Q: The average of 7 consecutive integers is 20. If these integers are taken in increasing order, what is the largest integer in this sequence?

Introduction / Context: This problem deals with consecutive integers and their average. For an odd number of consecutive terms, the average is exactly equal to the middle term. Once we identify the middle term, it is very easy to reconstruct the full sequence and find the largest value. Given Data / Assumptions: There are 7 consecutive integers. Their average (mean) is 20. We need to find the largest integer among these 7 consecutive numbers. Concept / Approach: For any set of consecutive integers with an odd count, the average is the middle element. If the 7 numbers are: n1, n2, n3, n4, n5, n6, n7 then n4 is the middle number because there are 3 numbers before it and 3 numbers after it. Hence the average equals n4. We then step up and down by 1 to get the remaining numbers, and identify the largest. Step-by-Step Solution: Step 1: Since there are 7 consecutive integers, the middle term is the 4th term. Step 2: Average = 20, so the middle term n4 = 20. Step 3: The sequence is: n1 = 20 - 3 = 17, n2 = 20 - 2 = 18, n3 = 20 - 1 = 19, n4 = 20, n5 = 20 + 1 = 21, n6 = 20 + 2 = 22, n7 = 20 + 3 = 23. Step 4: Largest integer in the sequence = n7 = 23. Verification / Alternative check: Check average: sum = 17 + 18 + 19 + 20 + 21 + 22 + 23. 17 + 18 = 35, +19 = 54, +20 = 74, +21 = 95, +22 = 117, +23 = 140. Average = 140 / 7 = 20, which matches the given condition. Why Other Options Are Wrong: Options A (21) and B (22) are values in the sequence but are not the largest term. Option D (24) and E (25) are larger but would force the average to be greater than 20 if they were part of a 7-term consecutive sequence centered at 20. Common Pitfalls: Confusing the average with one of the end terms rather than the middle term can lead to an incorrect sequence. Some students forget that there are 3 steps of +1 to go from the middle to the largest term when there are 7 consecutive numbers. Final Answer: The largest of the seven consecutive integers is 23.

Question 1

The average weight of three persons L, M and N taken together is 93 kg. If the average weight of L and M together is 89 kg and the average weight of M and N together is 96.5 kg, what is the weight of M in kilograms?

Accepted Answer

Introduction / Context: This is a straightforward averages and linear equations question involving the weights of three people. You are given the average of all three, as well as the averages of two different pairs. From this information, you must determine the individual weight of one person. Problems like this test your ability to convert averages to totals and then use simple algebra to isolate unknowns. Given Data / Assumptions: - Persons are L, M and N. - Average weight of L, M and N together = 93 kg. - Average weight of L and M together = 89 kg. - Average weight of M and N together = 96.5 kg. - We must find the weight of M in kilograms. Concept / Approach: The average of a group multiplied by the number of people gives the total weight of that group. We use this to find the total weights of the groups L + M + N, L + M and M + N. Then we subtract to find individual weights. Subtracting L + M from L + M + N gives N. Once N is known, we use M + N to solve for M. This is a clean and systematic way to handle such problems. Step-by-Step Solution: Step 1: Total weight of L, M and N = 3 * 93 = 279 kg.Step 2: Total weight of L and M = 2 * 89 = 178 kg.Step 3: Total weight of M and N = 2 * 96.5 = 193 kg.Step 4: From step 1 and step 2, find N: N = (L + M + N) - (L + M) = 279 - 178 = 101 kg.Step 5: Now use M + N = 193, so M + 101 = 193.Step 6: Solve for M: M = 193 - 101 = 92 kg.Step 7: Therefore, the weight of M is 92 kg. Verification / Alternative check: Check the calculation by reconstructing the averages. If M = 92 and N = 101, then L + M = 178 implies L = 178 - 92 = 86 kg. Now, L + M + N = 86 + 92 + 101 = 279 kg and 279 / 3 = 93 kg, matching the given average for all three. Also, L + M = 86 + 92 = 178 and 178 / 2 = 89, and M + N = 92 + 101 = 193 and 193 / 2 = 96.5, matching the pair averages. This confirms that the answer is consistent. Why Other Options Are Wrong: If M were 86, 88, 95 or 101 kg, the pair sums and averages would not match all three pieces of given information simultaneously. For example, if M were 86, then from L + M = 178 we would get L = 92, but this would not align with the value of N required to maintain the overall average of 93 kg and the given average for M and N. Common Pitfalls: Some students attempt to solve by guessing values rather than using totals and equations, which can be time consuming and error prone. Others confuse which totals correspond to which averages. Always compute the group totals first, subtract to get individual weights, and verify by checking all given averages at the end. Final Answer: The weight of M is 92 kg.

Question 2

Mahesh buys 3 shirts at an average price of Rs 1250 each. He then buys 2 more shirts at an average price of Rs 1450 each. What is the overall average price in rupees of all 5 shirts that he buys?

Accepted Answer

Introduction / Context: This question is a direct application of the concept of weighted average. Mahesh buys two sets of shirts at different average prices and we must determine the combined average price for all the shirts together. Such questions appear frequently in aptitude tests and are useful for understanding how averages combine across groups with different sizes and values. Given Data / Assumptions: - First purchase: 3 shirts at an average price of Rs 1250 each. - Second purchase: 2 shirts at an average price of Rs 1450 each. - Total number of shirts purchased altogether = 5. - We must find the overall average price per shirt for all 5 shirts. Concept / Approach: The overall average price is calculated by dividing the total amount spent by the total number of shirts. We first compute the cost of the first group and the cost of the second group. Then we add these costs to get the total expenditure and divide by the total number of shirts. This is an example of combining averages using weighted sums rather than simply averaging the two averages directly. Step-by-Step Solution: Step 1: Cost of the first 3 shirts = 3 * 1250 = Rs 3750.Step 2: Cost of the next 2 shirts = 2 * 1450 = Rs 2900.Step 3: Total cost for all 5 shirts = 3750 + 2900 = Rs 6650.Step 4: Total number of shirts = 3 + 2 = 5.Step 5: Overall average price per shirt = total cost / total number of shirts = 6650 / 5.Step 6: Compute 6650 / 5 = 1330.Step 7: Therefore, the average price of all 5 shirts is Rs 1330. Verification / Alternative check: The two group averages are 1250 and 1450. Since Mahesh bought slightly fewer shirts at the higher price (2 shirts) compared to the cheaper ones (3 shirts), the combined average should be closer to 1250 than to 1450, but still greater than 1250. The value 1330 fits this expectation, confirming that the result is reasonable as well as mathematically correct. Why Other Options Are Wrong: An average of 1370 or 1390 would place the combined average too close to or above the higher segment average of 1450, which is not possible when more shirts were bought at the lower price. Averages of 1310 or 1290 are lower than the true combined average 1330 and do not match the correct total expenditure when multiplied by 5. Common Pitfalls: Many students mistakenly take the average of 1250 and 1450 directly, that is (1250 + 1450) / 2 = 1350, ignoring the fact that the number of shirts in each price group is different. To avoid this, always compute the total expenditure for each group and then find the overall average from the grand total and total quantity. Final Answer: The overall average price of the 5 shirts is Rs 1330 per shirt.

Question 3

In a one day cricket match of 50 overs per innings, Team A scores at a run rate of 5.3 runs per over in its full 50 overs. Team B is chasing this score and, with 5 overs left to play, the required run rate to tie the match is 7.2 runs per over. What…

Accepted Answer

Introduction / Context: This question combines average rate concepts with a cricket run chase scenario. Team A has already batted and scored at a steady run rate. Team B is chasing and we are told the required run rate in the final overs to tie the match. From this information, we must find Team B current score after 45 overs. This type of question is common in quantitative aptitude sections involving sports contexts. Given Data / Assumptions: - Total overs in an innings = 50. - Team A run rate over 50 overs = 5.3 runs per over. - Team A therefore bats full 50 overs. - Team B is chasing Team A total. - After 45 overs, 5 overs remain for Team B. - Required run rate in the remaining 5 overs to tie the match = 7.2 runs per over. - We must find Team B score at the end of the 45th over. Concept / Approach: First, calculate Team A total runs from its run rate. This becomes the target to tie for Team B. Next, compute how many runs Team B needs to score in the final 5 overs using the required run rate. Subtract this required number of runs from the target to obtain Team B current score after 45 overs. This is a direct application of the relationship total runs = run rate * overs. Step-by-Step Solution: Step 1: Team A total runs = 5.3 runs per over * 50 overs.Step 2: Compute Team A total: 5.3 * 50 = 265 runs.Step 3: To tie the match, Team B must also score 265 runs.Step 4: Required run rate for the last 5 overs is 7.2 runs per over.Step 5: Runs needed in the last 5 overs = 7.2 * 5 = 36 runs.Step 6: Therefore, Team B current score after 45 overs must be 265 - 36 = 229 runs.Step 7: So Team B is on 229 runs at the end of the 45th over. Verification / Alternative check: If Team B is on 229 at 45 overs, and then scores 36 runs in the last 5 overs at 7.2 runs per over, their final total is 229 + 36 = 265, which exactly equals Team A total. This corresponds to a tie, matching the problem statement that 7.2 is the required rate to tie the match. Hence the current score of 229 is consistent. Why Other Options Are Wrong: If Team B score were 238, then to reach 265 they would need 27 runs from 5 overs, implying a required rate of 5.4 runs per over, not 7.2. If the score were 254, they would need only 11 runs from 5 overs, which is 2.2 runs per over. A score of 265 would already tie the match before the final 5 overs. The value 229 alone produces exactly 36 runs needed and a required run rate of 7.2. Common Pitfalls: Some students mistakenly treat 7.2 as the run rate over the entire inning instead of just the last 5 overs. Others forget that a tie means equal scores and not exceeding the target. Always work logically: compute the target, compute runs required in the remaining overs from the required rate, and then subtract from the target to obtain the current score. Final Answer: Team B score at the end of the 45th over is 229 runs.

Question 4

A batsman scores 98 runs in the 17th match of his career. As a result, his average runs per match increase by 2.5 runs. What was his batting average per match before playing the 17th match?

Accepted Answer

Introduction / Context: This is another cricket averages question. The batsman has a certain average over his first 16 matches. After scoring 98 runs in the 17th match, his overall average increases by 2.5 runs per match. We are asked to find his original average before the 17th match. This type of question tests your ability to convert average information into equations and solve for unknowns. Given Data / Assumptions: - Number of matches before the latest match = 16. - Let the old average be a runs per match. - In the 17th match, the batsman scores 98 runs. - New average after 17 matches is a + 2.5 runs per match. - We need to find the old average a. Concept / Approach: The total runs before the 17th match can be written as 16a. After the 17th match, the total runs become 16a + 98. The new average is this total divided by 17, and the problem tells us that this value is a + 2.5. Using this fact, we create an equation involving a and solve it. The process is very similar to other average change problems and relies on careful algebra. Step-by-Step Solution: Step 1: Old total runs in 16 matches = 16a.Step 2: After scoring 98 in the 17th match, new total runs = 16a + 98.Step 3: New average after 17 matches = (16a + 98) / 17.Step 4: We are told this equals a + 2.5, so (16a + 98) / 17 = a + 2.5.Step 5: Multiply both sides by 17: 16a + 98 = 17a + 42.5.Step 6: Rearrange: 16a + 98 - 17a - 42.5 = 0, giving -a + 55.5 = 0.Step 7: Hence a = 55.5.Step 8: Therefore, the batsman average before the 17th match was 55.5 runs per match. Verification / Alternative check: If a = 55.5, then total runs after 16 matches are 16 * 55.5 = 888 runs. Adding 98 from the 17th match gives new total runs of 986. New average after 17 matches is 986 / 17 = 58 runs per match. The increase in average is 58 - 55.5 = 2.5 runs, which matches the condition given. This confirms that the original average of 55.5 runs is correct. Why Other Options Are Wrong: If the old average had been 58, 60.5, 63 or 57, then using the same process would not produce an increase of exactly 2.5 runs after scoring 98. For instance, if a were 58, the new total after 17 matches would give a different new average increment than 2.5. Only a value of 55.5 satisfies the equation built from the given conditions. Common Pitfalls: Sometimes students mistakenly divide by 16 instead of 17 when computing the new average, or they set the new average equal to 2.5 rather than a + 2.5. Carefully label the old average, old total, new total and new average, and double check that the number of matches in the denominator matches the number of matches played at that stage. Final Answer: The batsman average before the 17th match was 55.5 runs per match.

Question 5

What is the average (arithmetic mean) of all integers between 100 and 200 inclusive that are exactly divisible by 13?

Accepted Answer

Introduction / Context: This question deals with averages of numbers in an arithmetic progression. You are asked to find the average of all integers between 100 and 200 that are divisible by 13. Because these numbers form a regular sequence with a fixed difference, there is a quick way to find the average without listing and adding all of them. Given Data / Assumptions: - We consider integers from 100 to 200 inclusive. - We only include those numbers that are divisible by 13. - We must find the average (arithmetic mean) of these selected numbers. Concept / Approach: All integers divisible by 13 form an arithmetic progression with common difference 13. The first multiple of 13 in the given range and the last multiple of 13 in the range determine the sequence. For any arithmetic progression, the average equals the midpoint of the first and last term, that is (first + last) / 2. So our main task is to identify the first and last multiples of 13 between 100 and 200 and then use this formula. Step-by-Step Solution: Step 1: Find the smallest multiple of 13 that is at least 100.Step 2: 13 * 7 = 91 (too small), 13 * 8 = 104, which lies within 100 to 200, so the first term is 104.Step 3: Find the largest multiple of 13 that is at most 200.Step 4: 13 * 15 = 195, and 13 * 16 = 208 (too large), so the last term in the range is 195.Step 5: The numbers divisible by 13 in the range form the sequence 104, 117, 130, ..., 195 with common difference 13.Step 6: Average of an arithmetic progression = (first term + last term) / 2.Step 7: Average = (104 + 195) / 2 = 299 / 2 = 149.5.Step 8: Thus, the required average is 149.5. Verification / Alternative check: You can list all multiples of 13 between 100 and 200 to confirm: 104, 117, 130, 143, 156, 169, 182, 195. There are 8 terms. Pairing from the ends, 104 + 195 = 299, 117 + 182 = 299, 130 + 169 = 299, 143 + 156 = 299. Each pair sums to 299, so the average of each pair is 299 / 2 = 149.5. Therefore the average of all 8 numbers is also 149.5, which agrees with our formula based method. Why Other Options Are Wrong: Values like 147.5, 145.5 or 143.5 correspond to midpoints of different ranges of numbers and do not match the specific first and last multiple of 13 in the 100 to 200 interval. The value 151.5 is higher than the midpoint of 104 and 195 and would correspond to a different set of end values. Only 149.5 is exactly the average of the true first and last multiples of 13 in the given range. Common Pitfalls: Some students forget to include the boundary values correctly or misidentify the first and last multiples of 13. Others attempt to compute the sum of all numbers individually, which is slower and prone to arithmetic mistakes. Remember that in any evenly spaced sequence, the average is simply the midpoint between the first and last term. Final Answer: The average of the numbers between 100 and 200 that are divisible by 13 is 149.5.

Question 6

The average marks of 18 students in an examination is 60. Later it is found that the marks of one student were wrongly entered as 63 instead of 36. What is the correct class average after the error is rectified?

Accepted Answer

Introduction / Context: This problem illustrates how a single incorrect data entry can change the average and how to correct the average once the mistake is discovered. The marks of one student were entered incorrectly, leading to an overstated class average. After identifying the error, we must recompute the correct average. This type of question is very common in aptitude tests dealing with averages and data correction. Given Data / Assumptions: - Number of students = 18. - Initially calculated average marks = 60. - Therefore, initial total marks based on the incorrect data = 18 * 60. - One student marks were wrongly entered as 63 instead of 36. - We must find the correct average after replacing 63 with 36. Concept / Approach: The idea is to first compute the incorrect total marks from the given incorrect average. Then we adjust this total by subtracting the wrong entry and adding the correct mark. After obtaining the corrected total, we divide by the number of students to get the corrected average. This is a simple and reliable correction method for average problems with misentered values. Step-by-Step Solution: Step 1: Incorrect total marks based on the wrong data = 18 * 60 = 1080.Step 2: We know that 63 was recorded instead of the true mark 36.Step 3: Difference between wrong and correct mark = 63 - 36 = 27.Step 4: The incorrect total is larger than the true total by 27 marks.Step 5: Correct total marks = 1080 - 27 = 1053.Step 6: Number of students remains 18.Step 7: Correct average = 1053 / 18.Step 8: Compute 1053 / 18. Since 18 * 58 = 1044 and 18 * 0.5 = 9, we have 1044 + 9 = 1053, so the average is 58.5.Step 9: Therefore, the correct class average is 58.5 marks. Verification / Alternative check: To verify, imagine the true total is 1053. If one student mark is 36 and the sum of the other 17 students marks is 1053 - 36 = 1017, adding them gives back 1053. Dividing by 18 gives 58.5, so the new average is consistent. Also, the incorrect mark 63 is 27 higher than 36, so using 63 instead of 36 inflated the total by 27 and the average by 27 / 18 = 1.5, which matches the difference between 60 and 58.5. Why Other Options Are Wrong: The options 59, 59.5 and 58 do not reflect the exact correction of 27 marks spread over 18 students. Option 60 is simply the original incorrect average and does not incorporate the correction. Only 58.5 matches the corrected total marks and the reduced average after replacing 63 with 36. Common Pitfalls: Some students mistakenly add 27 instead of subtracting it from the total, or they divide the difference 27 by the wrong number of students. Others forget that the number of students remains unchanged. To avoid these errors, always identify whether the wrong entry was too high or too low, adjust the total in the correct direction and divide by the correct count of data points. Final Answer: The correct average marks of the class is 58.5.

Question 7

The average of 44 consecutive odd numbers is 144. Based on this information, what is the largest odd number in this consecutive sequence?

Accepted Answer

Introduction / Context: This question uses the properties of arithmetic progressions and odd numbers. You are told that the average of 44 consecutive odd numbers is 144 and you must determine the largest number in the sequence. Because consecutive odd numbers differ by 2, they form a regular arithmetic progression, and the mean is directly related to the first and last terms. Given Data / Assumptions: - We have 44 consecutive odd numbers. - Average (mean) of these 44 numbers = 144. - The numbers are odd and consecutive, so the common difference is 2. - We need to find the largest odd number in this sequence. Concept / Approach: For any arithmetic progression, the average of all terms equals the average of the first and last terms, that is (first + last) / 2. For an odd count of terms, this is also the middle term, but here the count is even (44 terms). However, the relation between first and last still holds. If we denote the first term as a and the last term as l, then (a + l) / 2 = 144 and l can be expressed in terms of a and the number of terms using the common difference. Solving these equations yields both a and l. Step-by-Step Solution: Step 1: Let the first odd number in the sequence be a and the last (largest) number be l.Step 2: For an arithmetic progression, average = (a + l) / 2. Given average is 144, so (a + l) / 2 = 144.Step 3: Thus a + l = 288.Step 4: Since the numbers are consecutive odd numbers, the common difference is 2 and the number of terms n = 44.Step 5: For an arithmetic sequence, l = a + (n - 1) * d, where d is common difference.Step 6: Here l = a + (44 - 1) * 2 = a + 43 * 2 = a + 86.Step 7: Substitute l into a + l = 288: a + (a + 86) = 288.Step 8: This gives 2a + 86 = 288, so 2a = 202 and a = 101.Step 9: Now l = a + 86 = 101 + 86 = 187.Step 10: Therefore, the largest number in the sequence is 187. Verification / Alternative check: As a quick check, the first odd number is 101 and the last is 187. The midpoint between these two is (101 + 187) / 2 = 288 / 2 = 144, which matches the given average. The number of terms from 101 to 187 with step 2 is (187 - 101) / 2 + 1 = 86 / 2 + 1 = 43 + 1 = 44, so both the count and the average are consistent. Why Other Options Are Wrong: Values like 189, 191 or 193 would result in a different average when paired with the corresponding first term. For example, if the largest number were 189, the first number would adjust to keep the average 144, but the resulting sequence might not have exactly 44 terms. Only 187 produces both the required average and the correct number of consecutive odd terms. Common Pitfalls: Students sometimes confuse the formula for the nth term with the formula for average, or they accidentally use a common difference of 1 instead of 2 for odd numbers. Others forget that there are 44 terms, not 43. Keeping track of the common difference and the number of terms is essential for a correct answer. Final Answer: The largest odd number in the sequence is 187.

Question 8

A batsman makes 100 runs in the 25th match of his career, and this innings increases his average runs per match by 1.4 runs. What was his average runs per match before playing the 25th match?

Accepted Answer

Introduction / Context: This question is again about cricket averages and the effect of one innings on a batsman overall average. The batsman scores 100 runs in his 25th match, and this raises his average by 1.4 runs per match. From this information, we need to determine his average before this match. This is a standard type of average increment problem that appears in competitive exams. Given Data / Assumptions: - Before the 25th match, the batsman has played 24 matches. - Let his old average be a runs per match. - In the 25th match he scores 100 runs. - New average after 25 matches is a + 1.4 runs per match. - We must calculate the old average a. Concept / Approach: We compute the total runs before the 25th match as 24a. After the 25th match, the total runs become 24a + 100. The new average is this new total divided by 25 and is given to be a + 1.4. Setting up the equation (24a + 100) / 25 = a + 1.4 allows us to solve for a using simple algebra. This method is common for problems where a new data point changes the overall average. Step-by-Step Solution: Step 1: Old total runs in 24 matches = 24a.Step 2: After the 25th match, new total runs = 24a + 100.Step 3: New average after 25 matches = (24a + 100) / 25.Step 4: This new average is given to be a + 1.4, so (24a + 100) / 25 = a + 1.4.Step 5: Multiply both sides by 25: 24a + 100 = 25a + 35.Step 6: Rearrange: 24a + 100 - 25a - 35 = 0, which gives -a + 65 = 0.Step 7: Hence a = 65.Step 8: Therefore, the batsman average before the 25th match was 65 runs per match. Verification / Alternative check: If a = 65, then total runs after 24 matches are 24 * 65 = 1560 runs. After scoring 100 runs in the 25th match, new total runs = 1560 + 100 = 1660. New average = 1660 / 25 = 66.4 runs per match. The increase in average is 66.4 - 65 = 1.4 runs, which confirms that the old average of 65 is correct. Why Other Options Are Wrong: If the batsman original average were 55, 75, 45 or 70, the algebraic equation derived from the given condition would not hold, and the new average would not increase by exactly 1.4 runs after scoring 100. Only the value 65 satisfies both the relationship between old and new totals and the given increment in average. Common Pitfalls: Common errors include using 24 instead of 25 in the denominator for the new average or forgetting to add 100 to the old total when forming the new total. Some students also mistakenly assume that 100 divided by 25 is the increase in average without considering previous totals. Always form the equation carefully from total runs and the definition of average. Final Answer: The batsman average before the 25th match was 65 runs per match.

Question 9

A factory purchases 10 machines in total: 2 machines of type A costing Rs 95000 each, 3 machines of type B costing Rs 60000 each, and the remaining machines of type C costing Rs 50000 each. What is the average cost per machine in rupees for this ent…

Accepted Answer

Introduction / Context: This question applies the idea of weighted average to the cost of different categories of machines. A factory buys machines of three different types at different prices and you must find the overall average cost per machine. This is a direct application of total cost divided by total quantity, a fundamental concept in averages and cost analysis. Given Data / Assumptions: - Total machines purchased = 10. - Number of type A machines = 2, cost per type A machine = Rs 95000. - Number of type B machines = 3, cost per type B machine = Rs 60000. - Remaining machines are type C. Since 2 + 3 = 5, remaining = 10 - 5 = 5 type C machines. - Cost per type C machine = Rs 50000. - We must compute the average cost per machine for all 10 machines. Concept / Approach: The total cost is the sum of the cost of each group: type A, type B and type C. Once we have the total cost, the average cost per machine is simply total cost divided by total number of machines. This is a straightforward weighted average where each type of machine contributes according to its price and count. Step-by-Step Solution: Step 1: Cost of type A machines = 2 * 95000 = Rs 190000.Step 2: Cost of type B machines = 3 * 60000 = Rs 180000.Step 3: Cost of type C machines = 5 * 50000 = Rs 250000.Step 4: Total cost of all machines = 190000 + 180000 + 250000.Step 5: Compute total cost: 190000 + 180000 = 370000; 370000 + 250000 = 620000.Step 6: Total number of machines = 10.Step 7: Average cost per machine = total cost / total number = 620000 / 10 = Rs 62000.Step 8: Therefore, the average cost of each machine is Rs 62000. Verification / Alternative check: You can think about the distribution of prices: some machines are relatively expensive (Rs 95000), some are mid range (Rs 60000) and some are cheaper (Rs 50000). With 2 expensive, 3 mid range and 5 cheaper machines, it is reasonable that the average cost lies between 50000 and 95000, leaning closer to the lower side because more machines are cheaper. The answer Rs 62000 satisfies this intuition and matches the exact calculation. Why Other Options Are Wrong: Values like Rs 68333 or Rs 74666 would imply much higher total costs that cannot be obtained from the given individual prices and quantities. A figure like Rs 60500 is close but incorrect when multiplied by 10, and Rs 66000 is clearly larger than the true average. Only Rs 62000 matches the total of Rs 620000 divided by 10 machines. Common Pitfalls: Some students mistakenly average the three prices (95000, 60000 and 50000) without considering the different numbers of machines of each type. Others miscount the remaining type C machines. To avoid these errors, always compute group totals first and then divide by the overall quantity for a proper weighted average. Final Answer: The average cost per machine is Rs 62000.

Question 10

In coordinate geometry, a straight line cuts the x-axis at the point (3, 0) and the y-axis at the point (0, 6). What is the equation of this line in Cartesian form?

Accepted Answer

Introduction / Context: This question tests basic knowledge of straight line equations in coordinate geometry. A line is specified by its intercepts on the x axis and y axis, and you are required to find its algebraic equation. Recognizing how to compute the slope from two points and then forming the equation of a line is an essential skill in analytic geometry. Given Data / Assumptions: - The line crosses the x axis at point (3, 0). - The same line crosses the y axis at point (0, 6). - We assume a standard Cartesian coordinate system. - We need the equation of the line in the form y = mx + c or an equivalent linear form. Concept / Approach: A straight line can be determined uniquely by any two distinct points on it. Given two points (x1, y1) and (x2, y2), the slope m of the line is (y2 - y1) / (x2 - x1). Once the slope is known, we can write the equation in slope intercept form y = mx + c by substituting one of the points to find the intercept c. Alternatively, we can use the intercept form x/a + y/b = 1 when the intercepts a and b are known. Step-by-Step Solution: Step 1: The two given points are A(3, 0) on the x axis and B(0, 6) on the y axis.Step 2: Compute the slope m of line AB.Step 3: m = (y2 - y1) / (x2 - x1) = (6 - 0) / (0 - 3) = 6 / (-3) = -2.Step 4: Use the slope intercept form y = mx + c. We know m = -2.Step 5: Substitute point B(0, 6) to find c: 6 = -2 * 0 + c, so c = 6.Step 6: Therefore, the equation of the line is y = -2x + 6.Step 7: Check with point A(3, 0): y = -2 * 3 + 6 = -6 + 6 = 0, so the point lies on the line, confirming the equation. Verification / Alternative check: We can also use the intercept form of a line: x/a + y/b = 1, where a and b are x and y intercepts respectively. Here, a = 3 and b = 6, so x/3 + y/6 = 1. Multiplying by 6 gives 2x + y = 6, which can be rearranged to y = -2x + 6, matching the slope intercept form derived earlier. Since both approaches agree, the equation is confirmed. Why Other Options Are Wrong: The option y = 2x - 6 has a positive slope and passes through (3, 0) but would intersect the y axis at (0, -6), not (0, 6). The equation y = 2x + 6 has a positive slope and does not pass through (3, 0). The equation y = -2x - 6 has the correct slope magnitude but gives a y intercept of -6 instead of 6. The equation x + 2y = 6 rearranges to y = (6 - x) / 2, which has slope -0.5, not -2, and does not pass through the point (3, 0). Common Pitfalls: Some students invert the difference in x and y when calculating slope, leading to the wrong sign. Others forget that the x intercept has y coordinate zero and the y intercept has x coordinate zero. Always write down both points clearly, compute the slope carefully and verify that your final equation satisfies both given points. Final Answer: The equation of the line is y = -2x + 6.

Question 11

The average weight of three persons P, Q and R is 71 kg. If the average weight of P and Q together is 66 kg and the average weight of Q and R together is 76.5 kg, what is the weight of Q in kilograms?

Accepted Answer

Introduction / Context: This final question in the set is another averages and linear equations problem involving weights of three people. You know the average of all three, plus the averages of two different pairs. The goal is to determine the individual weight of Q. This structure mirrors many common aptitude questions where overlapping group averages lead to simple algebraic equations. Given Data / Assumptions: - Persons are P, Q and R. - Average weight of P, Q and R together = 71 kg. - Average weight of P and Q together = 66 kg. - Average weight of Q and R together = 76.5 kg. - We must find the weight of Q in kilograms. Concept / Approach: As in similar problems, we first convert each average to a total weight. We find totals for P + Q + R, P + Q and Q + R. By subtracting P + Q from the total, we get R, and by subtracting Q + R from the total we get P. Once we know P and R, we combine the relevant totals to isolate Q. The method is clean and uses basic algebra and arithmetic only. Step-by-Step Solution: Step 1: Total weight of P, Q and R = 3 * 71 = 213 kg.Step 2: Total weight of P and Q = 2 * 66 = 132 kg.Step 3: Total weight of Q and R = 2 * 76.5 = 153 kg.Step 4: From step 1 and step 2, compute R: R = (P + Q + R) - (P + Q) = 213 - 132 = 81 kg.Step 5: From step 1 and step 3, compute P: P = (P + Q + R) - (Q + R) = 213 - 153 = 60 kg.Step 6: Now use P + Q = 132 to find Q: 60 + Q = 132.Step 7: So Q = 132 - 60 = 72 kg.Step 8: Therefore, the weight of Q is 72 kg. Verification / Alternative check: Check all given averages. With P = 60, Q = 72 and R = 81, the total P + Q + R = 60 + 72 + 81 = 213, so the overall average is 213 / 3 = 71 kg, matching the first condition. For P and Q, total is 60 + 72 = 132 and average is 132 / 2 = 66 kg, matching the second condition. For Q and R, total is 72 + 81 = 153 and average is 153 / 2 = 76.5 kg, matching the third condition. All three given averages are consistent with these weights, confirming the answer. Why Other Options Are Wrong: If Q were 60, 75, 69 or 81, then at least one of the three average conditions would fail. For example, if Q were 75 and P + Q = 132, then P would be 57, and this combined with a suitable R would not satisfy the overall average of 71 and the given Q + R average at the same time. Only Q = 72 kg works with a consistent set of values for P and R. Common Pitfalls: Some learners attempt to guess Q without systematically converting averages to totals. Others mix up which totals correspond to which pairs. The safest approach is to write down the three equations for P + Q + R, P + Q and Q + R, express P and R in terms of these totals and then solve step by step. Finally, always verify your result by checking all given averages. Final Answer: The weight of Q is 72 kg.

Question 12

A batsman scores 87 runs in his 17th innings and thereby increases his batting average by 3 runs. What is his batting average after the 17th innings?

Accepted Answer

Introduction: This averages question comes from cricket statistics. When a batsman plays one more innings, his average can change depending on the new score. You are told his 17th innings score and how much his average increases, and must deduce his new average. This type of problem tests understanding of how totals and averages are related. Given Data / Assumptions: Before the 17th innings, he had completed 16 innings. Let his previous average be A runs per innings. In the 17th innings, he scores 87 runs. After this innings, his average increases by 3 runs, so his new average is A + 3. We are asked to find the new average after 17 innings. Concept / Approach: Average = (total runs) / (number of innings). Let the total runs before the 17th innings be 16A. After scoring 87 runs in the 17th innings, his new total runs become 16A + 87, and his new average over 17 innings is (16A + 87)/17. This new average is given to be A + 3. Equating these expressions leads to a linear equation in A. Step-by-Step Solution: Let previous average be A runs.Total runs after 16 innings = 16A.In the 17th innings he scores 87, so new total = 16A + 87.New average after 17 innings = (16A + 87) / 17.Given that this new average is 3 runs more than A, we have:(16A + 87) / 17 = A + 3.Multiply both sides by 17:16A + 87 = 17A + 51.Rearrange:87 − 51 = 17A − 16A ⇒ 36 = A.So previous average A = 36 runs.New average = A + 3 = 36 + 3 = 39 runs. Verification / Alternative check: Check with actual totals. If average after 16 innings is 36, then total runs = 16 * 36 = 576. Adding 87 gives new total = 663. Divide by 17: 663 / 17 = 39, which is indeed 3 runs more than 36. This confirms that the new average is 39 runs. Why Other Options Are Wrong: 40, 50 and 55 are too high and do not satisfy the equation (16A + 87)/17 = A + 3 for any integer previous average. The value 36 is the old average, not the new one, so it does not answer the question. Only 39 is consistent with all the given information. Common Pitfalls: Some students confuse the old and new averages and mistakenly report A instead of A + 3. Others may incorrectly set up the equation, such as writing 16A + 87 = 17(A + 3) incorrectly or forgetting to multiply the new average by the number of innings. Careful mapping of total runs before and after the new innings is crucial. Final Answer: The batsman's average after the 17th innings is 39 runs per innings.

Question 13

A cricket player has an average of 32 runs per innings over his first 10 innings. How many runs must he score in his 11th innings so that his overall batting average increases by 4 runs and becomes 36 runs per innings?

Accepted Answer

Introduction / Context: This problem tests the concept of arithmetic mean (average) and how it changes when additional data is added. We use the relationship between total runs, number of innings, and average score to determine how many runs the batsman must score in his next innings to raise his average by a fixed amount. Given Data / Assumptions: Number of initial innings = 10. Average runs for these 10 innings = 32 runs per innings. We add one more innings (the 11th innings). Required new average after 11 innings = 32 + 4 = 36 runs per innings. Let the runs scored in the 11th innings be R. Concept / Approach: Average is defined as: average = (sum of observations) / (number of observations). We first compute the total runs scored in the first 10 innings using the given average. Then, we express the new total runs after including the 11th innings in terms of R. By equating this new total to the required average multiplied by 11 innings, we can solve for R. Step-by-Step Solution: Step 1: Total runs in first 10 innings = 32 * 10 = 320. Step 2: Let runs in 11th innings be R. Then new total runs = 320 + R. Step 3: Required new average = 36 runs over 11 innings. So, (320 + R) / 11 = 36. Step 4: Multiply both sides by 11: 320 + R = 36 * 11. Step 5: Compute 36 * 11 = 396. Step 6: So 320 + R = 396 → R = 396 - 320 = 76. Verification / Alternative check: Check the final average: total runs = 320 + 76 = 396. Average over 11 innings = 396 / 11 = 36 runs per innings, which matches the required value. Why Other Options Are Wrong: Option B (79), C (85), D (87) and E (92) all produce totals different from 396, therefore the resulting averages would not equal 36 when divided by 11 innings. For example, if R = 79, new total = 399 and average = 399 / 11 = 36.27 (approx), not exactly 36. Common Pitfalls: A common error is to add 4 directly to the previous total or to the previous average without accounting for the change in the number of innings. Another mistake is to forget that the new average applies to all 11 innings, not just the latest one. Final Answer: The batsman must score 76 runs in his next innings to increase his average from 32 to 36 runs per innings.

Question 14

A grocer has sales of Rs 6435, Rs 6927, Rs 6855, Rs 7230 and Rs 6562 in five consecutive months. What should be his sales (in rupees) in the sixth month so that his average monthly sale over the six months becomes Rs 6500?

Accepted Answer

Introduction / Context: This is an average (mean) based aptitude question involving monthly sales figures. You are given the sales for five consecutive months and asked to find the sixth month's sales so that the overall average becomes a specified amount. Such questions are very common in competitive exams to test understanding of the mean formula and simple algebraic manipulation. Given Data / Assumptions: Sales for month 1 = Rs 6435. Sales for month 2 = Rs 6927. Sales for month 3 = Rs 6855. Sales for month 4 = Rs 7230. Sales for month 5 = Rs 6562. Desired average sale over 6 months = Rs 6500. Let the sale in month 6 be S rupees. Concept / Approach: Average (mean) is given by: average = (sum of all values) / (number of values). We first calculate the total sales of the first five months. Then we determine what the total sales must be over six months to get an average of Rs 6500. The difference between this required total and the already known five-month total gives us the required sale in the sixth month. Step-by-Step Solution: Step 1: Compute sum of sales for first 5 months. Sum = 6435 + 6927 + 6855 + 7230 + 6562. 6435 + 6927 = 13362. 13362 + 6855 = 20217. 20217 + 7230 = 27447. 27447 + 6562 = 34009. So total for 5 months = Rs 34009. Step 2: Required total for 6 months with average Rs 6500. Required total = 6500 * 6 = 39000. Step 3: Let sale in 6th month be S, then: 34009 + S = 39000. Step 4: Solve for S: S = 39000 - 34009 = 4991. Verification / Alternative check: Check the new average: New total = 34009 + 4991 = 39000. Average = 39000 / 6 = 6500, which matches the requirement. Why Other Options Are Wrong: Any value other than 4991 will make the total different from 39000, and the average will not be exactly Rs 6500. For example, if S = 5467, total = 39476 and average = 39476 / 6 ≈ 6579.33, which is higher than 6500. Common Pitfalls: Students sometimes take the average of the first five months and then adjust, instead of working directly with totals. Another mistake is to divide the difference in totals by 5 instead of 6, confusing the number of months. Final Answer: The grocer must have sales of Rs 4991 in the sixth month to achieve an average of Rs 6500 over six months.

Question 15

The average of 20 real numbers is zero. Out of these 20 numbers, what is the maximum possible number of terms that can be strictly greater than zero?

Accepted Answer

Introduction / Context: This question checks your conceptual understanding of averages and how positive and negative numbers can combine to give an average of zero. You are asked to determine the maximum possible count of numbers that can be greater than zero while the overall average of 20 numbers remains zero. It is a logical reasoning question based on the idea of balancing positive and negative values rather than heavy calculations. Given Data / Assumptions: Total number of numbers = 20. Average of the 20 numbers = 0. Numbers can be positive, negative, or zero. We want to maximize how many numbers are > 0. Concept / Approach: If the average of 20 numbers is zero, the sum of all 20 numbers must be zero. That is: sum of all 20 numbers = 0. To have many positive numbers, the remaining numbers must be sufficiently negative (or zero) to bring the total sum back to zero. We want to see how many numbers can be strictly positive while still allowing the sum to remain zero. Step-by-Step Solution: Step 1: Let k be the number of positive numbers. Step 2: Then the remaining 20 - k numbers are either negative or zero. Step 3: For the sum of all 20 numbers to be zero, the sum of the positive numbers must be balanced by the sum of the negatives. Step 4: To maximize k, we make exactly one number negative and all the others positive, with no zeros. So k can be 19, and the remaining 1 number is negative. Step 5: For example, suppose 19 numbers are +1 each, and the remaining number is -19. Then the sum is 19 * 1 + (-19) = 19 - 19 = 0 and the average is 0 / 20 = 0. Step 6: It is impossible to have all 20 numbers greater than zero because then the sum would be strictly positive and the average would also be positive, not zero. Verification / Alternative check: We have demonstrated one concrete example with 19 positive numbers and 1 negative number that yields an average of zero. This proves that 19 positive numbers are possible. We also argued that 20 positive numbers are impossible, so 19 is the maximum. Why Other Options Are Wrong: Option A (0): It is not necessary that all numbers be non-positive; we clearly constructed a case with many positive numbers. Option B (1): You can certainly have more than one positive number; our example has 19. Option C (19): This is the correct numeric count, but it is not listed as a separate option in the original set, so within the given options, the correct choice is "None of these". Option E (20): Having all 20 numbers positive would make the sum and thus the average strictly positive, contradicting the given average of zero. Common Pitfalls: Some students incorrectly assume that the numbers must be symmetrically distributed, such as 10 positive and 10 negative, which is not required. Others may overlook the possibility of using a single large negative number to balance many small positive numbers. Final Answer: The maximum possible number of positive numbers is 19, which is not explicitly listed, so the correct option is None of these.

Question 16

The average weight of 8 persons increases by 2.5 kg when a new person replaces one of the original 8, whose weight was 65 kg. What is the weight (in kilograms) of the new person who joined the group?

Accepted Answer

Introduction / Context: This question examines how the average of a group changes when one member is replaced by another person of a different weight. It combines average concepts with change in total value when one observation is removed and another is added. Understanding the relationship between total weight and average weight is key to solving such problems quickly in aptitude exams. Given Data / Assumptions: Number of persons initially = 8. Weight of the person who leaves = 65 kg. After replacement, the average weight of the 8 persons increases by 2.5 kg. Let the original average weight be A kg. Let the weight of the new person be W kg. Concept / Approach: If the average increases by 2.5 kg for 8 persons, the total weight of the group must increase by: increase in total = number of persons * increase in average. We compute this increase and relate it to the difference between the new person's weight and the old person's weight. This directly gives us the new weight without needing the original average explicitly. Step-by-Step Solution: Step 1: Original total weight = 8 * A. Step 2: New average = A + 2.5 kg. Step 3: New total weight = 8 * (A + 2.5) = 8A + 20. Step 4: When we replace the 65 kg person with the new person of weight W, new total weight = (original total) - 65 + W. So, 8A + 20 = 8A - 65 + W. Step 5: Cancel 8A on both sides: 20 = -65 + W. Step 6: Therefore, W = 20 + 65 = 85 kg. Verification / Alternative check: Increase in average = 2.5 kg for 8 persons, so total increase = 8 * 2.5 = 20 kg. This increase in total weight comes entirely from replacing 65 kg with W kg, so W - 65 = 20. Thus W = 85 kg, which confirms our earlier result. Why Other Options Are Wrong: Option A (70 kg), B (75 kg), C (80 kg) and E (90 kg) do not yield a net increase of exactly 20 kg when compared to 65 kg, so they would not produce a 2.5 kg increase in average for 8 persons. For example, if W = 80 kg, increase = 15 kg, total increase per person = 15 / 8 ≈ 1.875 kg, not 2.5 kg. Common Pitfalls: A common mistake is to try to compute the original average explicitly, which is unnecessary and may introduce errors. Another error is forgetting that the average change for the whole group scales the total change by the number of people. Final Answer: The weight of the new person who joins the group is 85 kg.

Question 17

The average of five consecutive odd numbers is 61. What is the difference between the largest and the smallest of these five odd numbers?

Accepted Answer

Introduction / Context: This problem uses the idea of consecutive odd numbers and their average. For any set of consecutive numbers (or consecutive odd or even numbers), the average is simply the middle term. Once we identify the sequence, finding the smallest and largest numbers and their difference becomes straightforward. Given Data / Assumptions: There are 5 consecutive odd numbers. The average (arithmetic mean) of these 5 numbers is 61. We are asked to find: largest number - smallest number. Concept / Approach: For an odd count of consecutive numbers, the average equals the middle term. So, in a sequence of 5 consecutive odd numbers: n1, n2, n3, n4, n5 where n1 < n2 < n3 < n4 < n5 and the numbers differ by 2 each. Then: average = n3. So, we can directly find the middle odd number, then move two steps backward and two steps forward to find the smallest and largest numbers respectively. Step-by-Step Solution: Step 1: Since there are 5 consecutive odd numbers, the middle number is the average. Step 2: Therefore, the middle odd number = 61. Step 3: Consecutive odd numbers differ by 2. Step 4: The five numbers are: n1 = 61 - 4 = 57, n2 = 61 - 2 = 59, n3 = 61, n4 = 61 + 2 = 63, n5 = 61 + 4 = 65. Step 5: Smallest number = 57, largest number = 65. Step 6: Difference = largest - smallest = 65 - 57 = 8. Verification / Alternative check: Check average: (57 + 59 + 61 + 63 + 65) / 5. Sum = 57 + 59 = 116, +61 = 177, +63 = 240, +65 = 305. Average = 305 / 5 = 61, matching the given average. Why Other Options Are Wrong: Option A (4), C (12), D (16), and E (20) correspond to either too small or too large gaps for five consecutive odd numbers centered at 61. With five consecutive odd numbers, the difference between smallest and largest must be 4 steps of 2, i.e. 8, and cannot be any other value. Common Pitfalls: Some students mistakenly treat the average as the first or last term instead of the middle term. Others may incorrectly use common difference 1 instead of 2 for odd numbers. Final Answer: The difference between the highest and lowest numbers is 8.

Question 18

The average weight of a class of 24 students is 35 kg. When the weight of the teacher is also included, the average weight increases by 400 grams (0.4 kg). What is the weight, in kilograms, of the teacher?

Accepted Answer

Introduction / Context: This question focuses on how the average (mean) changes when an additional data point (the teacher's weight) is added to a group. We use the relationship between total weight, number of persons, and average to determine the teacher's weight. Converting 400 grams into kilograms is an important small detail here. Given Data / Assumptions: Number of students = 24. Average weight of students = 35 kg. Teacher's weight is added to form a group of 25 people. New average weight = 35 kg + 0.4 kg = 35.4 kg. We need to find the teacher's weight in kg. Concept / Approach: Average (mean) is: average = total weight / number of persons. We first compute the total weight of the 24 students. Then we compute the total weight of all 25 persons using the new average. The difference between these totals directly gives the teacher's weight. Step-by-Step Solution: Step 1: Total weight of 24 students = 24 * 35 kg. 24 * 35 = 840, so total = 840 kg. Step 2: New average after including teacher = 35.4 kg. Number of persons now = 24 + 1 = 25. Step 3: Total weight of 25 persons = 25 * 35.4 kg. Compute 25 * 35.4 = (25 * 35) + (25 * 0.4) = 875 + 10 = 885 kg. Step 4: Teacher's weight = total weight of 25 persons - total weight of 24 students. Teacher's weight = 885 - 840 = 45 kg. Verification / Alternative check: If we add a 45 kg teacher to the 24 students whose total is 840 kg, new total = 840 + 45 = 885 kg. New average = 885 / 25 = 35.4 kg, which matches the required average (35 kg + 0.4 kg). Why Other Options Are Wrong: Weights 46 kg, 47 kg, 48 kg, or 50 kg would produce totals different from 885 kg, and so the average would not be 35.4 kg. For example, if the teacher weighed 48 kg, total = 840 + 48 = 888 kg and average = 888 / 25 = 35.52 kg, which is too high. Common Pitfalls: Students sometimes forget to convert 400 g to 0.4 kg and may mistakenly add 400 to 35. Another mistake is using 24 instead of 25 when multiplying the new average, which gives the wrong total. Final Answer: The weight of the teacher is 45 kg.

Question 19

The average of 7 consecutive integers is 20. If these integers are taken in increasing order, what is the largest integer in this sequence?

Accepted Answer

Introduction / Context: This problem deals with consecutive integers and their average. For an odd number of consecutive terms, the average is exactly equal to the middle term. Once we identify the middle term, it is very easy to reconstruct the full sequence and find the largest value. Given Data / Assumptions: There are 7 consecutive integers. Their average (mean) is 20. We need to find the largest integer among these 7 consecutive numbers. Concept / Approach: For any set of consecutive integers with an odd count, the average is the middle element. If the 7 numbers are: n1, n2, n3, n4, n5, n6, n7 then n4 is the middle number because there are 3 numbers before it and 3 numbers after it. Hence the average equals n4. We then step up and down by 1 to get the remaining numbers, and identify the largest. Step-by-Step Solution: Step 1: Since there are 7 consecutive integers, the middle term is the 4th term. Step 2: Average = 20, so the middle term n4 = 20. Step 3: The sequence is: n1 = 20 - 3 = 17, n2 = 20 - 2 = 18, n3 = 20 - 1 = 19, n4 = 20, n5 = 20 + 1 = 21, n6 = 20 + 2 = 22, n7 = 20 + 3 = 23. Step 4: Largest integer in the sequence = n7 = 23. Verification / Alternative check: Check average: sum = 17 + 18 + 19 + 20 + 21 + 22 + 23. 17 + 18 = 35, +19 = 54, +20 = 74, +21 = 95, +22 = 117, +23 = 140. Average = 140 / 7 = 20, which matches the given condition. Why Other Options Are Wrong: Options A (21) and B (22) are values in the sequence but are not the largest term. Option D (24) and E (25) are larger but would force the average to be greater than 20 if they were part of a 7-term consecutive sequence centered at 20. Common Pitfalls: Confusing the average with one of the end terms rather than the middle term can lead to an incorrect sequence. Some students forget that there are 3 steps of +1 to go from the middle to the largest term when there are 7 consecutive numbers. Final Answer: The largest of the seven consecutive integers is 23.

Question 20

The mean temperature from Monday to Wednesday is 37°C, and the mean temperature from Tuesday to Thursday is 34°C. If the temperature on Thursday is four-fifths of the temperature on Monday, what is the temperature (in degree Celsius) on Thursday?

Accepted Answer

Introduction / Context: This is a temperature and average based word problem involving four days: Monday, Tuesday, Wednesday and Thursday. You are given the average temperature over two overlapping three-day periods and a relationship between Monday's and Thursday's temperatures. The task is to find the temperature on Thursday using equations and basic algebra. Given Data / Assumptions: Let M, T, W, Th denote the temperatures on Monday, Tuesday, Wednesday and Thursday respectively (in °C). Mean temperature Monday to Wednesday is 37°C: (M + T + W) / 3 = 37. Mean temperature Tuesday to Thursday is 34°C: (T + W + Th) / 3 = 34. Thursday's temperature is four-fifths of Monday's temperature: Th = (4/5) * M. We must find the value of Th. Concept / Approach: We convert each average into an equation for the sum of the relevant temperatures. Then, we use subtraction to relate M and Th. Finally, we use the given ratio Th = (4/5)M to solve for M and then Th. It is a straightforward linear system of equations problem. Step-by-Step Solution: Step 1: From (M + T + W) / 3 = 37, we get M + T + W = 37 * 3 = 111. Step 2: From (T + W + Th) / 3 = 34, we get T + W + Th = 34 * 3 = 102. Step 3: Subtract the second equation from the first: (M + T + W) - (T + W + Th) = 111 - 102. This simplifies to M - Th = 9. Step 4: We also know Th = (4/5) * M. So M - (4/5)M = 9. Step 5: M * (1 - 4/5) = 9 → M * (1/5) = 9. Step 6: Therefore M = 9 * 5 = 45°C. Step 7: Now Th = (4/5) * M = (4/5) * 45 = 4 * 9 = 36°C. Verification / Alternative check: Check Monday to Wednesday: Only M matters here as we do not know T and W exactly, but our equations are consistent since M - Th = 9 is satisfied (45 - 36 = 9). Also Th is strictly less than M, matching the fraction 4/5. The equations are satisfied, so Th = 36°C is correct. Why Other Options Are Wrong: If Th = 38°C, then M - Th would not equal 9 and Th would not be 4/5 of M. The same inconsistency appears if we test 40°C, 39°C or 34°C for Thursday; they cannot satisfy both M - Th = 9 and Th = (4/5)M simultaneously. Common Pitfalls: Students sometimes confuse which equation to subtract from which and might incorrectly form Th - M = 9. Another common mistake is mixing up the ratio and writing M = (4/5) * Th instead of Th = (4/5) * M. Final Answer: The temperature on Thursday is 36 degree celsius.

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