Question 1

In a class of 78 students, 41 are taking French and 22 are taking German. Of the students taking French or German, 9 are taking both courses. How many students are not enrolled in either French or German?

Accepted Answer

Introduction / Context: This question is based on set theory and the principle of inclusion and exclusion. It uses a very common exam pattern where students are divided according to interests in two subjects. The goal is to find how many students are outside both groups. Understanding how to handle overlaps between sets is important for many aptitude and data interpretation problems. Given Data / Assumptions: Total number of students in the class is 78. 41 students are taking French. 22 students are taking German. 9 students are taking both French and German. We need the count of students who are not enrolled in either French or German. Concept / Approach: For two sets A and B, the number of elements in A union B is given by n(A ∪ B) = n(A) + n(B) - n(A ∩ B). Here, A is the set of students taking French and B is the set of students taking German. After finding how many students are taking at least one of the two languages, we subtract this count from the total number of students to get those who take neither language. Step-by-Step Solution: Step 1: Identify the sets. Let A be the set of students taking French and B be the set of students taking German. Step 2: Use the inclusion and exclusion principle. Number taking French or German = n(A ∪ B) = n(A) + n(B) - n(A ∩ B). Step 3: Substitute the given values. n(A ∪ B) = 41 + 22 - 9 = 63 - 9? Wait carefully: 41 + 22 = 63, and 63 - 9 = 54. So 54 students are taking at least one of the two languages. Step 4: Compute those taking neither language. Students taking neither French nor German = total students - students taking at least one language. So required number = 78 - 54 = 24. Verification / Alternative Check: We can mentally confirm that 24 students out of 78 not taking any language means 54 are taking at least one. Given 41 take French and 22 take German with 9 in the overlap, French only is 41 - 9 = 32, German only is 22 - 9 = 13, and both is 9. Total who take at least one language is 32 + 13 + 9 = 54, which matches our earlier calculation, so the answer is consistent. Why Other Options Are Wrong: Option 6 is far too small and would mean almost all students are in at least one language, inconsistent with the given numbers. Option 12 does not match the inclusion and exclusion calculation and would lead to an incorrect total when recomputed. Option 18 also fails when we reassemble the counts of French only, German only and both; it does not align with the total of 78 students. Common Pitfalls: One common mistake is to simply add 41 and 22 and ignore the overlap, which double counts the students taking both languages. Another error is subtracting the intersection twice. Some students also forget that the remainder after counting A ∪ B represents those who are in neither set, which is exactly what the question asks for. Final Answer: The number of students not enrolled in either course is 24.

Question 2

The average of 9 results is 50. The average of the first four results is 52 and the average of the last four results is 49. What is the value of the fifth result?

Accepted Answer

Introduction / Context: This is a classic averages question in which you are given the overall average of several results as well as the averages of subsets of those results. The aim is to find the missing central value. Problems like this test your ability to switch between averages and total sums and to carefully separate different groups within the data. Given Data / Assumptions: There are 9 results in total. The average of all 9 results is 50. The average of the first 4 results is 52. The average of the last 4 results is 49. We are required to find the fifth result, which lies between the first four and the last four results. Concept / Approach: The key idea is that average multiplied by the number of items gives the total sum of those items. We can compute the total of all 9 results, the total of the first 4, and the total of the last 4. The middle result is not counted in either of the two groups of four, so we subtract both partial totals from the overall total to obtain the fifth result. Step-by-Step Solution: Step 1: Compute the total of all 9 results. Total of 9 results = average * number of results = 50 * 9 = 450. Step 2: Compute the total of the first 4 results. Total of first 4 results = 52 * 4 = 208. Step 3: Compute the total of the last 4 results. Total of last 4 results = 49 * 4 = 196. Step 4: Use these totals to find the fifth result. Total of all 9 results = (first 4 results) + (fifth result) + (last 4 results). So, 450 = 208 + fifth result + 196. Add 208 and 196 to get 404. Therefore, fifth result = 450 - 404 = 46. Verification / Alternative Check: To verify, we can imagine example numbers that satisfy the given averages or simply trust the algebra. If the fifth result is 46, then the grand total 450 is exactly split into 208, 46 and 196 for each part. Recomputing the averages from these totals gives back 52 for the first four, 49 for the last four and 50 overall, which confirms our answer. Why Other Options Are Wrong: Option 48 would give a total of 208 + 48 + 196 = 452, which would lead to an overall average greater than 50, so it is wrong. Option 52 would produce 208 + 52 + 196 = 456, again giving a higher overall average than 50. Option 44 would give a total of 208 + 44 + 196 = 448, which would produce an overall average less than 50, so it is also incorrect. Common Pitfalls: A common error is to average 52 and 49 and assume that somehow produces the fifth result. Another mistake is forgetting that the middle result is counted only once in the total, while the groups of four do not include it at all. Students may also mistakenly treat 9 as if it is split into equal parts, which is not the case here because the central value is separate. Final Answer: The value of the fifth result is 46.

Question 3

The average price of three items of furniture is Rs. 15000. If their prices are in the ratio 3 : 5 : 7, what is the price of the cheapest item of furniture?

Accepted Answer

Introduction / Context: This problem uses the concepts of average and ratio together. You are given the average price of three items and the ratio in which their individual prices are divided. The task is to find the actual price of the cheapest item. Such questions are common in aptitude tests related to commerce, profit and loss, and basic arithmetic reasoning. Given Data / Assumptions: There are three items of furniture. The average price of the three items is Rs. 15000. The prices are in the ratio 3 : 5 : 7. We are required to find the price of the cheapest item, which corresponds to the smallest ratio term 3. Concept / Approach: First, from the average, we compute the total price of all three items. Then we use the ratio 3 : 5 : 7 to represent the three prices as 3k, 5k and 7k. The sum of these three ratio parts must equal the total price. So we solve for k and then multiply by the smallest term in the ratio to get the price of the cheapest item. Step-by-Step Solution: Step 1: Compute the total price of all three items. Total price = average price * number of items = 15000 * 3 = Rs. 45000. Step 2: Express the three prices using a common multiplier. Let the prices be 3k, 5k and 7k, where k is a positive constant. Step 3: Set up the equation using the total price. 3k + 5k + 7k = 45000. This gives 15k = 45000. Step 4: Solve for k. k = 45000 / 15 = 3000. Step 5: Compute the cheapest price. Cheapest item corresponds to 3k, so price = 3 * 3000 = Rs. 9000. Verification / Alternative Check: We can verify by computing all three prices: 3k = 9000, 5k = 15000 and 7k = 21000. Their sum is 9000 + 15000 + 21000 = 45000. Dividing this total by 3 gives an average of 15000, which matches the given average. This confirms that the distribution and the cheapest price are correct. Why Other Options Are Wrong: Option Rs. 6000 would correspond to 2k if k were 3000, but the ratio does not contain 2 as a term, so it is not valid. Option Rs. 7000 does not fit the ratio pattern because 7000 is not equal to 3k, 5k or 7k for integer k that gives total 45000. Option Rs. 8888 does not fit any sensible ratio here and does not give the correct total and average when combined with the other implied prices. Common Pitfalls: A common error is to directly divide the average by the smallest ratio term or to misinterpret average as relating to each ratio part rather than the total. Some students also forget to multiply the average by the number of items to get the total before applying the ratio, which leads to wrong equations and therefore incorrect answers. Final Answer: The price of the cheapest furniture item is Rs. 9000.

Question 4

In a mens hostel there were initially 95 students. After 30 more students were accommodated, the average expenditure per student decreased by Rs. 5, yet the total expenditure of the hostel increased by Rs. 500. What is the total expenditure of the h…

Accepted Answer

Introduction / Context: This question combines averages with total expenditure and a change in the number of members. It is a typical hostel or club expenditure problem that appears in aptitude exams to check your understanding of how per head average and total cost change when more people join and some financial adjustment occurs. The key is to form correct algebraic equations from the given statements. Given Data / Assumptions: Initially there are 95 students in the hostel. After some time 30 more students join, so the new strength becomes 125 students. When the new students join, the average expenditure per student decreases by Rs. 5. Despite the decrease in average, the total expenditure increases by Rs. 500. We need to find the total expenditure of the hostel after the increase in strength, that is, for 125 students. Concept / Approach: Let the initial average expenditure per student be x rupees. Then the initial total expenditure is 95 * x. After 30 more students join, the new average becomes (x - 5) and the number of students is 125. Therefore, the new total expenditure is 125 * (x - 5). We are told that this new total is Rs. 500 more than the old total, so we can form a linear equation and solve for x. Finally we compute the new total expenditure using the value of x. Step-by-Step Solution: Step 1: Let the initial average expenditure per student be x rupees. Initial total expenditure = 95 * x. Step 2: Express the new average and new total expenditure. After 30 more students join, number of students = 95 + 30 = 125. New average expenditure per student = x - 5. New total expenditure = 125 * (x - 5). Step 3: Use the condition about the increase in total expenditure. New total expenditure = old total expenditure + 500. So, 125 * (x - 5) = 95 * x + 500. Step 4: Solve the equation for x. 125x - 625 = 95x + 500. 125x - 95x = 500 + 625. 30x = 1125, so x = 1125 / 30 = 37.5 rupees. Step 5: Compute the new average and new total expenditure. New average = x - 5 = 37.5 - 5 = 32.5 rupees. New total expenditure = 125 * 32.5 = Rs. 4062.50. Verification / Alternative Check: Compute the old total expenditure: 95 * 37.5 = 3562.50 rupees. The new total is 4062.50 rupees. The difference is 4062.50 - 3562.50 = 500, which matches the given increase. The new average 32.5 is exactly Rs. 5 less than the old average 37.5, which also matches the statement. Therefore, the answer is fully consistent with all conditions. Why Other Options Are Wrong: Option Rs. 3825 gives an incorrect new average and does not differ from the original total by Rs. 500. Option Rs. 3937.50 also fails to produce the correct relation between old and new totals and the stated Rs. 5 drop in average. Option Rs. 4250 makes the increase in expenditure more than Rs. 500 and therefore does not satisfy the given condition. Common Pitfalls: Many students confuse the change in average with direct change in total without considering the new number of students. Another common mistake is to treat 500 as the change in average instead of the total expenditure. Errors in forming the equation 125 * (x - 5) = 95 * x + 500 are also frequent, especially in handling signs when moving terms from one side to the other. Final Answer: The total expenditure of the hostel after accommodating the new students is Rs. 4062.50.

Question 5

David obtained 76, 65, 82, 67 and 85 marks out of 100 in English, Mathematics, Physics, Chemistry and Biology respectively. What are his average marks across these five subjects?

Accepted Answer

Introduction / Context: This question is a straightforward average marks problem that checks whether you can correctly sum several values and then divide by the number of observations. It represents a typical scenario from school examinations where a student obtains different marks in different subjects and we need to compute the overall performance in terms of average marks. Given Data / Assumptions: The subjects are English, Mathematics, Physics, Chemistry and Biology. Marks obtained are 76, 65, 82, 67 and 85 respectively, each out of 100. We must find the average of these five marks. All marks are treated equally with no different weightage for any subject. Concept / Approach: The concept of average is simple: average = total sum of observations divided by the number of observations. Here we add the five subject marks and divide the total by 5. No complex formula is needed, but accuracy in addition and division is crucial to arrive at the correct answer. Step-by-Step Solution: Step 1: List the marks clearly. English = 76, Mathematics = 65, Physics = 82, Chemistry = 67, Biology = 85. Step 2: Add all the marks. Total marks = 76 + 65 + 82 + 67 + 85. Compute this sum: 76 + 65 = 141, 141 + 82 = 223, 223 + 67 = 290, 290 + 85 = 375. So, total marks = 375. Step 3: Divide by the number of subjects. Number of subjects = 5. Average marks = total marks / number of subjects = 375 / 5 = 75. Verification / Alternative Check: We can quickly verify the division: 5 * 75 = 375, which matches the total marks computed earlier. The average 75 is also reasonable given that most individual marks (76, 82, 85) are above 70 and only one mark is slightly lower at 65, so an overall average in the mid seventies is logically consistent. Why Other Options Are Wrong: Option 68 would give a total of 68 * 5 = 340, which is much less than the actual total of 375. Option 65 would correspond to 65 * 5 = 325, which is again below the true sum and does not reflect the higher marks in Physics and Biology. Option 60 would give a total of 300, which is far away from the actual total and therefore clearly incorrect. Common Pitfalls: Mistakes usually come from wrong addition of the five marks or from dividing by an incorrect number of subjects. Sometimes students also misread one of the marks or miss one subject entirely when adding. Double checking the sum and ensuring you divide by exactly 5 avoids these errors. Final Answer: David’s average marks across the five subjects are 75.

Question 6

The mean of 100 observations was originally calculated as 40. Later it was found that one observation was misread as 48 instead of 84. What is the correct mean of the 100 observations after this correction?

Accepted Answer

Introduction / Context: This problem deals with correction of a mean when an error is detected in one of the observations. Such questions commonly appear in statistics sections of aptitude tests. They test whether you understand that the mean is based on the total sum of observations and that any change in the data must be reflected by adjusting the total before recomputing the mean. Given Data / Assumptions: There are 100 observations in total. The originally calculated mean of these 100 observations is 40. One value was wrongly taken as 48 instead of the correct value 84. We must find the corrected mean after fixing this single error. All other observations are assumed to be correct. Concept / Approach: The mean is equal to the total sum divided by the number of observations. When one observation is misread, the original total is incorrectly low or high. To correct the mean, we first compute the original total from the original mean. Then we adjust this total by removing the wrong value and adding the correct value. Finally we divide the corrected total by the same number of observations, which is still 100 in this case. Step-by-Step Solution: Step 1: Compute the original total of all observations. Original mean = 40, number of observations = 100. Original total = mean * number of observations = 40 * 100 = 4000. Step 2: Adjust the total for the misread observation. The wrong value used was 48; the correct value should be 84. Corrected total = original total - wrong value + correct value. Corrected total = 4000 - 48 + 84. Compute 4000 - 48 = 3952, then 3952 + 84 = 4036. Step 3: Compute the new mean using the corrected total. Corrected mean = corrected total / number of observations = 4036 / 100 = 40.36. Verification / Alternative Check: The difference between the correct value and the wrong value is 84 - 48 = 36. Since there are 100 observations, the mean should change by 36 / 100 = 0.36. Adding 0.36 to the original mean 40 gives 40.36, which matches our detailed calculation. This quick method confirms that the corrected mean 40.36 is reliable and consistent. Why Other Options Are Wrong: Option 41.24 would correspond to an adjustment of 1.24 in the mean, which would mean a total correction of 124, not 36, so it is not correct. Option 41.92 represents an even larger change in the total and does not match the actual difference between 84 and 48. Option 42.05 would require a difference in total of 205, which is completely inconsistent with the true correction required. Common Pitfalls: A typical mistake is to divide the difference 84 - 48 by the wrong number of observations or to add and subtract in the wrong order. Some students also forget that only one value was corrected and incorrectly change the count of observations. Others recompute the mean without first recalculating or adjusting the total, which leads to inaccurate answers. Final Answer: The correct mean of the 100 observations after correction is 40.36.

Question 7

Without any stoppage, a person travels a certain distance at an average speed of 42 km/h. When stoppages are included, he covers the same distance at an average speed of 28 km/h. For how many minutes in each hour does he stop?

Accepted Answer

Introduction / Context: This question is about average speed with and without stoppages. It is a standard type of problem in time, speed and distance, where you are asked to relate the effective average speed including rest stops to the actual running speed. The goal is to understand how much time per hour is lost due to stoppages based on the reduction in average speed. Given Data / Assumptions: The person travels at an average speed of 42 km/h when there are no stoppages. Including stoppages, the average speed for the same distance reduces to 28 km/h. The running speed when in motion is assumed to remain 42 km/h. We must find how many minutes out of each hour are spent stopping. Concept / Approach: Average speed including stoppages is equal to total distance divided by total time, where total time includes both running time and stoppage time. Since the running speed is constant at 42 km/h, the ratio of effective average speed to running speed gives the fraction of time actually spent moving. The remaining fraction of each hour represents stoppage time. We express this fraction of an hour in minutes to get the final answer. Step-by-Step Solution: Step 1: Compare effective average speed with running speed. Running speed = 42 km/h, effective average speed including stops = 28 km/h. Fraction of time spent moving = effective average speed / running speed. So, fraction of time moving = 28 / 42 = 2 / 3. Step 2: Compute the fraction of time stopped. Total time in one hour is represented by 1. Fraction of time stopped = 1 - 2 / 3 = 1 / 3. Step 3: Convert this fraction of an hour into minutes. One hour = 60 minutes. Stoppage time per hour = (1 / 3) * 60 minutes = 20 minutes. Verification / Alternative Check: We can also reason using a simple distance, for example 42 km. If he runs without stopping, 42 km at 42 km/h takes 1 hour. With stoppages, his average is 28 km/h, so to cover 42 km he needs 42 / 28 = 1.5 hours. That means he spent 1 hour actually running and 0.5 hour stopping. Half an hour is 30 minutes, but that is total stoppage for 42 km. Per hour of clock time, the stoppage fraction is still one third of the time, which corresponds to 20 minutes per hour. The direct fraction method is cleaner and avoids confusion. Why Other Options Are Wrong: Option 15 minutes corresponds to a stoppage fraction of 1/4 hour, which would give a different effective speed ratio not equal to 28 / 42. Option 18 minutes is 0.3 hour, which again does not match the ratio 2 / 3 for moving time. Option 22 minutes is an awkward number and does not fit any simple fraction that would reproduce the correct speed reduction from 42 km/h to 28 km/h. Common Pitfalls: Students often confuse distance ratio with time ratio and may incorrectly try to subtract speeds. Another mistake is choosing an arbitrary distance and getting lost in calculation instead of directly using the simple ratio effective speed / running speed. Forgetting to convert the fraction of an hour into minutes correctly is another frequent source of error. Final Answer: The person stops for 20 minutes in each hour.

Question 8

In Arun's family, three people give their opinions about his weight. Arun thinks his weight is greater than 65 kg but less than 72 kg. His brother thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother believes that his weig…

Accepted Answer

Introduction / Context: This is a logical reasoning question involving ranges and inequalities rather than direct arithmetic only. Several family members estimate Arun's weight with different upper and lower bounds, and we are told that all of them are correct simultaneously. From this we must deduce the common range that satisfies all three statements and then find the average of the possible integer weights in that range. Given Data / Assumptions: Arun thinks his weight is greater than 65 kg and less than 72 kg. His brother thinks Arun's weight is greater than 60 kg and less than 70 kg. His mother believes Arun's weight is not greater than 68 kg. All three opinions are correct at the same time. Weights are assumed to be whole numbers in kilograms for counting different probable values. Concept / Approach: Each statement gives an interval of possible values. When all three are correct together, Arun's true weight must lie in the intersection of these three intervals. Once we find this intersection, we list all the different integer values of weight that satisfy it. The question then asks for the average of these different probable weights, which we compute by adding them and dividing by their count. Step-by-Step Solution: Step 1: Express each opinion as an inequality range. From Arun: weight is greater than 65 kg and less than 72 kg, so range (65, 72). From brother: weight is greater than 60 kg and less than 70 kg, so range (60, 70). From mother: weight cannot be greater than 68 kg, so weight is less than or equal to 68 kg. Step 2: Find the intersection of these ranges. Intersection of (65, 72) and (60, 70) is (65, 70). Applying the condition from mother (weight ≤ 68 kg) restricts this further to (65, 68]. Step 3: List the different probable integer weights in this intersection. The integer values strictly greater than 65 and less than or equal to 68 are 66 kg, 67 kg and 68 kg. Step 4: Compute the average of these possible weights. Sum of probable weights = 66 + 67 + 68 = 201 kg. Number of probable weights = 3. Average probable weight = 201 / 3 = 67 kg. Verification / Alternative Check: We can check that each of the three values 66, 67 and 68 fits all opinions. Each is greater than 65 and less than 72, so Arun's estimate is satisfied. Each is greater than 60 and less than 70, so the brother is also correct. None of them exceeds 68, so the mother is also correct. The average of these three values is clearly 67, confirming our calculation. Why Other Options Are Wrong: Option 65 kg is outside the allowed range because Arun claims his weight is greater than 65 kg, not equal to it. Option 66.5 kg is a reasonable looking decimal but the problem speaks of different probable weights typically interpreted as integer kilogram values, and the exact intersection average of integer possibilities is 67 kg, not 66.5 kg. Option 68.5 kg exceeds the mother's limit of 68 kg, and so it cannot be correct. Common Pitfalls: A frequent mistake is to average the upper and lower bounds of the largest range or to take an average of all boundary numbers from all three people. Another pitfall is to ignore the requirement that all three statements must be simultaneously true, so some students forget to compute the intersection of intervals. Finally, some candidates overlook whether to treat possible weights as integer values and misinterpret the phrase different probable weights. Final Answer: The average of the different probable weights of Arun is 67 kg.

Question 9

The average age of 15 students of a class is 15 years. Out of these, the average age of 5 students is 14 years and the average age of another 9 students is 16 years. What is the age of the remaining fifteenth student?

Accepted Answer

Introduction / Context: This question is another application of averages and total sums. The group is divided into two overlapping subgroups, and we are given the average age of each subgroup as well as the overall average. We need to find the age of the one student who is not counted in either of the two specified groups of 5 and 9 students. Given Data / Assumptions: There are 15 students in the class. The average age of all 15 students is 15 years. The average age of 5 selected students is 14 years. The average age of another 9 students is 16 years. The ages are all in years and we need the age of the remaining one student. Concept / Approach: We use the relation total age = average age * number of students for each group. We first compute the total age of all 15 students using the overall average. Then we compute the total ages of the groups of 5 and 9 students separately. The remaining student is the fifteenth one whose age is not included in those two sums. Therefore, age of the fifteenth student equals the overall total minus the sum of the totals of the two subgroups. Step-by-Step Solution: Step 1: Compute total age of all 15 students. Total age of 15 students = 15 * 15 = 225 years. Step 2: Compute total age of the 5 students. Total age of 5 students = 14 * 5 = 70 years. Step 3: Compute total age of the 9 students. Total age of 9 students = 16 * 9 = 144 years. Step 4: Use these totals to find the age of the fifteenth student. Sum of ages of the 5 and 9 students together = 70 + 144 = 214 years. Age of fifteenth student = total age of 15 students - 214 = 225 - 214 = 11 years. Verification / Alternative Check: We can quickly verify: if the single remaining student is 11 years old, then the three groups are consistent. The 5 students sum to 70, the 9 students to 144 and the last student 11 gives a total of 70 + 144 + 11 = 225. Dividing 225 by 15 yields an average of exactly 15 years, which matches the given information. Why Other Options Are Wrong: Option 12 would give a total of 70 + 144 + 12 = 226, which would produce an overall average higher than 15 years. Option 13 leads to a total of 227, again not matching the required total 225 for an average of 15. Option 14 would take the total to 228, making the average 228 / 15, which is not equal to 15 years, so it is incorrect. Common Pitfalls: Some students mistakenly average 14 and 16 or try to treat the 5 and 9 students as disjoint without considering the overall total properly. Another error is miscalculating the totals for the subgroups or forgetting that there is exactly one remaining student whose age must account for the difference between the partial sums and the whole class sum. Final Answer: The age of the fifteenth student is 11 years.

Question 10

The speeds of a car, a train and a bus are in the ratio 5 : 9 : 4. The average of their speeds taken together is 72 km/h. What is the average speed of the car and the train taken together?

Accepted Answer

Introduction / Context: This question combines averages with ratios of speeds. You are given the ratio of the speeds of three vehicles and the average of all three speeds. From this information, you must first determine each individual speed in actual km/h and then find the average speed of just two of the vehicles, the car and the train. Given Data / Assumptions: Speeds of car, train and bus are in the ratio 5 : 9 : 4. The average of the three speeds is 72 km/h. We need the average speed of the car and the train together. Speeds are assumed constant and measured in km/h. Concept / Approach: Let the speeds of the car, train and bus be 5k, 9k and 4k respectively. The average of the three speeds is the sum of the speeds divided by 3, and this is given as 72 km/h. Using this we can solve for k. Once we know k, we find the actual speed of the car and the train and then compute their average in the usual way by adding the two speeds and dividing by 2. Step-by-Step Solution: Step 1: Represent the speeds using a common multiplier. Let speed of car = 5k, train = 9k, bus = 4k. Step 2: Use the given average speed of all three vehicles. Average speed of three vehicles = (5k + 9k + 4k) / 3 = 72. So, (18k) / 3 = 72. This simplifies to 6k = 72. Step 3: Solve for k. k = 72 / 6 = 12. Step 4: Compute individual speeds. Car speed = 5k = 5 * 12 = 60 km/h. Train speed = 9k = 9 * 12 = 108 km/h. Bus speed = 4k = 4 * 12 = 48 km/h, which is not directly needed but confirms the ratio. Step 5: Find the average speed of car and train. Average speed of car and train = (60 + 108) / 2 = 168 / 2 = 84 km/h. Verification / Alternative Check: To verify, we can check that the average of all three speeds is still 72 km/h. Sum of speeds = 60 + 108 + 48 = 216 km/h. Average = 216 / 3 = 72 km/h, which matches the given condition. This ensures that the value of k and the derived speeds are correct, so the computed average of 84 km/h for the car and train is reliable. Why Other Options Are Wrong: Option 96 km/h would mean the sum of car and train speeds is 192 km/h, inconsistent with the correct individual speeds derived from the ratio and the overall average. Option 72 km/h is simply the average of all three vehicles, not of only the car and train. Option 60 km/h would correspond to the speed of the car alone, not the joint average of car and train. Common Pitfalls: Students may forget that the overall average is based on three speeds and incorrectly set 5k + 9k + 4k = 72 instead of dividing by 3. Another common error is to mis-handle the ratio or miscalculate k, which affects all subsequent speeds. Some may also attempt to directly average the ratio numbers 5 and 9 without scaling them properly, which is incorrect. Final Answer: The average speed of the car and the train together is 84 km/h.

Question 11

Out of three numbers, the first is twice the second and is half of the third. If the average of these three numbers is 56, what is the sum of the highest and the lowest numbers?

Accepted Answer

Introduction / Context: This question combines relationships between three unknown numbers with the concept of averages. The information about how the first number is related to the second and third allows us to express all three in terms of a single variable. The given average then helps us determine the exact values and finally compute the required sum of the largest and smallest numbers. Given Data / Assumptions: There are three numbers: first, second and third. The first number is twice the second number. The first number is half of the third number. The average of the three numbers is 56. We must find the sum of the highest and the lowest of these three numbers. Concept / Approach: We use algebra to express all three numbers in terms of one variable. If we let the second number be x, then the first number is 2x. Since the first number is half of the third, the third number is 4x. With these expressions we can compute the average in terms of x, set it equal to 56 and solve for x. Once we know the actual values of the three numbers, we identify the smallest and largest and add them together to get the final answer. Step-by-Step Solution: Step 1: Represent the numbers using one variable. Let the second number be x. Then the first number = 2x. Since the first number is half of the third, the third number = 4x. Step 2: Use the average condition. Average of the three numbers = (x + 2x + 4x) / 3 = 56. Simplify the numerator: x + 2x + 4x = 7x. So, 7x / 3 = 56. Step 3: Solve for x. Multiply both sides by 3: 7x = 56 * 3 = 168. Therefore, x = 168 / 7 = 24. Step 4: Find the three numbers. Second number = x = 24. First number = 2x = 48. Third number = 4x = 96. Step 5: Identify highest and lowest and find their sum. Lowest number = 24, highest number = 96. Required sum = 24 + 96 = 120. Verification / Alternative Check: We can verify by checking the average of 24, 48 and 96. Their sum is 24 + 48 + 96 = 168. Dividing by 3 gives 168 / 3 = 56, which matches the given average. The relationships are also satisfied: first number 48 is indeed twice the second number 24 and is half of the third number 96, confirming that our values are consistent. Why Other Options Are Wrong: Option 160 is greater than the sum of any two numbers from the derived set, so it cannot be correct. Option 80 would be 24 plus 56, but 56 is not one of the three numbers, so it does not match the actual numbers. Option 60 is 24 plus 36, but 36 is also not one of the actual numbers obtained from the average condition. Common Pitfalls: Students sometimes misinterpret the relationships and assign wrong expressions, for example making the third number 2x instead of 4x, which breaks the condition that the first is half of the third. Some also forget to divide by 3 when using the average or incorrectly simplify 7x / 3 = 56, leading to wrong values of x and consequently wrong final answers. Final Answer: The sum of the highest and lowest numbers is 120.

Question 12

The average age of a group of persons going for a picnic is 16 years. On the spot, 20 new persons with an average age of 15 years join the group, and the overall average age becomes 15.5 years. How many persons were there in the original picnic grou…

Accepted Answer

Introduction / Context: This is a classic mixture of averages problem where two groups with different average ages combine to form a larger group with a new average age. These types of questions appear widely in aptitude exams to test how well you handle the relationship between totals and averages when groups merge or split. Given Data / Assumptions: Initial group has an average age of 16 years. Let the number of persons in the initial group be n. On the spot, 20 new persons join, and these new persons have an average age of 15 years. After joining, the combined average age becomes 15.5 years. We need to find the original number of persons n in the picnic group. Concept / Approach: We treat the initial group and the new group as two sets with their own totals. The initial total age is 16 * n. The total age of the 20 new persons is 15 * 20. When they combine, the total number of persons is n + 20, and the total age is the sum of the two totals. The new combined average is 15.5, so total combined age = 15.5 * (n + 20). Setting the sum of the two group totals equal to this combined total yields an equation in n that we can solve. Step-by-Step Solution: Step 1: Express the total ages of the two groups. Initial total age = 16 * n. Total age of new 20 persons = 15 * 20 = 300. Step 2: Express the combined total age using the final average. Combined number of persons = n + 20. Combined total age = 15.5 * (n + 20). Step 3: Set up the equation. Initial total + new total = combined total. So, 16n + 300 = 15.5 * (n + 20). Step 4: Solve the equation for n. Expand the right side: 15.5 * (n + 20) = 15.5n + 310. So, 16n + 300 = 15.5n + 310. Bring terms involving n to one side: 16n - 15.5n = 310 - 300. 0.5n = 10, hence n = 10 / 0.5 = 20. Verification / Alternative Check: If the original group has 20 persons with average age 16, their total age is 20 * 16 = 320. The new 20 persons have total age 300. Combined total age = 320 + 300 = 620, and combined number of persons = 40. The combined average is 620 / 40 = 15.5 years, which matches the given final average, confirming that n = 20 is correct. Why Other Options Are Wrong: Option 15 would give an initial total of 240, leading to a different combined average not equal to 15.5 years. Option 23 or 24 similarly produce totals that do not result in the final average of 15.5 years when combined with the new group of 20 persons with average age 15 years. Common Pitfalls: A common mistake is to simply average 16 and 15 to get 15.5 and assume the groups are equal in size without checking the actual numbers. Another error is mishandling the algebraic equation, especially when multiplying the decimal 15.5. Some candidates also misinterpret the question and try to find the final number of persons instead of the original group size. Final Answer: The number of persons initially going for the picnic is 20.

Question 13

A motorist travels from one place to another 150 km away at an average speed of 50 km/h and returns to the starting point at an average speed of 30 km/h. What is his average speed for the entire round trip in km/h?

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Introduction / Context: This question is about average speed for a round trip where the distance is the same in both directions but the speeds are different. Many students go wrong by taking the simple average of the two speeds. However, for equal distances, the correct average speed is the harmonic mean of the two speeds, which accounts for different times taken at each speed. Given Data / Assumptions: One way distance between the two places is 150 km. Speed while going from start to destination is 50 km/h. Speed while returning is 30 km/h. The total journey is a round trip of 300 km. We must find the overall average speed for the entire journey. Concept / Approach: Average speed is defined as total distance divided by total time. For a two leg journey with equal distances but different speeds, we can either compute the total time for each leg and divide total distance by total time, or use the harmonic mean formula: average speed = 2 * v1 * v2 / (v1 + v2) where v1 and v2 are the two speeds. Both approaches yield the same result, but the harmonic mean formula is faster. Step-by-Step Solution: Step 1: Compute time taken for each leg of the journey. Time to go = distance / speed = 150 / 50 = 3 hours. Time to return = 150 / 30 = 5 hours. Step 2: Compute total distance and total time. Total distance for round trip = 150 + 150 = 300 km. Total time = 3 + 5 = 8 hours. Step 3: Compute average speed using total distance and total time. Average speed = total distance / total time = 300 / 8 = 37.5 km/h. Verification / Alternative Check: Using the harmonic mean formula for equal distances, average speed = 2 * v1 * v2 / (v1 + v2). Here v1 = 50 and v2 = 30. So, average speed = 2 * 50 * 30 / (50 + 30) = 3000 / 80 = 37.5 km/h. This matches our previous calculation, confirming the correctness of the answer. Why Other Options Are Wrong: Option 35 km/h would result from incorrect handling of the time or an arithmetic slip in division. Option 36 km/h may come from wrongly averaging times or using a rough estimate instead of exact calculation. Option 38.2 km/h does not correspond to any standard formula for combining speeds over equal distances. Common Pitfalls: The most common mistake is taking the arithmetic mean of the two speeds, that is (50 + 30) / 2 = 40 km/h, which is wrong because it ignores the different times spent at each speed. Another common error is forgetting to double the distance for the round trip or miscalculating the time for one of the legs. Always remember that average speed must be based on total distance divided by total time. Final Answer: The average speed for the entire round trip is 37.5 km/h.

Question 14

After replacing an old member by a new member, it is found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced member and the new member?

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Introduction / Context: This is a reasoning based average age question concerning a small group of club members. We are told that after a period of 3 years and the replacement of one member by another, the average age of the group is the same as it was 3 years earlier. This means that the effect of natural ageing has been exactly balanced by the change in membership. We must translate this information into a relationship between the ages of the old and new member. Given Data / Assumptions: There are 5 members in the club at any time. Consider the situation 3 years ago and the situation now after one replacement. After 3 years and replacing one old member with a new one, the average age of the 5 members now is equal to the average age 3 years ago. We must find the difference between the age of the replaced member and that of the new member. All ages are measured in years. Concept / Approach: If no one had been replaced, the age of each of the 5 members would have increased by 3 years over 3 years, so the total age of the group would have increased by 5 * 3 = 15 years. However, we are told that after replacement, the average age has not increased at all compared to 3 years ago, which means the total age is the same as it was 3 years earlier. Therefore the replacement must have effectively decreased the total age by exactly 15 years, and this difference comes from the difference between the ages of the old and new member. Step-by-Step Solution: Step 1: Consider what happens in 3 years with no replacement. If all 5 original members remained, each would be 3 years older after 3 years. Total age increase would be 5 * 3 = 15 years. Step 2: Use the given condition about average age. The question states that after 3 years and replacing one member, the average age now is the same as the average age 3 years ago. That means the total age of the group now is the same as the total age 3 years ago. Step 3: Relate this to the replacement. If no one had been replaced, the total would be old total + 15. But with replacement, the total is only old total. Therefore, the replacement reduced the total age by exactly 15 years. Step 4: Interpret the reduction. This reduction in total age is exactly the difference between the age of the replaced old member and the age of the new member. Hence, difference in ages = 15 years. Verification / Alternative Check: To verify, we can imagine a simple example. Suppose the total age of the 5 members 3 years ago was T. If everyone stayed, the total now would be T + 15. For the average to remain the same, after replacement the total must still be T. This means the replacement change has reduced the total by 15. Hence the old member must have been 15 years older than the new member, confirming the reasoning. Why Other Options Are Wrong: Option 16 years would mean the total age decreased by 16 years, which would make the total age now smaller than it was 3 years ago, contradicting the balance with ageing. Option 14 years or 13 years would not fully offset the natural 15 year total increase that would have occurred due to 5 members ageing for 3 years. Common Pitfalls: Some students mistakenly think the difference is 3 years or confuse the concept of average with individual changes. Others try to assign arbitrary ages instead of reasoning about total change over 3 years. Forgetting that 5 members each age by 3 years and therefore contribute a total increase of 15 years often leads to incorrect conclusions. Final Answer: The difference between the ages of the replaced and the new member is 15 years.

Question 15

The average monthly salary of 20 employees in an organisation is Rs. 1500. If the manager’s salary is added, the average salary increases by Rs. 100. What is the monthly salary of the manager?

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Introduction / Context: This is a simple salary average problem where the inclusion of an additional person changes the average. The question checks if you can correctly adjust totals when the average of a group changes due to one new member. Problems like these are common in basic quantitative aptitude relating to averages and money. Given Data / Assumptions: There are 20 employees initially. Their average monthly salary is Rs. 1500. When the manager joins, the average salary of the 21 people increases by Rs. 100 and becomes Rs. 1600. We must find the salary of the manager. All salaries are monthly and measured in rupees. Concept / Approach: The total salary of the initial group can be found by multiplying the average salary by the number of employees. Similarly, the total salary of the enlarged group including the manager is the new average multiplied by the new number of persons. The difference between these two totals is equal to the manager’s salary, since only that one person has been added. Step-by-Step Solution: Step 1: Compute total salary of the 20 employees. Total salary of 20 employees = 1500 * 20 = Rs. 30000. Step 2: Compute total salary after adding the manager. New average salary = 1500 + 100 = Rs. 1600. Number of persons now = 20 + 1 = 21. Total salary of 21 persons = 1600 * 21 = Rs. 33600. Step 3: Find the manager’s salary. Manager’s salary = total salary of 21 persons - total salary of 20 employees. Manager’s salary = 33600 - 30000 = Rs. 3600. Verification / Alternative Check: We can verify by taking the total salary including the manager, which is Rs. 33600, and dividing it by 21. This indeed gives 33600 / 21 = 1600, matching the new average. Removing the manager’s salary from this total leaves 30000 for 20 employees, which corresponds to the original average 1500. Thus all given conditions are satisfied by a manager’s salary of Rs. 3600. Why Other Options Are Wrong: Option Rs. 3700 would result in a new total of 30000 + 3700 = 33700, giving an average slightly more than Rs. 1600, which is not accepted by the question. Option Rs. 3800 or Rs. 3900 would make the increase in average more than Rs. 100, giving averages higher than Rs. 1600, so they are incorrect. Common Pitfalls: A common mistake is to think the manager’s salary is simply 1600 or 1500 plus 100 without considering how averages work for groups of different sizes. Some students also forget to multiply the average by the number of persons to get totals and instead attempt short cuts that do not hold in general. Final Answer: The monthly salary of the manager is Rs. 3600.

Question 16

In a mixture of three varieties of oil, the ratio of their weights is 4 : 5 : 8. If 5 kg of the first variety, 10 kg of the second variety, and some quantity of the third variety are added, the ratio of the three varieties becomes 5 : 7 : 9 in the f…

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Introduction / Context: This question is based on ratios and mixtures involving three components. We are given an initial ratio of three varieties of oil and then information about additional quantities added to each variety which results in a new ratio. Such problems are common in ratio and proportion sections and test your ability to convert ratios into algebraic expressions for actual quantities. Given Data / Assumptions: Initially, the ratio of the three varieties of oil is 4 : 5 : 8 by weight. Let the initial amounts be 4k, 5k and 8k kilograms respectively. Additional quantities added: 5 kg of the first variety, 10 kg of the second, and x kg of the third variety. After addition, the new ratio of the three varieties is 5 : 7 : 9. We must find x, the quantity of the third variety added, in kilograms. Concept / Approach: We convert the ratios into algebraic expressions using a common multiplier k. Then we express the new quantities after addition in terms of k and x. Since we are given that the final ratio is 5 : 7 : 9, the ratios of corresponding new quantities must be equal. This yields equations that we can solve to find k and x. Typically we equate two ratio pairs at a time to obtain a solvable system of equations. Step-by-Step Solution: Step 1: Represent initial amounts using a common multiplier. Initial amounts: first = 4k, second = 5k, third = 8k kilograms. Step 2: Write expressions for new amounts after addition. New amount of first variety = 4k + 5. New amount of second variety = 5k + 10. New amount of third variety = 8k + x. Step 3: Use the new ratio 5 : 7 : 9. We have (4k + 5) : (5k + 10) : (8k + x) = 5 : 7 : 9. First equate the ratio of first two types. (4k + 5) / (5k + 10) = 5 / 7. Step 4: Solve this equation for k. Cross multiply: 7 * (4k + 5) = 5 * (5k + 10). This gives 28k + 35 = 25k + 50. Rearrange: 28k - 25k = 50 - 35, so 3k = 15 and k = 5. Step 5: Use k to find x from the third ratio. Now the new amounts become: first = 4 * 5 + 5 = 25 kg, second = 5 * 5 + 10 = 35 kg, third = 8 * 5 + x = 40 + x. The final ratio is 5 : 7 : 9, so third : second = 9 : 7. Therefore, (40 + x) / 35 = 9 / 7. Cross multiply: 7 * (40 + x) = 9 * 35. This gives 280 + 7x = 315. So, 7x = 315 - 280 = 35, hence x = 35 / 7 = 5 kg. Verification / Alternative Check: With k = 5 and x = 5, the final amounts are 25, 35 and 45 kg. The ratio 25 : 35 : 45 simplifies by dividing each term by 5, giving 5 : 7 : 9, which matches the given final ratio exactly. This confirms that 5 kg of the third variety must have been added. Why Other Options Are Wrong: Option 15 kg would make the final third amount 55 kg, giving the ratio 25 : 35 : 55, which does not simplify to 5 : 7 : 9. Option 25 kg would yield 65 kg for the third variety, again not in the correct ratio with the first two types. Option 35 kg would give 75 kg as the third amount, resulting in 25 : 35 : 75, which cannot be simplified to 5 : 7 : 9. Common Pitfalls: Students sometimes forget to use a common multiplier k or mix up the order of the terms in the ratio. Another frequent error is to incorrectly set up the fraction equations or make algebraic mistakes while cross multiplying. It is also important to check the final ratio with the found values to ensure that no arithmetic slip occurred during calculations. Final Answer: The quantity of the third variety of oil added is 5 kg.

Question 17

What will be the total cost of building a fence around a square plot whose area is 289 square feet, if the cost of fencing is Rs. 58 per foot?

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Introduction / Context: This problem is a blend of basic geometry and unit cost calculation. You are given the area of a square plot and the cost per foot to build a fence around it. The task is to find the total fencing cost, which requires first finding the perimeter of the square from its area and then multiplying by the cost per unit length. Given Data / Assumptions: The plot is a square with area 289 square feet. The cost of building a fence is Rs. 58 per foot of fencing. We need to find the total cost of fencing the entire boundary of the square plot. All sides of the square are equal in length. Concept / Approach: For a square, area = side^2. From the given area, we find the side length by taking the square root. Once the side is known, the perimeter of the square is 4 times the side length. The total fencing cost is then perimeter multiplied by the cost per foot. Accuracy in computing the square root and multiplication is essential here. Step-by-Step Solution: Step 1: Find the side length of the square from its area. Area of square = side^2 = 289 square feet. So, side = square root of 289. 289 is 17^2, so side = 17 feet. Step 2: Compute the perimeter of the square. Perimeter of square = 4 * side = 4 * 17 = 68 feet. Step 3: Compute the total cost of fencing. Cost per foot = Rs. 58. Total cost = perimeter * cost per foot = 68 * 58. Compute 68 * 58 = 68 * (60 - 2) = 68 * 60 - 68 * 2 = 4080 - 136 = Rs. 3944. Verification / Alternative Check: We can quickly multiply in the standard way as well: 58 * 68. Doing the multiplication confirms the product 3944. Also, side 17 feet squared equals 289 square feet, which matches the given area, confirming that the side and perimeter values are correct and consistent with the problem statement. Why Other Options Are Wrong: Option Rs. 3748 would correspond to a different perimeter or cost per foot and does not result from 68 * 58. Option Rs. 3847 is not a multiple of 58 and therefore cannot represent the correct result for a whole number of feet. Option Rs. 3154 is significantly lower than the true cost and would come from an incorrect side length or perimeter calculation. Common Pitfalls: A common mistake is to miscalculate the square root of 289, sometimes confusing it with 19 or another nearby number. Others may incorrectly treat the given area as perimeter. Errors can also occur in the multiplication step if done too quickly. Carefully using area = side^2 and perimeter = 4 * side avoids these issues. Final Answer: The total cost of building the fence around the plot is Rs. 3944.

Question 18

It costs Rs p to send each of the first 1000 messages and Rs q to send each message thereafter. If r is greater than 1000, what is the total cost in rupees to send r messages in this messaging plan?

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Introduction: This question tests the concept of total cost when there are two different rates. The first part of the service charges one rate up to a certain limit, and any usage beyond that limit is charged at a different rate. Such problems are common in mobile plans, internet packages and messaging services, and they can be solved using simple algebra and careful counting of how many units are billed at each rate. Given Data / Assumptions: - Cost for each of the first 1000 messages is Rs p. - Cost for each message after the first 1000 is Rs q. - The total number of messages sent is r, and r is greater than 1000. - We need the total cost in rupees expressed in terms of p, q and r. Concept / Approach: The key idea is to split the total messages into two groups. The first group contains the first 1000 messages, all of which are charged at the rate p. The second group contains the remaining messages, which are charged at the rate q. The total cost is the cost of the first group plus the cost of the second group. This is a direct application of the idea: total cost = number of units * rate per unit for each slab. Step-by-Step Solution: Step 1: Number of messages in the first slab = 1000. Step 2: Cost for the first 1000 messages = 1000 * p = 1000p. Step 3: Since r is greater than 1000, the number of additional messages beyond 1000 is r - 1000. Step 4: Each additional message is charged Rs q, so cost for the extra messages = q * (r - 1000) = q(r - 1000). Step 5: Total cost = cost for first 1000 messages + cost for additional messages. Step 6: Therefore, total cost = 1000p + q(r - 1000). Verification / Alternative Check: If r = 1000, the number of additional messages is r - 1000 = 0, so the cost becomes 1000p + q(0) = 1000p, which matches the idea that only the first slab is used. If r = 1001, the cost should be 1000p for the first 1000 messages plus q for one extra message. The formula gives 1000p + q(1001 - 1000) = 1000p + q, which is correct. This confirms that the expression behaves correctly at the boundary. Why Other Options Are Wrong: Option b: 1000p + qr ignores that only r - 1000 messages are billed at rate q and therefore overcounts the cost of messages at rate q. Option c: pr + qr uses p for all r messages, which does not match the slab structure given in the question. Option d: 1000q + p(r - 1000) swaps the roles of p and q and so assigns the wrong rate to each slab. Option e: 1000(q - p) + qr does not simplify to the correct expression and gives an incorrect dependence on p and q. Common Pitfalls: A common mistake is to multiply r directly by either p or q without splitting the total messages into the two slabs. Another frequent error is to use qr for the second part instead of q(r - 1000). It is important to clearly separate the first 1000 messages and the remaining r - 1000 messages before writing the cost expression. Final Answer: The total cost in rupees to send r messages is 1000p + q(r - 1000).

Question 19

Three math classes A, B and C take an algebra test. The average score of class A is 83, the average of class B is 76 and the average of class C is 85. The average score of classes A and B together is 79 and the average score of classes B and C toget…

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Introduction: This question checks understanding of weighted averages and how class sizes affect an overall average. The averages of three different classes and the averages of pairs of classes are given. From these relations, we can find the relative sizes of the classes and then compute the combined average of all students in classes A, B and C together. Given Data / Assumptions: - Average score of class A = 83. - Average score of class B = 76. - Average score of class C = 85. - Average score of classes A and B together = 79. - Average score of classes B and C together = 81. - We are asked for the average score of all students in classes A, B and C together. Concept / Approach: Let the number of students in classes A, B and C be a, b and c respectively. Then the total scores of the three classes are 83a, 76b and 85c. Using the given combined averages of A and B, and B and C, we can set up equations involving a, b and c. From these equations we can find ratios between a, b and c. Once the ratios are known, the overall average is the total score divided by the total number of students. Step-by-Step Solution: Step 1: From the average of A and B, we have (83a + 76b) / (a + b) = 79. Step 2: Multiply both sides by (a + b): 83a + 76b = 79a + 79b. Step 3: Rearrange: 83a - 79a = 79b - 76b which gives 4a = 3b, so b = (4/3)a. Step 4: From the average of B and C, (76b + 85c) / (b + c) = 81. Step 5: Multiply both sides: 76b + 85c = 81b + 81c, so 85c - 81c = 81b - 76b which gives 4c = 5b, so c = (5/4)b. Step 6: Substitute b = (4/3)a into c = (5/4)b to get c = (5/4)*(4/3)a = (5/3)a. Step 7: Total score of all students = 83a + 76b + 85c = 83a + 76*(4/3)a + 85*(5/3)a. Step 8: Compute the coefficients: 76*(4/3) = 304/3 and 85*(5/3) = 425/3. So total score = 83a + (304/3)a + (425/3)a. Step 9: Combine terms: (304/3 + 425/3) = 729/3 = 243. Thus total score = 83a + 243a = 326a. Step 10: Total number of students = a + b + c = a + (4/3)a + (5/3)a = a + 3a = 4a. Step 11: Overall average = total score / total students = 326a / 4a = 81.5. Verification / Alternative Check: We can check that the derived ratios a : b : c = 3 : 4 : 5 are consistent. If a = 3, b = 4 and c = 5, total students = 12. Then total score = 83*3 + 76*4 + 85*5. This equals 249 + 304 + 425 = 978. Average = 978 / 12 = 81.5, which matches the result. The pairwise averages also work out correctly, confirming our solution. Why Other Options Are Wrong: Options 78, 79.5, 80 and 82 do not use the correct ratios between the class sizes and treat the averages as if each class had the same number of students or use incorrect algebra. They therefore lead to incorrect combined averages. Common Pitfalls: Students often assume that each class has the same number of students and simply average the three given averages. That method ignores the effect of class size on the combined average. Another common mistake is to mix up the equations when forming relations from the pairwise averages. Careful algebra and stepwise manipulation are essential. Final Answer: The average score of all students in classes A, B and C is 81.5.

Question 20

The average of the marks obtained in three subjects is 224. The marks in the first subject are twice the marks in the second subject and the marks in the second subject are twice the marks in the third subject. What are the marks in the second subje…

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Introduction: This problem involves an average of three subject marks and given relationships between those marks. It tests understanding of how to translate statements about relative sizes into algebraic equations and then use the average information to find individual values. Such questions are typical in aptitude tests on ratios, averages and basic algebra. Given Data / Assumptions: - There are three subjects with marks in each subject. - The average of the three subject marks is 224. - Marks in the first subject are twice the marks in the second subject. - Marks in the second subject are twice the marks in the third subject. - We must find the marks obtained in the second subject. Concept / Approach: If we let the third subject mark be a basic variable, then the other two marks can be written in terms of this variable using the given relationships. The average of three numbers is equal to their total sum divided by 3. Using the given average, we can find the total sum, then set up an equation in terms of the variable and solve for it. Finally, we compute the middle subject mark using the relationship. Step-by-Step Solution: Step 1: Let the marks in the third subject be x. Step 2: The second subject marks are twice the third subject, so second subject marks = 2x. Step 3: The first subject marks are twice the second subject, so first subject marks = 2 * 2x = 4x. Step 4: The three subject marks are therefore 4x, 2x and x. Step 5: Given that the average of these three marks is 224, total marks = 3 * 224 = 672. Step 6: Sum of the three marks in terms of x is 4x + 2x + x = 7x. Step 7: Set up the equation 7x = 672. Step 8: Solve for x: x = 672 / 7 = 96. Step 9: Second subject marks = 2x = 2 * 96 = 192. Verification / Alternative Check: If the third subject marks are 96, the second subject marks are 192 and the first subject marks are 384. Sum of marks = 96 + 192 + 384 = 672. The average is 672 / 3 = 224, which matches the given average. The relationship conditions are also satisfied: first subject 384 is twice second subject 192, and second subject 192 is twice third subject 96. This confirms that the calculation is correct. Why Other Options Are Wrong: Option 96 is the third subject mark, not the second subject mark. Option 128 and option 224 do not satisfy both relationships at the same time when used as the second subject mark. Option 256 leads to a total that does not give an average of 224 while maintaining the required ratios. Common Pitfalls: A common mistake is to confuse which subject is twice which and assign the ratios incorrectly. Another error is to forget to multiply the average by 3 to get the total sum. Keeping a clear representation such as 4x, 2x and x helps avoid these issues and makes the algebra straightforward. Final Answer: The marks obtained in the second subject are 192.

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