Question 1

Three workers with different capacities: “A” can complete the job in 2 hours. “B” can do three times this job in 8 hours (i.e., rate = 3 jobs in 8 h). “C” can complete the same job as A in 8 hours. If all three work together on one job, how long wil…

Accepted Answer

Introduction / Context: The key to this problem is to translate each statement into a clear unit rate relative to the same single job. Once the individual rates are aligned (jobs per hour), they add directly. The reciprocal of the total rate gives the completion time for one job together. Given Data / Assumptions: A alone: 2 hours ⇒ rate(A) = 1/2 job per hour. B: “thrice the work in 8 hours” ⇒ rate(B) = 3/8 job per hour. C: “same job as A in 8 hours” ⇒ rate(C) = 1/8 job per hour. All are working on one identical job together. Concept / Approach: Total rate = rate(A) + rate(B) + rate(C). Time = 1 / (total rate). Ensure all rates correspond to the same job, otherwise addition would be meaningless. Step-by-Step Solution: rate(A) = 1/2; rate(B) = 3/8; rate(C) = 1/8 jobs/hour.Total rate = 1/2 + 3/8 + 1/8 = 1/2 + 4/8 = 1/2 + 1/2 = 1 job/hour.Time = 1 / 1 = 1 hour. Verification / Alternative check: In 1 hour, A contributes 0.5, B contributes 0.375, C contributes 0.125; sum = 1.0 job, matching the requirement. Why Other Options Are Wrong: 2, 3, or 4 hours would imply total rates of 0.5, 0.333…, or 0.25 jobs/hour, which do not match the summed 1 job/hour capacity. Common Pitfalls: Misreading B’s statement as “B takes 8 hours for 1 job”; actually it is “3 jobs in 8 hours,” a faster rate than A. Final Answer: 1 hour

Question 2

Equivalent efficiencies for man, woman, and boy: One man or two women or three boys can complete a job in 88 days. If one man, one woman, and one boy work together, in how many days will they finish the job?

Accepted Answer

Introduction / Context: Statements like “one man or two women or three boys can finish in the same time” define fixed equivalence between individual rates. We convert to per-day rates for each category, then add the rates for one man, one woman, and one boy to get the combined rate and hence the time to complete the work together. Given Data / Assumptions: 1 man finishes in 88 days ⇒ man’s rate rm = 1/88 job/day. 2 women finish in 88 days ⇒ each woman’s rate rw = 1/(88*2) = 1/176 job/day. 3 boys finish in 88 days ⇒ each boy’s rate rb = 1/(88*3) = 1/264 job/day. Concept / Approach: Combined rate = rm + rw + rb. Then time = 1 / (combined rate). Keep all fractions exact to avoid rounding errors, and simplify to an integer if possible. Step-by-Step Solution: rm + rw + rb = 1/88 + 1/176 + 1/264.Common denominator 528: 1/88 = 6/528; 1/176 = 3/528; 1/264 = 2/528.Total = (6 + 3 + 2)/528 = 11/528 = 1/48 job/day.Therefore, time = 48 days. Verification / Alternative check: The equivalence implies 1 man ≡ 2 women ≡ 3 boys. So 1 man + 1 woman + 1 boy = 1 man + 0.5 man + 0.333… man = 1.833… man. Since 1 man takes 88 days, 1.833… man would take 88 / 1.833… = 48 days, consistent. Why Other Options Are Wrong: 46, 54, 44, and 52 days do not reflect the exact combined rate 11/528. Common Pitfalls: Treating 1 man, 1 woman, 1 boy as equal contributors or miscomputing individual rates from the shared 88-day statement. Final Answer: 48 days

Question 3

Time relations relative to the team: A alone takes 25 days more than (A + B) together, and B alone takes 49 days more than (A + B) together. How many days do A and B together take to finish the work?

Accepted Answer

Introduction / Context: This problem defines solo times in terms of the team time. Let t be the time taken by A + B together; then A alone is t + 25 and B alone is t + 49. Setting the sum of solo rates equal to the team rate yields a solvable equation for t. Given Data / Assumptions: (A + B) time = t days. A alone time = t + 25 days. B alone time = t + 49 days. Work size = 1 unit. Concept / Approach: Rates add: 1/(t + 25) + 1/(t + 49) = 1/t. Clear denominators to form a quadratic in t, then solve for the positive root to find the team time directly. Step-by-Step Solution: 1/(t+25) + 1/(t+49) = 1/t((t+49) + (t+25))/((t+25)(t+49)) = 1/t(2t + 74) / ((t+25)(t+49)) = 1/tt(2t + 74) = (t+25)(t+49)2t^2 + 74t = t^2 + 74t + 1225 ⇒ t^2 = 1225 ⇒ t = 35. Verification / Alternative check: A alone = 60 days, B alone = 84 days. Check: 1/60 + 1/84 = (7 + 5)/420 = 12/420 = 1/35, matching the team time of 35 days. Why Other Options Are Wrong: 25, 15, or 45 days do not satisfy the relation between solo and team times defined in the equation; only 35 days does. Common Pitfalls: Adding times instead of rates or making algebraic errors while cross-multiplying. Keep denominators intact until the final quadratic emerges. Final Answer: 35 days

Question 4

Who is fastest given partial-work times: X can do 1/4 of the work in 10 days, Y can do 40% of the work in 40 days, and Z can do 1/3 of the work in 13 days. If they work individually, who will finish the whole work first?

Accepted Answer

Introduction / Context: Partial-work information can be scaled to full-work times by proportion. The one with the smallest full-work time is the fastest. Convert each worker’s given partial completion into a per-day rate and then into the full time required for one job to objectively compare speeds. Given Data / Assumptions: X does 1/4 work in 10 days ⇒ rate = (1/4)/10 = 1/40 job/day ⇒ full time = 40 days. Y does 40% = 2/5 in 40 days ⇒ rate = (2/5)/40 = 1/100 job/day ⇒ full time = 100 days. Z does 1/3 in 13 days ⇒ rate = (1/3)/13 = 1/39 job/day ⇒ full time ≈ 39 days. Concept / Approach: For each: full time = (given time) * (1 / given fraction). Compare 40, 100, and 39 days. The shortest time indicates the fastest worker in isolation. Step-by-Step Solution: X full time = 10 * 4 = 40 days.Y full time = 40 * (5/2) = 100 days.Z full time = 13 * 3 = 39 days. Verification / Alternative check: Rate comparison: 1/39 (Z) > 1/40 (X) > 1/100 (Y), so Z is indeed the fastest. Why Other Options Are Wrong: X is slightly slower than Z; Y is much slower. “Both X and Z” is incorrect because their times differ by 1 day. Common Pitfalls: Averaging days or fractions incorrectly. Always convert to either common rates or full times before comparing. Final Answer: Z

Question 5

One leaves after partial teamwork: P and Q together can finish a work in 30 days. They work together for 10 days and then Q leaves. If P finishes the remaining work alone in 20 more days, how long would P take to complete the whole work alone?

Accepted Answer

Introduction / Context: This is a standard “dropout” work problem. Use the team phase to find how much work remains, then use P’s solo phase duration to deduce P’s solo rate. Finally invert to find P’s solo time for the entire job from start to finish. Given Data / Assumptions: (P + Q) time = 30 days ⇒ rate(P+Q) = 1/30 job/day. Together phase = 10 days ⇒ work done = 10 * 1/30 = 1/3. Remaining work = 2/3 completed by P alone in 20 days. Concept / Approach: P’s solo rate = (remaining work) / (time taken) = (2/3) / 20 = 1/30 job/day. Therefore P’s solo time for 1 job is 30 days. This also implies that Q’s contribution during the first 10 days matched P’s, since the total combined rate was 1/30 and P’s rate is 1/30 as well, making Q’s rate effectively zero across the later phase (as Q left). Step-by-Step Solution: Work in first 10 days = 10/30 = 1/3.Remaining = 2/3 completed by P in 20 days ⇒ P’s rate = (2/3)/20 = 1/30.Therefore P alone would take 30 days for the whole work. Verification / Alternative check: If P’s rate is 1/30, then (P+Q) = 1/30 ⇒ Q’s rate must be 0 across those 10 days on average. Practically, this signals that the data force P’s solo rate to equal the combined rate, leaving Q’s effective rate negligible or indicating the team rate was driven by P alone—a common stylized outcome in such problems. Why Other Options Are Wrong: 20, 50, 40, and 60 days do not reproduce the observed split of 1/3 and 2/3 with the given calendar durations. Common Pitfalls: Subtracting times instead of rates or miscomputing the fraction of work done in the team phase. Always track work fractions precisely. Final Answer: 30 days

Question 6

One worker leaves midway; other completes: A can do a job in 60 days and B in 75 days. They begin together, but after some days A leaves and B finishes the remaining work in 30 days. After how many days from the start did A leave?

Accepted Answer

Introduction / Context: When one worker departs partway through, split the timeline into the joint phase and the solo phase. Convert solo phase duration into completed fraction of work, then deduce how much was done during the joint phase. From that, compute the length of the joint phase by dividing its fraction by the combined rate. Given Data / Assumptions: A alone time = 60 days ⇒ rate(A) = 1/60. B alone time = 75 days ⇒ rate(B) = 1/75. B finishes the remainder in 30 days after A leaves. Concept / Approach: Work done by B alone in 30 days = 30 * 1/75 = 2/5. Hence, the fraction completed during the joint phase is 1 − 2/5 = 3/5. Since the combined rate is 1/60 + 1/75 = 9/300 = 3/100, the joint-phase duration t satisfies (3/100)*t = 3/5. Solve for t to get when A left. Step-by-Step Solution: Work by B alone = 30/75 = 2/5.Work in joint phase = 1 − 2/5 = 3/5.Combined rate = 1/60 + 1/75 = (5 + 4)/300 = 9/300 = 3/100.(3/100) * t = 3/5 ⇒ t = (3/5) / (3/100) = 20 days. Verification / Alternative check: In 20 days together they do (3/100)*20 = 3/5 of the work; remaining 2/5 is done by B in 30 days, consistent. Why Other Options Are Wrong: 25, 21, 24, or 18 days would not split the work into 3/5 and 2/5 consistent with B’s 30-day solo effort. Common Pitfalls: Forgetting to convert the solo phase into a fraction of work or misadding the combined rate of A and B. Final Answer: 20 days

Question 7

More workers reduce the days: 20 women can complete a piece of work in 7 days. If 8 more women are added (i.e., total 28 women), in how many days will they finish the work?

Accepted Answer

Introduction / Context: For constant-efficiency workers, total work measured in woman-days remains invariant. If the number of workers increases, the days required decrease proportionally. The product of workers and days equals the fixed total work for the job in this model. Given Data / Assumptions: Total work W = (20 women) * (7 days) = 140 woman-days. New team size = 28 women (20 + 8 added). Concept / Approach: With the same type of workers, W = workers * days. Solve for days = W / workers with the new team size. No change in efficiency per worker is assumed across scenarios. Step-by-Step Solution: W = 20 * 7 = 140 woman-days.Days with 28 women = 140 / 28 = 5 days. Verification / Alternative check: If 28 women work 5 days, total = 28 * 5 = 140 woman-days, matching the original job’s workload. Why Other Options Are Wrong: 4.5 or 5.5 days misrepresent the exact proportional decrease; 6 or 4 days also do not maintain the required woman-day total. Common Pitfalls: Assuming non-linear effects or changing efficiencies when the problem states identical workers. Here, proportional scaling applies directly. Final Answer: 5 days

Question 8

Changing the workforce size: 12 men can finish a piece of work in 24 days. If only 8 men are assigned, how many days will they need to complete the same work (assume equal efficiency)?

Accepted Answer

Introduction / Context: Workload is constant and equals workers times days when all workers have equal efficiency. Decreasing the number of workers increases the required days proportionally to keep the product equal to the same total work. Given Data / Assumptions: Total work W in man-days = 12 * 24 = 288 man-days. New team = 8 men, identical efficiency. Concept / Approach: Days = W / workers = 288 / 8. This directly gives the revised duration to finish the job when worker count changes and productivity per worker remains unchanged. Step-by-Step Solution: W = 288 man-days.Days with 8 men = 288 / 8 = 36 days. Verification / Alternative check: Check proportionality: Reducing workers from 12 to 8 is a factor of 2/3; time increases by reciprocal 3/2: 24 * 1.5 = 36 days, consistent. Why Other Options Are Wrong: 28, 48, and 52 days do not preserve the 288 man-day total; 32 is also inconsistent with the proportional scale-up. Common Pitfalls: Averaging or guessing instead of preserving worker-days. Always keep total man-days invariant for identical efficiency conditions. Final Answer: 36

Question 9

Output changes with manpower and time: 20 men can cut 30 trees in 4 hours. If 4 men leave (so 16 men remain), how many trees will be cut in 12 hours by the remaining men (assume proportional output)?

Accepted Answer

Introduction / Context: When output is proportional to the product of (number of workers) * (time), we can find a per man-hour productivity and scale it to any new combination of workers and hours. This is a direct application of the “work is proportional to man-hours” model. Given Data / Assumptions: With 20 men in 4 hours, trees cut = 30. Assume constant productivity per man-hour across scenarios. New scenario: 16 men for 12 hours. Concept / Approach: Compute trees per man-hour first: trees / (men * hours). Then multiply by the new man-hour total to get the output for the changed team and duration. Step-by-Step Solution: Productivity = 30 / (20 * 4) = 30 / 80 = 3/8 tree per man-hour.New man-hours = 16 * 12 = 192.Trees cut = 192 * (3/8) = 24 * 3 = 72. Verification / Alternative check: Proportional scaling: Men reduced by factor 16/20 = 0.8, hours increased by 12/4 = 3; overall scale = 0.8 * 3 = 2.4; thus trees = 30 * 2.4 = 72, same result. Why Other Options Are Wrong: 80 and 79 exceed the proportional output; 68 and 64 are too low given the large increase in total man-hours. Common Pitfalls: Forgetting to scale both men and hours, or assuming linearity in one but not the other. Output is proportional to their product under constant productivity. Final Answer: 72

Question 10

Scaling wall construction with men, length, and hours: 40 men build a 200 m wall in 12 days working 8 h/day. How many days will 30 men take to build a similar 300 m wall working 6 h/day (assume proportional productivity)?

Accepted Answer

Introduction / Context: In uniform construction tasks, output is proportional to the product of (number of men) * (days) * (hours per day). First deduce the man-hour requirement per meter from the base scenario, then use it to compute the required man-days when the workforce and daily hours change and the length increases to 300 m. Given Data / Assumptions: Base: 40 men, 12 days, 8 h/day build 200 m. New: 30 men, 6 h/day, length = 300 m. Productivity assumed constant (similar wall, same conditions). Concept / Approach: Compute man-hours per meter from the base case. Then total man-hours needed for 300 m equals (man-hours/m) * 300. Divide by daily man-hours (men * hours/day) to obtain the number of days for the new scenario. Step-by-Step Solution: Base man-hours = 40 * 12 * 8 = 3840 mh for 200 m.Man-hours per meter = 3840 / 200 = 19.2 mh/m.For 300 m: required mh = 300 * 19.2 = 5760 mh.Daily mh with 30 men at 6 h/day = 30 * 6 = 180 mh/day.Days required = 5760 / 180 = 32 days. Verification / Alternative check: Ratio method: Scale by length (300/200 = 1.5), inverse by men (40/30 = 4/3), and inverse by hours (8/6 = 4/3). Days_new = 12 * 1.5 * (40/30) * (8/6) = 12 * 1.5 * 4/3 * 4/3 = 32, consistent. Why Other Options Are Wrong: 18, 36, 9, or 28 days do not satisfy the man-hour balance and proportional scaling across all factors. Common Pitfalls: Ignoring the change in daily working hours or scaling only by men and length without adjusting for hours per day. Final Answer: 32

Question 11

If m men, working m hours per day, can complete exactly m units of work in m days, then in n days how many units of work will be completed by n men working n hours per day? (Assume work ∝ men * hours/day * days and the same job type.)

Accepted Answer

Introduction / Context:This question tests direct proportion reasoning in time and work. When task type and productivity conditions remain identical, total work completed varies directly with the product of number of workers, effective hours per day, and the number of days. Given Data / Assumptions: Case 1: m men, m hours/day, m days produce m units. Case 2: n men, n hours/day, n days produce X units (unknown). Same job environment and linear productivity with respect to men and hours. Concept / Approach:Let k be the proportionality constant so that Work = k * (men * hours/day * days). Use Case 1 to determine k, then apply to Case 2. Step-by-Step Solution: m = k * (m * m * m) = k * m^3k = m / m^3 = 1 / m^2X = k * (n * n * n) = (1 / m^2) * n^3 = n^3 / m^2 Verification / Alternative check:If n = m, then result gives m units, matching the first scenario, so the expression is consistent. Why Other Options Are Wrong: m^3 / n^2: Inverts dependence and breaks symmetry when n = m. m^4 / n^2: Has incorrect powers and dimensions. n^4 / m^3: Incorrect exponent pattern; fails dimensionally with proportional reasoning. None of these: Incorrect because a correct closed form exists. Common Pitfalls:Confusing units or assuming work depends on men + hours + days instead of the product men * hours * days; forgetting to solve for the proportionality constant before substitution. Final Answer:n^3 / m^2

Question 12

Twenty men complete one third of a job in 20 days. How many additional men must be employed so that the remaining two thirds of the job is finished in 25 more days?

Accepted Answer

Introduction / Context:Time and work problems often assume constant individual productivity. Total work equals rate times time. When more workers are added, the group rate scales linearly with the number of workers, assuming identical efficiency. Given Data / Assumptions: 20 men do 1/3 of the job in 20 days. Remaining work is 2/3 of the job. Goal: finish the remaining 2/3 in 25 days with a larger team. Concept / Approach:Let r be the daily work of one man. Work done = men * r * days. First use the first phase to compute r, then determine how many men are needed for the second phase, and finally compute how many extra men must be added to the initial 20. Step-by-Step Solution: 20 * r * 20 = 1/3 ⇒ r = 1 / 1200Let x be the required number of men for the second phase.x * r * 25 = 2/3x * (1/1200) * 25 = 2/3 ⇒ x = (2/3) * 1200 / 25 = 32Additional men needed = 32 − 20 = 12 Verification / Alternative check:With 32 men for 25 days, total work = 32 * (1/1200) * 25 = 2/3, which exactly meets the remaining requirement. Why Other Options Are Wrong: 15, 18, 25, 10 give total men ≠ 32, leading to over or under completion. Common Pitfalls:Subtracting men instead of adding; calculating total men as the answer rather than the number of additional men; mixing up 1/3 and 2/3 portions. Final Answer:12

Question 13

Forty five people dig a pond in 18 days. If the pond must be dug in 15 days, how many people are required?

Accepted Answer

Introduction / Context:For uniform work rates, the number of worker days (people * days) needed is constant for the same job. If the deadline shortens, more workers are needed proportionally. Given Data / Assumptions: Original team: 45 people. Original duration: 18 days. New duration: 15 days. Equal productivity and identical job. Concept / Approach:Total person days for the job are fixed. Compute the total person days from the original plan and divide by the new number of days to get the required team size. Step-by-Step Solution: Person days needed = 45 * 18 = 810Required workers for 15 days = 810 / 15 = 54 Verification / Alternative check:Check proportionality: time drops from 18 to 15 (factor 5/6), so workers must rise by factor 6/5: 45 * (6/5) = 54, which matches the computed value. Why Other Options Are Wrong:50, 60, 72, 48 do not meet the exact person day requirement 810 and would either leave work incomplete or waste capacity. Common Pitfalls:Using an additive rather than multiplicative adjustment; rounding mid calculation; forgetting that person days must remain constant for the same job. Final Answer:54

Question 14

Three men or four women can build a wall in 43 days. In how many days can seven men and five women build the same wall?

Accepted Answer

Introduction / Context:This is a combined workforce problem with different worker categories. We convert each category to a consistent daily rate and then add rates for mixed teams. Given Data / Assumptions: 3 men complete 1 wall in 43 days. 4 women complete 1 wall in 43 days. Find days for 7 men and 5 women together. Concept / Approach:Compute the per person daily rate for men and women using the equivalences given. Sum the rates for the mixed team, then take the reciprocal to get the time to complete 1 wall. Step-by-Step Solution: Men rate: 3 * r_m * 43 = 1 ⇒ r_m = 1 / 129 per man per dayWomen rate: 4 * r_w * 43 = 1 ⇒ r_w = 1 / 172 per woman per dayTeam rate = 7 * (1/129) + 5 * (1/172)Compute: 7/129 + 5/172 = 1/12Time = 1 / (1/12) = 12 days Verification / Alternative check:Proportions are consistent with the given single category completions. The exact fractional arithmetic yields a clean 1/12 rate, confirming 12 days. Why Other Options Are Wrong:16, 25, 21, 18 days correspond to incorrect rate sums or arithmetic slips when converting single category times to per person rates. Common Pitfalls:Assuming 1 man equals 1 woman; ignoring that both categories have different per person rates; averaging days instead of adding rates. Final Answer:12 days

Question 15

A and B together take 18 days to finish a job, B and C take 24 days, and C and A take 36 days. In how many days can A, B, and C together finish the job?

Accepted Answer

Introduction / Context:When pairwise completion times are given, we can derive individual rates and then sum them to get the combined time. Given Data / Assumptions: (A + B) time = 18 days ⇒ rate = 1/18 (B + C) time = 24 days ⇒ rate = 1/24 (C + A) time = 36 days ⇒ rate = 1/36 Concept / Approach:Let a, b, c be daily rates. Then a + b = 1/18, b + c = 1/24, c + a = 1/36. Add all three to eliminate pairs and solve for a + b + c. Step-by-Step Solution: (a + b) + (b + c) + (c + a) = 1/18 + 1/24 + 1/362(a + b + c) = 1/18 + 1/24 + 1/36 = 1/8a + b + c = 1/16Therefore, all three together finish in 16 days. Verification / Alternative check:Individual rates computed from pair sums are consistent, but the problem only requires the total sum, which we obtained directly. Why Other Options Are Wrong:8, 24, 36, 12 come from misadding reciprocals or forgetting to divide the pairwise sum by 2 to get the triple sum. Common Pitfalls:Adding times instead of rates; not halving the total of pair rates; arithmetic mistakes with fractions 1/18, 1/24, 1/36. Final Answer:16

Question 16

A and B can do a job in 72 days, B and C in 120 days, and A and C in 90 days. In how many days can A alone finish the job?

Accepted Answer

Introduction / Context:With three workers and pairwise times, sum rates to get the combined rate, then back out individual rates. The target is the time for A alone. Given Data / Assumptions: A + B: 72 days ⇒ 1/72 B + C: 120 days ⇒ 1/120 A + C: 90 days ⇒ 1/90 Concept / Approach:Let a, b, c be daily rates. Then a + b + b + c + c + a = 2(a + b + c) equals the sum of given pair rates. Once a + b + c is known, use a + c and subtract c to solve for a, or use any pair. Step-by-Step Solution: 2(a + b + c) = 1/72 + 1/120 + 1/90 = 1/30 + 1/120 = 1/24? Compute exactly: total = 1/72 + 1/120 + 1/90 = 1/30a + b + c = 1/60From A + C = 1/90 ⇒ a + c = 1/90c = (a + b + c) − (a + b) but quicker: c = (1/60) − (1/72) = 1/360a = (a + c) − c = 1/90 − 1/360 = 1/120Time for A alone = 1 / (1/120) = 120 days Verification / Alternative check:Plug a = 1/120 into A + B = 1/72 to get b, and into A + C = 1/90 to get c. The values are consistent with B + C = 1/120. Why Other Options Are Wrong:80, 100, 150, 90 do not satisfy all three pair equations when back substituted. Common Pitfalls:Arithmetic slips when summing reciprocals; forgetting to divide by 2 after adding pair rates; inconsistent rounding mid calculation. Final Answer:120 days

Question 17

Twenty five men can reap a field in 20 days. If 15 men leave after some days and the field must then be finished in 37.5 days from that leaving point, after how many days should the 15 men leave?

Accepted Answer

Introduction / Context:This staged-work problem has two phases with different team sizes. The total work remains constant, but the team rate changes after workers leave. We compute the partial work done before departure and equate the remainder to the plan after departure. Given Data / Assumptions: Full job by 25 men takes 20 days. After some t days, 15 men leave, leaving 10 men to finish. From leaving time, completion must occur in 37.5 days. Concept / Approach:Let r be one man’s daily work. Total work W = 25 * r * 20. Before leaving: work done = 25 * r * t. After leaving: remaining work is completed by 10 men in 37.5 days, that is 10 * r * 37.5. Sum equals W. Step-by-Step Solution: W = 25 * r * 2025 * r * t + 10 * r * 37.5 = 25 * r * 20Divide by r: 25t + 375 = 50025t = 125 ⇒ t = 5 days Verification / Alternative check:Before leaving, 25 men for 5 days do 125 man days. Remaining is 500 − 125 = 375 man days, exactly equal to 10 men for 37.5 days. Why Other Options Are Wrong:6, 4, 7 days do not split the work into the required 375 man days remainder after departure. Common Pitfalls:Interpreting 37.5 as total project time; forgetting that 37.5 is measured after the leaving moment; dropping the factor of 25t in the first phase. Final Answer:5 days

Question 18

A can do a job in 5 hours, B in 9 hours, and C in 15 hours. If C works with A and B for only 1 hour, how much additional time will A and B together take to finish the remaining work?

Accepted Answer

Introduction / Context:Mixed participation problems require computing the work done during a short collaborative interval, then finishing with the subset at their combined rate. Given Data / Assumptions: A alone: 5 hours ⇒ rate 1/5 per hour. B alone: 9 hours ⇒ rate 1/9 per hour. C alone: 15 hours ⇒ rate 1/15 per hour. All three work together for the first 1 hour, then only A and B continue. Concept / Approach:Compute work completed in the first hour by A, B, and C. Subtract from 1 to find the remaining work. Divide the remainder by the combined rate of A and B to get the extra time. Step-by-Step Solution: First hour output = 1/5 + 1/9 + 1/15Common denominator 45: 9 + 5 + 3 = 17 ⇒ 17/45Remaining work = 1 − 17/45 = 28/45A + B rate = 1/5 + 1/9 = 14/45Time needed = (28/45) / (14/45) = 2 hours Verification / Alternative check:Compute total time: 1 hour with all three plus 2 more hours with A and B equals 3 hours overall; check that total work equals 1 by summing hourly outputs. Why Other Options Are Wrong:3, 3.5, 4, 2.5 hours result from arithmetic slips in the fractional sums or dividing remaining work by only one worker’s rate. Common Pitfalls:Forgetting that only A and B continue after 1 hour; mixing up addition of times versus addition of rates; reducing 17/45 incorrectly. Final Answer:2 hours

Question 19

A alone can finish a job in 24 days, and B alone in 16 days. Working together with C, they finish the job in 8 days. In how many days can C alone finish the job?

Accepted Answer

Introduction / Context:This is a standard rate extraction problem: use individual times for A and B and the joint time for all three to isolate the third individual rate. Given Data / Assumptions: A: 24 days ⇒ rate 1/24 per day. B: 16 days ⇒ rate 1/16 per day. A + B + C: 8 days ⇒ rate 1/8 per day. Concept / Approach:Sum A and B rates, subtract from the combined rate to get C’s rate, then invert to get C’s time. Step-by-Step Solution: A + B rate = 1/24 + 1/16 = 5/48C rate = 1/8 − 5/48 = 6/48 − 5/48 = 1/48C alone time = 1 / (1/48) = 48 days Verification / Alternative check:Check: 1/24 + 1/16 + 1/48 = 2/48 + 3/48 + 1/48 = 6/48 = 1/8, consistent with the given joint time. Why Other Options Are Wrong:32, 36, 40, 24 do not match the extracted rate from the combined data and lead to incorrect total rates when recombined. Common Pitfalls:Arithmetic mistakes with denominators 24, 16, and 8; forgetting to subtract A + B from the total rather than averaging times. Final Answer:48 days

Question 20

A can do a job in 10 days. B is 60 percent more efficient than A. In how many days can B alone complete the same job?

Accepted Answer

Introduction / Context:Efficiency comparisons translate into rate multipliers. If B is 60 percent more efficient than A, B’s work rate is 1.6 times A’s rate. Time is the reciprocal of rate for one whole job. Given Data / Assumptions: A time = 10 days ⇒ A rate = 1/10 per day. B efficiency = 160 percent of A. Concept / Approach:Multiply A’s rate by 1.6 to get B’s rate, then take the reciprocal to obtain B’s time to finish one job. Step-by-Step Solution: B rate = 1.6 * (1/10) = 0.16 per dayB time = 1 / 0.16 = 6.25 days = 6 1/4 days Verification / Alternative check:If A does 1 unit in 10 days, in 6.25 days A would do 0.625 units. Since B is 1.6 times as fast, B’s output in 6.25 days is 1 unit, confirming the calculation. Why Other Options Are Wrong:6 and 6 2/3 arise from rounding or misinterpreting 60 percent more as 60 percent of A; 8 days assumes B is slower than reality; 5 days overestimates B’s efficiency. Common Pitfalls:Confusing 60 percent more with 60 percent of; converting 0.25 to fractional days incorrectly; mixing time and rate comparisons directly. Final Answer:6 1/4

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