Question 1

Teacher’s weight from a slight average increase: The average weight of 40 students is 40 kg. Including the teacher increases the average by 0.5 kg. What is the teacher’s weight?

Accepted Answer

Introduction / Context: A small change in average across many people implies a significant change in total weight. Use totals before and after including the teacher to isolate their weight. Given Data / Assumptions: Students: 40 with average 40 kg. With teacher: 41 persons with average 40.5 kg. Concept / Approach: Compute totals from averages. Teacher’s weight = new total − old total. Step-by-Step Solution: Students’ total = 40 * 40 = 1600 kgWith teacher total = 41 * 40.5 = 1660.5 kgTeacher’s weight = 1660.5 − 1600 = 60.5 kg Verification / Alternative check: Check average difference: 0.5 * 41 = 20.5 extra total. Students at 1600; adding a 60.5 kg teacher indeed gives +60.5, which results in +0.5 average across 41. Why Other Options Are Wrong: 40.5 kg: Too light; would hardly change average. 60 kg or 62 kg: Do not match the exact totals. Common Pitfalls: Using 40 (not 41) when multiplying the increased average or forgetting that the count changes to 41. Final Answer: 60.5 kg

Question 2

Coach added raises the average by 20%: The average age of four players is 18.5 years. If the coach is included, the average increases by 20%. What is the coach’s age?

Accepted Answer

Introduction / Context: An average increased by a percentage indicates a multiplicative change to the mean. Convert to totals before and after adding the coach to find their age. Given Data / Assumptions: Players: 4 with average 18.5 years ⇒ total 74. Including coach: 5 people; new average = 18.5 * 1.20 = 22.2 years. Concept / Approach: Coach’s age = (new average * 5) − old total. Step-by-Step Solution: Old total (players) = 4 * 18.5 = 74New total with coach = 5 * 22.2 = 111Coach’s age = 111 − 74 = 37 years Verification / Alternative check: Average check with ages consistent: (74 + 37) / 5 = 111/5 = 22.2 years, which is 20% higher than 18.5. Why Other Options Are Wrong: 28, 31, 34 years: Too small to raise the average by the stated 20% with count increasing to 5. Common Pitfalls: Adding 20% of total instead of 20% of average, or forgetting that headcount increases from 4 to 5. Final Answer: 37 years

Question 3

Four children spaced by 4 years: The average age is 12 years. If consecutive age gaps are 4 years, what is the age of the youngest child?

Accepted Answer

Introduction / Context: An arithmetic progression model fits siblings with equal spacing. Represent ages in terms of the youngest and sum to match the given average. Given Data / Assumptions: Four children; average = 12 ⇒ total = 48. Spacing between consecutive ages = 4 years. Concept / Approach: Let youngest = a. Then ages are a, a + 4, a + 8, a + 12. Sum equals 4a + 24; set equal to 48 and solve for a. Step-by-Step Solution: 4a + 24 = 484a = 24 ⇒ a = 6 Verification / Alternative check: Ages become 6, 10, 14, 18 with sum 48; average is 12, consistent with the prompt. Why Other Options Are Wrong: 7, 8, 9 years: Do not satisfy the sum required by the given average and spacing. Common Pitfalls: Using the average as the youngest directly or ignoring that four terms are spread symmetrically around the mean. Final Answer: 6 years

Question 4

Production dip and monthly average: A TV factory produces on average 60 sets/day for the first 25 days of a month. Illness lowers output over the last 5 days so that the monthly average becomes 58. What is the average production per day during the l…

Accepted Answer

Introduction / Context: Block averages (first 25 days, last 5 days) can be combined via totals to reach the monthly average. Subtract the known block to get the unknown block's total and divide by its length. Given Data / Assumptions: First 25 days average = 60 ⇒ total = 1500. Monthly average = 58 across 30 days ⇒ total = 1740. Need average over last 5 days. Concept / Approach: Last 5 days total = monthly total − first 25 days total. Then divide by 5. Step-by-Step Solution: Monthly total = 30 * 58 = 1740First-25 total = 25 * 60 = 1500Last-5 total = 1740 − 1500 = 240Average last 5 = 240 / 5 = 48 Verification / Alternative check: Weighted average check: (1500 + 240)/30 = 1740/30 = 58. Works. Why Other Options Are Wrong: 45 or 52: Would yield monthly totals different from 1740. 58: That is the monthly average, not the last 5-day average after a dip. Common Pitfalls: Averaging 60 and 58 directly instead of using totals and correct day counts. Final Answer: 48

Question 5

Passed candidates from overall average: Among 120 candidates, the overall average marks are 35. Passed candidates average 39, and failed candidates average 15. How many candidates passed?

Accepted Answer

Introduction / Context: Classic two-group mixture: express grand total via overall average and equate it to the sum of group totals from their averages to solve the passed count. Given Data / Assumptions: Total candidates = 120; overall average = 35 ⇒ total marks = 4200. Passed average = 39; Failed average = 15. Concept / Approach: Let p be the number passed. Then 39p + 15(120 − p) = 4200. Solve for p. Step-by-Step Solution: 39p + 1800 − 15p = 420024p = 2400 ⇒ p = 100 Verification / Alternative check: Totals: pass = 3900; fail = 20*15 = 300; sum = 4200; average = 4200/120 = 35. All consistent. Why Other Options Are Wrong: 110, 120, 150: Do not satisfy the weighted-average equation with given group means. Common Pitfalls: Attempting to average the group averages without weighting by counts leads to wrong answers. Final Answer: 100

Question 6

Grouped averages within a class: The average age of 30 students is 12 years. One group of 5 averages 10 years and another group of 5 averages 14 years. What is the average age of the remaining students?

Accepted Answer

Introduction / Context: Partitioning a population into subgroups with known averages allows computing the remainder's average by subtracting subgroup totals from the grand total. Given Data / Assumptions: Total students = 30; overall average = 12 ⇒ total age = 360. Group A (5 students) average = 10 ⇒ total = 50. Group B (5 students) average = 14 ⇒ total = 70. Concept / Approach: Remaining students = 30 − 10 = 20. Their total age = 360 − (50 + 70) = 240. Average = total / count. Step-by-Step Solution: Remaining total = 360 − 120 = 240Remaining count = 20Average = 240 / 20 = 12 years Verification / Alternative check: Weighted average check: (50 + 70 + 240)/30 = 360/30 = 12, matching the given overall average. Why Other Options Are Wrong: 8 or 10 or 14 years: Do not balance to the grand total of 360 when combined with the two groups. Common Pitfalls: Dividing by 30 instead of 20 for the remainder or misadding subgroup totals. Final Answer: 12 years

Question 7

Average for remaining days: A shopkeeper earns Rs 504 in 12 days. His average for the first 4 days is Rs 40/day. What is his average for the remaining days?

Accepted Answer

Introduction / Context: Totals across time blocks with a known partial average let us isolate the remaining block’s total and compute its average. Given Data / Assumptions: Total for 12 days = Rs 504. First 4 days average = Rs 40 ⇒ total = Rs 160. Remaining days = 8 days. Concept / Approach: Remaining total = grand total − first block total. Then divide by days remaining. Step-by-Step Solution: Remaining total = 504 − 160 = 344Average for remaining 8 days = 344 / 8 = Rs 43 Verification / Alternative check: Add back: 160 + 344 = 504; overall average = 504/12 = 42, consistent with a heavier latter block average of 43 balancing with 40 early on. Why Other Options Are Wrong: Rs 40 or Rs 42 or Rs 45: Do not equal the exact computed 344/8. Common Pitfalls: Dividing by the full 12 instead of the remaining 8 or miscomputing the first block total. Final Answer: Rs 43

Question 8

Chain replacement and averages: A, B, C average 84 kg. Adding D yields a new average 80 kg. If E (weighing 3 kg more than D) replaces A, then the average of B, C, D, E is 79 kg. What is A’s weight?

Accepted Answer

Introduction / Context: This problem combines two average transformations: first adding a member, then replacing a member with another whose weight is defined relative to the added member. Solving sequentially with totals yields A's weight. Given Data / Assumptions: Average(A,B,C) = 84 ⇒ A + B + C = 252. Average(A,B,C,D) = 80 ⇒ A + B + C + D = 320 ⇒ D = 68. E = D + 3 = 71. Average(B,C,D,E) = 79 ⇒ B + C + D + E = 316. Concept / Approach: Use totals to isolate the unknowns: obtain D, then E, then compute B + C, and finally deduce A from A + (B + C) = 252. Step-by-Step Solution: From 4-person average: D = 320 − 252 = 68E = 68 + 3 = 71B + C = 316 − (D + E) = 316 − (68 + 71) = 177A = 252 − (B + C) = 252 − 177 = 75 Verification / Alternative check: Check the second condition: B + C + D + E = 177 + 68 + 71 = 316 ⇒ average 316/4 = 79. Consistent. Why Other Options Are Wrong: 70, 72, 80 kg: Do not satisfy all given averages simultaneously. Common Pitfalls: Forgetting to recompute totals after the replacement step or mixing the 3 kg offset direction for E relative to D. Final Answer: 75 kg

Question 9

Family average at birth of youngest: The average age of a family of 6 is 22 years now. The youngest member is 7 years old. What was the family’s average age at the time of the youngest member’s birth?

Accepted Answer

Introduction / Context: To find the past average at the birth of the youngest, roll back ages by the youngest’s current age for all older family members and set the infant’s age to 0 at birth. Then compute the average across all six members (including the newborn). Given Data / Assumptions: Family members = 6. Present average = 22 years ⇒ present total = 132. Youngest's present age = 7 years. Concept / Approach: At the birth of the youngest (7 years ago), the five older members were each 7 years younger. Subtract 7*5 from the sum of the older five; the youngest contributes 0. Divide by 6 for the past average. Step-by-Step Solution: Present total = 6 * 22 = 132Sum of five elders now = 132 − 7 = 125Sum of five elders 7 years earlier = 125 − 5*7 = 125 − 35 = 90Youngest at birth = 0 ⇒ total at birth = 90Average at birth = 90 / 6 = 15 years Verification / Alternative check: Forward check by adding 7 years to each of the six members’ birth-time ages adds 42 to the total (7*6), raising 90 to 132, which matches the present total. Why Other Options Are Wrong: 17 or 17.5 or 18 years: Overestimate the rolled-back ages for the older five. Common Pitfalls: Subtracting 7 years from the group total six times (including the newborn) instead of only for the five older members. Final Answer: 15 years

Question 10

Recovering absent students’ heights from class averages: The average height of 40 students is 163 cm. On a particular day, three students A, B, and C were absent, and the average of the remaining 37 students was 162 cm. If A and B have equal height …

Accepted Answer

Introduction / Context: This problem uses the definition of the arithmetic mean to reconstruct the combined height of absent students from two given averages. Once we know the total of the absent trio, the stated relations between A, B, and C allow us to solve for A’s exact height. Given Data / Assumptions: Total students = 40; class average = 163 cm Present students = 37; present average = 162 cm A and B have equal height; let A = B = h C is 2 cm less than A → C = h - 2 Concept / Approach: Average = total / count. Hence, total height of all 40 and total of the 37 present can be found. Their difference is the combined height of A, B, and C. Then solve the linear equation from h + h + (h - 2). Step-by-Step Solution: Total (40) = 40 * 163 = 6520 cmTotal (37) = 37 * 162 = 5994 cmSum(A,B,C) = 6520 - 5994 = 526 cmLet A = B = h, C = h - 2 → 3h - 2 = 526 → 3h = 528 → h = 176 cm Verification / Alternative check: Heights would be 176, 176, and 174. Their sum is 526, which matches the computed absent total. Substituting back confirms the two given averages. Why Other Options Are Wrong: 166, 180, 186, and 172 cm do not satisfy the linear relation 3h - 2 = 526, hence they cannot produce the correct totals. Common Pitfalls: Forgetting that C is exactly 2 cm less than A (not B), or using 40 instead of 37 when recomputing the present group’s total. Keep totals and counts consistent. Final Answer: 176 cm

Question 11

Average speed over a multi-leg trip with different speeds: Columbus travels 200 km from Lucknow to Kolkata at 40 km/h, then 300 km to Bengaluru at 20 km/h, and finally 500 km to Ahmedabad at 10 km/h. What is his overall average speed for the full 10…

Accepted Answer

Introduction / Context: Average speed for a complete trip equals total distance divided by total time, not the average of speeds. When each leg covers a fixed distance at a different speed, we must compute each leg’s time and sum them. Given Data / Assumptions: D1 = 200 km at v1 = 40 km/h D2 = 300 km at v2 = 20 km/h D3 = 500 km at v3 = 10 km/h Total distance D = 1000 km Concept / Approach: Total time T = D1/v1 + D2/v2 + D3/v3. Overall average speed v_avg = D / T. Avoid taking a naive mean of 40, 20, and 10 because distances differ and time is the true averaging weight. Step-by-Step Solution: T1 = 200 / 40 = 5 hT2 = 300 / 20 = 15 hT3 = 500 / 10 = 50 hTotal T = 5 + 15 + 50 = 70 hv_avg = 1000 / 70 = 14.2857… km/h = 14 2/7 km/h Verification / Alternative check: Because the slowest leg (10 km/h) covers the largest distance, the average must be closer to 10 than to 20 or 40; 14 2/7 fits this intuition. Why Other Options Are Wrong: 14 5/7 and 15.6 km/h are too high; “none of these” is incorrect because 14 2/7 exactly matches the computed value; 13 6/7 is too low. Common Pitfalls: Averaging speeds arithmetically or weighting by distances incorrectly. Always compute total time explicitly. Final Answer: 14 2/7 KM/hr

Question 12

Mean from deviations about an arbitrary value: The algebraic sum of deviations of 20 observations measured from 23 is 70. What is the mean of these observations?

Accepted Answer

Introduction / Context: When deviations are measured from an arbitrary number a, the sum of observations is sum(xi) = n * a + sum(xi - a). Knowing the deviation sum lets us recover the true mean quickly. Given Data / Assumptions: n = 20 observations Reference a = 23 Sum of deviations = Σ(xi - 23) = 70 Concept / Approach: Compute Σxi = n * a + Σ(xi - a) = 20 * 23 + 70, then divide by n to obtain the mean. Step-by-Step Solution: Σxi = 20 * 23 + 70 = 460 + 70 = 530Mean = 530 / 20 = 26 Verification / Alternative check: Equivalently, the mean equals 23 + (70 / 20) = 23 + 3.5 = 26.5? Careful: 70/20 = 3.5, but we must use Σxi = 460 + 70 = 530, and 530/20 = 26; direct “add 70/20 to 23” only works if 23 is the previous mean, not merely a reference. Hence the formula via totals is the robust path and yields 26. Why Other Options Are Wrong: 24, 25, and 27 do not match Σxi / n; 23 is just the reference about which deviations were computed, not the mean. Common Pitfalls: Adding deviation average to the reference as if the reference were the prior mean. Always compute Σxi correctly first. Final Answer: 26

Question 13

Adjusting an average after removing two values: The average of 50 numbers is 38. If two values, 45 and 55, are excluded, what is the average of the remaining 48 numbers?

Accepted Answer

Introduction / Context: Average adjustments are often easiest by working with totals. Removing some values reduces both the total and the count; recompute the mean from the new total and count. Given Data / Assumptions: Initial count = 50; average = 38 Removing values 45 and 55 Concept / Approach: Compute original total, subtract the removed numbers, then divide by the new count (48). Step-by-Step Solution: Original total = 50 * 38 = 1900New total = 1900 - (45 + 55) = 1900 - 100 = 1800New average = 1800 / 48 = 37.5 Verification / Alternative check: Because 45 and 55 average to 50 (above 38), removing them should decrease the mean, which aligns with 37.5. Why Other Options Are Wrong: 36.5 and 37 are too low; 37.52 is a rounding-like distractor; 38 ignores the removal of above-average values. Common Pitfalls: Dividing by 50 after removal or forgetting to subtract both removed values from the total. Final Answer: 37.5

Question 14

Correcting class average after two marking errors: The average marks of 30 students are 45. After rechecking, one student gains 45 more marks and another loses 15 marks. What is the adjusted class average?

Accepted Answer

Introduction / Context: Error correction in a dataset changes the total, which in turn changes the average. Here two opposite corrections occur, and their net effect adjusts the mean for the same class size. Given Data / Assumptions: n = 30 students; initial average = 45 Corrections: +45 to one student, −15 to another Concept / Approach: Adjusted total = original total + net change. Adjusted average = adjusted total / n. Net change is (+45) + (−15) = +30. Step-by-Step Solution: Original total = 30 * 45 = 1350Adjusted total = 1350 + 30 = 1380Adjusted average = 1380 / 30 = 46 Verification / Alternative check: Average increases by net change / n = 30 / 30 = 1, so 45 + 1 = 46, confirming the result. Why Other Options Are Wrong: 45 ignores the net increase; 44 decreases incorrectly; 47 and 46.5 overstate the impact of the +30 net correction. Common Pitfalls: Adding absolute changes without considering their signs or re-dividing by a wrong class size. Final Answer: 46

Question 15

Finding a missing mark from the group mean: The mean marks of a group of 7 students is 226. If six marks are 340, 180, 260, 56, 275, and 307, what is the seventh student’s mark?

Accepted Answer

Introduction / Context: Given a group mean and all but one values, we can compute the missing value from totals. This is a direct application of the definition of mean. Given Data / Assumptions: n = 7 students; mean = 226 Known six marks: 340, 180, 260, 56, 275, 307 Concept / Approach: Total = n * mean. Subtract the sum of known marks from the total to get the missing mark. Step-by-Step Solution: Total = 7 * 226 = 1582Sum of known six = 340 + 180 + 260 + 56 + 275 + 307 = 1418Seventh mark = 1582 − 1418 = 164 Verification / Alternative check: Plugging 164 back, the seven marks sum to 1582, and 1582 / 7 = 226 exactly. Why Other Options Are Wrong: 226 and 340 are existing or distractor numbers; “cannot be determined” is false because totals suffice; 188 does not satisfy the mean requirement. Common Pitfalls: Arithmetic slips when adding multiple numbers—group them carefully to avoid mistakes. Final Answer: 164

Question 16

Average age rises by 20% when adding the head of family: The average age of 4 family members is 25 years. When the head of the family is included (making 5 people), the average increases by 20%. Find the age of the head.

Accepted Answer

Introduction / Context: Adding one member to a group changes the average depending on how that member compares to the current mean. Here the mean increases by a percentage, letting us work with totals cleanly. Given Data / Assumptions: Average of 4 = 25 → current total = 100 New average rises by 20% → new average = 25 * 1.20 = 30 New count = 5 Concept / Approach: Compute the new total from the new average and count, then subtract the old total to get the head’s age. Step-by-Step Solution: New total = 5 * 30 = 150Head’s age = 150 − 100 = 50 years Verification / Alternative check: Check average with 5 people: (100 + 50) / 5 = 150 / 5 = 30, exactly 20% above 25. Why Other Options Are Wrong: 45, 55, and 60 would not give a 20% rise to 30; 40 keeps the average too low. Common Pitfalls: Applying 20% to the total incorrectly or to just one person’s age; the percentage applies to the average (hence to the total proportionally). Final Answer: 50 yr

Question 17

Average age increases after replacing three men with three women: A committee has 11 persons. Replacing men aged 32, 34, and 33 by three women increases the committee’s average age by 2 years. What is the average age of the three women?

Accepted Answer

Introduction / Context: When a group’s average increases after replacement without changing the headcount, the total age increases by (increase in average) * (group size). That increase must come from the difference between the incoming and outgoing members’ totals. Given Data / Assumptions: Total persons remain 11 Average increases by 2 years → total age increases by 11 * 2 = 22 years Outgoing men: 32, 34, 33 (sum = 99) Concept / Approach: Let W be the sum of the three women’s ages. Then W − 99 = 22 → W = 121. The average of the three women is W / 3. Step-by-Step Solution: Increase in total = 22W − 99 = 22 → W = 121Average (women) = 121 / 3 = 40 1/3 years Verification / Alternative check: Replacing 99 by 121 raises the committee total by 22, which across 11 members raises the mean by exactly 2 years. Why Other Options Are Wrong: 40, 41, and 39 2/3 do not produce the exact +22 total change; 411/3 is a misread of the mixed fraction needed here (correct is 401/3). Common Pitfalls: Confusing “sum increases by 22” with “each woman is 22 years older” or misreading mixed fractions. Always compute totals first. Final Answer: 401/3 yr

Question 18

Average weight rises when two women are replaced by two men: The average weight of 10 women increases by 1 kg when two women weighing 42 kg and 38 kg are replaced by two men. What is the average weight of the two men (in kg)?

Accepted Answer

Introduction / Context: Replacing members in an average changes the total. If the average for 10 people rises by 1 kg, the total rises by 10 kg. That total rise must equal the difference between the men’s combined weight and the two women’s combined weight. Given Data / Assumptions: n = 10; average increases by 1 kg → total increases by 10 kg Women replaced: 42 kg and 38 kg (sum = 80 kg) Let men’s combined weight = M Concept / Approach: M − 80 = 10 → M = 90. Average weight of the two men is M / 2. Step-by-Step Solution: Increase in total = 10 kgM − 80 = 10 → M = 90 kgMen’s average = 90 / 2 = 45 kg Verification / Alternative check: Replacing 80 kg by 90 kg increases the 10-person total by 10, making the average rise by 1 kg, exactly as stated. Why Other Options Are Wrong: 46, 44, 42, 47 kg do not match the required 10 kg increase in the 10-person total. Common Pitfalls: Using the original average (unknown) or dividing the 10 kg rise by 2 erroneously without first adding back the replaced weights. Compute combined totals first. Final Answer: 45kg

Question 19

Finding boys from overall and subgroup means: In a class of 150 students, the overall mean weight is 60 kg. The mean weight of boys is 70 kg and that of girls is 55 kg. How many boys are there?

Accepted Answer

Introduction / Context: This is a standard two-group weighted average problem. The overall mean equals the weighted sum of subgroup means divided by the total count; solving yields the subgroup size. Given Data / Assumptions: Total n = 150; overall mean = 60 Boys’ mean = 70; Girls’ mean = 55 Let number of boys = b, girls = 150 − b Concept / Approach: (70b + 55(150 − b)) / 150 = 60. Solve for b. Step-by-Step Solution: (70b + 8250 − 55b) / 150 = 60(15b + 8250) = 900015b = 750 → b = 50 Verification / Alternative check: Totals: boys 50 * 70 = 3500; girls 100 * 55 = 5500; combined 9000 / 150 = 60 matches overall mean. Why Other Options Are Wrong: 60, 75, 90, 100 do not satisfy the weighted-average equation. Common Pitfalls: Swapping subgroup sizes or forgetting to weight means by their counts. Final Answer: 50

Question 20

Recovering a deceased member’s income from changing family averages: The average monthly income of 4 earning members of a family is ₹7350. After one member passes away, the average monthly income becomes ₹6500 for the remaining 3 members. What was t…

Accepted Answer

Introduction / Context: Averages convert to totals seamlessly. With the original four-member average and the new three-member average, we can find the lost member’s contribution as the difference between totals. Given Data / Assumptions: 4-member average = ₹7350 → total = ₹29400 3-member average = ₹6500 → total = ₹19500 Concept / Approach: Income of the deceased = original total − new total. Step-by-Step Solution: Original total = 4 * 7350 = ₹29400New total = 3 * 6500 = ₹19500Missing income = ₹29400 − ₹19500 = ₹9900 Verification / Alternative check: Re-add ₹9900 to the remaining totals: (₹19500 + ₹9900) / 4 = ₹29400 / 4 = ₹7350, consistent. Why Other Options Are Wrong: Other values do not reconcile the two totals to the given averages. Common Pitfalls: Mistaking an average change as a per-person change; always work with totals first. Final Answer: ₹ 9900

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