Question 1

Monthly average and a half-average month: A business has total annual sales of ₹132000. If June's sales were half of the monthly average, what were the sales in June?

Accepted Answer

Introduction / Context:Find monthly average from annual total, then take half for June. Given Data / Assumptions: Annual sales = ₹132000 Months = 12 June = half of monthly average Concept / Approach:Monthly average = total / 12. Half of that is the required June sales. Step-by-Step Solution:Monthly average = 132000 / 12 = ₹11000June sales = 11000 / 2 = ₹5500 Verification / Alternative check:If one month is at half-average, another could be above average to keep the annual sum correct; the question asks only for June's value, which is ₹5500. Why Other Options Are Wrong:₹7200, ₹6300, ₹5600, ₹8400 are not half of ₹11000. Common Pitfalls:Dividing by 6 instead of 12; or halving the annual total directly by mistake. Final Answer:₹ 5500

Question 2

Effect of an annual raise on monthly per-person average: In a family of 5 members, the average monthly income per person is ₹1000. If one person's income increases by ₹12000 per year, what is the new monthly average income per person for the family?

Accepted Answer

Introduction / Context:The question converts an annual increment to a monthly figure and distributes it across all family members to find the new per-person monthly average. Given Data / Assumptions: Members = 5 Current per-person monthly average = ₹1000 One person's annual increase = ₹12000 Concept / Approach:₹12000 per year equals ₹1000 per month. This increases the family's total monthly income by ₹1000; divide by 5 to get the per-person increase. Step-by-Step Solution:Annual increase = ₹12000 → Monthly increase = ₹12000 / 12 = ₹1000Per-person monthly increment = ₹1000 / 5 = ₹200New per-person monthly average = ₹1000 + ₹200 = ₹1200 Verification / Alternative check:Old family monthly total = ₹1000 * 5 = ₹5000. New total = ₹6000. New per-person = ₹6000 / 5 = ₹1200. Why Other Options Are Wrong:₹1600, ₹2000, ₹3400, ₹1100 are not consistent with the increment distribution across 5 people. Common Pitfalls:Forgetting to convert annual to monthly before averaging; distributing the increment to just one person when asked for per-person average. Final Answer:₹ 1200

Question 3

Weekly temperature average with a missing day: The average temperature of the first three days of a week is 25° C and that of the next three days is 27° C. If the average for the whole week is 26.5° C, what is the temperature on the seventh day?

Accepted Answer

Introduction / Context:Find the single missing value from grouped averages within a 7-day week. Given Data / Assumptions: Average days 1–3 = 25° C Average days 4–6 = 27° C Average days 1–7 = 26.5° C Concept / Approach:Compute totals from averages for both 3-day blocks and for all 7 days, then subtract to get day 7. Step-by-Step Solution:Sum days 1–3 = 3 * 25 = 75Sum days 4–6 = 3 * 27 = 81Sum days 1–7 = 7 * 26.5 = 185.5Day 7 = 185.5 - (75 + 81) = 29.5° C Verification / Alternative check:Check that the combined totals match the weekly average after inserting 29.5 for day 7; they do. Why Other Options Are Wrong:31.5° C is too high; 21° C too low; 27.5° C and 26.0° C do not satisfy the weekly average constraint. Common Pitfalls:Using 6 rather than 7 days for the weekly total; not converting averages back to totals first. Final Answer:29.5° C

Question 4

Average age shift after five years and a new entrant: Five years ago, the average age of four men was 48 years. Now, a new man joins and the current average increases by 2 years compared to what it would be otherwise. What is the present age of the …

Accepted Answer

Introduction / Context:This problem combines time progression of averages with the addition of a new member. Clarifying the timeline is crucial: the original average is given five years ago; then, now, someone joins, and the resulting average is higher by 2 years than it otherwise would have been. Given Data / Assumptions: Five years ago: average of 4 men = 48 → total then = 4 * 48 = 192 Now, each of the 4 is 5 years older → total now (without new man) = 192 + 4 * 5 = 212 With the new man joining now, the current average becomes 2 years higher than 53 (the average the 4 would now have), i.e., new average = 55 Concept / Approach:Compute the current total for the 4, then use the target 5-person average to deduce the new man’s age. Step-by-Step Solution:Average of original 4 now = 48 + 5 = 53Stated increase by joining = +2 → new 5-person average = 55Total for 5 at 55 = 5 * 55 = 275Age of new man = 275 - 212 = 63 years Verification / Alternative check:If the new man were 63, the new average is exactly 55, which is 2 more than the 53 that would have prevailed without him. Why Other Options Are Wrong:58, 42, 41, 38 do not produce the required +2 rise in the current average when added to the existing group. Common Pitfalls:Adding 2 to 48 instead of to 53; forgetting that 5 years have passed for each of the original members before the new man joins. Final Answer:63 yr

Question 5

Combining multiple average relations: Average of (a, b, c) is 11; average of (c, d, e) is 17; average of (e, f) is 22; and average of (e, c) is 17. What is the average of (a, b, c, d, e, f)?

Accepted Answer

Introduction / Context:We are given several overlapping averages and asked for the combined average. Eliminating variables through sums is effective. Given Data / Assumptions: (a + b + c) / 3 = 11 → a + b + c = 33 (c + d + e) / 3 = 17 → c + d + e = 51 (e + f) / 2 = 22 → e + f = 44 (e + c) / 2 = 17 → e + c = 34 Concept / Approach:Compute S = a + b + c + d + e + f using the equations. Note that (c + d + e) + (a + b + c) + (e + f) - (e + c) subtracts the overlap to yield S. Step-by-Step Solution:S = (a + b + c) + (c + d + e) + (e + f) - (e + c)S = 33 + 51 + 44 - 34 = 94Average of six = S / 6 = 94 / 6 = 47 / 3 = 15 2/3 Verification / Alternative check:Express variables in terms of parameters and show total remains 94 regardless, confirming invariance. Why Other Options Are Wrong:151⁄3, 162⁄3, 171⁄2 are not equal to 47/3; only 152⁄3 matches 15 2/3. Common Pitfalls:Double-counting c and e; forgetting to subtract the shared pair once. Final Answer:152⁄3

Question 6

Weighted average purchase price per kg: Mumbai procures wheat as follows: 1200 kg at ₹30/kg from Nasik, 1500 kg at ₹25/kg from Ratnagiri, and 2000 kg at ₹32/kg from Kolhapur. What is the average cost per kg?

Accepted Answer

Introduction / Context:This is a weighted average of prices with different quantities bought at different rates. Given Data / Assumptions: 1200 kg @ ₹30/kg 1500 kg @ ₹25/kg 2000 kg @ ₹32/kg Concept / Approach:Weighted average price = (sum of (quantity * price)) / (total quantity). Step-by-Step Solution:Total cost = 1200*30 + 1500*25 + 2000*32= 36000 + 37500 + 64000 = 137500Total quantity = 1200 + 1500 + 2000 = 4700 kgAverage price = 137500 / 4700 ≈ ₹29.25 per kg Verification / Alternative check:The result is between ₹25 and ₹32 and slightly biased upward due to a large 2000 kg at ₹32; ₹29.25 is consistent. Why Other Options Are Wrong:₹29, ₹29.50, ₹29.75, ₹28.90 are not the exact weighted average from the given quantities and rates. Common Pitfalls:Averaging the prices (simple mean) instead of using quantity weights. Final Answer:₹ 29.25 per kg

Question 7

Average of three ages given a ratio: The average age of three boys is 15 years. If their ages are in the ratio 3 : 5 : 7, what is the age of the youngest boy?

Accepted Answer

Introduction / Context:We translate the ratio to actual ages using the total derived from the average. Given Data / Assumptions: Average of 3 boys = 15 → sum = 45 Ratio of ages = 3 : 5 : 7 → total parts = 15 Concept / Approach:If total parts = 15 correspond to total age 45, then each part = 45 / 15 = 3 years. Multiply the youngest’s parts by 3 to get his age. Step-by-Step Solution:One part = 45 / 15 = 3 yearsYoungest (3 parts) = 3 * 3 = 9 years Verification / Alternative check:The ages are 9, 15, 21; average is (9 + 15 + 21) / 3 = 45 / 3 = 15, consistent. Why Other Options Are Wrong:12, 15, 18, 21 do not match the youngest boy determined from the ratio and total sum. Common Pitfalls:Using 3 as the answer (the ratio number) instead of converting to years using the total sum. Final Answer:9 year

Question 8

Average of odd numbers up to 100: What is the average of all odd numbers from 1 to 99 (i.e., all odd numbers up to 100)?

Accepted Answer

Introduction / Context:The average of an arithmetic sequence equals the mean of its first and last terms. Given Data / Assumptions: Odd numbers from 1 to 99 form an arithmetic sequence with common difference 2 First term = 1, last term = 99 Concept / Approach:Average of any arithmetic sequence = (first + last) / 2. The count (50 odds) is not necessary to compute the average itself. Step-by-Step Solution:Average = (1 + 99) / 2 = 100 / 2 = 50 Verification / Alternative check:Pair 1 with 99, 3 with 97, etc.; each pair averages to 50 and there are equal pairs, confirming 50. Why Other Options Are Wrong:51, 49.5, 49, 50.5 are inconsistent with the symmetry of the odd sequence around 50. Common Pitfalls:Confusing with the average of integers 1 to 100 (50.5) instead of odd numbers 1 to 99 (50). Final Answer:50

Question 9

Average speed over a trip with different speeds by distance share: On a journey across Delhi, a taxi travels 60% of the distance at 30 km/h, 20% at 20 km/h, and the remaining 20% at 10 km/h. What is the average speed for the whole journey?

Accepted Answer

Introduction / Context:When different speeds apply to different fractions of the distance, the correct average speed is the harmonic-mean-like combination weighted by distance fractions, not the simple arithmetic mean. Given Data / Assumptions: Fractions of distance: 0.6, 0.2, 0.2 Speeds: 30 km/h, 20 km/h, 10 km/h respectively Concept / Approach:If fractions of distance are f1, f2, f3 at speeds v1, v2, v3, then 1 / v_avg = f1 / v1 + f2 / v2 + f3 / v3. Step-by-Step Solution:1 / v_avg = 0.6 / 30 + 0.2 / 20 + 0.2 / 10= 0.02 + 0.01 + 0.02 = 0.05v_avg = 1 / 0.05 = 20 km/h Verification / Alternative check:Let total distance be 1 unit; compute total time as sum of (fi * distance)/vi and divide distance 1 by total time; it yields 20 km/h. Why Other Options Are Wrong:22.5, 24.625, 25, 18 km/h do not satisfy the harmonic sum of times given the stated fractions and speeds. Common Pitfalls:Taking the arithmetic mean of speeds or weighting by time instead of distance when the fractions refer to distance parts. Final Answer:20 km/hr

Question 10

A train reaches on time when it travels at an average speed of 40 km/h. If it travels at an average speed of 35 km/h, it reaches 15 minutes late. What is the total journey distance (in km)?

Accepted Answer

Introduction: This average speed question tests the relationship between distance, speed, and time. If a journey is on time at one speed but late at a lower speed, the difference in travel times for the same distance equals the lateness. By translating the 15-minute delay into hours and equating the two time expressions, we can solve for the journey time and distance. Given Data / Assumptions: On-time average speed = 40 km/h Reduced average speed = 35 km/h Lateness at 35 km/h = 15 minutes = 0.25 hours Distance is constant in both cases Concept / Approach: Let the scheduled travel time be T hours and distance be D km. Then D = 40 * T for the on-time case, and D = 35 * (T + 0.25) for the delayed case. Equate these two expressions for D and solve for T. Finally compute D using D = 40 * T. Step-by-Step Solution: 40 * T = 35 * (T + 0.25) 40T = 35T + 8.75 → 5T = 8.75 → T = 1.75 h Distance D = 40 * 1.75 = 70 km Verification / Alternative check: Time at 35 km/h: 70 / 35 = 2.00 h. Scheduled time: 1.75 h. Difference = 0.25 h = 15 minutes, which matches the given delay. Why Other Options Are Wrong: 80, 60, 50, and 30 km do not satisfy the 15-minute difference when checked against the two speeds; only 70 km yields exactly 0.25 h extra at 35 km/h. Common Pitfalls: Using average of speeds or subtracting speeds directly. The key is equating times for the same distance. Final Answer: 70 km

Question 11

The average of 10 numbers was calculated as 15. Later it was found that one value 36 was wrongly read as 26. What is the correct average of the 10 numbers?

Accepted Answer

Introduction: This problem checks error correction in averages. When one observed value is misread, we adjust the total sum by the difference between the correct and incorrect readings, then recompute the average using the same count of observations. Given Data / Assumptions: n = 10 numbers Originally computed average = 15 → original computed sum = 15 * 10 = 150 Misread: 36 was read as 26 (under-recorded by 10) Concept / Approach: Correct sum = computed sum + (true value - misread value) = 150 + (36 - 26) = 160. Then correct average = 160 / 10. Step-by-Step Solution: Computed sum = 150 Correction = +10 (because 36 - 26 = 10) Correct sum = 160 Correct average = 160 / 10 = 16 Verification / Alternative check: Replace the misread 26 with 36 in any concrete 10-number set that produces the same initial average; the average rises by 1 exactly because +10 spread across 10 items adds +1 to the average. Why Other Options Are Wrong: 14, 12.4, 18.6, and 15.5 do not reflect the exact +10 adjustment distributed over 10 items. Common Pitfalls: Dividing the correction by 9 or adding 10 directly to the average. Always adjust the sum first, then divide by the original count. Final Answer: 16

Question 12

The average of 13 results is 68. The average of the first 7 results is 63 and that of the last 7 results is 70. What is the seventh (overlapping) result?

Accepted Answer

Introduction: This average puzzle uses overlapping groups. The “first seven” and “last seven” sets overlap in exactly one element (the seventh result). Sums from the two groups can be combined and compared with the total sum to isolate the overlapped value. Given Data / Assumptions: Total results = 13; overall average = 68 → total sum = 68 * 13 = 884 First 7 average = 63 → sum = 63 * 7 = 441 Last 7 average = 70 → sum = 70 * 7 = 490 Concept / Approach: The two seven-element blocks together count every element once except the central (7th) element, which is counted twice. Hence: (sum first 7 + sum last 7) = total sum + seventh result. Solve for the seventh result. Step-by-Step Solution: 441 + 490 = 931 931 = 884 + (seventh) seventh = 931 - 884 = 47 Verification / Alternative check: Visualize the 13 results in order. The first seven include positions 1–7; the last seven include 7–13. Position 7 is double-counted; subtract the total once to isolate it. Why Other Options Are Wrong: 65.5, 73.5, 94, and 52 do not satisfy the overlap arithmetic when plugged back into the sums. Common Pitfalls: Adding averages instead of sums or forgetting that the seventh item is counted twice in the two partial sums. Final Answer: 47

Question 13

The average of 25 results is 18. The average of the first 12 is 14 and the average of the last 12 is 17. What is the 13th (middle) result?

Accepted Answer

Introduction: Here, two disjoint groups (first twelve and last twelve) exclude the central 13th result. By comparing their combined sum with the total sum of all 25 results, we can directly extract the middle value without reconstructing the full list. Given Data / Assumptions: Total count = 25; overall average = 18 → total sum = 18 * 25 = 450 First 12 average = 14 → sum = 14 * 12 = 168 Last 12 average = 17 → sum = 17 * 12 = 204 Concept / Approach: The first 12 and last 12 are disjoint and together contain 24 results (excluding the 13th). Therefore, (sum first 12 + sum last 12) = total sum − 13th. Rearranging yields the 13th value. Step-by-Step Solution: Sum(first 12) + Sum(last 12) = 168 + 204 = 372 Total sum = 450 13th result = 450 − 372 = 78 Verification / Alternative check: Summing all 25 as (first 12 + 13th + last 12) reproduces the total 450, confirming that the 13th must be 78. Why Other Options Are Wrong: 28, 72, 85, and 62 do not reconcile the three averages simultaneously when plugged back into the partition sums. Common Pitfalls: Mistaking the setup for an overlapping case (like 7-and-7 over 13). Here, the groups are disjoint; no item is double-counted. Final Answer: 78

Question 14

The average temperature of Monday–Thursday is 38 °C and of Tuesday–Friday is 40 °C. If Monday was 30 °C, what was the temperature on Friday?

Accepted Answer

Introduction: This moving-average style problem compares two consecutive 4-day averages that share three common days. Subtracting the shared sums isolates the difference between Friday and Monday, allowing direct computation of Friday’s temperature from Monday’s value and the two averages. Given Data / Assumptions: Average(Mon–Thu) = 38 °C → M + T + W + Th = 152 Average(Tue–Fri) = 40 °C → T + W + Th + F = 160 M = 30 °C Concept / Approach: Subtract the first 4-day sum from the second: (T + W + Th + F) − (M + T + W + Th) = 160 − 152 = 8, giving F − M = 8. Then substitute M = 30 °C to find F. Step-by-Step Solution: F − M = 8 F = M + 8 = 30 + 8 = 38 °C Verification / Alternative check: Compute T + W + Th from the first equation: T + W + Th = 152 − 30 = 122. Then from the second, F = 160 − 122 = 38 °C, confirming the result. Why Other Options Are Wrong: 39° or 40° would imply different 4-day sums; 30° repeats Monday and contradicts the 8-degree difference; 37° does not satisfy both averages simultaneously. Common Pitfalls: Attempting to solve for each daily temperature separately; only the difference F − M is needed thanks to overlapping windows. Final Answer: 38°

Question 15

The average of 5 consecutive integers is n. If the next two consecutive integers are also included, what will be the new average of all 7 integers?

Accepted Answer

Introduction: Averages of consecutive integers track the middle value. For an odd count, the average equals the exact middle term. Adding two next higher consecutive numbers shifts the middle position upward by one, thus increasing the average by exactly 1, irrespective of n. Given Data / Assumptions: Five consecutive integers centered at some integer m Average of these five = n, so m = n Add the next two consecutive integers (m + 3 and m + 4) Concept / Approach: For 5 numbers, the middle is the 3rd term = m = n. For 7 numbers, the middle is the 4th term, which is m + 1. Hence the average increases by exactly 1. No computation of actual values is necessary beyond understanding the symmetry of consecutive integers. Step-by-Step Solution: Original set (5): m − 2, m − 1, m, m + 1, m + 2 → average = m = n After adding two next numbers: m − 2, …, m + 2, m + 3, m + 4 → middle = 4th = m + 1 New average = m + 1 = n + 1 → increase by 1 Verification / Alternative check: Test with m = 10: original average of 8, 9, 10, 11, 12 is 10. Adding 13 and 14, the 7-number average is 11. Increase = 1. Why Other Options Are Wrong: Increases by 2 or 1.4 contradict the fixed shift of the middle; “Remains the same” is false; “Decrease by 1” is impossible when only larger values are added symmetrically above. Common Pitfalls: Averaging endpoints or miscounting positions; always use the middle term logic for consecutive sequences. Final Answer: increase by 1

Question 16

A bowler’s average is 12.4 runs per wicket. In his last match he took 5 wickets for 26 runs and his overall bowling average decreased by 0.4. How many wickets had he taken before his last match?

Accepted Answer

Introduction: Bowling average = total runs conceded / total wickets. When a new performance is added, the updated average can be expressed in terms of the prior totals plus the last match’s runs and wickets. This yields a linear equation in the unknown prior wickets. Given Data / Assumptions: Prior average = 12.4 runs/wicket Last match: 5 wickets for 26 runs New average = 12.4 − 0.4 = 12.0 runs/wicket Concept / Approach: Let prior wickets = x and prior runs = 12.4x. After the last match: wickets = x + 5; runs = 12.4x + 26. Set new average = (12.4x + 26) / (x + 5) = 12 and solve for x. Step-by-Step Solution: (12.4x + 26) / (x + 5) = 12 12.4x + 26 = 12x + 60 0.4x = 34 → x = 85 Verification / Alternative check: New totals: runs = 12.4 * 85 + 26 = 1054 + 26 = 1080; wickets = 85 + 5 = 90; 1080 / 90 = 12.0, confirming the drop by 0.4. Why Other Options Are Wrong: 78, 72, and 64 do not satisfy the equation; 90 would give an average different from 12.0 after adding the last match’s performance. Common Pitfalls: Averaging averages or mixing up the order of addition; always work with total runs and total wickets. Final Answer: 85

Question 17

A hostel has 35 students. When 7 more students join (making 42), the daily mess expense increases by Rs 42, yet the average expense per head decreases by Re 1. What was the original total daily mess expense?

Accepted Answer

Introduction: This is a classic average cost problem. Increasing the headcount changes both the total and per-head costs. We set up two equations: one for the new total expense and another linking new per-head cost to the old per-head cost minus Re 1, then solve for the original total expense. Given Data / Assumptions: Original students = 35; new students = 42 Increase in total expense = Rs 42 Decrease in per-head expense = Re 1 Concept / Approach: Let original total expense be E. Then original per-head = E / 35. New total = E + 42, so new per-head = (E + 42) / 42. Given (E + 42) / 42 = (E / 35) − 1. Solve for E. Step-by-Step Solution: (E + 42) / 42 = E / 35 − 1 Multiply by 210: 5(E + 42) = 6E − 210 5E + 210 = 6E − 210 → E = 420 Verification / Alternative check: Original per-head = 420 / 35 = 12. New total = 462; new per-head = 462 / 42 = 11, exactly Re 1 less. Why Other Options Are Wrong: 432, 442, 40, and 410 fail the per-head check when substituted back into the relationships. Common Pitfalls: Subtracting Rs 42 from per-head instead of adjusting totals; mixing up “increase” and “decrease”. Final Answer: Rs. 420

Question 18

Five years ago, the average age of a husband and wife was 23 years. Now, the average age of the husband, the wife, and their child (born in the interim) is 20 years. How old is the child now?

Accepted Answer

Introduction: Average age problems with shifting family sizes are best handled via total ages. Convert averages to totals at each time point, account for aging across years, and compare to deduce the newcomer’s current age. Given Data / Assumptions: 5 years ago: average of 2 people = 23 → total then = 46 Now: average of 3 people = 20 → total now = 60 Husband and wife each aged 5 years over the interval Concept / Approach: Present total of the couple = past total + 2 * 5 = 46 + 10 = 56. Present family total (including the child) is 60, so the child’s current age is the difference 60 − 56. Step-by-Step Solution: Couple now = 46 + 10 = 56 Family now = 60 Child age now = 60 − 56 = 4 years Verification / Alternative check: Five years ago child did not exist or was newly born later; totals align with averages exactly when using 4 years as the child’s age now. Why Other Options Are Wrong: 3, 1, less than 1, and 5 years lead to totals that do not match the given averages simultaneously. Common Pitfalls: Averaging ages again after adding years or forgetting both adults gain 5 years each, not 5 years in total. Final Answer: 4 year

Question 19

Ten years ago, the average age of a family of 4 members was 24 years. Since then two children were born, and the average age of the family is the same (24 years) today. If the two children differ in age by 2 years, what is the present age of the you…

Accepted Answer

Introduction: Family average age questions are solved by converting averages to totals and tracking how totals change as years pass and members are added. The constancy of the average despite added members reveals the combined current age of the new children, which can be split using their given age gap. Given Data / Assumptions: 10 years ago: 4 members, average 24 → total then = 24 * 4 = 96 Now: 6 members, average still 24 → total now = 24 * 6 = 144 Two new children; their ages differ by 2 years Concept / Approach: If the original 4 members aged 10 years each, their combined total increased by 40. Therefore current total from the original 4 is 96 + 40 = 136. Since the family total is 144, the two children together sum to 8 years now. Let the younger be x and the older be x + 2. Then 2x + 2 = 8 → x = 3. Step-by-Step Solution: Original four now = 96 + 40 = 136 Children total now = 144 − 136 = 8 x + (x + 2) = 8 → x = 3 (younger), older = 5 Verification / Alternative check: With children aged 3 and 5, the current average returns to 24 for 6 members exactly. Why Other Options Are Wrong: 1, 2, 4, or 5 years for the youngest break the 8-year total constraint or the 2-year gap. Common Pitfalls: Forgetting that each of the initial 4 members adds 10 years to the total; mixing up “difference” and “sum”. Final Answer: 3 years

Question 20

The mean temperature of Monday to Wednesday is 37 °C and of Tuesday to Thursday is 34 °C. If Thursday’s temperature is 4/5 of Monday’s temperature, what was Thursday’s temperature?

Accepted Answer

Introduction: This problem combines overlapping averages with a proportion condition between two specific days. By turning averages into equations for 3-day sums and using the 4/5 relation, we can solve for Monday and then obtain Thursday directly. Given Data / Assumptions: (M + T + W) / 3 = 37 → M + T + W = 111 (T + W + Th) / 3 = 34 → T + W + Th = 102 Th = (4/5) * M Concept / Approach: Subtract the two 3-day sums: (T + W + Th) − (M + T + W) = 102 − 111 = −9 → Th − M = −9 → Th = M − 9. Combine with Th = (4/5)M to solve for M, then compute Th. Step-by-Step Solution: M − 9 = (4/5)M M − (4/5)M = 9 → (1/5)M = 9 → M = 45 Th = (4/5)*45 = 36 °C Verification / Alternative check: With M = 45 and Th = 36, the difference Th − M = −9 matches the subtraction of the sums (−9), confirming consistency. Why Other Options Are Wrong: 35.5°, 34°, and 36.5° fail the system of equations. 33° does not satisfy the 4/5 proportion when paired with any integer Monday consistent with the sums. Common Pitfalls: Averaging the averages or ignoring the given proportion; always convert to linear equations first. Final Answer: 36° C

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