Question 1

Building vs. breaking: net progress with an interrupter A can build a structure in 8 days, while B can demolish the same structure in 3 days. A works alone for 4 days, then B joins and they work together for 2 days. After that, in how many days will…

Accepted Answer

Introduction / Context:This problem involves a positive builder and a negative (demolisher) worker. When they work together, the net effect can reduce prior progress, so careful fraction accounting is essential. Given Data / Assumptions: A builds in 8 days ⇒ +1/8 per day. B breaks in 3 days ⇒ −1/3 per day. A works 4 days alone; then A and B work together for 2 days; finally A works alone again. Concept / Approach:Track cumulative progress: first A alone, then the net effect with B, then compute the remaining fraction for A alone. Step-by-Step Solution: After first 4 days (A alone): 4 * (1/8) = 1/2 builtA + B per day: 1/8 − 1/3 = (3 − 8)/24 = −5/24Two days together: 2 * (−5/24) = −5/12Progress after 6 days: 1/2 − 5/12 = 1/12 builtRemaining = 1 − 1/12 = 11/12Time for A alone = (11/12) / (1/8) = 88/12 = 22/3 days Verification / Alternative check:Multiply A’s rate by 22/3 to confirm: (1/8)*(22/3) = 22/24 = 11/12, exactly the remainder. Why Other Options Are Wrong:7, 8, 9 days are integer guesses; only 22/3 satisfies the exact net accounting. Common Pitfalls:Ignoring the negative contribution when B is demolishing or forgetting that net can decrease progress. Final Answer:22/3 days

Question 2

Catching up to a wall-length target with added manpower A 1000-meter wall must be built in 50 days. 56 men are hired, but after 27 days only 448 meters are completed. How many additional men must be employed to finish the remaining work on time?

Accepted Answer

Introduction / Context:This is a throughput calibration question. We infer the per-man productivity from the initial phase, then size the workforce needed to hit the deadline for the remaining work. Given Data / Assumptions: Total wall length = 1000 m; deadline = 50 days. Initial crew = 56 men for 27 days; output = 448 m. Assume constant per-man productivity throughout. Concept / Approach:Compute per-man rate from realized output, then compute how many men are needed to complete the remaining meters in the remaining days. Finally subtract current men to get “additional men.” Step-by-Step Solution: Per-man rate p satisfies 56 * 27 * p = 448 ⇒ p = 448 / (56 * 27) = 8/27 m per man-dayRemaining meters = 1000 − 448 = 552Remaining days = 50 − 27 = 23Men needed m: m * p * 23 = 552 ⇒ m = 552 / ( (8/27) * 23 ) = 81 men (total)Additional men required = 81 − 56 = 25 Verification / Alternative check:Check capacity: 81 men for 23 days at 8/27 gives 81 * 23 * 8/27 = 552 m remaining, so the target is hit exactly. Why Other Options Are Wrong:23, 27, 31 do not equal the computed increment from 56 to 81; any other total misses 552 m in 23 days. Common Pitfalls:Using the planned average rate (20 m/day) instead of inferring the true p from realized output. Final Answer:25

Question 3

Inferring individual copy speed from a joint throughput Mr. Modi can copy 40 pages in 10 minutes. Together, Mr. Xerox and Mr. Modi copy 250 pages in 25 minutes. How many minutes will Mr. Xerox alone take to copy 36 pages?

Accepted Answer

Introduction / Context:This problem asks you to back out an unknown worker’s rate from the combined rate and a known co-worker’s rate. Once Mr. Xerox’s pages-per-minute rate is known, time equals pages divided by rate. Given Data / Assumptions: Mr. Modi: 40 pages in 10 minutes ⇒ 4 pages/min. Mr. Modi + Mr. Xerox: 250 pages in 25 minutes ⇒ 10 pages/min combined. Rates are constant during the interval. Concept / Approach:Combined rate = sum of individual rates. Subtract Mr. Modi’s rate to find Mr. Xerox’s rate, then compute time for 36 pages. Step-by-Step Solution: Combined rate = 250 / 25 = 10 pages/minMr. Modi’s rate = 4 pages/minMr. Xerox’s rate = 10 − 4 = 6 pages/minTime for 36 pages = 36 / 6 = 6 minutes Verification / Alternative check:If both worked 25 minutes at their rates (4 and 6), output is (4 + 6) * 25 = 250 pages, which matches. Why Other Options Are Wrong:5 minutes implies 7.2 pages/min (too high); 3 is far too high; 12 is half the needed speed. Common Pitfalls:Dividing 250 by 10 first (correct) but then forgetting to subtract Modi’s rate to isolate Xerox’s rate. Final Answer:6 minutes

Question 4

Mixed-team productivity then women-only completion Twenty-five men and fifteen women can complete a job in 12 days. All begin together; after 8 days the women stop. The remaining work is finished by 25 men in 6 days. How long would it take if only 1…

Accepted Answer

Introduction / Context:This is a two-phase workforce problem used to infer relative productivities of men and women. After learning the ratio, compute the total work in man-rate units and then find the women-only duration. Given Data / Assumptions: Phase 1: (25 men + 15 women) for 8 days, then women leave. Phase 2: 25 men alone finish the remainder in 6 days. Productivity is linear and constant. Concept / Approach:Let man rate = m and woman rate = w (jobs per day). Equate Phase-2 remainder to solve for w in terms of m, then compute total job in m-units and divide by 15w. Step-by-Step Solution: Work done in first 8 days = (25m + 15w)*8Remaining work = total W − (25m + 15w)*8Given: 25m * 6 = remaining ⇒ 150m = W − 8(25m + 15w)Also, W = (25m + 15w)*12 (from the 12-day completion by the mixed team)Substitute: 150m = (25m + 15w)*12 − 8(25m + 15w) = 4(25m + 15w)150m = 100m + 60w ⇒ 50m = 60w ⇒ m/w = 6/5 ⇒ w = (5/6)mTotal work W in m-units = (25m + 15*(5/6)m)*12 = (25m + 12.5m)*12 = 450mWomen-only daily rate (15 women) = 15w = 15*(5/6)m = 12.5mTime with 15 women alone = 450m / 12.5m = 36 days Verification / Alternative check:36 days is consistent with the inferred m:w ratio 6:5 and total job size 450m. Why Other Options Are Wrong:60, 88, 94 do not match the derived 36 days. Hence the correct choice within the given set is “None of these.” Common Pitfalls:Assuming the total work is 12 days of the mixed team without converting to rates; or forgetting to scale by 15w for the women-only case. Final Answer:None of these

Question 5

AB work 3 days; C is faster than (A+B) together by a factor of time A can do a job in 12 days, B in 8 days. C can do the same job in four-fifths of the time required by (A and B together). A and B work together for 3 days, then C alone completes the…

Accepted Answer

Introduction / Context:The original stem was broken. By the Recovery-First Policy, we minimally repair it to reflect a common variant: C’s solo time is four-fifths of the joint time of A and B. We then compute how long C needs to finish after A and B have worked some initial days. Given Data / Assumptions: A = 12 days ⇒ rate 1/12 per day. B = 8 days ⇒ rate 1/8 per day. (A+B) rate = 1/12 + 1/8 = 5/24 per day ⇒ time = 24/5 days. C’s time = (4/5)*(24/5) = 96/25 days ⇒ C’s rate = 25/96 per day. A and B work jointly for 3 days first. Concept / Approach:Compute initial work by A and B, find the leftover fraction, then divide by C’s rate. Step-by-Step Solution: Work by A+B in 3 days = 3 * (5/24) = 15/24 = 5/8Remaining work = 1 − 5/8 = 3/8C’s time = (3/8) / (25/96) = (3/8) * (96/25) = 36/25 = 1.44 days Verification / Alternative check:Multiply C’s rate by 36/25: (25/96)*(36/25) = 36/96 = 3/8, which matches the remainder. Why Other Options Are Wrong:8, 6, 3 are far from 1.44 days; the only valid choice is “None of these.” Common Pitfalls:Interpreting “four-fifths” against A or B instead of the joint A+B time; or adding times rather than rates. Final Answer:None of these

Question 6

One-day head start by A, then both finish the job A completes a work in 15 days and B in 20 days. A works alone for 1 day, then A and B work together to finish. How many additional days are needed after the first day?

Accepted Answer

Introduction / Context:Problems with an initial solo day followed by joint work are solved by subtracting the first day’s contribution and then dividing the remainder by the combined daily rate. Given Data / Assumptions: A alone: 15 days ⇒ 1/15 per day. B alone: 20 days ⇒ 1/20 per day. A works alone for 1 day; then both work together. Concept / Approach:Remaining fraction = 1 − 1/15. Combined rate = 1/15 + 1/20. Step-by-Step Solution: After day 1 by A: completed = 1/15; remaining = 14/15Joint rate = 1/15 + 1/20 = (4 + 3)/60 = 7/60 per dayDays needed = (14/15) / (7/60) = (14/15) * (60/7) = 8 days Verification / Alternative check:Total time = 1 + 8 = 9 days; check: 9 * (1/15) + 8 * (1/20) = 0.6 + 0.4 = 1 job. Why Other Options Are Wrong:7 or 6 days leave unfinished work; “None of these” is unnecessary because 8 is exact. Common Pitfalls:Forgetting that the question asks for additional days after the first solo day. Final Answer:8 days

Question 7

A leaves mid-way; B finishes the remainder A can do a job in 10 days and B in 20 days. They start together, but after 5 days A leaves. In how many days will B complete the remaining work?

Accepted Answer

Introduction / Context:When one worker leaves part-way, compute the completed fraction so far, then divide the remainder by the staying worker’s rate. Given Data / Assumptions: A: 10 days ⇒ 1/10 per day. B: 20 days ⇒ 1/20 per day. They work together for the first 5 days. Concept / Approach:Together rate = 1/10 + 1/20 = 3/20 per day. Remaining = 1 − (3/20)*5. Step-by-Step Solution: Work in first 5 days = (3/20) * 5 = 15/20 = 3/4Remaining = 1 − 3/4 = 1/4B’s rate = 1/20 ⇒ time = (1/4) / (1/20) = 5 days Verification / Alternative check:Total job = 3/4 + (1/4) = 1. Timings are consistent. Why Other Options Are Wrong:6, 8, 10 contradict B’s fixed rate of 1/20 for a quarter job. Common Pitfalls:Adding 5 directly to an option instead of calculating the leftover fraction properly. Final Answer:5 days

Question 8

Revised crew after day 3: how long to finish? Twelve men can complete a work in 8 days. After 3 days, three more men join the team. In how many days will they complete the remaining work?

Accepted Answer

Introduction / Context:This scenario uses man-days to track total effort. Once some work is done by the original team, add the new hires and compute how many days are necessary to deliver the remainder. Given Data / Assumptions: 12 men finish in 8 days ⇒ total work = 96 man-days. First 3 days: 12 men work ⇒ 36 man-days completed. Then 3 more men join (total 15 men). Concept / Approach:Remaining work = 96 − 36 = 60 man-days. With 15 men, time = remaining / 15. Step-by-Step Solution: Total work = 12 * 8 = 96 man-daysDone in first 3 days = 12 * 3 = 36 man-daysRemaining = 96 − 36 = 60 man-daysWith 15 men, time = 60 / 15 = 4 days Verification / Alternative check:Overall duration = 3 + 4 = 7 days, which is less than 8 because of extra manpower. Why Other Options Are Wrong:2, 5, 6 do not correspond to 60/15. Common Pitfalls:Forgetting to compute the first 3 days of progress before adding the newcomers. Final Answer:4

Question 9

B and C start; A finishes after they leave A can complete a job in 9 days, B in 10 days, and C in 15 days. B and C work together for 2 days and then leave. How long will A take to finish the remaining work?

Accepted Answer

Introduction / Context:We first determine how much B and C accomplish together, then hand off to A for the remainder. Converting each time into a daily rate keeps the arithmetic straightforward. Given Data / Assumptions: A: 9 days ⇒ 1/9 per day. B: 10 days ⇒ 1/10 per day. C: 15 days ⇒ 1/15 per day. B and C work 2 days, then stop. Concept / Approach:Combined rate(B+C) = 1/10 + 1/15 = 1/6 per day. Remainder = 1 − 2*(1/6). Then divide by A’s rate. Step-by-Step Solution: Work by B and C in 2 days = 2 * (1/6) = 1/3Remaining = 1 − 1/3 = 2/3A’s rate = 1/9 ⇒ time = (2/3) / (1/9) = 6 days Verification / Alternative check:In 6 days, A completes 6*(1/9) = 2/3, matching the remaining fraction. Why Other Options Are Wrong:9, 10, 13 contradict the exact fraction 2/3 at rate 1/9. Common Pitfalls:Adding 2 days to an option rather than recomputing with A’s solo rate. Final Answer:6 days

Question 10

Convert women to man-equivalents to size the final push Twenty-four men complete a job in 16 days. Thirty-two women can complete the same job in 24 days. Sixteen men and sixteen women work together for 12 days. How many more men must be added to fin…

Accepted Answer

Introduction / Context:Mixed-gender crew problems can be simplified by converting all productivity into a single “man-equivalent” or “woman-equivalent” unit. Once equivalents are established, remaining work and required headcount follow from linear rate arithmetic. Given Data / Assumptions: 24 men → 16 days ⇒ total work = 384 man-days. 32 women → 24 days ⇒ 768 woman-days = the same work ⇒ 1 man-day = 2 woman-days. Current crew for first 12 days: 16 men + 16 women. Women continue during the final 2-day sprint. Concept / Approach:Convert women to man-equivalents, compute done vs. remaining, then solve for extra men x so that the 2-day capacity equals the remaining man-days. Step-by-Step Solution: Total work W = 384 man-daysMan-equivalent of 16 women = 8 men (since 1 man = 2 women)Crew man-equivalent/day = 16 + 8 = 24 menWork done in 12 days = 24 * 12 = 288 man-daysRemaining = 384 − 288 = 96 man-daysFinal phase duration = 2 days with (16 + x) men + 16 women = (16 + x) + 8 = (24 + x) man-equivalents/dayCapacity over 2 days = 2 * (24 + x) = 96 ⇒ 24 + x = 48 ⇒ x = 24 extra men Verification / Alternative check:With 24 extra, the final daily man-equivalent = 48; over 2 days = 96, exactly the remainder. Why Other Options Are Wrong:48 or 36 overshoot; “None of these” is unnecessary because 24 fits perfectly. Common Pitfalls:Forgetting to convert women to man-equivalents before summing crew productivity. Final Answer:24

Question 11

Three typists A, B, and C work on the same report. A alone can finish the report in 16 hours; B alone in 20 hours; C alone in 24 hours. All three start typing together at 9:00 a.m. At exactly 1:00 p.m. (after 4 hours), typist A stops and leaves the …

Accepted Answer

Introduction / Context:Time-and-work problems often require converting individual times to rates, adding rates for simultaneous work, and then tracking partial progress across phases. Here, three typists start together; one leaves after 4 hours, and the others finish. We must compute the exact finish time from the start at 9:00 a.m. Given Data / Assumptions: A's time = 16 h ⇒ rate = 1/16 report per hour. B's time = 20 h ⇒ rate = 1/20 per hour. C's time = 24 h ⇒ rate = 1/24 per hour. All start at 9:00 a.m.; A stops at 1:00 p.m. (4 hours). Concept / Approach:Use unit-work method. Total work W = 1 report. Sum rates when people work together; multiply by time to get completed fraction. Remainder divided by the later combined rate gives the additional time needed. Step-by-Step Solution: Combined rate for first 4 h = 1/16 + 1/20 + 1/24.Compute LCM(16,20,24) = 240; sum = 15/240 + 12/240 + 10/240 = 37/240 per hour.Work done in first 4 h = 4 * (37/240) = 148/240 = 37/60.Remaining work = 1 - 37/60 = 23/60.After 1:00 p.m., only B and C work: rate = 1/20 + 1/24 = 6/120 + 5/120 = 11/120 per hour.Time to finish remainder = (23/60) / (11/120) = (23/60) * (120/11) = 46/11 h ≈ 4.1818 h.0.1818 h ≈ 0.1818 * 60 ≈ 10.9 min ≈ 11 min.Finish time = 1:00 p.m. + 4 h 11 min ≈ 5:11 p.m. Verification / Alternative check:As a quick reasonableness check, A contributes 4*(1/16)=1/4. B and C together at 11/120 per hour need about (3/4)/(11/120) ≈ 8.18 hours if A never came; but A contributed more than 1/4 because everyone worked initially; our precise arithmetic above gives 5:11 p.m., which is consistent. Why Other Options Are Wrong:04:10 p.m.: underestimates remaining time after A leaves.05:45 p.m.: overestimates B+C time.06:15 p.m.: significantly longer than required by computed remainder and rate. Common Pitfalls:Mixing average time with sum of times; not converting to per-hour rates; rounding too early; or assuming integer minutes. Keep exact fractions until the end. Final Answer:05:11 p.m.

Question 12

A team of 14 persons can complete a job in 16 days (uniform productivity). Initially, 8 persons start the work and continue for 12 days. After 12 days, 8 more persons join (so the team becomes 16 persons). How many days will the team take to complet…

Accepted Answer

Introduction / Context:This is a classic manpower–time problem with a change in crew size mid-project. We translate total work into person-days, account for work already done, and then compute the duration for the remainder with the increased team. Given Data / Assumptions: 14 persons complete the work in 16 days ⇒ total work W = 14 * 16 person-days = 224 person-days. Phase 1: 8 persons work for 12 days. Phase 2: 8 more join ⇒ 16 persons for the remainder. Uniform productivity per person per day. Concept / Approach:Compute the total work in person-days. Subtract the contribution of the first phase (already completed). Divide leftover person-days by the new team size to get the additional days required after the joining event. Step-by-Step Solution: Total work W = 224 person-days.Work done in 12 days by 8 persons = 8 * 12 = 96 person-days.Remaining work = 224 - 96 = 128 person-days.After 12 days, team size = 16 persons.Days to finish remainder = 128 / 16 = 8 days. Verification / Alternative check:Total elapsed time from the start would be 12 + 8 = 20 days. The question specifically asks for the time to complete the remaining work after the join, which is 8 days. Since 8 days is not among the listed numeric options, the correct choice is “None of these.” Why Other Options Are Wrong:12 days: Ignores increased manpower; would overshoot.7 days: Underestimates the remaining person-days for 16 workers.9 days: Overestimates remainder time compared with exact division (128/16). Common Pitfalls:Confusing total duration with post-joining duration; not converting to person-days; or rounding prematurely. Always calculate remainder precisely. Final Answer:None of these (the required time for the remainder is 8 days)

Question 13

Rates of men and boys in reaping hectares are different. If 5 men and 3 boys together reap 23 hectares in 4 days, and 3 men and 2 boys reap 7 hectares in 2 days, then how many boys must assist 7 men so that 45 hectares are reaped in 6 days?

Accepted Answer

Introduction / Context:We have two groups with men and boys working at different rates. By forming linear equations for daily output, we can determine individual rates and then scale to the target combination (7 men plus x boys) to meet a specified area in a given time. Given Data / Assumptions: (5 men + 3 boys) working 4 days ⇒ 23 hectares. (3 men + 2 boys) working 2 days ⇒ 7 hectares. Goal: (7 men + x boys) working 6 days ⇒ 45 hectares. Uniform rates per person per day; constant productivity. Concept / Approach:Let m = hectares per man per day, b = hectares per boy per day. Translate both statements into equations, solve for m and b, then find x such that (7m + x*b) * 6 = 45. Step-by-Step Solution: (5m + 3b) * 4 = 23 ⇒ 5m + 3b = 23/4 = 5.75.(3m + 2b) * 2 = 7 ⇒ 3m + 2b = 7/2 = 3.5.Solving gives m = 1 hectare/day, b = 1/4 hectare/day.For 7 men + x boys over 6 days: (7*1 + x*(1/4)) * 6 = 45.(7 + 0.25x) * 6 = 45 ⇒ 42 + 1.5x = 45 ⇒ 1.5x = 3 ⇒ x = 2. Verification / Alternative check:Plug m = 1, b = 0.25 back into the original equations to confirm they satisfy both, ensuring consistency. They do, so x = 2 is valid. Why Other Options Are Wrong:6 boys: Would exceed 45 hectares within 6 days.4 boys / 5 boys: Produce totals different from 45 hectares when evaluated; they do not match the target exactly.3 boys: Still short of 45 hectares. Common Pitfalls:Forgetting to divide by days; mixing area with rate; or assuming men and boys contribute equally. Always solve for individual rates first. Final Answer:2 boys

Question 14

A piece of work was planned to finish in 40 days. The initial team managed to complete only half the work in 24 days. Then, 16 more men were added to the team, and the whole work was still finished exactly in the scheduled 40 days. How many men were…

Accepted Answer

Introduction / Context:When team size changes mid-project, convert to rate equations relative to the total work W. Here half the work is done by the initial team in 24 days; the rest is done by a larger team in the final 16 days. We solve for the original team size. Given Data / Assumptions: Total planned duration = 40 days. Half of W completed in first 24 days by x men. 16 more men added after day 24; remaining half finished in 16 days. Each man works at the same constant rate r (work units per day). Concept / Approach:Let W be total work. Then x*r*24 = W/2 and (x + 16)*r*16 = W/2. Eliminate W and r to solve for x, the initial number of men. Step-by-Step Solution: From half-work equations: x*r*24 = (x + 16)*r*16.Cancel r: 24x = 16x + 256.8x = 256 ⇒ x = 32. Verification / Alternative check:First half: 32*r*24 = W/2. Second half: 48*r*16 = W/2. Both equal W/2 ⇒ consistent. Total time = 24 + 16 = 40 as required. Why Other Options Are Wrong:16/24/48 men yield equations that do not balance to halves over 24 and 16 days. Only 32 satisfies the equal half-work condition under a constant per-man rate. Common Pitfalls:Assuming linearity without setting equations; mixing total duration with per-phase work; or forgetting both phases are exactly half of W. Final Answer:32 men

Question 15

Workers A, B, and C can complete a job at different individual times. A alone takes 10 days; B alone takes 24 days. When A, B, and C work together, they finish the job in 6 days. In how many days can C alone complete the job?

Accepted Answer

Introduction / Context:Given individual completion times for A and B, and the combined time for A+B+C, we find C's solo time by subtracting known rates from the combined rate. This is a standard rate decomposition problem. Given Data / Assumptions: A's time = 10 days ⇒ rate = 1/10 job/day. B's time = 24 days ⇒ rate = 1/24 job/day. (A+B+C) time = 6 days ⇒ combined rate = 1/6 job/day. Concept / Approach:Let rC be C's rate. Then rC = 1/6 - (1/10 + 1/24). Convert to a single fraction and then invert to get C's required days (1/rC). Step-by-Step Solution: 1/10 + 1/24 = (12/120) + (5/120) = 17/120.Combined rate = 1/6 = 20/120.C's rate rC = 20/120 - 17/120 = 3/120 = 1/40.Therefore, C's time alone = 1 / (1/40) = 40 days. Verification / Alternative check:Check: 1/10 + 1/24 + 1/40 = 12/120 + 5/120 + 3/120 = 20/120 = 1/6. Valid. Why Other Options Are Wrong:25, 50, 75, 60 do not satisfy the exact rate balance when combined with A and B to produce 1/6 job/day. Common Pitfalls:Arithmetic slips when summing fractions or inverting the final rate; not using a common denominator can lead to errors. Keep precise fractions to avoid rounding mistakes. Final Answer:40

Question 16

Production is proportional to men × hours × days. If 15 men make 12 toys by working 5 hours per day for 16 days, in how many days can 10 men make 15 toys by working 8 hours per day (same productivity per man-hour)?

Accepted Answer

Introduction / Context:Output in such problems scales directly with the product of number of men, hours per day, and days, assuming a constant per man-hour productivity. We set up a proportionality to relate the two scenarios. Given Data / Assumptions: Scenario 1: 15 men, 5 h/day, 16 days ⇒ 12 toys. Scenario 2: 10 men, 8 h/day, D days ⇒ 15 toys. Uniform productivity k toys per man-hour. Concept / Approach:If k is toys per man-hour, then toys = men * hours/day * days * k. Use Scenario 1 to find k, then plug into Scenario 2 to solve for D. Step-by-Step Solution: Scenario 1 man-hours = 15 * 5 * 16 = 1200 man-hours.Thus k = 12 / 1200 = 0.01 toy per man-hour.Scenario 2 equation: 10 * 8 * D * 0.01 = 15.0.8D = 15 ⇒ D = 15 / 0.8 = 18.75 days.18.75 days = 18 3/4 days. Verification / Alternative check:Proportion check: Desired toys 15 is 1.25 times 12, but man-hours per day changed; explicit calculation above is safest and consistent. Why Other Options Are Wrong:18 2/3 and 18 4/5 are near but do not match the exact decimal 18.75; 18 5/7 and 19 deviate further from the computed value. Common Pitfalls:Confusing total hours with hours per day; forgetting to include all three factors (men, hours, days) in proportion; rounding too early. Final Answer:18 3/4 days

Question 17

In a garrison, food was stocked for 1000 soldiers for one month (assume 30 days of rations). After 10 days, 1000 more soldiers joined (so total becomes 2000). For how many additional days will the remaining food last after the reinforcements arrive?

Accepted Answer

Introduction / Context:Conservation of total rations (measured in soldier-days) is a common approach. We first compute the total available soldier-days, subtract what has been consumed, and then divide by the new daily consumption after reinforcements. Given Data / Assumptions: Initial stock = enough for 1000 soldiers for 30 days. First phase = 10 days with 1000 soldiers. After day 10, soldiers = 2000. Uniform daily consumption per soldier. Concept / Approach:Use the soldier-day model. Total stock in soldier-days is fixed. Consumption reduces it. Remaining soldier-days divided by current soldiers gives remaining days. Step-by-Step Solution: Total soldier-days = 1000 * 30 = 30000.Consumed in first 10 days = 1000 * 10 = 10000.Remaining soldier-days = 30000 - 10000 = 20000.After reinforcements: daily consumption = 2000 soldier-days per day.Remaining days = 20000 / 2000 = 10 days. Verification / Alternative check:Total elapsed time until exhaustion = 10 (first) + 10 (remaining) = 20 days from the start. The question asks for duration after the join, hence 10 days is correct. Why Other Options Are Wrong:25/20/15 days conflate total-from-start with remaining or miscompute consumption. Only 10 days matches the soldier-day arithmetic post-join. Common Pitfalls:Not converting to soldier-days; assuming “one month” as 31; or forgetting that the question asks for the remaining duration after the reinforcements arrive. Final Answer:10 days

Question 18

Workers A and B can complete a job individually in 8 hours and 12 hours, respectively. If they work in alternate one-hour turns, starting with A, at what time (duration) will the job be completed?

Accepted Answer

Introduction / Context:Alternate-work problems require tracking the cumulative fraction completed after each segment. With 1-hour alternations, we can bundle two successive hours (one by A, one by B) into a 2-hour cycle to simplify computation. Given Data / Assumptions: A's solo time = 8 h ⇒ rate = 1/8 per hour. B's solo time = 12 h ⇒ rate = 1/12 per hour. Work proceeds in 1-hour blocks with A starting. Concept / Approach:In each 2-hour cycle (A then B), the completed fraction equals 1/8 + 1/12. Repeatedly add cycle output until the remaining fraction can be finished by a partial final hour if needed. Step-by-Step Solution: Two-hour cycle output = 1/8 + 1/12 = 5/24 of the job.After 4 full cycles (8 hours), work done = 4 * (5/24) = 20/24 = 5/6; remaining = 1/6.Next hour A works: adds 1/8; remaining = 1/6 - 1/8 = 1/24.B's rate is 1/12 per hour, so time to finish the last 1/24 = (1/24) / (1/12) = 1/2 hour.Total duration = 8 h + 1 h + 0.5 h = 9.5 h = 9 1/2 h. Verification / Alternative check:Breaking into hour-by-hour sums yields the same cumulative fraction; cycle grouping just accelerates calculation. Why Other Options Are Wrong:9 h: stops too early (1/24 remains). 8 1/2 h and 8 h: underestimate combined progress. 10 h: overshoots by an extra half-hour. Common Pitfalls:Forgetting partial hours are allowed; mis-adding fractions in cycles; assuming a whole-hour completion. Final Answer:9 1/2 h

Question 19

A contractor planned to complete a job in 9 days. He hired a certain number of men. However, 6 of them were absent from day one, and the remaining men finished the job in 15 days. How many men had the contractor originally planned to employ?

Accepted Answer

Introduction / Context:This question applies the person-days (or rate) method. If a contractor expects n men to finish in 9 days, the total work can be written as n * 9 in man-days. When only (n − 6) men actually work and the job takes 15 days, the same work is (n − 6) * 15 man-days. Equating these allows us to solve for n. Given Data / Assumptions: Planned workforce = n men. Planned duration = 9 days. Actual workforce = n − 6 men (from day one). Actual duration = 15 days. Per-man productivity is constant throughout. Concept / Approach:Total work is invariant. Express it both ways and solve for n. Using person-days avoids needing the per-man rate explicitly; it cancels out naturally. Step-by-Step Solution: Planned work = n * 9.Actual work = (n − 6) * 15.Equate: n * 9 = (n − 6) * 15.Expand: 9n = 15n − 90 ⇒ 6n = 90 ⇒ n = 15. Verification / Alternative check:If 15 men could finish in 9 days, but only 9 men (15 − 6) worked, the duration factor increases by 15/9 = 5/3; 9 * (5/3) = 15 days, consistent. Why Other Options Are Wrong:6, 9, 12, 13 do not satisfy 9n = 15(n − 6); only n = 15 balances both sides. Common Pitfalls:Confusing total work with rate; forgetting that fewer workers increase time proportionally; attempting to assign arbitrary per-man rates (unnecessary). Final Answer:15

Question 20

Two inlet pipes A and B can fill a tank in 40 minutes and 60 minutes, respectively. An outlet pipe C empties the tank. With all three pipes open together, the tank fills in 30 minutes. How long would C alone take to empty a full tank?

Accepted Answer

Introduction / Context:Pipe-and-cistern problems use additive rates: inlets add to the fill rate while outlets subtract from it. Knowing the net fill time with all pipes open, we can isolate the outlet’s rate and invert to get its emptying time. Given Data / Assumptions: A fills in 40 min ⇒ rate = 1/40 tank/min. B fills in 60 min ⇒ rate = 1/60 tank/min. With A, B, C open together, fill time = 30 min ⇒ net rate = 1/30 tank/min. C is an outlet ⇒ its rate is subtractive. Concept / Approach:Let rC be C’s emptying rate (tank/min). Then 1/30 = 1/40 + 1/60 − rC. Solve for rC and invert to get time. Step-by-Step Solution: 1/40 + 1/60 = 3/120 + 2/120 = 5/120 = 1/24.Equation: 1/30 = 1/24 − rC ⇒ rC = 1/24 − 1/30.Compute rC: LCM(24,30)=120 ⇒ rC = (5 − 4)/120 = 1/120 tank/min.Time for C alone = 1 / (1/120) = 120 min = 2 h. Verification / Alternative check:Net with all three: 1/24 − 1/120 = (5 − 1)/120 = 4/120 = 1/30, matches the given net time. Why Other Options Are Wrong:1 h/3 h/4 h/90 min lead to net rates different from 1/30 when recombined with A and B. Common Pitfalls:Adding rather than subtracting the outlet, or mishandling fractions. Final Answer:2 h

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