Question 1

Section-wise pass averages: Section A has 20 students with an average of 80%. Section B has 30 students with an average of 70%. Find the combined passing average of all 50 students.

Accepted Answer

Introduction / Context: Combining two groups with different averages requires a weighted mean using their sizes. Larger groups influence the result more than smaller groups. Given Data / Assumptions: Section A: 20 students, average 80% Section B: 30 students, average 70% Concept / Approach: Combined average = (sum of group totals) / (total students). Compute each total as average * size and divide by 50. Step-by-Step Solution: Total A = 20 * 80 = 1600Total B = 30 * 70 = 2100Combined total = 1600 + 2100 = 3700Combined average = 3700 / 50 = 74% Verification / Alternative check: The combined mean must lie between 70% and 80% and be closer to 70% since the larger group (30) is at 70%. 74% fits that intuition. Why Other Options Are Wrong: 72%, 75%, 77%, 73%: do not equal 3700/50. Common Pitfalls: Averaging 80 and 70 directly to 75% ignores group sizes and leads to an incorrect result here. Final Answer: 74%

Question 2

Backdating family averages: The present average age of a 4-member family is 20 years. The youngest is 4 years old now. What was the average age of the family at the time of the youngest's birth (i.e., 4 years ago, before the baby was counted)?

Accepted Answer

Introduction / Context: This problem asks for the average at the time of the youngest child’s birth, when only the three older members were counted. We must backdate ages by 4 years for those three and exclude the newborn (age 0 then). Given Data / Assumptions: Current average (4 members) = 20 → current total = 80 Youngest current age = 4 → youngest was 0 at birth At birth time: only 3 members are counted Concept / Approach: Four years ago, the three older members were each 4 years younger. Subtract 4 for each of those three from the current total, exclude the youngest (who was 0 then), and divide by 3. Step-by-Step Solution: Current total (all 4) = 80Remove youngest now to isolate three older = 80 - 4 = 76Backdate 4 years for each of the three: subtract 3 * 4 = 12Total of three at birth time = 76 - 12 = 64Average then = 64 / 3 = 21 1/3 years Verification / Alternative check: If we included the newborn at age 0 and divided by 4, we would get 16, but the question clarifies “before the baby was counted,” so use 3 members, giving 21 1/3. Why Other Options Are Wrong: 27, 22: do not match the correctly backdated total for three members. 23 1/3: overestimates the average. 16: corresponds to including the newborn at age 0, which the prompt excludes. Common Pitfalls: Including the newborn in the average four years ago (age 0) or forgetting to reduce each older member by 4 years when backdating. Final Answer: 21 1/3 yr

Question 3

Average unchanged after birth: Three years ago, a 5-member family averaged 17 years. Today, even after a child is born, the family’s average is still 17. Find the present age of the child.

Accepted Answer

Introduction / Context: Similar to earlier average-over-time problems, we compare the current total with the projected total of the original members. The difference yields the child’s present age. Given Data / Assumptions: 3 years ago: 5 members, average = 17 → total = 85 Today: 6 members after birth, average still 17 → total = 6 * 17 = 102 Concept / Approach: Increase in total for original five due to 3 years = 5 * 3 = 15. Add this to 85 to get the original-five total today, then compare with 102 to get the child’s age now. Step-by-Step Solution: Original five today would total = 85 + 15 = 100Actual total today = 102Child’s present age = 102 - 100 = 2 years Verification / Alternative check: Total rise in the five members over 3 years is 15 years. The child must be 2 so that the six-person average remains 17 today. Why Other Options Are Wrong: 1, 3, 2.5, 4: do not equate the present total to 102. Common Pitfalls: Forgetting to add the 3-year increment for each of the original five or miscomputing the current grand total for six members. Final Answer: 2 yr

Question 4

Equal-expenditure purchases: A person spends ₹36 at each of five markets, where prices per orange are ₹1.00, ₹1.50, ₹1.80, ₹2.00, and ₹2.25. Find the average price per orange across all oranges bought.

Accepted Answer

Introduction / Context: When spending equal amounts at different prices, the number of items purchased varies inversely with price. The overall average price equals (total spent) / (total items), which simplifies using reciprocals of the prices (a harmonic-mean style setup weighted by equal expenditure). Given Data / Assumptions: Spend per market = ₹36 (five markets) Prices: ₹1.00, ₹1.50, ₹1.80, ₹2.00, ₹2.25 per orange Concept / Approach: Total oranges = Σ(36 / price). Total spend = 5 * 36. Average price = (total spend) / (total oranges) = 5 / Σ(1 / price). Step-by-Step Solution: Σ(1/price) = 1/1 + 1/1.5 + 1/1.8 + 1/2 + 1/2.25= 1 + 0.666666... + 0.555555... + 0.5 + 0.444444... = 19/6Average price = 5 / (19/6) = 30/19 = ₹1.5789... ≈ ₹1.58 Verification / Alternative check: Compute total oranges explicitly: 36, 24, 20, 18, 16 → total 114 oranges for ₹180 → 180/114 = ₹1.5789..., same result. Why Other Options Are Wrong: ₹1.91, ₹2.00, ₹1.80, ₹1.70: do not equal 30/19 to two decimal places. Common Pitfalls: Averaging prices directly (arithmetic mean) without accounting for different quantities obtained at each price due to equal spending. Final Answer: ₹ 1.58

Question 5

Linked subgroup averages: avg(a,b,c)=11; avg(c,d,e)=17; avg(e,f)=22; avg(e,c)=17. Find the average of a, b, c, d, e, f.

Accepted Answer

Introduction / Context: We are given several overlapping averages. Converting each to totals yields linear relations among the variables. The goal is the grand total divided by 6, which can be expressed using the provided sums and the known e + c relation. Given Data / Assumptions: a + b + c = 33 c + d + e = 51 e + f = 44 e + c = 34 Concept / Approach: Express T = a + b + c + d + e + f using known sums. Replace a + b by 33 - c, d by 51 - c - e, and f by 44 - e. Substitute and simplify, then divide by 6 for the average. Step-by-Step Solution: T = (33 - c) + c + (51 - c - e) + e + (44 - e)= 33 + 51 + 44 - c - eBut c + e = 34 ⇒ T = 128 - 34 = 94Average = T / 6 = 94 / 6 = 47 / 3 = 15 2/3 Verification / Alternative check: Plugging back: If average is 47/3, total is 94. All intermediate sums remain satisfied since we used identities only. Why Other Options Are Wrong: 18 1/2, 21 1/3, 16 1/2, 17 1/6: do not equal 94/6. Common Pitfalls: Algebra slips when substituting or forgetting to use e + c = 34, leading to an expression still in terms of c and e. Final Answer: 15 2/3

Question 6

Average correction after data error: The average age of 54 girls was initially computed as 14 years. One girl's actual age is 10.5 years, but it was taken as 13. Find the corrected average (approximate to the nearest whole year).

Accepted Answer

Introduction / Context: When a recording error is found, adjust the total by removing the wrong entry and adding the correct one. The new average is the corrected total divided by the unchanged count. The prompt asks for an approximate whole-year answer. Given Data / Assumptions: Count = 54 girls Old average = 14 years → old total = 54 * 14 = 756 Wrong entry used = 13 years; correct value = 10.5 years Concept / Approach: Corrected total = old total - wrong value + correct value. Corrected average = corrected total / 54. Round to the nearest year as requested. Step-by-Step Solution: Corrected total = 756 - 13 + 10.5 = 753.5Corrected average = 753.5 / 54 ≈ 13.9537...Rounded to nearest whole year ≈ 14 Verification / Alternative check: Because the correction lowers the total by only 2.5 across 54 students, the average hardly changes and remains approximately 14 when rounded to the nearest year. Why Other Options Are Wrong: 10.50, 12.50, 12: far from the computed 13.95. 13: this is closer than others but the prompt asks for rounding to a whole year; 13.95 rounds to 14. Common Pitfalls: Dividing by 53 after removing the incorrect entry; the headcount remains 54 because we replaced a value, not removed a student. Final Answer: 14 yr

Question 7

Father and twin sons: The average age of a father and his two twin sons is 30 years. If the ratio of the father’s age to one son’s age is 5 : 2, find the father’s current age.

Accepted Answer

Introduction / Context: We have one parent and two children with an average and a ratio. Transform the ratio into algebraic expressions and use the average to get the exact ages. Given Data / Assumptions: Average of three = 30 → total = 90 Father : one son = 5 : 2 Sons are twins, so both have the same age Concept / Approach: Let one son be S and father be (5/2)S. With two sons, total = (5/2)S + 2S = (9/2)S = 90. Solve for S, then compute the father’s age. Step-by-Step Solution: (9/2)S = 90 ⇒ S = 20Father’s age = (5/2) * 20 = 50 years Verification / Alternative check: Totals: father 50, sons 20 and 20 → sum 90, average 30, ratio 50 : 20 = 5 : 2, all conditions met. Why Other Options Are Wrong: 30, 45, 20, 40: inconsistent with both the ratio and the total 90. Common Pitfalls: Treating the 5 : 2 ratio as 5 : 4 for two sons; the ratio is stated for one son, then doubled separately when summing ages. Final Answer: 50 yr

Question 8

Inventory ratio and mean: In a stationery shop, the ratio of pens to pencils is 3 : 2, and the average of the two counts is 180 (i.e., (pens + pencils) / 2 = 180). What is 30% of the number of pencils?

Accepted Answer

Introduction / Context: The phrase “average of the two counts is 180” means (pens + pencils)/2 = 180. With the ratio of 3 : 2, we can find each count and then compute 30% of the pencil count (which may be fractional if not an integer multiple of 10). Given Data / Assumptions: Pens : Pencils = 3 : 2 (Pens + Pencils) / 2 = 180 ⇒ Pens + Pencils = 360 Concept / Approach: Let pens = 3k and pencils = 2k. Then 5k = 360 → k = 72. Thus pencils = 2k = 144. Compute 30% of 144 as 0.30 * 144. Step-by-Step Solution: k = 360 / 5 = 72Pencils = 2k = 14430% of pencils = 0.30 * 144 = 43.2 Verification / Alternative check: Pens = 3k = 216. Check average: (216 + 144)/2 = 360/2 = 180, consistent with the prompt. Why Other Options Are Wrong: 144, 72, 57.6, 86.4: these are counts or other percentages, not exactly 30% of 144. Common Pitfalls: Misreading “average” as 180 each or misapplying the ratio to the average rather than to the actual counts. Final Answer: 43.2

Question 9

Pairwise averages of three people: If avg(Rakesh, Mohan) = 15, avg(Mohan, Ramesh) = 12, and avg(Rakesh, Ramesh) = 13, find Mohan’s age.

Accepted Answer

Introduction / Context: We are given three pairwise averages among Rakesh (Rk), Mohan (Mo), and Ramesh (Ra). Converting to pairwise sums allows elimination and solution for an individual age directly. Given Data / Assumptions: Rk + Mo = 30 Mo + Ra = 24 Rk + Ra = 26 Concept / Approach: Add the first two equations to obtain Rk + 2Mo + Ra. Then substitute Rk + Ra from the third equation to isolate Mo. Step-by-Step Solution: (Rk + Mo) + (Mo + Ra) = 30 + 24 = 54Rk + 2Mo + Ra = 54But Rk + Ra = 26 ⇒ 26 + 2Mo = 542Mo = 28 ⇒ Mo = 14 Verification / Alternative check: Compute individuals: From Rk + Mo = 30, Rk = 16; from Mo + Ra = 24, Ra = 10. Check Rk + Ra = 26, consistent. Why Other Options Are Wrong: 16, 13, 12, 15: do not satisfy all three pairwise sums simultaneously. Common Pitfalls: Attempting to solve using three variables directly without first converting to pairwise sums and applying elimination succinctly. Final Answer: 14

Question 10

Family ages with youngest at birth: The average age of five sisters is 20 years at present. The youngest sister is currently 4 years old. What was the average age of the full group (all five, counting the newborn as age 0) at the exact time of the y…

Accepted Answer

Introduction / Context: Average age problems often require translating “average now” into a total sum and then rolling ages backward by a fixed number of years. Here, we compute the group average at the moment the youngest was born by deducting elapsed years from the older siblings and setting the newborn's age to 0. Given Data / Assumptions: Five sisters in total. Present average age = 20 years. Youngest sister's present age = 4 years. At the time of birth of the youngest (4 years ago), the other four were each 4 years younger; the youngest was 0. Concept / Approach: Convert the present average to a present total. Move 4 years back in time: subtract 4 years for each of the four elder sisters; the youngest contributes 0 at birth. Finally, divide by 5 to get the past average. Step-by-Step Solution: Present total = 5 * 20 = 100Sum of four elders now = 100 - youngest(4) = 96Four elders 4 years earlier: 96 - 4*4 = 96 - 16 = 80Youngest at birth = 0Total at birth = 80 + 0 = 80Average at birth = 80 / 5 = 16 years Verification / Alternative check: Forward check: Add 4 years to each of the five to get from birth to now adds 20 years to total, consistent with the present totals computed above. Why Other Options Are Wrong: 15 years: Would imply a total of 75 at birth, inconsistent with present data. 17 or 18 years: Overestimates the elders’ contribution after rolling back ages. Common Pitfalls: Forgetting to reduce each elder’s age individually by 4 years or treating the newborn as already aged 4 at birth. Final Answer: 16 years

Question 11

Targeting an overall percentage: Three annual exams are each out of 500 marks. A student averaged 45% in the first and 55% in the second. To achieve an overall average of 60% across all three, how many marks are needed in the third exam?

Accepted Answer

Introduction / Context: Weighted averages across equal-maximum exams reduce to totals. The overall target percentage dictates a target total; subtract known totals to find the remaining requirement. Given Data / Assumptions: Three exams, each with 500 marks. Exam 1 average = 45% of 500. Exam 2 average = 55% of 500. Desired overall average = 60% across 1500 marks total. Concept / Approach: Total required = target percentage * grand total marks. Marks needed in exam 3 = required total minus (marks from exam 1 + exam 2). Step-by-Step Solution: Exam 1 = 0.45 * 500 = 225Exam 2 = 0.55 * 500 = 275Target total for 60% = 0.60 * 1500 = 900Needed in exam 3 = 900 - (225 + 275) = 900 - 500 = 400 Verification / Alternative check: Totals become 225 + 275 + 400 = 900; 900/1500 = 0.60 = 60%. Why Other Options Are Wrong: 450: Would give 950 total (63.33%). 350: Only 850 total (56.67%). 300: Just 800 total (53.33%). Common Pitfalls: Averaging the percentages directly without converting to totals can mislead when paper maxima differ; here they are equal but totals still keep logic clean. Final Answer: 400

Question 12

Women and children workforce: The combined average age is 15 years. The 16 children average 8 years and the women average 22 years. If 10 of the women are married, how many women workers are unmarried?

Accepted Answer

Introduction / Context: Mixture (weighted average) problems with two groups can be solved by equating total ages computed from averages. Once the number of women is known, subtract married to get unmarried. Given Data / Assumptions: Average (all) = 15 years. Children: 16 members, average 8 years. Women: average 22 years, count unknown. Married women = 10; need unmarried women count. Concept / Approach: Total age = average * count. If w is women count, set 16*8 + 22*w = 15*(16 + w) and solve for w. Then unmarried = w - married. Step-by-Step Solution: Children total = 16 * 8 = 128Women total = 22 * wOverall total = 15 * (16 + w) = 240 + 15wEquation: 128 + 22w = 240 + 15w7w = 112 ⇒ w = 16Unmarried women = 16 - 10 = 6 Verification / Alternative check: Totals: 128 + 22*16 = 128 + 352 = 480; group size = 32; 480/32 = 15, consistent. Why Other Options Are Wrong: 16 or 12 or 8: Do not align with the solved women count 16 and 10 married. Common Pitfalls: Forgetting to multiply averages by group sizes or mixing up the subtraction of married from total women. Final Answer: 6

Question 13

Pairwise averages: If (Dinesh + Mahesh)/2 = 16, (Mahesh + Rahul)/2 = 13, and (Dinesh + Rahul)/2 = 14, what is Mahesh’s age?

Accepted Answer

Introduction / Context: With three pairwise averages, linear equations allow recovery of each individual age. Adding and subtracting strategically eliminates variables. Given Data / Assumptions: (D + M)/2 = 16 ⇒ D + M = 32 (M + R)/2 = 13 ⇒ M + R = 26 (D + R)/2 = 14 ⇒ D + R = 28 Concept / Approach: Add equations to collect terms and compare with a third to isolate one variable. Here, summing the first two and subtracting the third isolates M. Step-by-Step Solution: (D + M) + (M + R) = 32 + 26 = 58So D + 2M + R = 58But D + R = 28Subtract: (D + 2M + R) - (D + R) = 58 - 282M = 30 ⇒ M = 15 Verification / Alternative check: Compute D = 32 - M = 17 and R = 26 - M = 11; check D + R = 28. Works. Why Other Options Are Wrong: 20 yr, 58 yr, 16 yr: Inconsistent with the three equations. Common Pitfalls: Trying to average the three averages directly without solving the system or mishandling subtraction. Final Answer: 15 yr

Question 14

Pass–fail split from overall average: In a class of 80 students, overall average marks are 40. Passed students average 60 and failed students average 35. How many students failed?

Accepted Answer

Introduction / Context: Mixture-averages with two groups can be solved by equating weighted totals. Let p pass and 80 - p fail; use the given group averages to match the overall total. Given Data / Assumptions: Total students = 80; overall average = 40 ⇒ total marks = 3200. Passed average = 60; Failed average = 35. Concept / Approach: Total = (pass count * pass avg) + (fail count * fail avg). Solve for p, then compute failed = 80 - p. Step-by-Step Solution: Let p = passed ⇒ failed = 80 - p60p + 35(80 - p) = 320060p + 2800 - 35p = 320025p = 400 ⇒ p = 16Failed = 80 - 16 = 64 Verification / Alternative check: Totals: 16*60 = 960; 64*35 = 2240; sum = 3200, average = 3200/80 = 40. Correct. Why Other Options Are Wrong: 63, 60, 72: Do not satisfy the weighted-total equation with the given averages. Common Pitfalls: Confusing pass and fail counts or forgetting to multiply averages by their respective counts. Final Answer: 64

Question 15

Recession effect on counts: Sahara, Tata, Singhania, and Birla each had 250 industries last year. This year each sold 10 factories. What is their average number of industries now?

Accepted Answer

Introduction / Context: When each member of a set changes by the same amount, the average changes by that same amount. Here, a uniform decrease occurs for all four companies. Given Data / Assumptions: Four companies, each had 250 industries last year. Each sold 10 factories this year. Concept / Approach: If every element decreases by 10, the average also decreases by 10. So new average = old average - 10. Step-by-Step Solution: Old average = 250Decrease across the board = 10New average = 250 - 10 = 240 Verification / Alternative check: Totals: old total = 4*250 = 1000; new total = 4*(250 - 10) = 4*240 = 960; average = 960/4 = 240. Why Other Options Are Wrong: 260: Would require an increase, not a decrease. 25: Unit error (should be industries, not tens). None of these: Incorrect since 240 is attainable and correct. Common Pitfalls: Forgetting the property that uniform shifts translate directly to average shifts. Final Answer: 240

Question 16

Father–daughter ages: Present ratio is 3:1. Four years ago it was 4:1. What will be the average age of the father and the daughter two years hence?

Accepted Answer

Introduction / Context: Age-ratio problems combine present ratios with past ratios to recover exact ages. After solving the present ages, projecting 2 years forward and averaging is straightforward. Given Data / Assumptions: Current ratio father:daughter = 3:1 ⇒ Father = 3x, Daughter = x Four years ago ratio = 4:1 ⇒ (3x - 4):(x - 4) = 4:1 Find the average 2 years hence. Concept / Approach: Set up and solve a linear equation from the historical ratio, then add 2 to each age and compute the mean. Step-by-Step Solution: (3x - 4) / (x - 4) = 43x - 4 = 4x - 16x = 12 ⇒ Daughter now = 12, Father now = 36Two years hence: Daughter = 14, Father = 38Average in 2 years = (38 + 14) / 2 = 26 Verification / Alternative check: Check past ratio: 36 - 4 = 32 and 12 - 4 = 8 ⇒ 32:8 = 4:1, as given. Why Other Options Are Wrong: 24 or 25: Underestimates the forward ages. 36: That is the father’s present age, not the required average 2 years later. Common Pitfalls: Mixing present and past ages or averaging the ratios instead of solving the equation. Final Answer: 26

Question 17

Linked triples: The average of three numbers is 42. The first is twice the second and the second is twice the third. What is the difference between the largest and the smallest number?

Accepted Answer

Introduction / Context: Ratio chaining across three numbers allows expressing all in one variable. With the average known, compute each and then the required difference (range). Given Data / Assumptions: Average of three = 42 ⇒ sum = 126. First = 2 * second; Second = 2 * third. Concept / Approach: Let third = t ⇒ second = 2t ⇒ first = 4t. Then 4t + 2t + t = 7t = 126, so t = 18. Largest is 4t, smallest is t. Step-by-Step Solution: 7t = 126 ⇒ t = 18Numbers: 72, 36, 18Difference largest - smallest = 72 - 18 = 54 Verification / Alternative check: Average check: (72 + 36 + 18) / 3 = 126/3 = 42, consistent. Why Other Options Are Wrong: 18 and 36: Intermediate values from the ratio, not the range. 72: Largest value itself, not the difference. Common Pitfalls: Reversing the doubling chain or forgetting that average fixes the sum first. Final Answer: 54

Question 18

Overlapping four-day averages: A mechanic’s average earning for the first four days of a week is Rs 18, and for the last four days is Rs 22. If he earned Rs 20 on the fourth day, what is his average earning for the entire seven-day week?

Accepted Answer

Introduction / Context: Two overlapping four-day averages (days 1–4 and days 4–7) allow recovery of the seven-day total if the overlapping day’s value is known. Subtract the overlap once to avoid double-counting. Given Data / Assumptions: Average days 1–4 = Rs 18 ⇒ sum1 = 72. Average days 4–7 = Rs 22 ⇒ sum2 = 88. Day 4 earning = Rs 20. Concept / Approach: Sum over the week equals sum1 + sum2 minus the overlap day once (because day 4 appears in both sums). Then divide by 7 for the weekly average. Step-by-Step Solution: sum1 + sum2 = 72 + 88 = 160Weekly total = 160 - day4 = 160 - 20 = 140Weekly average = 140 / 7 = Rs 20 Verification / Alternative check: Construct a small example with variables for the unknown days and verify the same relation; algebra reduces exactly to subtracting the overlap once. Why Other Options Are Wrong: Rs 18.95, Rs 16, Rs 25.71: Do not match the computed weekly average from correct total 140. Common Pitfalls: Forgetting day 4 is counted twice in the two four-day blocks; failing to subtract it once leads to an inflated total. Final Answer: Rs 20

Question 19

Two-group salary mix: The average salary across all workers is Rs 850. Seven technicians average Rs 1000 each. The rest average Rs 780 each. How many total workers are there in the workshop?

Accepted Answer

Introduction / Context: Another two-group mixture. With the overall average and one subgroup known, solve for total headcount by equating totals and using the other subgroup’s average. Given Data / Assumptions: Overall average = Rs 850. Technicians: 7 people, average = Rs 1000. Others: average = Rs 780, count unknown. Concept / Approach: Let total workers be n. Then total payroll = 850*n. Also equals 7*1000 + (n - 7)*780. Solve for n. Step-by-Step Solution: 850n = 7000 + 780(n - 7)850n = 7000 + 780n - 5460850n - 780n = 1540 ⇒ 70n = 1540 ⇒ n = 22 Verification / Alternative check: Totals: technicians = 7000; others = 15*780 = 11700; grand = 18700; 18700/22 = 850.
 Why Other Options Are Wrong: 18, 20, 24: Do not satisfy the weighted-average equality. Common Pitfalls: Misplacing parentheses in (n - 7)*780 or equating averages instead of totals. Final Answer: 22

Question 20

Replacement effect on average weight: The average weight of 8 men rises by 2 kg when a 50 kg man is replaced by a newcomer. What is the newcomer’s weight?

Accepted Answer

Introduction / Context: When a replacement raises the average by a known amount, the total for the group rises by (group size * rise in average). Add this increase to the departing member’s weight to get the newcomer’s weight. Given Data / Assumptions: Group size = 8 men. Average increase = 2 kg. Departing weight = 50 kg. Concept / Approach: Total increase = group size * increase in average. Newcomer = departing + total increase (since only one person changed). Step-by-Step Solution: Total increase = 8 * 2 = 16 kgNewcomer weight = 50 + 16 = 66 kg Verification / Alternative check: Let old total be T. New total must be T - 50 + 66 = T + 16; average rises by 16/8 = 2, as stated. Why Other Options Are Wrong: 52, 58: Too small; would not raise the average by 2. 68: Would raise the total by 18, hence average by 2.25, not 2. Common Pitfalls: Multiplying the average increase by 7 (excluding the replaced person) instead of the full group count 8. Final Answer: 66 kg

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