Question 1

The average weight of 100 students in a school is 32 kg. The average weight of the first 49 students is 30 kg and the average weight of the last 50 students is 34 kg. What is the weight of the 50th student in kilograms?

Accepted Answer

Introduction / Context:This question checks your understanding of averages and overlapping groups. One hundred students have a known average weight. You are told the average weight of the first 49 students and the average of the last 50 students. The 50th student does not belong to either of these two groups, and you must use the total information to find that student weight. This pattern appears frequently in aptitude tests. Given Data / Assumptions:- Total number of students = 100.- Average weight of all 100 students = 32 kg.- Average weight of the first 49 students = 30 kg.- Average weight of the last 50 students = 34 kg.- We are asked to find the weight of the 50th student alone. Concept / Approach:The key idea is that average multiplied by the number of students gives the total weight. We first compute the total weight of all 100 students. Then we compute the total weight of the first 49 students and the total of the last 50 students. The 50th student is not part of these two groups, so if we add the sums of the first 49 and last 50 we have accounted for all students except the 50th. This observation allows us to isolate the 50th student weight by simple subtraction. Step-by-Step Solution:Step 1: Total weight of all 100 students = 100 * 32 = 3200 kg.Step 2: Total weight of the first 49 students = 49 * 30 = 1470 kg.Step 3: Total weight of the last 50 students = 50 * 34 = 1700 kg.Step 4: Notice that the last 50 students are students numbered 51 to 100. The 50th student is not included in this group.Step 5: If we add the total weights of the first 49 and the last 50 students, we get 1470 + 1700 = 3170 kg.Step 6: The total weight of all 100 students is 3200 kg, which includes the 50th student, but the sum 3170 kg does not.Step 7: Therefore, weight of the 50th student = 3200 - 3170 = 30 kg. Verification / Alternative check:We can quickly verify consistency. If the 50th student weighs 30 kg, then the total of the first 49 plus the 50th is 1470 + 30 = 1500 kg for the first 50 students, giving an average of 1500 / 50 = 30 kg. The last 50 students have a total weight of 1700 kg and an average of 34 kg as given. Combined, 1500 + 1700 = 3200 kg, and the overall average remains 3200 / 100 = 32 kg, matching the question statement. Why Other Options Are Wrong:Weights like 25 kg, 32 kg or 33 kg for the 50th student lead to total weights that no longer produce an overall average of 32 kg when combined with the given subgroup totals. Simple substitution and recomputation of the total would show that the resulting averages do not match what the problem describes. Common Pitfalls:Some learners mistakenly assume that the 50th student is included in both subgroups or in neither and attempt more complicated setups. The key is to recognize that the first group covers students 1 to 49 and the second group covers 51 to 100, so the 50th student is missing from both. Once this is clear, the subtraction approach becomes straightforward and reliable. Final Answer:The weight of the 50th student is 30 kg.

Question 2

The average annual revenue of a company over 9 consecutive years is Rs. 76 lakhs. The average revenue for the first 5 years is Rs. 71 lakhs and the average revenue for the last 5 years is Rs. 83 lakhs. What is the revenue of the 5th year in lakhs?

Accepted Answer

Introduction / Context:This problem involves overlapping averages over several years of company revenue. You are given the average revenue over 9 years, the average over the first 5 years, and the average over the last 5 years. The 5th year is common to both the first group and the last group, and you must find the revenue in that particular year. This is a textbook example of how to handle overlapping intervals in average calculations for business mathematics. Given Data / Assumptions:- The company has revenue figures for 9 consecutive years.- Average revenue for all 9 years = Rs. 76 lakhs per year.- Average revenue for the first 5 years = Rs. 71 lakhs per year.- Average revenue for the last 5 years = Rs. 83 lakhs per year.- Year numbering is assumed as year 1 to year 9, and the last 5 years are years 5 to 9.- We must find the revenue for year 5 in lakhs. Concept / Approach:Let the revenues in years 1 to 9 be R1, R2, ..., R9. The first 5 year average gives the sum R1 + R2 + R3 + R4 + R5 and the last 5 year average gives R5 + R6 + R7 + R8 + R9. Notice that year 5 appears in both sums. The total over 9 years is R1 + ... + R9. If we add the first 5 year sum and the last 5 year sum, we count R5 twice, while all other years are counted once. Therefore the sum of the two 5 year totals equals the 9 year total plus R5. This leads to a simple equation for R5. Step-by-Step Solution:Step 1: Total revenue for all 9 years = 9 * 76 = Rs. 684 lakhs.Step 2: Total revenue for the first 5 years = 5 * 71 = Rs. 355 lakhs.Step 3: Total revenue for the last 5 years = 5 * 83 = Rs. 415 lakhs.Step 4: Let the revenue of the 5th year be R5 lakhs.Step 5: The sum of the first 5 years plus the sum of the last 5 years is (R1 + R2 + R3 + R4 + R5) + (R5 + R6 + R7 + R8 + R9).Step 6: This combined sum equals the total of all 9 years plus R5, because year 5 is counted twice: 355 + 415 = 684 + R5.Step 7: Add the left side: 355 + 415 = 770.Step 8: Set up the equation 770 = 684 + R5, so R5 = 770 - 684 = 86.Step 9: Therefore, revenue in the 5th year = Rs. 86 lakhs. Verification / Alternative check:Check the logic. If year 5 revenue is 86 lakhs, then the total for nine years is still 684 lakhs. The first 5 year sum must be 355, which includes R5, and the last 5 year sum must be 415, also including R5. Adding 355 and 415 produces 770, and subtracting the 9 year total of 684 leaves 86, agreeing with our computed value for R5. This confirms that the overlapping sum reasoning is correct. Why Other Options Are Wrong:Values like Rs. 88 lakhs or Rs. 82 lakhs do not satisfy the relation 355 + 415 = 684 + R5. Substituting any of those figures leads to a mismatch between the combined 5 year totals and the 9 year total. Only Rs. 86 lakhs keeps all averages and totals consistent with the information given in the problem. Common Pitfalls:Some students incorrectly think the last 5 years are years 4 to 8 instead of 5 to 9, while others forget that year 5 is double counted when adding the two 5 year sums. Another frequent error is trying to guess R5 directly from the three averages without using the total sums. Always convert averages to totals and use the double counting idea for overlapping intervals. Final Answer:The revenue for the 5th year is Rs. 86 lakhs.

Question 3

What is the average of all natural numbers from 21 to 39 inclusive?

Accepted Answer

Introduction / Context:This question is about finding the average of a sequence of consecutive natural numbers. The numbers run from 21 to 39 inclusive, and you must compute their arithmetic mean. This is a basic concept in aptitude tests and helps build familiarity with arithmetic progressions and averages. Given Data / Assumptions:- We are dealing with natural numbers from 21 to 39 inclusive.- The numbers form an arithmetic progression with common difference 1.- We must find the arithmetic mean of all these numbers. Concept / Approach:For any arithmetic progression with equally spaced terms, the average or arithmetic mean is simply the mean of the first and last terms. This works because the sequence is symmetric: the smallest and largest terms balance each other around the center, and similarly for the inner pairs. Therefore we do not need to list or add all terms; we can use the simple formula average = (first + last) / 2. Step-by-Step Solution:Step 1: Identify the first term of the sequence: 21.Step 2: Identify the last term of the sequence: 39.Step 3: Use the formula for the average of an arithmetic progression: average = (first term + last term) / 2.Step 4: Substitute the values: average = (21 + 39) / 2.Step 5: Compute the sum: 21 + 39 = 60.Step 6: Divide by 2: 60 / 2 = 30.Step 7: Therefore the required average is 30. Verification / Alternative check:If you want to check manually, you can write the numbers as pairs that add up to 60: (21, 39), (22, 38), (23, 37), (24, 36), (25, 35), (26, 34), (27, 33), (28, 32), (29, 31). The middle term is 30, and for each pair the average is 30. Because all pairs and the central value 30 balance around 30, the overall average is also 30. This confirms the formula based method. Why Other Options Are Wrong:Options like 29, 31 or 28 arise if someone mistakenly counts the number of terms or excludes an endpoint. However, the start and end points 21 and 39 are clearly specified and must be included. Any average different from (21 + 39) / 2 is inconsistent with the structure of this arithmetic progression. Common Pitfalls:A common mistake is to add all numbers one by one which is time consuming and prone to arithmetic errors, especially under exam pressure. Another error is to misread the range and stop at 38 instead of 39. Remember that for consecutive numbers, the average is always halfway between the smallest and largest term, which is a very quick and reliable shortcut. Final Answer:The average of all natural numbers from 21 to 39 is 30.

Question 4

What is the average of the first 39 even natural numbers?

Accepted Answer

Introduction / Context:This problem asks for the average of the first 39 even natural numbers. Even numbers follow a very regular pattern, so this is a straightforward application of the concept of averages in an arithmetic progression. Such questions are common in competitive exams to test quick calculation skills and understanding of sequences. Given Data / Assumptions:- We consider the first 39 even natural numbers.- The sequence of even natural numbers is 2, 4, 6, 8, and so on, with common difference 2.- We need to find the arithmetic mean of these 39 even numbers. Concept / Approach:The first 39 even natural numbers form an arithmetic progression where the first term is 2 and the common difference is 2. The nth term of this progression is 2n. For n = 39, the last term is 2 * 39 = 78. For any arithmetic progression, the average equals the mean of the first and last terms, because the terms can be paired symmetrically around the center. Step-by-Step Solution:Step 1: Identify the first even number in the list: 2.Step 2: Identify the last even number in the list, which is the 39th even number.Step 3: The nth even number is 2n, so the 39th even number is 2 * 39 = 78.Step 4: Use the arithmetic mean formula for an arithmetic progression: average = (first term + last term) / 2.Step 5: Substitute the values: average = (2 + 78) / 2.Step 6: Compute the sum: 2 + 78 = 80.Step 7: Divide by 2: 80 / 2 = 40.Step 8: Therefore, the average of the first 39 even natural numbers is 40. Verification / Alternative check:You can pair the numbers from opposite ends: (2, 78), (4, 76), (6, 74), and so on. Each pair sums to 80, so the average of each pair is 80 / 2 = 40. Since all pairs share the same average and the list is made entirely from such pairs, the overall average of the full set must also be 40. This further supports the calculated answer. Why Other Options Are Wrong:Values such as 39 or 20 might be chosen if someone splits or counts the numbers incorrectly, while 68 or 78 are closer to the higher end of the range and cannot possibly represent the average of numbers that start from 2. Only 40 lies exactly in the middle of the first and last term, which is the correct mean for this arithmetic progression. Common Pitfalls:Some students mistakenly think that the average of the first 39 even numbers should be equal to the 20th term or the common difference. Others forget to identify the last term correctly as 78. Always remember the simple rule: for evenly spaced numbers, the average is the midpoint between the smallest and largest term. Final Answer:The average of the first 39 even natural numbers is 40.

Question 5

A batsman scores 81 runs in his 16th match and thereby increases his average runs per match by 3 runs. What is his batting average after the 16th match?

Accepted Answer

Introduction / Context:This cricket based aptitude question involves averages and how they change when a new performance is added. A batsman has some unknown average for his first 15 matches. After scoring 81 runs in his 16th match, his average increases by 3 runs. You must determine his new average after this 16th match. This is a common pattern in sports related average problems in competitive exams. Given Data / Assumptions:- The batsman has already played 15 matches before the 16th match.- Let his average runs per match before the 16th match be a runs.- Total runs in 15 matches before the 16th match = 15a.- In the 16th match he scores 81 runs.- As a result, his average increases by 3 runs, so new average after 16 matches is a + 3.- We must find his average after the 16th match in runs per match. Concept / Approach:The idea is to translate the given information into an equation using the relationship average = total runs / number of matches. We express the new average after the 16th match in terms of the old total and the new score. Then we set this equal to the increased average a + 3. Solving this equation gives us the old average a, and therefore the new average a + 3. Step-by-Step Solution:Step 1: Let a be the batsman average runs per match for the first 15 matches.Step 2: Total runs in the first 15 matches = 15a.Step 3: In the 16th match he scores 81 runs, so new total runs after 16 matches = 15a + 81.Step 4: New average after 16 matches = (15a + 81) / 16.Step 5: We are told that this new average is 3 runs more than the old average a, so (15a + 81) / 16 = a + 3.Step 6: Multiply both sides by 16: 15a + 81 = 16a + 48.Step 7: Rearrange: 15a + 81 - 16a - 48 = 0 which simplifies to -a + 33 = 0.Step 8: Therefore a = 33.Step 9: New average after the 16th match = a + 3 = 33 + 3 = 36. Verification / Alternative check:If the old average is 33, the batsman total after 15 matches is 15 * 33 = 495 runs. After scoring 81 runs in the 16th match, the new total is 495 + 81 = 576 runs. The new average is 576 / 16 = 36 runs per match. The increase in average is 36 - 33 = 3 runs, which matches the condition in the question. This confirms that the new average is correct. Why Other Options Are Wrong:Values like 35, 34 or 33 do not produce the required 3 run increase when you back calculate the old average and total runs. Only an average of 36 is consistent with a score of 81 in the 16th match and the specified increase in average. You can verify by plugging each option back into the total and average formulas and checking whether the increase matches 3 runs. Common Pitfalls:Some learners mistakenly set (15a + 81) / 16 equal to 3 instead of a + 3, or they forget that the new average applies to 16 matches, not 15. Another common mistake is trying to guess the answer by trial without forming the equation, which can be slow and error prone. The clean algebraic method is fast, accurate and easy to reuse in similar problems. Final Answer:The batsman average after the 16th match is 36 runs per match.

Question 6

A batsman scores 111 runs in his 10th match and thereby increases his average runs per match by 5 runs. What is his batting average after the 10th match?

Accepted Answer

Introduction / Context:This is another cricket average question involving a change in average after a new innings. The batsman has an unknown average over his first 9 matches. After scoring 111 runs in the 10th match, his average increases by 5 runs. You must determine his new average after the 10th match. This question reinforces the same idea of average and total runs used in many sports and exam scenarios. Given Data / Assumptions:- The batsman has played 9 matches before the 10th match.- Let his average runs per match before the 10th match be a runs.- Total runs in 9 matches before the 10th match = 9a.- In the 10th match he scores 111 runs.- His average increases by 5 runs, so new average after 10 matches is a + 5.- We are asked to find this new average after 10 matches. Concept / Approach:As before, average is total runs divided by number of matches. Write the expression for the new average in terms of the old total and the new score, and set it equal to a + 5. Solving this simple linear equation provides the old average a, and adding 5 gives the required new average. This clear algebraic approach is the standard way to tackle such problems. Step-by-Step Solution:Step 1: Let the old average across 9 matches be a runs per match.Step 2: Total runs in the first 9 matches = 9a.Step 3: In the 10th match he scores 111 runs, so the new total runs after 10 matches = 9a + 111.Step 4: New average after 10 matches = (9a + 111) / 10.Step 5: The problem tells us that this new average is 5 runs more than the old average, so (9a + 111) / 10 = a + 5.Step 6: Multiply both sides by 10: 9a + 111 = 10a + 50.Step 7: Rearrange: 9a + 111 - 10a - 50 = 0 which simplifies to -a + 61 = 0.Step 8: Therefore a = 61.Step 9: New average after the 10th match = a + 5 = 61 + 5 = 66 runs per match. Verification / Alternative check:If the old average was 61, the batsman total after 9 matches was 9 * 61 = 549 runs. After scoring 111 runs in the 10th match, his new total becomes 549 + 111 = 660 runs. Dividing by 10 matches gives a new average of 660 / 10 = 66 runs per match. The increase in average is 66 - 61 = 5 runs, exactly as the question describes. This verifies the correctness of the calculation. Why Other Options Are Wrong:A new average of 61, 62, 64 or 60 does not satisfy the condition of a 5 run increase after scoring 111. For each of these, back calculating the old average and checking total runs leads to contradictions. Only the value 66 produces an old average of 61 and the correct total of 660 runs after 10 matches. Common Pitfalls:Common mistakes include using 9 instead of 10 in the denominator when computing the new average or thinking that the new average is 111 / 10. Another error is to confuse the increase in total runs with the increase in average. Always keep track of how many matches the average applies to, and build your equation from totals and averages carefully. Final Answer:The batsman average after the 10th match is 66 runs per match.

Question 7

The average weight of two football players P and Q is 38 kg. The average weight of P, Q and their coach T is 49 kg. What is the weight of the coach in kilograms?

Accepted Answer

Introduction / Context:This problem is a simple application of averages with a small group. You are given the average weight of two players and then the average weight when their coach is also included. Using these two pieces of information, you must determine the coach weight. This style of question is frequently seen in school level aptitude tests and entrance exams. Given Data / Assumptions:- Players P and Q have an average weight of 38 kg.- The average weight of P and Q together is 38 kg, so their combined weight is 2 * 38 kg.- When coach T is included with P and Q, the new average becomes 49 kg for the three persons.- We must find the coach T weight in kilograms. Concept / Approach:The key idea is again that average multiplied by the number of people gives the total weight. First, calculate the combined weight of P and Q from their average. Next, compute the combined weight of P, Q and T from the new average. The difference between these two totals must be the coach weight. This is a simple and reliable technique for small group average problems. Step-by-Step Solution:Step 1: Average weight of P and Q is 38 kg.Step 2: Total weight of P and Q together = 2 * 38 = 76 kg.Step 3: When coach T joins, there are 3 persons (P, Q and T) with average weight 49 kg.Step 4: Total weight of P, Q and T = 3 * 49 = 147 kg.Step 5: Coach weight T = total of three persons minus total of players P and Q.Step 6: T = 147 - 76 = 71 kg.Step 7: Therefore, the coach weight is 71 kg. Verification / Alternative check:If P and Q together weigh 76 kg and the coach weighs 71 kg, then the total for all three is 76 + 71 = 147 kg. The average for three persons is 147 / 3 = 49 kg, which matches the given average. The original average of P and Q alone is 76 / 2 = 38 kg, which also matches the problem statement. This confirms that the calculation is consistent and correct. Why Other Options Are Wrong:Values such as 46 kg, 76 kg or 91 kg do not produce a total of 147 kg when added to 76 kg, and so they do not yield an average of 49 kg for three persons. A quick recomputation of total and average with these alternative values will show that the resulting averages are not 49 kg. Hence they cannot be correct. Common Pitfalls:Students sometimes confuse whether the average is given for two persons or three persons, or they mistakenly divide by 2 instead of 3 when computing the new average. Another pitfall is to think that the coach weight is simply 49 - 38, ignoring the effect of the number of persons. Always convert averages to total sums and work with those totals to avoid such errors. Final Answer:The coach weight is 71 kg.

Question 8

Find the median of the following numbers: 14, 12, 12, 16, 13 and 18.

Accepted Answer

Introduction / Context:This question belongs to the topic of measures of central tendency in statistics, specifically the median. Unlike the mean, which is an arithmetic average, the median is the middle value when data are arranged in ascending or descending order. You are given six numbers and asked to find their median. Such simple median questions regularly appear in school exams and basic aptitude tests. Given Data / Assumptions:- The given numbers are 14, 12, 12, 16, 13 and 18.- We assume that each value represents one observation and that no value has extra weight beyond its appearance in the list.- We must compute the median of these six numbers. Concept / Approach:The median of a data set is the middle value when the observations are arranged in order. For an odd number of observations, the median is the single middle term. For an even number of observations, as in this case with six values, the median is the average of the two central terms after sorting. Therefore, our approach is to sort the data, identify the two middle values and then compute their mean. Step-by-Step Solution:Step 1: List the data: 14, 12, 12, 16, 13 and 18.Step 2: Arrange these values in ascending order: 12, 12, 13, 14, 16, 18.Step 3: Count the number of observations. There are 6 observations, which is an even number.Step 4: For an even number of observations, the median is the average of the 3rd and 4th values in the ordered list.Step 5: Identify the 3rd value: 13.Step 6: Identify the 4th value: 14.Step 7: Compute the median = (13 + 14) / 2 = 27 / 2 = 13.5.Step 8: Therefore, the median of the given data set is 13.5. Verification / Alternative check:You can think of the median as the value that has an equal number of observations on either side when sorted. In the ordered list 12, 12, 13, 14, 16, 18, the two central positions are between 13 and 14. There are two values below 13 and two values above 14, so the midpoint between 13 and 14 is the correct median. That midpoint is 13.5, confirming our earlier calculation. Why Other Options Are Wrong:Values such as 13 or 14 correspond to one of the central terms but ignore the fact that the sample size is even. The median must be the average of the two central values, not just one of them. Values like 14.5 or 15 do not represent the midpoint between 13 and 14 and therefore cannot be the correct median for this particular data set. Common Pitfalls:A frequent mistake is to forget to sort the data before finding the median. Another common error is to pick either the 3rd or 4th observation alone when there are an even number of data points, instead of averaging the two central ones. Always remember to order the data and, for an even count, compute the mean of the two middle observations. Final Answer:The median of the numbers 14, 12, 12, 16, 13 and 18 is 13.5.

Question 9

The simple interest on a sum of money for 20 years is equal to five fourths of the principal. What is the per annum rate of interest in percentage?

Accepted Answer

Introduction / Context:This problem involves the basic formula for simple interest in quantitative aptitude. You are told that over 20 years, the simple interest earned on a principal amount is equal to five fourths of the principal itself. The task is to find the annual rate of interest in percentage terms. This type of question helps reinforce the relationship between principal, rate, time and interest in financial mathematics. Given Data / Assumptions:- Let the principal amount be P rupees. - Time period t = 20 years. - Simple interest over 20 years is equal to five fourths of the principal, that is SI = (5 / 4) * P. - We need to find the rate of interest per annum r in percent. Concept / Approach:The simple interest formula is SI = (P * r * t) / 100, where P is principal, r is the annual rate in percent and t is the time in years. The problem directly gives SI as (5 / 4) * P. By equating the formula expression to this given value and simplifying, we can solve for r. Since P appears on both sides of the equation, it cancels out, making the calculation straightforward. Step-by-Step Solution:Step 1: Write the formula for simple interest: SI = (P * r * t) / 100.Step 2: Substitute t = 20 years into the formula: SI = (P * r * 20) / 100.Step 3: According to the question, SI is also equal to (5 / 4) * P.Step 4: Set the two expressions equal: (P * r * 20) / 100 = (5 / 4) * P.Step 5: Cancel P from both sides, assuming P is not zero.Step 6: This gives (r * 20) / 100 = 5 / 4.Step 7: Simplify left side: r * 20 / 100 = r / 5.Step 8: So r / 5 = 5 / 4.Step 9: Multiply both sides by 5: r = (5 / 4) * 5 = 25 / 4 = 6.25.Step 10: Therefore, the rate of interest per annum is 6.25 percent. Verification / Alternative check:Take a simple principal, for example P = 4 rupees for easy fractional calculation. If r = 6.25 percent and t = 20 years, then SI = (4 * 6.25 * 20) / 100 = (4 * 125) / 100 = 500 / 100 = 5 rupees. Five rupees is equal to five fourths of 4 rupees, so SI = (5 / 4) * P is satisfied. This numerical check confirms that the derived rate of 6.25 percent is consistent with the problem statement. Why Other Options Are Wrong:If r were 12.5, 2.5 or 5 percent, substituting into the formula SI = (P * r * 20) / 100 would not yield (5 / 4) * P. For example, with r = 5 percent, SI over 20 years would be (P * 5 * 20) / 100 = P, which is only equal to the principal, not five fourths of it. Only 6.25 percent produces the required interest to principal ratio. Common Pitfalls:Some students confuse simple interest with compound interest and attempt more complex formulas. Others misinterpret five fourths of the principal as a rate instead of an amount. Always read carefully whether the statement is about the interest amount or the rate. Here the interest amount over 20 years is compared directly to the principal, which fits perfectly into the simple interest formula. Final Answer:The rate of interest is 6.25 percent per annum.

Question 10

The average weight of 50 boys in a class is 45 kg. When one boy leaves the class, the average weight reduces by 0.1 kg. What is the weight of the boy who left the class, in kilograms?

Accepted Answer

Introduction / Context:This aptitude problem explores how removing one member from a group affects the average. You are given the average weight of 50 boys and told that when one boy leaves, the average decreases by 0.1 kg. Using this information you must determine the weight of the boy who left. This type of question checks understanding of how totals and averages interact when elements are added or removed. Given Data / Assumptions:- There are initially 50 boys in the class.- Average weight of these 50 boys is 45 kg.- After one boy leaves, the class has 49 boys.- The new average weight is 0.1 kg lower, that is 44.9 kg.- We must find the weight of the boy who left the class. Concept / Approach:The total weight of the group is always equal to average multiplied by the number of boys. We first calculate the total weight of all 50 boys. Then we compute the total weight after one boy leaves using the new average and 49 boys. The difference between the original total and the new total gives the weight of the boy who left. This approach is straightforward and widely used in average problems involving entry or exit of members. Step-by-Step Solution:Step 1: Initial average weight of 50 boys = 45 kg.Step 2: Initial total weight of all 50 boys = 50 * 45 = 2250 kg.Step 3: After one boy leaves, the new average is 45 - 0.1 = 44.9 kg.Step 4: Number of boys left in the class = 49.Step 5: New total weight after the boy leaves = 49 * 44.9 kg.Step 6: Compute the new total: 49 * 44.9 = 2200.1 kg.Step 7: Weight of the boy who left = initial total weight - new total weight.Step 8: Weight of the boy who left = 2250 - 2200.1 = 49.9 kg. Verification / Alternative check:We can verify by reasoning about the changed average. Removing a boy slightly lighter than the original average would increase the average, while removing a boy heavier than the original average would decrease it. The average dropped from 45 to 44.9, so the removed boy must be slightly heavier than the new average but slightly lighter than some of the others. Computing the exact difference between totals gives 49.9 kg, and recomputing the new average 2200.1 / 49 confirms the value of 44.9 kg exactly. Why Other Options Are Wrong:Weights like 50 kg or 49 kg would produce different new totals and would not reduce the average by exactly 0.1 kg. For example, if the boy weighed 50 kg, the new total would be 2250 - 50 = 2200 kg and the new average would be 2200 / 49 which is slightly less than 44.9, not exactly equal. Similar inconsistencies arise with the other options. Only 49.9 kg produces the correct new average of 44.9 kg. Common Pitfalls:Students sometimes subtract 0.1 kg directly from 45 and assume that this is the boy weight, which is incorrect because that value is an average, not an individual weight. Others mistakenly treat 49 as the divisor instead of 50 when computing the original total. Always compute the original and new totals, then use their difference to find the individual weight that left the group. Final Answer:The weight of the boy who left the class is 49.9 kg.

Question 11

If a + 1/a = 2 for a nonzero real number a, what is the value of a^5 + 1/a^5?

Accepted Answer

Introduction / Context:This is an algebraic identity problem that often appears in quantitative aptitude and algebra sections. You are given a relation involving a variable a and its reciprocal 1/a, and you must compute a higher power expression a^5 + 1/a^5. Such questions test understanding of algebraic manipulation and recognition that the given equation may uniquely determine the value of a. Given Data / Assumptions:- a is a nonzero real number so that 1/a is defined.- It is given that a + 1/a = 2.- We are required to find the value of a^5 + 1/a^5.- All calculations are in real numbers. Concept / Approach:One way to handle expressions involving a and 1/a is to use algebraic identities and successive squaring. However, in this particular problem, the equation a + 1/a = 2 is very special. If we multiply both sides by a and rearrange terms, we can obtain a quadratic equation in a. Solving that quadratic shows that a has a single real value, which allows us to compute a^5 and 1/a^5 directly without heavy algebraic expansion. Step-by-Step Solution:Step 1: Start from the given equation a + 1/a = 2.Step 2: Multiply both sides by a to remove the denominator: a^2 + 1 = 2a.Step 3: Rearrange to standard quadratic form: a^2 - 2a + 1 = 0.Step 4: Recognize this as a perfect square: a^2 - 2a + 1 = (a - 1)^2.Step 5: Set (a - 1)^2 = 0, so a - 1 = 0 and a = 1.Step 6: Since a is nonzero and the equation has only one real root, a must be 1.Step 7: Now compute a^5 + 1/a^5 for a = 1.Step 8: a^5 = 1^5 = 1 and 1/a^5 = 1/1^5 = 1.Step 9: Therefore, a^5 + 1/a^5 = 1 + 1 = 2. Verification / Alternative check:Substitute a = 1 back into the original relation: a + 1/a = 1 + 1 = 2, which satisfies the given condition exactly. Since the quadratic equation derived from the original relation has only one real root, there is no other possible real value for a. Thus, any expression in a and 1/a must evaluate consistently when we use a = 1, confirming that a^5 + 1/a^5 equals 2. Why Other Options Are Wrong:Values such as 0, 1, 3 or 4 would arise only if a had some other value. For example, if a were 0 or -1, then a + 1/a either is not defined or does not equal 2. Since the equation a + 1/a = 2 forces a to be exactly 1, any value other than 2 for a^5 + 1/a^5 would contradict the algebraic steps performed above. Common Pitfalls:One mistake is to attempt long expansions using identities such as (a + 1/a)^2 and (a + 1/a)^3 without first checking whether a itself can be determined. This leads to unnecessary work. Another error is to ignore the possibility that the given relation might produce a unique solution for a. When you see simple equations like a + 1/a = 2, always consider solving directly for a before doing power expansions. Final Answer:The value of a^5 + 1/a^5 is 2.

Question 12

In the first 39 overs of a cricket match, the run rate of a team was 4.6 runs per over. What run rate is required in the remaining 11 overs so that the team reaches a target of 252 runs?

Accepted Answer

Introduction / Context:This is a run rate calculation problem from cricket, a favourite topic in quantitative aptitude. The team has already played 39 overs at a known run rate and now needs to chase a target of 252 runs. You must determine the required run rate in the remaining 11 overs so that the team can reach the target. This tests basic arithmetic with averages and a clear understanding of how run rate relates to total runs. Given Data / Assumptions:- Total overs in the innings = 50 overs (39 already played and 11 remaining, though the exact full quota is not strictly needed). - Run rate in the first 39 overs = 4.6 runs per over. - Target score = 252 runs. - Overs already completed = 39. - Overs remaining = 11. - We must find the required average run rate in the last 11 overs. Concept / Approach:The total runs scored so far can be calculated by multiplying the run rate by the overs already bowled. The remaining runs to be scored are equal to the target minus runs already scored. Once we know how many runs remain and how many overs are left, the required run rate is simply remaining runs divided by remaining overs. This is exactly the same concept as average speed or average marks, but applied to cricket runs. Step-by-Step Solution:Step 1: Runs scored in the first 39 overs = 39 * 4.6.Step 2: Compute 39 * 4.6 = 39 * 46 / 10 = 1794 / 10 = 179.4 runs.Step 3: Target runs = 252, so remaining runs to be scored = 252 - 179.4.Step 4: Calculate the remaining runs: 252 - 179.4 = 72.6 runs.Step 5: Overs remaining = 50 - 39 = 11 overs.Step 6: Required run rate in the remaining overs = remaining runs / remaining overs = 72.6 / 11.Step 7: Perform the division: 72.6 / 11 = 6.6 runs per over.Step 8: Therefore, the team must score at an average of 6.6 runs per over in the last 11 overs to reach 252 runs. Verification / Alternative check:Multiply the required run rate back by the remaining overs: 6.6 * 11 = 72.6. Adding this to the runs already scored gives 179.4 + 72.6 = 252. This exactly matches the target score, confirming that the run rate of 6.6 runs per over is correct. Any small difference would show up immediately when you recompute the final total. Why Other Options Are Wrong:If the required rate were 7.2, then runs in the last 11 overs would be 7.2 * 11 = 79.2 and the final total would be 179.4 + 79.2 = 258.6, which overshoots the target. With 5.7 or 6.3 runs per over, the total runs would fall short of 252. Only 6.6 runs per over produces exactly the target score. Common Pitfalls:One common error is to ignore the decimal part of the run rate 4.6 and approximate it as 5, which significantly changes the calculation. Another is to forget that the target includes all runs, not just those after a certain over. Always work with the exact run rate, calculate the precise runs scored and the precise remaining runs, and then divide by the remaining overs for the required rate. Final Answer:The required run rate in the last 11 overs is 6.6 runs per over.

Question 13

In a 50 over one day cricket match, Team A scores at an average of 4.5 runs per over for the first 30 overs and 5.5 runs per over for the remaining 20 overs. Team B plays the full 50 overs while chasing and loses by 10 runs. What is the average runs…

Accepted Answer

Introduction / Context:This is a cricket run rate comparison problem. Team A scores at two different average rates in different phases of its innings. Team B then bats the full 50 overs while chasing this total and falls short by 10 runs. You must calculate Team B average scoring rate. Such questions combine average calculations, total runs and a comparison between two teams performance. Given Data / Assumptions:- Total overs in the match for each team = 50 overs.- Team A scores at 4.5 runs per over in the first 30 overs.- Team A scores at 5.5 runs per over in the remaining 20 overs.- Team B plays all 50 overs and loses by 10 runs.- We must find the average runs per over scored by Team B. Concept / Approach:The first step is to determine Team A total score using the given run rates for the two segments of its innings. Once we know this total, Team B score must be 10 runs less because it loses by 10 runs. Dividing Team B total runs by 50 overs gives the average runs per over scored by Team B. The entire solution uses only basic average and multiplication operations. Step-by-Step Solution:Step 1: Runs scored by Team A in the first 30 overs = 30 * 4.5.Step 2: Calculate this: 30 * 4.5 = 135 runs.Step 3: Runs scored by Team A in the remaining 20 overs = 20 * 5.5.Step 4: Calculate this: 20 * 5.5 = 110 runs.Step 5: Total runs scored by Team A = 135 + 110 = 245 runs.Step 6: Team B loses by 10 runs, so Team B total score = 245 - 10 = 235 runs.Step 7: Team B plays all 50 overs, so average runs per over for Team B = 235 / 50.Step 8: Compute the average: 235 / 50 = 4.7 runs per over.Step 9: Therefore, Team B average scoring rate is 4.7 runs per over. Verification / Alternative check:To check, multiply Team B average by total overs: 4.7 * 50 = 235. Since Team A scored 245 runs, Team B is indeed 10 runs short, confirming the match result described. Also note that Team B average is slightly higher than Team A early run rate of 4.5 but lower than Team A later run rate of 5.5, which is reasonable for a chasing effort that falls slightly short. Why Other Options Are Wrong:An average of 5.2 runs per over would give Team B a total of 5.2 * 50 = 260 runs, which would win the match instead of losing by 10 runs. Similarly, 4.4 or 4.6 runs per over would produce totals of 220 or 230 runs, which do not match the given margin of defeat. Only 4.7 runs per over corresponds exactly to a total of 235 runs and a 10 run loss against 245. Common Pitfalls:Some students mistakenly average the two run rates of Team A without considering the number of overs in each segment, or they try to average Team A and Team B run rates directly. Others forget to subtract 10 from Team A total to get Team B score. Always compute exact totals from the given averages and then apply the stated winning margin to determine the chasing team total. Final Answer:Team B scored at an average rate of 4.7 runs per over.

Question 14

In a class of 60 students there are 30 girls. The average weight of these 30 girls is 58 kg and the average weight of the full class is 63 kg. What is the average weight of the boys in the class, in kilograms?

Accepted Answer

Introduction / Context:This averages question involves two subgroups within a class, girls and boys. You are told the size and average weight of the girls group, as well as the overall average weight of the entire class. From this information, you must determine the average weight of the boys. This is a classic weighted average problem that appears frequently in aptitude tests. Given Data / Assumptions:- Total number of students in the class = 60. - Number of girls = 30. - Therefore, number of boys = 60 - 30 = 30. - Average weight of the girls = 58 kg. - Average weight of the full class = 63 kg. - We must find the average weight of the boys in kilograms. Concept / Approach:The combined total weight of all students can be computed using the overall average. The total weight of the girls group can be calculated from their average and count. The difference between the combined total and the girls total gives the total weight of the boys. Dividing this by the number of boys yields the average weight of the boys. This approach is an example of decomposing a weighted average into component groups. Step-by-Step Solution:Step 1: Compute total weight of all 60 students using the overall average.Step 2: Total weight of class = 60 * 63 = 3780 kg.Step 3: Compute total weight of the 30 girls using their average.Step 4: Total weight of girls = 30 * 58 = 1740 kg.Step 5: Total weight of boys = total class weight - total girls weight = 3780 - 1740 = 2040 kg.Step 6: Number of boys = 30.Step 7: Average weight of boys = total weight of boys / number of boys = 2040 / 30.Step 8: Compute 2040 / 30 = 68 kg.Step 9: Therefore, the average weight of the boys is 68 kg. Verification / Alternative check:We can confirm the result by recomputing the combined average. If boys average 68 kg and girls average 58 kg, with 30 of each, then total boys weight is 30 * 68 = 2040 kg and total girls weight is 30 * 58 = 1740 kg. Combined total is 2040 + 1740 = 3780 kg. Dividing 3780 by 60 students gives an overall average of 63 kg, which matches the given data. This confirms that the boys average weight must be 68 kg. Why Other Options Are Wrong:If the boys average were 67, 66 or 65 kg, the total combined weight would differ and the overall average would not remain 63 kg. For example, with an average of 66 kg for boys, total boys weight would be 1980 kg, and the class total would be 1980 + 1740 = 3720 kg, which gives an overall average of 3720 / 60 = 62 kg, not 63 kg. Hence these alternatives do not satisfy the conditions. Common Pitfalls:Some students attempt to average the two given averages 58 and 63 directly without considering that the group sizes are equal or different. Other students miscalculate the total weights due to simple multiplication errors. Always convert each group average into a group total, sum or subtract as required, and then divide by the correct group size to obtain the missing average. Final Answer:The average weight of the boys in the class is 68 kg.

Question 15

The mean of 21 observations, all of them distinct, is 40. If each observation greater than the median is increased by 21, what will be the new mean of all observations?

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Introduction / Context:This statistics question links the mean and median of a data set and describes a transformation applied only to the values greater than the median. You are told the mean of 21 distinct observations. Then each observation greater than the median is increased by 21, and you must find the new mean. This problem is useful for understanding how selective changes in a data set affect the overall mean. Given Data / Assumptions:- There are 21 observations, all different from one another.- The mean (average) of these 21 observations is 40. - This means the total of all 21 observations is 21 * 40. - Afterward, each observation greater than the median is increased by 21. - For 21 ordered observations, the median is the 11th observation. - Therefore, there are 10 observations greater than the median. - We must find the new mean after each of those 10 observations is increased by 21. Concept / Approach:The mean is total sum divided by the number of observations. If we know how much the total sum changes, we can easily calculate the new mean. The key observation here is that only the 10 largest values change, and each of them increases by 21. This means the total increase in the sum is 10 * 21. Since the number of observations remains 21, we divide the new total by 21 to get the new mean. We do not need to know the individual values or even the median itself. Step-by-Step Solution:Step 1: Original mean = 40.Step 2: Number of observations = 21.Step 3: Original total sum of all observations = 21 * 40 = 840.Step 4: For 21 ordered observations, the median is the 11th observation.Step 5: Observations greater than the median are the top 10 observations (positions 12 to 21).Step 6: Each of these 10 observations is increased by 21, so the total increase in the sum = 10 * 21 = 210.Step 7: New total sum after the change = original total + increase = 840 + 210 = 1050.Step 8: The number of observations remains 21.Step 9: New mean = new total sum / number of observations = 1050 / 21.Step 10: Compute 1050 / 21 = 50.Step 11: Therefore, the new mean of the observations is 50. Verification / Alternative check:You can imagine a simple illustrative set of 21 values arranged in increasing order. Increasing each of the 10 largest values by 21 clearly adds 210 to the total sum. Since the number of observations stays the same, the mean must increase by 210 / 21 = 10. Starting from an original mean of 40, the new mean becomes 40 + 10 = 50. This reasoning perfectly matches the calculated result. Why Other Options Are Wrong:A new mean of 30 or 45 would correspond to a decrease or a smaller increase in the total, which contradicts the fact that several values have been raised by a positive amount. A mean of 40 would imply no change in total, which is not true because 10 numbers are increased by 21 each. The value 50.5 might appear close, but it corresponds to a total sum that is not an integer multiple of the increased increments. Only 50 is consistent with adding 210 to the total and dividing by 21. Common Pitfalls:Some learners mistakenly think that only the median increases or that the number of modified observations is 11 instead of 10. Others attempt to find the actual median value, which is unnecessary. The key is to understand that there are exactly 10 values greater than the median in a set of 21 distinct ordered observations and that each of those 10 values increases by 21, so the effect on the mean is straightforward to compute. Final Answer:The new mean of the observations will be 50.

Question 16

In a temple, the average number of fruits offered in a week (7 days) is 75. The average number of fruits offered in six days excluding Tuesday is 72. How many fruits were offered on Tuesday?

Accepted Answer

Introduction / Context:This problem is a straightforward average and total calculation scenario involving daily offerings at a temple. You are given the average number of fruits offered over a full week and a separate average over six days excluding Tuesday. From this, you must determine how many fruits were offered on Tuesday. Such problems help solidify comfort with translating between averages and totals in day wise contexts. Given Data / Assumptions:- The time period considered is a full week of 7 days. - Average number of fruits offered per day over the full week = 75. - So the total number of fruits offered over 7 days is 7 * 75. - Average number of fruits offered over six days excluding Tuesday = 72. - So the total number of fruits offered in those six days is 6 * 72. - We must find the number of fruits offered on Tuesday alone. Concept / Approach:The guiding idea is that average equals total divided by number of days. First, compute the total fruits for the entire week using the weekly average. Next, compute the total for the six non Tuesday days using their average. The difference between the weekly total and the six day total gives the fruits offered on Tuesday, because Tuesday is the only day not included in the six day group. Step-by-Step Solution:Step 1: Total fruits offered over 7 days = weekly average * 7 = 75 * 7.Step 2: Compute 75 * 7 = 525 fruits.Step 3: Total fruits offered in six days excluding Tuesday = 6 * 72.Step 4: Compute 6 * 72 = 432 fruits.Step 5: The only day not included in this second total is Tuesday.Step 6: Fruits offered on Tuesday = total for full week - total for the six non Tuesday days.Step 7: Fruits offered on Tuesday = 525 - 432 = 93 fruits.Step 8: Therefore, 93 fruits were offered on Tuesday. Verification / Alternative check:We can verify by recalculating the averages. Suppose fruits offered on Tuesday are 93. The total for the entire week is 432 plus 93 = 525, and the average is 525 / 7 = 75, which matches the weekly average. The total for the six days excluding Tuesday is 432, and the average for those six days is 432 / 6 = 72, which matches the given six day average. This confirms that 93 is the correct number for Tuesday. Why Other Options Are Wrong:If Tuesday had 90, 92 or 72 fruits, then the recalculated weekly and six day averages would not match the given values simultaneously. For instance, with 90 fruits on Tuesday, the weekly total becomes 522 and the weekly average becomes 522 / 7 which is not exactly 75. Only 93 satisfies both average conditions together. Common Pitfalls:Some students mistakenly divide the difference of the averages by the number of days or miscount the number of days in the week. Others attempt to average the two given averages directly. The correct method is always to convert each average to a total using the corresponding number of days, then subtract to isolate the unknown day total. Final Answer:The number of fruits offered on Tuesday is 93.

Question 17

Nineteen friends go to a restaurant for dinner. Thirteen of them spend Rs 79 each, and each of the remaining six spends Rs 4 more than the overall average expenditure of all nineteen. What is the total amount spent by the group?

Accepted Answer

Introduction: (Reserved duplicate to ensure clarity that the computed total is not among the numerical choices.) This mirrors the earlier hotel bill structure and reinforces solving with an unknown overall average A. Given Data / Assumptions: 13 spend Rs 79 each; 6 spend A + 4 each. Total = 19A. Concept / Approach: Set 13*79 + 6*(A+4) = 19A and solve for A, then compute 19A. Step-by-Step Solution: 1051 + 6A = 19A ⇒ 13A = 1051 ⇒ A ≈ 80.8461538. Total ≈ 19 * 80.8461538 ≈ Rs 1,536.08. Final Answer: Since Rs 1,536.08 is not listed numerically among options a–d, the correct choice remains None of these.

Question 18

Seven years ago, the average age of A, B, and C was 51 years. Today A is 3 years older than B, and B is 3 years older than C. What are their present ages (in years) for A, B, and C?

Accepted Answer

Introduction: Age-averages often require translating averages into totals and then applying given relationships between people. Here, we know a past average and fixed differences among the three persons' ages. We compute the current total and solve using linear relations. Given Data / Assumptions: Seven years ago: average of A, B, C = 51 years ⇒ past total = 3 * 51 = 153. Each person aged 7 years since then ⇒ current total adds 3 * 7 = 21. A = B + 3; B = C + 3 ⇒ A = C + 6. Concept / Approach: Compute the present total first, then express all ages in terms of the youngest (C). Solve the single-variable equation for C, then back-calculate B and A using the given differences. Step-by-Step Solution: Present total = 153 + 21 = 174. Let C = x. Then B = x + 3, A = x + 6. Sum = x + (x + 3) + (x + 6) = 3x + 9. Set 3x + 9 = 174 ⇒ 3x = 165 ⇒ x = 55. So C = 55, B = 58, A = 61. Verification / Alternative check: Seven years ago the ages would be 54, 51, and 48. Average then = (54 + 51 + 48) / 3 = 153 / 3 = 51, which matches the given value. Why Other Options Are Wrong: 54, 51, and 48: These are the ages seven years ago, not the present ages. 55, 58, and 61 or 48, 51, and 54 or 62, 59, and 56: These violate the required order A = B + 3 and B = C + 3 with the correct present total 174. Common Pitfalls: Using 51 as the present average, or forgetting to add 7 years to each person before solving. Always update the total when time passes for multiple people. Final Answer: The present ages are 61, 58, and 55 years for A, B, and C respectively.

Question 19

At a marriage party, 32% are women, 54% are men, and there are 196 children. How many men are present at the party?

Accepted Answer

Introduction: This question mixes percentages with an absolute headcount to find the total and then the number of men. Recognizing that all groups are percentages of the same total lets us back out the total from the known count of children. Given Data / Assumptions: Women = 32% of total. Men = 54% of total. Children = remainder = 100% − (32% + 54%) = 14% of total. Children count = 196 (absolute). Concept / Approach: If 14% corresponds to 196, then total attendees T satisfy 0.14 * T = 196. Solve for T, then compute 54% of T to get the number of men. Step-by-Step Solution: 0.14 * T = 196 ⇒ T = 196 / 0.14 = 1400. Men = 54% of 1400 = 0.54 * 1400 = 756. Verification / Alternative check: Women = 32% of 1400 = 448. Children = 14% of 1400 = 196. Check: 448 + 756 + 196 = 1400, which matches the total. Why Other Options Are Wrong: 448: That is the number of women (32% of 1400), not men. 332 and 324: Do not equal 54% of the computed total. 672: Corresponds to 48% of 1400, not the given 54% for men. Common Pitfalls: Using 196 as 32% or 54% mistakenly, or adding percentages incorrectly. Always ensure percentages sum to 100% and identify which share matches the given absolute count. Final Answer: The number of men present is 756.

Question 20

A school has 200 students, of which 7/10 are boys. How many girls are in the school?

Accepted Answer

Introduction: A simple fraction and complement problem: if a fraction of the total are boys, the rest are girls. Convert the fraction to a count and subtract from the total. Given Data / Assumptions: Total students = 200. Boys = 7/10 of total. Girls = total − boys. Concept / Approach: Compute boys as (7/10) * 200, then subtract from 200 to get girls. This is a direct application of fractions and complements. Step-by-Step Solution: Boys = (7/10) * 200 = 140. Girls = 200 − 140 = 60. Verification / Alternative check: Girls are 3/10 of the total. (3/10) * 200 = 60, confirming the result. Why Other Options Are Wrong: 80 and 120: These correspond to 40% and 60% and do not match the remaining 30%. 40: Too small (20%). 140: That is the number of boys, not girls. Common Pitfalls: Confusing the fraction for boys with the fraction for girls or miscomputing (7/10) * 200. Always compute one and subtract from the total. Final Answer: The number of girls is 60.

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