Question 1

What is the average of all even numbers between 222 and 250?

Accepted Answer

Introduction / Context: This question tests your understanding of averages for a sequence of evenly spaced numbers, in this case all even numbers in a given interval. Such sequences form an arithmetic progression, and there is a simple shortcut to find their average using only the first and last term of the sequence. Given Data / Assumptions: We are concerned with even numbers between 222 and 250. All numbers are integers and differ by 2, since they are consecutive even numbers. The term “between” here still yields the same answer whether we treat endpoints as included or excluded, due to symmetry. We must compute the arithmetic mean of all these even numbers. Concept / Approach: Even numbers in order form an arithmetic progression (AP) with common difference 2. For any AP, the average of all terms is equal to (first term + last term) / 2. Therefore, once we know the smallest even number at or above 222 and the largest even number at or below 250, we can directly use this property instead of summing many terms. Step-by-Step Solution: Step 1: Identify the first even number in the range. 222 is itself an even number, so the smallest even number is 222. Step 2: Identify the last even number in the range. 250 is also even, so the largest even number is 250. Step 3: Use the AP average property. Average of all even numbers from 222 to 250 = (222 + 250) / 2. Compute: 222 + 250 = 472, and 472 / 2 = 236. Verification / Alternative Check: If we consider exclusive bounds (224 to 248), the first even number is 224 and the last is 248. Their average is (224 + 248) / 2 = 472 / 2 = 236 again. Because the even numbers are symmetrically placed around the center, both interpretations lead to the same average of 236, confirming that our result is robust. Why Other Options Are Wrong: 232 and 234 are smaller than the true central value and would correspond to a range whose endpoints do not match 222 and 250 appropriately. 230 is even farther from the midpoint and clearly not the center of an evenly spaced set running from 222 to 250. Common Pitfalls: A frequent error is to assume that the average is always the middle term; this is only true when you know the exact count and list. Others may try to sum all terms manually, which is time consuming and prone to arithmetic mistakes. Remember that for any arithmetic progression, the average is simply half the sum of the first and last terms. Final Answer: The average of all even numbers between 222 and 250 is 236.

Question 2

In a class of 60 students there are 20 girls who scored an average of 40 marks in a test. If the average marks of the whole class is 60, what are the average marks of the boys in the class?

Accepted Answer

Introduction / Context: This question deals with averages of two subgroups (boys and girls) inside a larger group (the whole class). You are given the class average and the girls' average and asked to find the boys' average. This is a very typical weighted average problem seen in basic quantitative aptitude tests. Given Data / Assumptions: Total students in the class = 60. Number of girls = 20, so number of boys = 60 - 20 = 40. Average marks of the girls = 40 marks. Average marks of the entire class = 60 marks. Marks are all from the same test and on the same scale. Concept / Approach: Average marks multiplied by the number of students gives total marks. So we first compute the total marks of the entire class using the overall average, and the total marks of the girls using their average. Subtracting the girls' total from the class total yields the total marks scored by the boys. Dividing that total by the number of boys produces the boys' average marks. Step-by-Step Solution: Step 1: Total marks scored by all 60 students. Total class marks = 60 * 60 = 3600. Step 2: Total marks scored by 20 girls. Girls' total marks = 20 * 40 = 800. Step 3: Total marks scored by 40 boys. Boys' total marks = 3600 - 800 = 2800. Step 4: Average marks of the boys. Average boys' marks = 2800 / 40 = 70. Verification / Alternative Check: We can quickly verify by recombining the subgroup totals. Girls contribute 800 marks and boys contribute 2800 marks. Combined total = 3600 marks. Dividing by the total of 60 students, we get 3600 / 60 = 60 marks, which matches the original overall average. This confirms that the boys' average of 70 marks is consistent with the entire scenario. Why Other Options Are Wrong: 60 would imply that boys and class have the same average, which is impossible because girls score lower on average (40), so boys must score higher to pull the class average up to 60. 50 would pull the overall class average down below 60 when combined with girls averaging 40. 80 would make the class average higher than 60 because boys are a majority (40 out of 60) and such a high average would dominate the combined score. Common Pitfalls: A common error is to directly average 40 and 60, ignoring that there are more boys than girls. Some learners also forget to compute totals and rely on intuition about how averages combine. Always translate average information into totals first when dealing with subgroups, then work backwards to find the missing average. Final Answer: The average marks of the boys in the class is 70.

Question 3

An oil refinery buys crude oil at Rs. 3600 per barrel. During processing there is a 10% wastage. If the refinery wants to earn a 5% profit on its effective cost, then at what price per barrel should it sell the oil so that the selling price includes…

Accepted Answer

Introduction / Context: This question combines several commercial arithmetic ideas: wastage in quantity, desired profit percentage, and tax on the selling price. Such problems are very common in business mathematics and aptitude tests because they require careful step-by-step reasoning and accurate percentage calculations. Given Data / Assumptions: Cost price of crude oil for the refinery = Rs. 3600 per barrel purchased. There is 10% wastage in processing, so only 90% of the purchased oil is actually saleable. The refinery wants to earn a 5% profit on its effective cost per saleable barrel. An 8% tax is charged on the selling price, and the final quoted price must include this tax. We need to find the final selling price per barrel including tax. Concept / Approach: Because of the 10% wastage, one barrel purchased does not correspond to one barrel sold. For every barrel sold, more crude must be bought. First we compute the effective cost per saleable barrel by dividing the purchase cost by 0.9. Then we add 5% profit to get the base selling price before tax. Finally, we add 8% tax on this base price to get the final price that the customer pays. Each adjustment must be done multiplicatively and in the correct order. Step-by-Step Solution: Step 1: Compute effective cost per saleable barrel. Saleable quantity is 90% of purchased quantity, so cost per saleable barrel = 3600 / 0.9 = Rs. 4000. Step 2: Add 5% profit on this effective cost. Base selling price before tax = 4000 * 1.05 = Rs. 4200. Step 3: Add 8% tax on the selling price. Final selling price including tax = 4200 * 1.08 = 4536. So the refinery should charge Rs. 4536 per barrel (inclusive of tax). Verification / Alternative Check: We can check profits: effective cost per saleable barrel is Rs. 4000 and base selling price before tax is Rs. 4200. Profit per barrel before tax = 4200 - 4000 = 200, which is 200 / 4000 = 5%, as required. The tax of 8% on 4200 is 336, and 4200 + 336 = 4536, confirming that the final price correctly includes tax at the right rate. Why Other Options Are Wrong: Rs. 3674 and Rs. 3711 are too low; they do not even recover a proper 5% profit on the effective cost once wastage is considered. Rs. 4219 is closer but effectively ignores or miscalculates the combined effect of wastage, profit, and tax, and does not match the correct multiplicative sequence 3600 → 4000 → 4200 → 4536. Common Pitfalls: One common mistake is to take a 5% profit on the original cost of Rs. 3600 without accounting for wastage, which underestimates the needed selling price. Another error is to treat tax as part of profit or to incorrectly apply percentages additively (for example, adding 5% and 8% directly) instead of multiplying step by step. Keeping a clear order—first adjust for wastage, then add profit, then add tax—prevents these errors. Final Answer: The refinery should sell the processed oil at Rs. 4536 per barrel including 8% tax.

Question 4

Two students appeared for an examination. One of them secured 26 marks more than the other, and his marks were 75% of the sum of their marks. What are the marks obtained by the two students?

Accepted Answer

Introduction / Context: This problem is a simple algebra word problem involving two unknown quantities (the marks of two students) and two equations derived from the statements. It tests your ability to translate verbal conditions into algebraic equations and then solve them systematically. Given Data / Assumptions: There are two students; let their marks be x and y. One student scored 26 marks more than the other. The higher mark is equal to 75% of the sum of both students' marks. Marks are measured on the same scale (for example, out of 100). We must find the actual marks obtained by both students. Concept / Approach: Let the higher scoring student have marks x and the other student have marks y. Then x = y + 26. Also, x is 75% of (x + y), so x = 0.75 * (x + y). These two equations in x and y can be solved simultaneously. Solving them gives the unique pair of marks that satisfies both conditions. Step-by-Step Solution: Step 1: Express the relationships using equations. From the difference condition: x = y + 26. From the percentage condition: x = 0.75 * (x + y). Step 2: Use fractional form for clarity. Write 0.75 as 3/4: x = (3/4) * (x + y). Multiply both sides by 4: 4x = 3x + 3y, so x = 3y. Step 3: Combine with the first equation. We now have x = 3y and also x = y + 26. So 3y = y + 26, which gives 2y = 26 and hence y = 13. Step 4: Find x. x = 3y = 3 * 13 = 39. Thus, the two students scored 39 marks and 13 marks. Verification / Alternative Check: Check the conditions: the difference between their marks is 39 - 13 = 26, which matches the first condition. The sum of their marks is 39 + 13 = 52. Seventy five percent of this sum is 0.75 * 52 = 39, which is exactly the higher mark. Both conditions are satisfied, so the solution is correct. Why Other Options Are Wrong: 53 and 27 differ by 26, but 53 is not 75% of (53 + 27). 60 and 34 also differ by 26, yet 60 is not 75% of their sum. 78 and 52 keep the same difference, but 78 is not 75% of (78 + 52); the percentage condition fails. Common Pitfalls: A common mistake is to mix up which mark is 75% of the sum or to mis-handle the percentage as 75 / 100 in equations. Some learners also forget to convert 0.75 into a fraction before clearing the denominator, which can cause arithmetic slips. Always write both conditions clearly in algebraic form and solve systematically to avoid guesswork. Final Answer: The two students obtained marks of 39 and 13 respectively.

Question 5

In the first 32 overs of a cricket match, the scoring rate was 7.2 runs per over. What run rate is required in the remaining 18 overs so that the team reaches a total of 297 runs?

Accepted Answer

Introduction / Context: Run rate problems are very common in cricket-based aptitude questions. They test your understanding of averages and remaining requirement calculations. Here, you are given the scoring rate for an initial set of overs and a target total for the whole innings, and you must determine the required scoring rate in the remaining overs to reach that target. Given Data / Assumptions: First 32 overs have been completed. Run rate in these 32 overs = 7.2 runs per over. Total target score = 297 runs. Remaining overs = 18 (since a typical limit is 50 overs, but here only the remaining 18 are relevant). We must find the required run rate (runs per over) in these 18 overs to reach 297 runs. Concept / Approach: Run rate is just average runs per over. First we compute the runs already scored using the run rate in the first 32 overs. Then we subtract this from the target to find how many more runs are needed. Finally, we divide the required runs by the remaining overs to get the required run rate. Although in real matches runs are whole numbers, in aptitude problems a decimal run rate is acceptable for calculation purposes. Step-by-Step Solution: Step 1: Calculate runs scored in the first 32 overs. Runs scored = run rate * overs bowled = 7.2 * 32. 7.2 * 32 = 230.4 runs (mathematically, this is the product used for further calculation). Step 2: Find runs still required. Target score = 297 runs. Required runs = 297 - 230.4 = 66.6 runs. Step 3: Compute required run rate in the remaining 18 overs. Required run rate = required runs / remaining overs. Required run rate = 66.6 / 18 ≈ 3.7 runs per over. Verification / Alternative Check: If the team scores at 3.7 runs per over for 18 overs, it will score approximately 3.7 * 18 = 66.6 runs in those overs. Adding this to 230.4 gives total runs of about 297. The small decimal remainder is due to the decimal nature of the initial run rate, but for exam purposes, 3.7 is the appropriate required rate and matches the option given. Why Other Options Are Wrong: 4.3 or 4.9 runs per over would yield totals well above 297 when added to the initial runs, overshooting the target. 3.1 runs per over is too low and would not allow the team to reach 297; it would leave them short of the target. Common Pitfalls: One frequent mistake is to round intermediate values too early or to assume that 7.2 runs per over for 32 overs must give a whole number of runs. Another error is to average the initial run rate and some guess of the final run rate instead of calculating the required rate from remaining runs and overs. Always calculate runs already scored, subtract from target, then divide by remaining overs. Final Answer: The team must score at a rate of approximately 3.7 runs per over in the remaining 18 overs.

Question 6

The mean height of 25 boys in a class is 150 cm, and the mean height of 35 girls in the same class is 145 cm. What is the approximate combined mean height of all 60 students in the class?

Accepted Answer

Introduction / Context: This question illustrates a classic case of combining averages of two different groups into a single overall average. Boys and girls have different mean heights, and you are asked to find the combined mean height of the entire class. This is a typical weighted average problem in basic statistics and aptitude tests. Given Data / Assumptions: Number of boys = 25; their mean height = 150 cm. Number of girls = 35; their mean height = 145 cm. Total number of students = 25 + 35 = 60. Heights are measured in centimetres. We need the combined mean height of all 60 students, approximately. Concept / Approach: The combined mean height is found by dividing the total height of all students by the total number of students. To do this, we first find the total height of the boys from their mean and count, and similarly for the girls. Adding these totals gives the total height for the entire class. Dividing this by 60 yields the combined average height. Because the resulting value may not be an integer, we choose the closest option provided. Step-by-Step Solution: Step 1: Compute total height of the boys. Total height of boys = 25 * 150 = 3750 cm. Step 2: Compute total height of the girls. Total height of girls = 35 * 145 = 5075 cm. Step 3: Compute total height of all 60 students. Combined total height = 3750 + 5075 = 8825 cm. Step 4: Compute combined mean height. Combined mean height = 8825 / 60 ≈ 147.0833 cm. So the mean height is approximately 147 cm. Verification / Alternative Check: Notice that the boys' average (150 cm) is higher than the girls' average (145 cm), and there are more girls than boys. Therefore, the combined mean height must lie between 145 cm and 150 cm and closer to 145 cm but not too close, because 25 boys still influence the mean significantly. The approximate value 147 cm fits this reasoning and matches our calculation of 147.08 cm. Why Other Options Are Wrong: 145 cm would be equal to the girls' average and would ignore the taller boys; with 25 boys at 150 cm average, the combined mean cannot be as low as 145. 143 cm is below both subgroup averages, which is impossible for a combined mean of only these two groups. 146 cm is slightly lower than the true value and does not match the precise weighted calculation, which is closer to 147 cm. Common Pitfalls: A common mistake is to take a simple average of 150 and 145, giving 147.5, without weighting by the different numbers of boys and girls. Others may miscalculate totals or divide by the wrong total number of students. Always multiply each group average by its group size, add those totals, and then divide by the combined group size to get the correct weighted mean. Final Answer: The combined mean height of all 60 students is approximately 147 cm.

Question 7

Rakesh has marks in five subjects. The average of his marks in the first three subjects is 79, and the average of his marks in the last three subjects is 86. If his mark in the third subject is 80, what is his average mark across all five subjects?

Accepted Answer

Introduction / Context: This is a clever averages question that involves overlapping groups of subjects. You are given the average of the first three subjects, the average of the last three subjects, and the actual mark in the overlapping subject. The task is to determine the overall average in all five subjects. It tests your ability to manipulate sums and averages with overlapping data. Given Data / Assumptions: Rakesh has marks in 5 subjects, call them S1, S2, S3, S4 and S5. Average of first three subjects (S1, S2, S3) = 79. Average of last three subjects (S3, S4, S5) = 86. Mark in the third subject S3 = 80. We need the average of all five subjects combined. Concept / Approach: Average = total marks / number of subjects. From the given averages of first three and last three subjects, we can find the sums S1 + S2 + S3 and S3 + S4 + S5. Since S3 is common to both groups and its value is known, we can separate out sums S1 + S2 and S4 + S5. Adding these with S3 gives the total marks in all five subjects, and dividing by 5 yields the required overall average. Step-by-Step Solution: Step 1: Convert averages into totals. S1 + S2 + S3 = 79 * 3 = 237. S3 + S4 + S5 = 86 * 3 = 258. Step 2: Use the given value of S3. S3 = 80. So S1 + S2 = 237 - 80 = 157. And S4 + S5 = 258 - 80 = 178. Step 3: Find total marks in all five subjects. Total = (S1 + S2) + S3 + (S4 + S5) = 157 + 80 + 178 = 415. Step 4: Compute overall average. Average over 5 subjects = 415 / 5 = 83. Verification / Alternative Check: We can check consistency by reconstructing plausible marks or simply reusing the sums. S1 + S2 + S3 = 237, S3 + S4 + S5 = 258, and S3 = 80 give S1 + S2 = 157 and S4 + S5 = 178 as before. Adding all five subjects again gives 415, so 415 / 5 = 83, confirming the calculation. The value 83 also lies between 79 and 86, which is reasonable because the full set of five subjects includes both groups. Why Other Options Are Wrong: 81 and 82 are both lower than the correct overall mean and would not yield the correct totals when multiplied by 5. 85 is too high and would require the total to be 425, which contradicts the sums computed from the given averages. Common Pitfalls: One common mistake is to simply average 79 and 86, which ignores the overlapping subject and yields 82.5, not an integer and not correct. Another issue is forgetting to subtract the third subject when transitioning between the sum of three subjects and the sum of the other two. Keeping track of the overlapping subject S3 and working systematically with sums avoids these errors. Final Answer: Rakesh's average mark across all five subjects is 83.

Question 8

Consider the data set: 12, 1, 10, 1, 9, 3, 4, 9, 7, 9. What is the mode of this data?

Accepted Answer

Introduction / Context: This question is taken from basic statistics and tests your understanding of the concept of mode. The mode is the value that occurs most frequently in a data set. Unlike the mean and median, which are measures of central tendency based on position or arithmetic combinations, the mode is purely frequency based. Given Data / Assumptions: The data set is: 12, 1, 10, 1, 9, 3, 4, 9, 7, 9. All values are treated as distinct observations in a simple list (no grouping or class intervals). We assume that we are asked to find the unique mode if it exists, that is, the single most frequent value. Concept / Approach: The mode of a data set is defined as the value that appears the greatest number of times. To find it, we count the frequency of each distinct number in the list. The number with the highest frequency is the mode. If two or more numbers are tied for highest frequency, the data set is multimodal, but in this problem one value clearly appears most often. Step-by-Step Solution: Step 1: List each distinct value and count frequencies. Values present: 1, 3, 4, 7, 9, 10, 12. Count for 1: appears twice. Count for 3: appears once. Count for 4: appears once. Count for 7: appears once. Count for 9: appears three times. Count for 10: appears once. Count for 12: appears once. Step 2: Identify the highest frequency. The highest frequency is 3, corresponding to the value 9. Therefore, the mode of the data set is 9. Verification / Alternative Check: We can quickly verify by re-scanning the original list: 12, 1, 10, 1, 9, 3, 4, 9, 7, 9. Counting again, 9 appears at three positions, 1 at two positions, and all other numbers appear only once. Since 9 clearly has the maximum frequency, the mode is unambiguously 9. No other value matches its frequency. Why Other Options Are Wrong: 12 appears only once, so it cannot be the mode. 1 appears twice, but this is fewer than the three occurrences of 9. 7 also appears only once and therefore is not the most frequent value. Common Pitfalls: Learners sometimes confuse mode with mean or median and try to perform arithmetic operations instead of frequency counting. Another common error is to miscount the occurrences of each value, especially in longer lists. Taking a systematic approach—listing each distinct value and tallying occurrences—helps ensure that the correct mode is identified. Final Answer: The mode of the given data set is 9.

Question 9

The average of the marks of 17 students in an examination was first calculated as 71. Later it was found that one mark had been wrongly entered as 65 instead of 56, and another mark as 24 instead of 50. What is the correct average of the marks?

Accepted Answer

Introduction / Context: This is a typical “correction of average” problem in statistics. It shows how an initially computed mean can change when errors in individual data values are detected and corrected. You are given the originally reported average and details of the mistakes, and you must recompute the correct average by adjusting the total appropriately. Given Data / Assumptions: Number of students = 17. Originally calculated average = 71 marks. One student's mark was entered as 65 instead of the true value 56. Another student's mark was entered as 24 instead of the true value 50. All other marks are assumed to have been recorded correctly. We must find the correct average after adjusting for both errors. Concept / Approach: Average = total marks / number of students. First, we calculate the original total marks using the original average and number of students. Then we remove the wrong entries and add the correct ones. This yields the true total marks. Finally, we divide the true total by 17 again to get the corrected average. The key is to adjust the total by the net effect of correcting the two errors. Step-by-Step Solution: Step 1: Compute original total marks. Original total = 71 * 17 = 1207. Step 2: Adjust for the first mistake. Wrongly entered as 65 instead of 56, so the recorded total is 9 marks too high. Correcting this reduces the total by 65 - 56 = 9. Step 3: Adjust for the second mistake. Wrongly entered as 24 instead of 50, so the recorded total is 26 marks too low. Correcting this increases the total by 50 - 24 = 26. Step 4: Compute corrected total. Corrected total = 1207 - 9 + 26 = 1207 + 17 = 1224. Step 5: Compute corrected average. Corrected average = 1224 / 17 = 72. Verification / Alternative Check: Note that the net adjustment to the total is +17 marks (subtract 9, add 26). Increasing the total by 17 marks over 17 students increases the average by exactly 1 mark. So the corrected average must be 71 + 1 = 72, which matches the detailed calculation. This provides a quick mental check that our result is consistent. Why Other Options Are Wrong: 70 would imply that the corrected total is less than the original total, which contradicts the net increase of 17 marks after fixing the entries. 71 is the original incorrect average and does not reflect the corrected mark entries. 73 would require a total increase of 34 marks across 17 students, but our corrections only add a net 17 marks. Common Pitfalls: Learners sometimes forget to subtract the wrong values before adding the correct ones, or they confuse which wrong entry was too high and which was too low. Others try to re-average only the two corrected marks without considering the total. Remember that the average is based on the total of all observations, so corrections must always be applied to the total sum first. Final Answer: The correct average of the marks is 72.

Question 10

The average of the runs scored by a cricketer in his first 99 innings is 99. How many runs must he score in his 100th innings so that his average over 100 innings becomes exactly 100?

Accepted Answer

Introduction / Context: This is a classic cricket and averages question that appears frequently in aptitude exams. It deals with how a single new observation (the 100th innings) affects the overall average after many innings. The problem highlights that even a large additional score may change the average by only a small amount when many prior innings are already included. Given Data / Assumptions: Number of innings already played = 99. Average runs over these 99 innings = 99 runs per innings. We want the average over 100 innings to be 100 runs per innings. Runs in the 100th innings are unknown and need to be found. Concept / Approach: Average runs = total runs / number of innings. We first find the total runs scored in the first 99 innings using the existing average. Then we compute the total runs required after 100 innings to have an average of 100. The difference between the required total and the existing total gives the runs needed in the 100th innings. This is a straightforward but important application of average and total relationships. Step-by-Step Solution: Step 1: Compute total runs in the first 99 innings. Total runs after 99 innings = 99 * 99 = 9801. Step 2: Compute the total runs required for an average of 100 after 100 innings. Required total after 100 innings = 100 * 100 = 10000. Step 3: Find the runs needed in the 100th innings. Runs in 100th innings = required total - current total. Runs needed = 10000 - 9801 = 199. Verification / Alternative Check: If the batsman scores 199 runs in his 100th innings, then his total runs become 9801 + 199 = 10000. Dividing by 100 innings gives 10000 / 100 = 100 runs per innings, which matches the required new average. This confirms that 199 is indeed the correct score needed. Why Other Options Are Wrong: 99 or 100 runs would give totals of 9900 or 9901 respectively, leading to averages less than 100. 101 runs would give a total of 9902, yielding an average of only 99.02, still below 100. Common Pitfalls: Some students make the mistake of thinking that adding 1 run to the average means scoring only a little more in the next innings, such as 101 instead of 99. They forget that the new innings must pull up the average of all 100 innings, not just itself. The correct approach is always to work through total runs: multiply the average by the number of innings to get totals, then find the difference required to reach the new total. Final Answer: He must score 199 runs in his 100th innings to raise his average to 100.

Question 11

In a class of 65 students there are 39 girls. The average weight of these girls is 60 kg and the average weight of the whole class is 64 kg. What is the average weight of the boys in the class?

Accepted Answer

Introduction / Context: This question is another example of a weighted average and subgroup problem. The class is divided into girls and boys with different average weights, and the average weight of the entire class is also given. The goal is to use this information to determine the boys' average weight. Given Data / Assumptions: Total number of students = 65. Number of girls = 39, so number of boys = 65 - 39 = 26. Average weight of the girls = 60 kg. Average weight of the whole class = 64 kg. We must find the average weight of the boys, in kilograms. Concept / Approach: Average weight multiplied by the number of students gives total weight. We first compute the total weight of all 65 students using the class average. Next, we compute the total weight of the 39 girls using their average. Subtracting the girls' total weight from the class total gives the total weight of the 26 boys. Dividing this by 26 yields the average weight of the boys. Step-by-Step Solution: Step 1: Compute total weight of the full class. Total class weight = 64 * 65 = 4160 kg. Step 2: Compute total weight of the girls. Total girls' weight = 60 * 39 = 2340 kg. Step 3: Compute total weight of the boys. Total boys' weight = total class weight - total girls' weight. Total boys' weight = 4160 - 2340 = 1820 kg. Step 4: Compute average weight of the boys. Number of boys = 26. Average boys' weight = 1820 / 26 = 70 kg. Verification / Alternative Check: We can verify by recomputing the overall average. Girls contribute 2340 kg, boys contribute 1820 kg, giving a total of 4160 kg. Dividing by 65 students gives 4160 / 65 = 64 kg, which matches the given class average. This confirms that 70 kg is the consistent average for the boys. Why Other Options Are Wrong: 66 kg or 68 kg would give a smaller total weight for the boys, which would pull the overall class average below 64 kg, contradicting the question. 69 kg would be slightly low and would still not produce the exact total required for a 64 kg class average when combined with the girls' total weight. Common Pitfalls: Some learners try to find the boys' average by simply taking a difference between the two averages (for example, 64 and 60) without considering the number of students in each subgroup. Others miscalculate totals or forget to use the correct counts of boys and girls. Always work with totals: class total, girls' total, then boys' total, and finally compute the boys' average from their total and count. Final Answer: The average weight of the boys in the class is 70 kg.

Question 12

What is the average of the first seven multiples of 7?

Accepted Answer

Introduction / Context: This question asks about the average of the first several multiples of a given number, in this case 7. The multiples of any fixed number form an arithmetic progression. Knowing this allows you to use a simple formula or property to find the average without adding all the terms one by one. Given Data / Assumptions: We are dealing with multiples of 7. The first seven multiples of 7 are 7, 14, 21, 28, 35, 42 and 49. We need to find the average (arithmetic mean) of these seven numbers. All numbers are positive integers and equally spaced, with a common difference of 7. Concept / Approach: The multiples of 7 form an arithmetic progression (AP) with first term a = 7 and common difference d = 7. In any AP, the average of all its terms is equal to the average of the first and last terms, that is, (first term + last term) / 2. This shortcut saves time compared with adding all the terms. Alternatively, we could add the terms and divide by 7 to double check. Step-by-Step Solution: Step 1: Write the first seven multiples of 7 explicitly. 7, 14, 21, 28, 35, 42, 49. Step 2: Use the AP average property. First term = 7, last term = 49. Average = (first + last) / 2 = (7 + 49) / 2 = 56 / 2 = 28. So the required average is 28. Verification / Alternative Check: We can verify by direct summation: 7 + 14 + 21 + 28 + 35 + 42 + 49. Pairing terms symmetrically, (7 + 49) + (14 + 42) + (21 + 35) + 28 = 56 + 56 + 56 + 28 = 196. Dividing this total by 7 terms gives 196 / 7 = 28, which agrees with our shortcut method. Why Other Options Are Wrong: 7 and 14 are just single multiples; they are not central enough in the list of seven multiples to represent the overall average. 21 is one of the interior terms but is below the true central tendency of the full seven-term sequence. Common Pitfalls: Some learners mistakenly think the average of the first n multiples of 7 is always 7n / 2, which is not correct without checking the sequence. Others add the numbers but make arithmetic mistakes. Remember the simple rule: for any arithmetic progression, the average equals (first term + last term) / 2, which here simplifies the work greatly. Final Answer: The average of the first seven multiples of 7 is 28.

Question 13

In a class there are 28 boys and some girls. In a test, the mean score obtained by the boys is 12.5, while the mean score obtained by the girls is 14.5. If the overall average score of all the students is 13.1, what is the total number of students i…

Accepted Answer

Introduction / Context: This problem is another weighted average question where you know the averages of two subgroups (boys and girls) and the combined average of the class. The number of boys is known, but the number of girls is unknown. You must use the relationship between group totals and the overall average to find the total class size. Given Data / Assumptions: Number of boys = 28. Mean score of boys = 12.5 marks. Mean score of girls = 14.5 marks. Overall mean score of all students = 13.1 marks. Let the number of girls be g, and total students be 28 + g. Concept / Approach: The total marks of boys plus the total marks of girls must equal the overall mean multiplied by the total number of students. We compute the total marks of the boys using their mean and number. Then we represent the total marks of the girls using g and their mean. Setting the sum equal to 13.1 * (28 + g) gives an equation in g, which we solve. Once we find g, we add it to 28 to get the total class strength. Step-by-Step Solution: Step 1: Compute total marks of the boys. Total boys' marks = 28 * 12.5 = 350. Step 2: Express total marks of the girls in terms of g. Mean score of girls = 14.5, so total girls' marks = 14.5 * g. Step 3: Use the overall mean condition. Overall mean = 13.1 marks. Total class marks = 13.1 * (28 + g). Equation: 350 + 14.5g = 13.1 * (28 + g). Step 4: Solve for g. 13.1 * (28 + g) = 13.1 * 28 + 13.1g = 366.8 + 13.1g. So 350 + 14.5g = 366.8 + 13.1g. 14.5g - 13.1g = 366.8 - 350 → 1.4g = 16.8. g = 16.8 / 1.4 = 12. Step 5: Find total number of students. Total students = 28 + 12 = 40. Verification / Alternative Check: With 12 girls, total girls' marks = 14.5 * 12 = 174. Total marks for the class = 350 + 174 = 524. Dividing this by 40 students gives 524 / 40 = 13.1, which matches the given overall average, confirming that the class has 40 students. Why Other Options Are Wrong: 38 or 42 or 44 would correspond to different values of g, and substituting those values back into the equation would not produce an overall average of 13.1. Only g = 12 gives a consistent solution. Common Pitfalls: Some learners incorrectly attempt to average 12.5 and 14.5 directly, ignoring the different numbers of boys and girls. Others make algebraic errors when expanding and simplifying the equation, especially with decimals. Carefully computing the totals for each subgroup and then solving the linear equation step by step helps avoid these mistakes. Final Answer: The total number of students in the class is 40.

Question 14

Two students appeared for an examination. One student secured 21 marks more than the other, and his marks were equal to 80% of the sum of their marks. What were the marks obtained by the two students?

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Introduction / Context: This aptitude question tests linear equation skills and understanding of percentages in word problems. You must translate the relation between two students marks into algebra and then solve for their individual scores. Given Data / Assumptions: - Two students appeared for the same examination. - Let the lower score be y marks. - The higher score is 21 marks more, so it is y + 21. - The higher score is 80% of the sum of their marks. - All marks are assumed to be non negative and realistic for a test. Concept / Approach: We convert the statement into an equation. The sum of their marks is (y + y + 21). The higher score equals 0.8 times this sum. That equation can be solved for y, after which the higher score is obtained by adding 21. Finally we check which option matches the pair of marks. Step-by-Step Solution: Step 1: Let the lower score be y and the higher score be y + 21. Step 2: Sum of marks = y + (y + 21) = 2y + 21. Step 3: Given that higher score = 80% of sum, so y + 21 = 0.8(2y + 21). Step 4: Expand right side: y + 21 = 1.6y + 16.8. Step 5: Rearrange: 21 - 16.8 = 1.6y - y gives 4.2 = 0.6y. Step 6: Hence y = 4.2 / 0.6 = 7. Step 7: Higher score = y + 21 = 7 + 21 = 28. The marks are 28 and 7. Verification / Alternative check: With marks 28 and 7, the sum is 35. Eighty percent of 35 is 0.8 * 35 = 28, which matches the higher score. The difference between the scores is 28 - 7 = 21, which also matches the given condition. So the pair 28 and 7 satisfies both conditions exactly. Why Other Options Are Wrong: - 88 and 67 differ by 21 but 80% of their sum is not 88. - 89 and 68 also differ by 21 but do not match the 80% relation. - 98 and 77 differ by 21 but again fail the 80% condition. - None of these is a generic distractor and is not required since one option already fits perfectly. Common Pitfalls: Many learners misuse percentage and set 80% of the larger mark equal to the sum instead of 80% of the sum equal to the larger mark. Others forget to represent both conditions simultaneously and try to guess values. Writing a clear variable equation avoids confusion and makes the solution systematic. Final Answer: The two students obtained 28 and 7 marks respectively.

Question 15

In a class there are 9 boys and some girls. In a test the mean score of the boys is 12 and the mean score of the girls is 14. If the overall average score of all the students is 13.1, what is the total number of students in the class?

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Introduction / Context: This question checks understanding of weighted averages where two groups have different averages and sizes. You must infer the number of girls from the overall class average and then find the total strength of the class. Given Data / Assumptions: - Number of boys = 9. - Average score of boys = 12. - Let number of girls be g. - Average score of girls = 14. - Overall average of all students = 13.1. Concept / Approach: Average equals total score divided by number of students. We find the total score of boys and of girls in terms of g. The sum of these gives total class score. Dividing that sum by total students (9 + g) and equating it to 13.1 leads to a solvable linear equation for g. Then we add boys and girls to find the total number of students. Step-by-Step Solution: Step 1: Total score of boys = 9 * 12 = 108. Step 2: Total score of girls = 14g. Step 3: Total students = 9 + g. Step 4: Overall average = (108 + 14g) / (9 + g) = 13.1. Step 5: Write equation (108 + 14g) = 13.1(9 + g). Step 6: Expand right side: 108 + 14g = 117.9 + 13.1g. Step 7: Rearrange: 14g - 13.1g = 117.9 - 108 gives 0.9g = 9.9, so g = 11. Step 8: Total students = 9 + 11 = 20. Verification / Alternative check: With 9 boys and 11 girls, total score = 108 + 14 * 11 = 108 + 154 = 262. Overall average = 262 / 20 = 13.1, which matches the given value. Thus the solution is consistent with all data in the question. Why Other Options Are Wrong: - 19, 21, 22 and 24 all correspond to different numbers of girls and lead to overall averages that are not 13.1. - Quick mental guesses usually ignore the weighted nature of the average, so they do not satisfy the precise equation. Common Pitfalls: Learners sometimes take the simple average of 12 and 14, which is 13, and try to match 13.1 by trial and error without equations. Others forget that more girls pull the average closer to 14 while more boys pull it toward 12. Writing an equation using totals is the most reliable method. Final Answer: The total number of students in the class is 20.

Question 16

The average age of 5 members of a family is 24 years. The youngest member is 8 years old. What was the average age of the family, in years, at the time of the birth of the youngest member?

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Introduction / Context: Age and average problems are very common in aptitude tests. Here you must understand that when time moves backward, everyone except the newborn becomes younger, and the newborn has age zero at birth. Using this idea, we recalculate the earlier average of the same family. Given Data / Assumptions: - Number of family members now = 5. - Present average age = 24 years. - Youngest member is now 8 years old. - We assume that all 5 members were present at the time of birth of the youngest, with the youngest having age zero then. Concept / Approach: The present total age can be found by multiplying average by number of members. To go back 8 years, we subtract 8 years from the age of each of the 5 members, which subtracts 5 * 8 from the total age. At that time, the youngest had age zero but is still counted as a member, so we divide the earlier total by 5 to get the required earlier average. Step-by-Step Solution: Step 1: Present total age = 5 * 24 = 120 years. Step 2: We go back 8 years to the time of the youngest child birth. Step 3: Each of the 5 members was 8 years younger then. Step 4: Total age decreases by 5 * 8 = 40 years. Step 5: Earlier total age = 120 - 40 = 80 years. Step 6: Number of members then is still 5, including the newborn with age zero. Step 7: Earlier average age = 80 / 5 = 16 years. Verification / Alternative check: Move forward from birth to now. Starting from total 80, add 8 years for each of 5 members, which gives 80 + 40 = 120. Average then becomes 120 / 5 = 24, matching the given current average. This confirms that the earlier average of 16 years is consistent with the situation described. Why Other Options Are Wrong: - 20, 24 and 30 ignore the effect of subtracting 8 years from all members. - 18 does not lead back to the current average when 8 years are added to each member age. Common Pitfalls: A frequent mistake is to subtract 8 only once from the average instead of from each member. Another error is to assume that at the time of birth there were only 4 older members and forget that the newborn is also part of the family, with age zero. Always work with total ages rather than guessing averages directly. Final Answer: The average age of the family at the time of the youngest member birth was 16 years.

Question 17

Nine students of a class contribute money towards a common fund. Seven of them contribute Rs 50 each, and the remaining two contribute Rs 50 and Rs 90 more than this amount respectively. What is the average contribution of the class of 9 students?

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Introduction / Context: This question involves average contribution when different students contribute different amounts. Seven students pay the same fixed amount and two students pay larger amounts. The task is to compute the average contribution over all nine students. Given Data / Assumptions: - There are 9 students in total. - Seven students contribute Rs 50 each. - One student contributes Rs 50 more than Rs 50, that is Rs 100. - Another student contributes Rs 90 more than Rs 50, that is Rs 140. - We must find the average contribution per student. Concept / Approach: Average contribution equals total amount contributed divided by number of students. We first find the sum contributed by the seven equal contributors and then add the two higher contributions. Dividing this total by 9 gives the required average, which will be a non integer value. Step-by-Step Solution: Step 1: Contribution of each of the seven students = Rs 50. Step 2: Total contribution of these seven students = 7 * 50 = Rs 350. Step 3: Eighth student contributes 50 more than 50, so contributes Rs 100. Step 4: Ninth student contributes 90 more than 50, so contributes Rs 140. Step 5: Total contribution of all nine students = 350 + 100 + 140 = Rs 590. Step 6: Average contribution = total contribution / number of students = 590 / 9. Step 7: 590 / 9 is approximately equal to 65.56 rupees. Verification / Alternative check: We can express 590 / 9 as a mixed fraction. Dividing 590 by 9 gives quotient 65 and remainder 5, so the exact average is 65 5/9 rupees. Converting 5/9 to decimal gives about 0.56, so 65.56 is a good approximation. Multiplying this approximate average back by 9 returns about 590, confirming consistency within rounding limits. Why Other Options Are Wrong: - 60 is lower than the base contribution of 50 plus the effect of higher contributors. - 70, 75 and 80 are too high and do not match the precise total of Rs 590. - Only approximately 65.56 matches total 590 when multiplied by 9. Common Pitfalls: Some learners mistakenly add 50 and 90 directly to the average or misinterpret the phrase more than this amount. Another common error is averaging the three distinct values 50, 100 and 140 without considering how many students pay each amount. Always calculate the total rupees first, then divide by the total number of contributors. Final Answer: The average contribution of the 9 students is approximately Rs 65.56.

Question 18

The average age of 120 members of a society is 60.7 years. After 30 new members join, the average age of the whole group becomes 56.3 years. What is the average age, in years, of the newly joined members?

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Introduction / Context: This is a classic weighted average question involving an existing group and a group of new members. The overall average age drops after new members join, so the new members must be younger on average than the original members. The task is to compute that new average. Given Data / Assumptions: - Original number of members = 120. - Original average age = 60.7 years. - New members joining = 30. - New average age of all 150 members = 56.3 years. - We must find the average age of the 30 new members. Concept / Approach: The total age of the original group is original average multiplied by 120. The new total age after addition is new average multiplied by 150. The difference between these totals is exactly the combined age of the 30 new members. Dividing that difference by 30 gives the required average age for the newcomers. Step-by-Step Solution: Step 1: Original total age = 120 * 60.7 = 7284 years. Step 2: New total number of members = 120 + 30 = 150. Step 3: New total age = 150 * 56.3 = 8445 years. Step 4: Total age of new members = new total age minus original total age. Step 5: So, age of new members = 8445 - 7284 = 1161 years. Step 6: Number of new members = 30. Step 7: Average age of new members = 1161 / 30 = 38.7 years. Verification / Alternative check: Check the direction of change. Original average was 60.7; after adding members with average 38.7, overall average falls to 56.3, which lies between 38.7 and 60.7 and is closer to 60.7 because there are more original members. This confirms that the computed value is reasonable and consistent with average behaviour. Why Other Options Are Wrong: - 36.5, 37.2 and 38.3 would lead to a different combined total and not give an overall average of 56.3. - 40.1 is too high and would not reduce the average as much as required by the question. Common Pitfalls: Learners sometimes average 60.7 and 56.3 directly or forget to multiply by the number of members. Another mistake is to divide the change in average by the change in count without using totals. Always go back to total age = average * number of members for each group and then compare. Final Answer: The average age of the newly joined members is 38.7 years.

Question 19

A company starts with 8000 workers. The number of workers increases by 5% at the end of the first year, by 10% at the end of the second year and by 20% at the end of the third year. How many workers are there at the beginning of the fourth year?

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Introduction / Context: This percentage growth question is about successive increases. The number of workers grows each year by a given percentage, so the new workforce becomes the base for the next percentage increase. You must apply percentage multipliers in sequence rather than simply adding percentages. Given Data / Assumptions: - Initial number of workers = 8000. - Increase after first year = 5% of current number. - Increase after second year = 10% of current number. - Increase after third year = 20% of current number. - All increases are compounded year on year. Concept / Approach: A percentage increase of p% means the new quantity is multiplied by (1 + p/100). Because increases are successive, we multiply the original number by each factor in turn: 1.05, 1.10 and 1.20. The final product gives the number of workers at the start of the fourth year. Step-by-Step Solution: Step 1: After first year, workers = 8000 * 1.05. Step 2: 8000 * 1.05 = 8400 workers. Step 3: After second year, workers = 8400 * 1.10 = 9240. Step 4: After third year, workers = 9240 * 1.20. Step 5: 9240 * 1.20 = 11088. Step 6: This is the number of workers present at the beginning of the fourth year. Verification / Alternative check: Combine the multipliers directly: overall factor = 1.05 * 1.10 * 1.20. This equals 1.386. Multiplying 8000 by 1.386 gives 11088, which matches the detailed stepwise calculation. This confirms the result and shows that we cannot replace the combined growth with a simple 5 + 10 + 20 = 35% increase. Why Other Options Are Wrong: - 10188 and 11008 are values that could come from arithmetic mistakes or rounding. - 11808 and 12000 assume larger or linear percentage increases and ignore compounding details. Common Pitfalls: Many candidates add percentages and compute 8000 * 1.35, which gives 10800, not found here but still incorrect. Others apply percentages to the wrong base, such as always using 8000 rather than the updated count each year. Always treat each year change as a fresh percentage of the current number of workers. Final Answer: The company has 11088 workers at the beginning of the fourth year.

Question 20

The average of 11 numbers is 7. If every number is doubled, what will be the new average of these 11 numbers?

Accepted Answer

Introduction / Context: This is a basic concept check on how averages behave under scaling operations. The question asks what happens to the average if each number in the set is multiplied by a constant factor, in this case doubled. Given Data / Assumptions: - There are 11 numbers. - Their average is 7. - Each of the 11 numbers is doubled. - We must find the new average after doubling every number. Concept / Approach: Average equals total sum divided by number of terms. When each term is multiplied by a constant k, the total sum is also multiplied by k, while the number of terms stays the same. Therefore, the new average is simply k times the old average. Here k is 2 because we double each number. Step-by-Step Solution: Step 1: Let the 11 numbers be n1, n2, ..., n11. Step 2: Their sum S satisfies S / 11 = 7, so S = 7 * 11 = 77. Step 3: After doubling, each number becomes 2n1, 2n2, ..., 2n11. Step 4: New total sum S new = 2n1 + 2n2 + ... + 2n11 = 2(n1 + ... + n11) = 2S. Step 5: Hence S new = 2 * 77 = 154. Step 6: New average = S new / 11 = 154 / 11 = 14. Verification / Alternative check: Use the scaling property directly. Since each number is multiplied by 2, the average must also be multiplied by 2. Old average = 7, so new average = 2 * 7 = 14. This agrees with the detailed calculation and confirms the reasoning. Why Other Options Are Wrong: - 3.5 is obtained by halving instead of doubling. - 7 is the original average and ignores the change. - 10.5 and 21 come from incorrect or random multipliers applied to the original average. Common Pitfalls: Learners sometimes confuse operations on each term with operations on the number of terms, or they try to recalculate with assumed values. Remember that scaling every term by the same constant scales the average by exactly that constant. This rule saves time in many aptitude questions. Final Answer: The new average of the numbers is 14.

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