept / Approach: The average of a group of numbers is the total sum divided by the number of numbers. When one number is incorrect, the total is off by the difference between the wrong and correct values. After adjusting the total, you divide by the same count to get the corrected average. Step-by-Step Solution: Step 1: Let S be the total sum based on the incorrect data.Step 2: Since the average with the mistake is 50, S = 10 * 50 = 500.Step 3: The incorrect value is 64 and the correct value is 46.Step 4: Correct total sum = S - incorrect value + correct value = 500 - 64 + 46.Step 5: Compute 500 - 64 = 436, then 436 + 46 = 482.Step 6: Correct average = corrected total sum / number of values = 482 / 10.Step 7: 482 / 10 = 48.2. Verification / Alternative check: The difference between the correct and incorrect value is 46 - 64 = -18. This means the true total is 18 less than the original total. The average must therefore decrease by 18 / 10 = 1.8. Subtracting 1.8 from the incorrect average of 50 gives 48.2, confirming our calculation. Why Other Options Are Wrong: 48: This would correspond to a total change of 20 instead of 18.48.1: This is slightly low and reflects incorrect handling of the decimal change.49: This is too high and does not account fully for the reduced total after correction. Common Pitfalls: Many students forget to adjust the sum by both removing the incorrect entry and adding the correct one. Others may incorrectly divide the difference or misplace the decimal point. Carefully tracking the change in total and dividing by the correct number of observations is essential. Final Answer: The correct average of the 10 numbers is 48.2.

Question 11

The average cost of 4 items in a shopping list is Rs 1,250 per item. If one more item costing Rs 2,000 is added to the list, what will be the new average cost (in Rs) per item for all 5 items together?

Accepted Answer

Introduction / Context: This question deals with updating an average when a new item is added to an existing group. It helps to understand how the inclusion of a more expensive or cheaper item affects the overall average cost of all items combined. Given Data / Assumptions: • Number of items initially = 4.• Average cost of these 4 items = Rs 1,250.• Cost of the new 5th item = Rs 2,000.• All costs are in rupees. Concept / Approach: To update an average when a new value is added, we first compute the total cost of the original items using the initial average. Then we add the cost of the new item to get the new total and divide by the new number of items to get the new average. Since the new item is more expensive than the original average, we expect the overall average to increase. Step-by-Step Solution: Step 1: Let the total cost of the initial 4 items be S.Step 2: Since the average cost is 1,250, S = 4 * 1,250.Step 3: Compute S = 5,000.Step 4: Add the cost of the 5th item, which is Rs 2,000.Step 5: New total cost = 5,000 + 2,000 = 7,000.Step 6: New number of items = 5.Step 7: New average cost = new total cost / number of items = 7,000 / 5.Step 8: 7,000 / 5 = 1,400. Verification / Alternative check: The new item costs Rs 750 more than the original average (2,000 minus 1,250). When you spread this extra 750 over 5 items, the average should increase by 750 / 5 = 150. So the new average should be 1,250 + 150 = 1,400, which matches our detailed calculation. Why Other Options Are Wrong: 1100: This is lower than the original average and cannot happen when a more expensive item is added.1500: This overestimates the impact of the costlier item.1250: This is the original average, before adding the more expensive fifth item. Common Pitfalls: Some students forget to multiply the original average by the number of items to find the total, or they mistakenly average the old average and the new value directly without considering the differing counts. Always convert averages into totals before combining or updating groups. Final Answer: The new average cost per item is Rs 1,400.

Question 12

The average marks of 10 students in an examination was calculated as 25. Later, it was found that the marks of one student had been wrongly entered as 34 instead of 43. What is the correct average of the marks of the 10 students after this correctio…

Accepted Answer

Introduction / Context: This problem focuses on correcting an average after discovering a data entry error. It is a common exam style question where a single wrong entry needs to be fixed and the new average recalculated correctly. Given Data / Assumptions: • Number of students = 10.• Initially calculated average marks = 25.• One student's marks were recorded as 34.• The correct marks of that student are 43.• All other marks are assumed correct. Concept / Approach: The average marks are the total of all marks divided by the number of students. If one mark is incorrect, we need to correct the total sum. This is done by subtracting the wrong value and adding the correct one. Then we divide by the same number of students to obtain the corrected average. Step-by-Step Solution: Step 1: Let S be the total marks based on the incorrect data.Step 2: Since the initial average is 25, S = 10 * 25 = 250.Step 3: Incorrect marks used = 34, correct marks = 43.Step 4: Correct total = S - incorrect value + correct value = 250 - 34 + 43.Step 5: Compute 250 - 34 = 216, then 216 + 43 = 259.Step 6: Correct average = corrected total / number of students = 259 / 10.Step 7: 259 / 10 = 25.9. Verification / Alternative check: The difference between the correct and incorrect marks is 43 - 34 = 9. Hence, the correct total marks should be 9 more than the original total. Therefore, the average should increase by 9 / 10 = 0.9 marks. Starting from the original average of 25, we get 25 + 0.9 = 25.9, which supports our detailed calculation. Why Other Options Are Wrong: 25.5: This assumes an increase of only 0.5 marks in the average, not supported by the 9 mark difference.24.1 and 24.5: These are lower than the original average, while the correction actually increases the total and the mean. Common Pitfalls: Students may forget that the total must be adjusted by the full difference between the correct and incorrect values. Another mistake is to change the number of students, which remains the same. Always recompute the total properly and divide by the unchanged count of observations. Final Answer: The correct average marks of the 10 students is 25.9.

Question 13

Three numbers have an average of 112. The first number is equal to one sixth of the sum of the other two numbers. What is the value of the first number?

Accepted Answer

Introduction / Context: This question combines averages with a simple algebraic condition relating three numbers. You are given the overall average of the three numbers and a relationship between the first number and the sum of the other two numbers. Using algebra, you can determine the actual value of the first number. Given Data / Assumptions: • There are three numbers.• Average of the three numbers = 112.• First number is equal to one sixth of the sum of the other two numbers.• All numbers are real numbers, typically expected to be positive in such aptitude questions. Concept / Approach: The average of three numbers equals their sum divided by three. Thus, knowing the average gives us their total. Let us denote the first number as a and the other two numbers as b and c. The relation a = (b + c) / 6 provides a second equation. Combining both expressions, we can solve for a directly using substitution. Step-by-Step Solution: Step 1: Let the three numbers be a, b, and c.Step 2: Given that their average is 112, so (a + b + c) / 3 = 112.Step 3: Multiply both sides by 3: a + b + c = 336.Step 4: The first number a is one sixth of the sum of the other two: a = (b + c) / 6.Step 5: From this, b + c = 6a.Step 6: Substitute b + c = 6a into the total sum: a + 6a = 336.Step 7: This gives 7a = 336.Step 8: Therefore, a = 336 / 7 = 48. Verification / Alternative check: Once we have a = 48, we can compute b + c = 6a = 6 * 48 = 288. Then a + b + c = 48 + 288 = 336, and the average is 336 / 3 = 112, matching the given average. Also, a is indeed one sixth of the sum of the other two, since 48 = 288 / 6. Everything is consistent, confirming the correctness of the solution. Why Other Options Are Wrong: 45, 30, 15: None of these values satisfy both conditions simultaneously. If you plug any of them into a = (b + c) / 6 and a + b + c = 336, you will not get consistent values for b and c. Common Pitfalls: Some students misinterpret the condition and write 6a = b + c incorrectly or forget to multiply the average by the number of values to get the total sum. Others may confuse average with one of the numbers. Correctly setting up the algebraic equations is the key step here. Final Answer: The first number is 48.

Question 14

The average weight of P, Q and R is 45 kg. The average weight of P and Q is 36.5 kg and the average weight of Q and R is 52 kg. What is the weight (in kg) of Q?

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Introduction / Context: This problem deals with averages of overlapping groups. You are given the average weight of three persons together and the averages of two different pairs. The goal is to find the weight of the common person in both pairs, which requires forming and solving linear equations. Given Data / Assumptions: • Persons are P, Q, and R.• Average weight of P, Q, and R together = 45 kg.• Average weight of P and Q together = 36.5 kg.• Average weight of Q and R together = 52 kg.• All weights are in kilograms. Concept / Approach: The average of a group multiplied by the number of members gives the total weight. Using this property, we can write equations for P + Q + R, P + Q, and Q + R. From these, we can isolate individual weights. The main idea is to subtract equations to find R and then back substitute to find Q. Step-by-Step Solution: Step 1: Let the weights be P, Q, and R.Step 2: From the given average of P, Q, R: (P + Q + R) / 3 = 45, so P + Q + R = 3 * 45 = 135.Step 3: Given average of P and Q: (P + Q) / 2 = 36.5, so P + Q = 2 * 36.5 = 73.Step 4: Given average of Q and R: (Q + R) / 2 = 52, so Q + R = 2 * 52 = 104.Step 5: From P + Q + R = 135 and P + Q = 73, subtract to find R: R = 135 - 73 = 62.Step 6: Use Q + R = 104 to find Q: Q + 62 = 104.Step 7: Therefore, Q = 104 - 62 = 42. Verification / Alternative check: Check the computed values. With Q = 42 and R = 62, then Q + R = 42 + 62 = 104, so the pair average is 104 / 2 = 52 as given. Also, P + Q = 73 gives P = 73 - 42 = 31. Now P + Q + R = 31 + 42 + 62 = 135, and 135 / 3 = 45, which matches the given average for all three. This confirms that Q = 42 kg is correct. Why Other Options Are Wrong: 44, 46, 48: Substituting any of these values for Q will break at least one of the given average relationships. The sums P + Q, Q + R, or P + Q + R will not match the required totals. Common Pitfalls: Students may treat the averages independently without converting them to sums, or they may overlook that Q appears in both pair averages. Another mistake is incorrect arithmetic when subtracting totals. Writing the equations clearly and checking each step prevents such errors. Final Answer: The weight of Q is 42 kg.

Question 15

The average age of 20 students in a group is 21 years. Two students leave the group and one new student joins, after which the average age becomes 20 years. If the age of one student who left the group is 26 years and the age of the student who join…

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Introduction / Context: This question explores how changes in group membership affect the average age. When some members leave and others join, the total sum of ages changes. Using averages before and after the change, we can determine the age of an unknown student who left the group. Given Data / Assumptions: • Initial number of students = 20.• Initial average age = 21 years.• Two students leave and one new student joins.• New number of students = 19.• New average age = 20 years.• Age of one student who left = 26 years.• Age of the student who joined = 20 years.• Age of the second student who left is unknown and needs to be found. Concept / Approach: Average age multiplied by the number of students gives the total sum of ages. We use the initial average to find the initial total sum. After two students leave and one joins, we use the new average to find the new total sum. By subtracting the known leaving and entering ages from the totals, we can isolate and compute the unknown age of the other student who left. Step-by-Step Solution: Step 1: Initial total age of 20 students = 20 * 21 = 420 years.Step 2: After two students leave and one joins, the group has 19 students with average age 20 years.Step 3: New total age of 19 students = 19 * 20 = 380 years.Step 4: Let the age of the second student who left be x years.Step 5: The new total age can also be written as: initial total age minus ages of two students who left plus age of new student.Step 6: So 380 = 420 - 26 - x + 20.Step 7: Simplify the right side: 420 - 26 = 394, then 394 + 20 = 414.Step 8: Therefore, 380 = 414 - x.Step 9: Rearranging, x = 414 - 380 = 34. Verification / Alternative check: Check with the found value x = 34. Total age of the two students who left = 26 + 34 = 60. So from the original total 420, after removing 60 we get 360. Adding the age of the new student (20) gives 380, which equals 19 * 20, matching the new average. This confirms that the age of the second student who left is 34 years. Why Other Options Are Wrong: 20, 22, 16: Substituting any of these values for x will not produce the correct new total of 380. The resulting average will differ from 20 years, so they do not satisfy the conditions. Common Pitfalls: Students sometimes confuse the direction of change in total age or forget that the group size changes from 20 to 19. Others may incorrectly subtract or add the wrong ages. Carefully organizing the equation for the new total age prevents these mistakes. Final Answer: The age of the other student who left the group is 34 years.

Question 16

Each member of a club contributes as many rupees and as many paise as the total number of members in the club. If the total contribution collected from all members is Rs 2525, how many members are there in the club?

Accepted Answer

Introduction / Context: This question involves understanding money in rupees and paise and setting up a quadratic equation. Each member contributes an amount that depends on the total number of members, and you are asked to find that number given the total contribution. Given Data / Assumptions: • Let the number of members in the club be n.• Each member contributes n rupees and n paise.• Total contribution from all members is Rs 2525.• We use 100 paise = 1 rupee. Concept / Approach: If there are n members, and each gives n rupees and n paise, then each member contributes n + n/100 rupees. The total contribution is n times this amount. This leads to a quadratic equation in n, which we solve to find the number of members. We also check that the solution is a positive integer that makes sense in the context of the problem. Step-by-Step Solution: Step 1: Let the number of members be n.Step 2: Each member contributes n rupees and n paise.Step 3: Convert n paise to rupees: n paise = n / 100 rupees.Step 4: So contribution per member in rupees = n + n / 100 = (100n + n) / 100 = 101n / 100.Step 5: Total contribution from n members = n * (101n / 100) rupees.Step 6: Given total contribution = Rs 2525, so n * (101n / 100) = 2525.Step 7: Multiply both sides by 100: 101n^2 = 2525 * 100 = 252500.Step 8: Divide both sides by 101: n^2 = 252500 / 101.Step 9: Note that 101 * 2500 = 252500, so n^2 = 2500.Step 10: Therefore, n = square root of 2500 = 50 (reject negative value). Verification / Alternative check: If n = 50, then each member contributes 50 rupees and 50 paise. In rupees, that is 50.50 or 50 + 50 / 100. Total contribution = 50 * 50.50 = 50 * 50.5 = 2525, which matches the given total. This confirms that n = 50 is correct. Why Other Options Are Wrong: 60, 45, 55: Substituting these values into the expression n * (101n / 100) does not yield 2525. The resulting totals differ from the given amount, so these cannot be the correct number of members. Common Pitfalls: Some students forget to convert paise into rupees or misinterpret the phrase as n rupees or n paise instead of both. Others may make algebraic mistakes when forming or solving the quadratic equation. Careful unit conversion and stepwise algebra help avoid errors. Final Answer: The number of members in the club is 50.

Question 17

The arithmetic mean (average) amount of savings of ten students is Rs 600. Three students have no savings, and each of the remaining seven students has at least Rs 250, including Nihar, who has exactly Rs 1300. Under these conditions, what is the la…

Accepted Answer

Introduction / Context: This problem is about averages and optimization with constraints. You know the overall average savings, some students have zero savings, others have at least a minimum amount, and one student has a fixed amount. You are asked to find the maximum amount that any one student could have while still satisfying all these conditions. Given Data / Assumptions: • There are 10 students in total.• Average savings of all 10 students = Rs 600.• Three students have no savings at all (Rs 0 each).• The remaining 7 students each have at least Rs 250.• Nihar is one of the seven and has exactly Rs 1300.• We want the maximum possible savings of any one student. Concept / Approach: The total savings of all students can be found from the average. To maximize the savings of one student, we minimize the savings of the remaining students subject to the given constraints. Since the three students already have zero, we keep them as they are. Among the remaining six students apart from the maximum saver, five should be set at the minimum allowed level (Rs 250) while Nihar is fixed at Rs 1300. The leftover amount then becomes the maximum possible for the remaining student. Step-by-Step Solution: Step 1: Total savings of all 10 students = 10 * 600 = Rs 6000.Step 2: Three students have zero savings, so they contribute Rs 0.Step 3: Out of the remaining 7 students, one is Nihar with Rs 1300. Let the student with maximum savings have M rupees.Step 4: The other 5 students each have at least Rs 250. To maximize M, set each of these 5 students at exactly Rs 250.Step 5: Total savings of these 5 students = 5 * 250 = Rs 1250.Step 6: Now total savings = 0 (three students) + 1300 (Nihar) + 1250 (five students) + M (one student) = 6000.Step 7: So 1300 + 1250 + M = 6000.Step 8: Add 1300 and 1250 to get 2550, so 2550 + M = 6000.Step 9: Therefore, M = 6000 - 2550 = Rs 3450. Verification / Alternative check: Check that M = 3450 satisfies all constraints. The seven students other than the three with zero have savings: 1300 (Nihar), 3450, and five times 250. Their total is 1300 + 3450 + 1250 = 6000, which matches the overall total. Each of the seven has at least 250, the average is 600, and the three zeros are included. Thus the conditions are satisfied. Why Other Options Are Wrong: 3250: This is less than the value obtained after minimizing the others, so it is not the maximum possible.3650 and 3850: These are too large; if any student had these amounts while others met the constraints, the total would exceed Rs 6000. Common Pitfalls: Students may forget that to maximize one value, the other variable values must be minimized according to the constraints. Another mistake is ignoring the fixed amount for Nihar or wrongly changing the zero savings for the three students. Careful organization of the constraints leads to the correct maximum value. Final Answer: The largest possible savings any one student could have is Rs 3450.

Question 18

A batsman has a certain average of runs for 12 innings. In the 13th inning he scores 96 runs, and his average runs per inning increases by 5 runs. What will be his new batting average after the 13th inning?

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Introduction / Context: This problem is about averages in the context of cricket scores. When a batsman plays another inning and scores runs significantly higher than his current average, his average increases. By relating the original and new totals through the change in average, we can find the new average directly. Given Data / Assumptions: • The batsman has played 12 innings initially.• His average after these 12 innings is some unknown value A.• In the 13th inning he scores 96 runs.• After the 13th inning, his average increases by 5 runs, becoming A + 5.• We need to find the value of A + 5, the new average. Concept / Approach: The total runs scored in 12 innings is 12A. After scoring 96 in the 13th inning, the total becomes 12A + 96. This new total divided by 13 equals the new average A + 5. This gives an equation in A which can be solved algebraically. Once A is known, A + 5 gives the required average after 13 innings. Step-by-Step Solution: Step 1: Let the original average after 12 innings be A runs.Step 2: Total runs scored in 12 innings = 12A.Step 3: In the 13th inning he scores 96 runs, so new total runs = 12A + 96.Step 4: New average after 13 innings is given as A + 5.Step 5: Therefore, (12A + 96) / 13 = A + 5.Step 6: Multiply both sides by 13: 12A + 96 = 13(A + 5).Step 7: Expand right side: 13(A + 5) = 13A + 65.Step 8: So 12A + 96 = 13A + 65.Step 9: Rearranging, 96 - 65 = 13A - 12A gives 31 = A.Step 10: New average = A + 5 = 31 + 5 = 36. Verification / Alternative check: With A = 31, total after 12 innings is 12 * 31 = 372. Adding the 13th inning score, total runs = 372 + 96 = 468. Average after 13 innings = 468 / 13 = 36, which is indeed 5 more than 31. So the result is consistent with the problem statement. Why Other Options Are Wrong: 28, 32, 42: These do not produce a 5 run increase from a consistent original average when 96 is added as the 13th inning score. Substituting them back into the equation leads to contradictions. Common Pitfalls: Students may incorrectly assume the original average or forget that averages depend on total runs and total innings. Another common error is to equate the 13th inning score directly with the increase in average without forming the algebraic equation. Using the proper relation between old total, new total, and new average avoids these issues. Final Answer: The batsman's average after the 13th inning is 36 runs per inning.

Question 19

A team of 8 persons takes part in a shooting competition. The best marksman in the team scores 85 points. If he had scored 92 points instead, the average score of the team would have been 84 points. What was the actual total number of points scored …

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Introduction / Context: This problem involves averages and a hypothetical change in one person's score. You are told what the team average would have been if the best marksman had scored slightly more. Using this, you can deduce the team's actual total score before that hypothetical change. Given Data / Assumptions: • Number of team members = 8.• Best marksman's actual score = 85 points.• If he had scored 92 points, the team average would have been 84 points.• We are asked to find the actual total points scored by the team. Concept / Approach: The key idea is to use the hypothetical scenario to compute the team total if the best marksman had scored 92. Then we recognize that this hypothetical total is exactly 7 points more than the real total, because the only difference is his score changing from 85 to 92. Subtracting 7 from the hypothetical total gives the actual total score. Step-by-Step Solution: Step 1: Let T be the actual total score of the team with the best marksman scoring 85.Step 2: If he had scored 92 instead of 85, his score would increase by 7 points.Step 3: Then the hypothetical total score would be T + 7.Step 4: In that hypothetical case, the average score per person would be 84.Step 5: Therefore, (T + 7) / 8 = 84.Step 6: Multiply both sides by 8: T + 7 = 84 * 8.Step 7: Compute 84 * 8 = 672.Step 8: So T + 7 = 672, which gives T = 672 - 7 = 665. Verification / Alternative check: With T = 665, the original average is 665 / 8 = 83.125, which is slightly below 84. If we increase only one person's score by 7 points, new total becomes 665 + 7 = 672. New average = 672 / 8 = 84, exactly as stated in the problem. Thus, the total score of 665 is consistent with the hypothetical condition. Why Other Options Are Wrong: 672: This is the hypothetical total after adding 7 points, not the actual total.645, 588: These totals do not satisfy the average condition when 7 is added and divided by 8. The average does not become 84 in those cases. Common Pitfalls: Students sometimes confuse the actual and hypothetical totals, directly using 672 as the answer. Another issue is forgetting that only one person's score changes, so the difference is exactly 7 points. Always distinguish between the real scenario and the imagined one used for calculation. Final Answer: The total number of points actually scored by the team is 665.

Question 20

A librarian purchased 60 story books for his library at a certain average price per book. Later, he found that by spending Rs 336 more in total, he could have bought 4 extra books (making 64 books in all) and the overall average price per book would…

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Introduction / Context: This question combines averages with a change in quantity and total cost. The librarian considers an alternative purchase plan in which he spends a bit more money, buys more books, and ends up with a lower average price. From these relationships, we can deduce the original average price per book. Given Data / Assumptions: • Number of books actually purchased = 60.• Original average price per book = x rupees (unknown).• So original total cost = 60x rupees.• If he spent Rs 336 more, total cost would be 60x + 336 rupees.• With this extra money, he could have bought 4 more books, that is 64 books.• In that case, the new average price per book would be (x - 1) rupees, that is Re 1 less. Concept / Approach: The key is to equate the total cost in the hypothetical case to the number of books multiplied by the new average price. This leads to a linear equation in x. Once we solve for x, we get the original average price per book. Such word problems are solved systematically by translating verbal relationships into algebraic equations. Step-by-Step Solution: Step 1: Let the original average price per book be x rupees.Step 2: Original total cost for 60 books = 60x.Step 3: In the hypothetical scenario, the librarian spends 336 rupees more, so new total cost = 60x + 336.Step 4: In that scenario, he could buy 64 books with average price (x - 1) rupees.Step 5: Therefore, new total cost can also be written as 64 * (x - 1).Step 6: Set the two expressions for new total cost equal: 60x + 336 = 64(x - 1).Step 7: Expand the right side: 64(x - 1) = 64x - 64.Step 8: So 60x + 336 = 64x - 64.Step 9: Rearrange to collect x terms: 336 + 64 = 64x - 60x.Step 10: 400 = 4x, so x = 400 / 4 = 100. Verification / Alternative check: With x = 100, original total cost is 60 * 100 = Rs 6000. If he spent 336 more, total becomes 6000 + 336 = Rs 6336. With 64 books, the new average price would be 6336 / 64 = 99 rupees, which is exactly Re 1 less than 100. This confirms that 100 rupees was the original average price per book. Why Other Options Are Wrong: Rs 84, Rs 83, Rs 68: If any of these were used as the original average, the equations would not balance. The total cost with 336 extra rupees would not equal the cost of 64 books at a reduced average price of x - 1. Common Pitfalls: Students sometimes confuse which average is larger or forget that the second average is lower by Re 1. Others may misinterpret the increase in number of books. Writing down the algebraic equation carefully and following the steps avoids such confusion. Final Answer: The previous average price of each book was Rs 100.

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