Question 1

If 10 spiders can catch 10 flies in 10 minutes (all working at the same constant rate), how many flies can 200 spiders catch in 200 minutes?

Accepted Answer

Introduction: Catching flies here scales with the number of spiders and with time, assuming a constant per-spider catch rate. We infer the rate from the base scenario and then scale to the larger numbers using direct proportionality (spiders × minutes × rate per spider-minute). Given Data / Assumptions: 10 spiders catch 10 flies in 10 minutes. All spiders have the same constant efficiency. Find flies caught by 200 spiders in 200 minutes. Concept / Approach: Let r be the number of flies caught by one spider per minute. Then total flies = spiders × minutes × r. Using the base case we find r, then apply it to the new case by multiplication without rounding since values are exact here. Step-by-Step Solution: Base: 10 = 10 * 10 * r → r = 10 / 100 = 0.1 flies per spider-minute New: Flies = 200 * 200 * 0.1 = 4000 Verification / Alternative check: Proportion method: New/Base = (200/10) * (200/10) = 20 * 20 = 400. Then 10 * 400 = 4000, same result. Why Other Options Are Wrong: 2000, 3000, 5000, 2500 are not consistent with the direct quadratic scaling in spiders and minutes based on the stated rate. Common Pitfalls: Treating the relationship as linear in one factor only (either spiders or minutes). The catch count scales with both. Final Answer: 4000

Question 2

Two thousand soldiers in a fort have food for 20 days. Some soldiers are transferred out and the same food then lasts for 25 days. How many soldiers were transferred?

Accepted Answer

Introduction: This is another man-days conservation question. With fixed food, men × days is constant. When the duration rises from 20 to 25 days, it must be because the number of soldiers decreased proportionally. We can solve for the new headcount and then find how many left. Given Data / Assumptions: Initial soldiers = 2000; initial duration = 20 days. After transfer, same stock lasts 25 days. Per-soldier daily ration unchanged. Concept / Approach: Total food = 2000 * 20 = 40000 soldier-days. Let remaining soldiers be S. Then S * 25 = 40000 → compute S and subtract from 2000 to get the transferred number. Step-by-Step Solution: S * 25 = 40000 → S = 40000 / 25 = 1600 Transferred = 2000 − 1600 = 400 soldiers Verification / Alternative check: Check: With 1600 soldiers, duration = 40000 / 1600 = 25 days, exactly as stated. Why Other Options Are Wrong: 450, 500, 525, 600 give remaining soldiers that do not multiply to 40000 soldier-days over 25 days. Common Pitfalls: Averaging 20 and 25 to guess, or thinking duration increased by 5 days implies only 5 men left; the relationship is multiplicative via man-days, not additive. Final Answer: 400

Question 3

In a hostel, food is enough for 50 students for 45 days. If 75 students share the same stock (no wastage, same ration), for how many days will the food last?

Accepted Answer

Introduction: Identical to other stock-sharing problems, this uses the constancy of student-days (total food). Increasing the number of students reduces the number of days inversely, with the per-student ration unchanged. The arithmetic is straightforward and exact here. Given Data / Assumptions: Total food supports 50 students for 45 days. New student count = 75. Consumption per student per day remains the same. Concept / Approach: Student-days = 50 * 45. For 75 students, new days = (50 * 45) / 75. This is the standard unitary-method structure for such problems. Step-by-Step Solution: Student-days = 2250 New duration D = 2250 / 75 = 30 days Verification / Alternative check: Ratio approach: D scales by 50/75 = 2/3 → 45 * (2/3) = 30 days. Same result, faster mental math. Why Other Options Are Wrong: 25, 28, 34, and 40 days violate the fixed student-days constraint. Common Pitfalls: Forgetting inverse proportion: with more students and the same stock, the duration must go down. Final Answer: 30 days

Question 4

Rations are provided to feed 72 soldiers for 54 days. If the same stock is used to feed 90 soldiers but the individual ration is reduced by 10%, how long will the food last?

Accepted Answer

Introduction: When headcount changes and per-person ration also changes, convert everything to “soldier-equivalent” days. A 10% reduction means each soldier consumes 0.9 of the original daily ration, so 90 soldiers consume the equivalent of 81 “full-ration” soldiers per day. Apply man-day conservation to compute the new duration. Given Data / Assumptions: Original: 72 soldiers for 54 days → stock = 72 * 54 full-ration soldier-days. New: 90 soldiers at 90% ration → effective daily consumption = 90 * 0.9 = 81 full-ration soldier-days/day. Consumption is uniform; no wastage. Concept / Approach: Let S be the total stock in full-ration soldier-days. Then duration D satisfies S = 81 * D. Compute S from the original configuration and divide by 81 to find D in days. Step-by-Step Solution: S = 72 * 54 = 3888 full-ration soldier-days New effective daily consumption = 81 D = 3888 / 81 = 48 days Verification / Alternative check: Simplify first: 72/81 = 8/9; thus D = (8/9)*54 = 8*6 = 48 days, confirming without large numbers. Why Other Options Are Wrong: 72 or 54 days ignore the ration cut effect; 126 and 36 are inconsistent with the effective 81-soldier-per-day consumption. Common Pitfalls: Treating 90 soldiers at 90% ration as 90 full-ration soldiers (it is effectively 81), or forgetting to convert to full-ration equivalents. Final Answer: 48 days

Question 5

Three men can finish a piece of work in 18 days. Six boys can also finish the same work in 18 days. Working together at the same constant rates, in how many days will 4 men and 4 boys finish the work?

Accepted Answer

Introduction: When two types of workers can each complete the same job in the same number of days but with different group sizes, we can infer individual rates for a “man” and a “boy.” Summing their daily rates allows us to compute the combined time to finish the work together. Given Data / Assumptions: 3 men finish in 18 days. 6 boys finish in 18 days. Rates are constant and additive; no interference effects. Concept / Approach: Let work = 1 unit. Then daily rate of 3 men = 1/18 → per man rate = (1/18)/3 = 1/54. Similarly, daily rate of 6 boys = 1/18 → per boy rate = (1/18)/6 = 1/108. Combine 4 men and 4 boys to get total daily rate; invert to get total time in days. Step-by-Step Solution: Per man rate = 1/54 job/day Per boy rate = 1/108 job/day 4 men rate = 4/54 = 2/27 job/day 4 boys rate = 4/108 = 1/27 job/day Combined rate = 2/27 + 1/27 = 3/27 = 1/9 job/day Time = 1 / (1/9) = 9 days Verification / Alternative check: Check that 3 men alone need 18 days (3 * 1/54 = 1/18) and 6 boys alone need 18 days (6 * 1/108 = 1/18). The combined rate arithmetic is consistent with both references. Why Other Options Are Wrong: 6 and 8 days are too short (would require a higher daily rate); 10 and 12 days are too long given the additive rates computed. Common Pitfalls: Averaging days directly or treating “man” and “boy” as equal-rate workers. Compute and add rates, not times, when workers collaborate. Final Answer: 9 days

Question 6

Work-rate with unequal efficiencies: 10 men and 8 women together finish a job in 5 days. Given that one woman's daily work equals half of one man's daily work, in how many days will 4 men and 6 women complete the same job?

Accepted Answer

Introduction / Context:This unitary-method problem tests conversion of mixed-efficiency teams to a common unit (e.g., “man-equivalents”) and then using rate * time = work. By expressing every worker's contribution in comparable terms, we can scale the time for any new team composition. Given Data / Assumptions: 10 men + 8 women finish the work in 5 days. 1 woman does half the daily work of 1 man. Find days needed by 4 men + 6 women to finish the same job. Concept / Approach:Let a man's daily rate be m units/day. Then a woman's daily rate is m/2. Convert each team to man-equivalents, compute the total work once, and reuse it to find the time for the second team. Step-by-Step Solution: Daily rate of (10 men + 8 women) = 10m + 8*(m/2) = 10m + 4m = 14m Total work W = (rate) * (time) = 14m * 5 = 70m Daily rate of (4 men + 6 women) = 4m + 6*(m/2) = 4m + 3m = 7m Required days = W / (new rate) = 70m / 7m = 10 days Verification / Alternative check:Because 4 men + 6 women is exactly half of the original team in man-equivalents (7m vs 14m), their time must double from 5 to 10 days. This proportionality check confirms the result. Why Other Options Are Wrong:9, 8 2/3, 11, 12 days do not preserve the rate-time inverse relationship between a 14m team and a 7m team for the same job size. Common Pitfalls:Forgetting that a woman works at m/2, or adding women as if they worked at rate m like men. Mixing up proportional reasoning is another frequent error. Final Answer:10 days

Question 7

Mixed workforce rates: 8 men finish the job in 12 days; 4 women finish it in 48 days; 10 children finish it in 24 days. In how many days will 10 men, 4 women, and 10 children working together complete the same job?

Accepted Answer

Introduction / Context:Different categories of workers often have different speeds. Convert each category to a per-person daily rate, add rates to get a team rate, and then compute time = work / rate. The unitary-method approach keeps the arithmetic clean and scalable. Given Data / Assumptions: 8 men complete in 12 days ⇒ 1 man's rate = 1 / (8*12) = 1/96 job/day. 4 women complete in 48 days ⇒ 1 woman's rate = 1 / (4*48) = 1/192 job/day. 10 children complete in 24 days ⇒ 1 child's rate = 1 / (10*24) = 1/240 job/day. Team: 10 men + 4 women + 10 children. Concept / Approach:Team rate = sum of individual rates, then time = 1 / (team rate). Be precise with fractions to avoid compounding rounding errors. Step-by-Step Solution: Men: 10 * (1/96) = 10/96 = 5/48 Women: 4 * (1/192) = 4/192 = 1/48 Children: 10 * (1/240) = 1/24 Team rate = 5/48 + 1/48 + 1/24 = 6/48 + 2/48 = 8/48 = 1/6 job/day Time = 1 / (1/6) = 6 days Verification / Alternative check:Since the combined rate is exactly 1/6 job/day, finishing in 6 days is exact (no rounding). Why Other Options Are Wrong:5, 7, 8, 10 days contradict the exact combined-rate computation 1/6 job/day. Common Pitfalls:Adding times instead of rates, or using proportional reasoning on days rather than on daily work rates leads to wrong results. Final Answer:6 days

Question 8

Scaling manpower to meet a deadline: 20 men complete one-third of a job in 20 days. How many additional men are required to finish the remaining work in 25 more days (assume constant individual rates)?

Accepted Answer

Introduction / Context:Use man-days as the unit of work. First compute the total job in man-days from a known partial completion. Then compute how many men are needed to deliver the remaining man-days within the new deadline, and subtract the existing workforce to find the extra men required. Given Data / Assumptions: 20 men finish 1/3 of the job in 20 days. Remaining work = 2/3 of the job. Deadline for the remainder = 25 days. Concept / Approach:Let r be one man’s daily work and W be total work. From the partial completion, determine r or W/r. Then use rate * time = work to solve for the required team size to finish the remainder in the allotted time. Step-by-Step Solution: Given: 20 men * 20 days * r = W/3 ⇒ 400r = W/3 ⇒ r = W/1200 Let M be men needed for the remaining 2W/3 in 25 days: M * 25 * r = 2W/3 M = (2W/3) / (25r) = (2/3) * (1200/25) = 32 men (total) Extra men required = 32 − 20 = 12 Verification / Alternative check:Man-days remaining = (2/3)W = (2/3)*(1200r) = 800r. In 25 days, daily manpower = 800r / 25 = 32r ⇒ 32 total men. This reconfirms the calculation. Why Other Options Are Wrong:10, 15, 20, 8 do not satisfy the man-day balance for 25 remaining days. Common Pitfalls:Adding or averaging days instead of using man-days, or forgetting to subtract the existing 20 men from the total required. Final Answer:12

Question 9

Scaling manpower to meet a deadline: 40 men complete one-third of a job in 40 days. How many additional men are needed to finish the remaining work in 50 more days (constant rates)?

Accepted Answer

Introduction / Context:As with all man-day problems, compute total work units from the partial completion, determine the required manpower for the remaining work within the new deadline, and then find the additional men beyond the current team size. Given Data / Assumptions: 40 men finish 1/3 of the job in 40 days. Remaining work = 2/3. Remaining time allowed = 50 days. Concept / Approach:Let r be a man’s daily rate, W total work. From 40*40*r = W/3, determine r in terms of W. Then compute the men needed to do 2W/3 in 50 days, and subtract 40 to get the additional men. Step-by-Step Solution: 40*40*r = W/3 ⇒ 1600r = W/3 ⇒ r = W/4800 Men needed M: M*50*r = 2W/3 ⇒ M = (2W/3) / (50r) = (2/3)*(4800/50) = 64 Additional men = 64 − 40 = 24 Verification / Alternative check:Man-days remaining = (2/3)W = (2/3)*(4800r) = 3200r. In 50 days ⇒ men/day = 3200r/50 = 64r ⇒ 64 men, consistent with the above. Why Other Options Are Wrong:12, 18, 20, 28 do not match the required man-day balance for a 50-day completion window. Common Pitfalls:Not subtracting the existing workforce from total men required, or mixing up the one-third and two-thirds portions of the job. Final Answer:24

Question 10

Basic unitary method (constant rate): If 7 spiders make 7 webs in 7 days, how many days does 1 spider need to make 1 web?

Accepted Answer

Introduction / Context:This is a direct unitary-method question. If multiple identical workers complete a proportional number of identical tasks in a given time, the rate per worker per day can be extracted and used to answer any equivalent scaling question. Given Data / Assumptions: 7 spiders produce 7 webs in 7 days. All spiders work at the same constant rate. Find the time for 1 spider to make 1 web. Concept / Approach:Let r be web/spider/day. Total webs = (number of spiders) * (days) * r. Solve for r from the given data, then compute the time for a single spider to make one web as 1/r days. Step-by-Step Solution: 7 spiders * 7 days * r = 7 webs ⇒ 49r = 7 ⇒ r = 7/49 = 1/7 web per spider per day Time for 1 spider to make 1 web = 1 / (1/7) = 7 days Verification / Alternative check:Proportionality check: If one spider makes 1 web in 7 days, then 7 spiders make 7 webs in the same 7 days, matching the premise. Why Other Options Are Wrong:3, 4, 6, 10 contradict the constant-rate calculation of r = 1/7 web/day per spider. Common Pitfalls:Assuming 1 day by incorrectly cancelling numbers, or thinking 7 spiders * 7 days = 49 webs (confusing rate extraction). Final Answer:7

Question 11

Men–hours–days scaling: 16 men working 7 hours per day plough a field in 48 days. In how many days will 14 men working 12 hours per day plough the same field?

Accepted Answer

Introduction / Context:Total work is proportional to men * hours/day * days when productivity per man-hour is constant. Equate total man-hours across scenarios to find the unknown time for the changed team and schedule. Given Data / Assumptions: Scenario 1: 16 men, 7 h/day, 48 days. Scenario 2: 14 men, 12 h/day, D days. Productivity per man-hour is constant. Concept / Approach:Set man-hours equal: 16*7*48 = 14*12*D, then solve for D. This is a classic men–hours–days proportionality problem under the unitary method. Step-by-Step Solution: 16*7*48 = 14*12*D Reduce by 14: (16*(7/14))*48 = 12*D ⇒ (16*0.5)*48 = 12D ⇒ 8*48 = 12D 384 = 12D ⇒ D = 32 days Verification / Alternative check:Man-hours in scenario 1 = 16*7*48 = 5376; scenario 2 = 14*12*32 = 5376; equality confirmed. Why Other Options Are Wrong:56, 38, 30, 28 do not satisfy the equal-man-hours relation for the same field. Common Pitfalls:Multiplying or dividing the wrong factors or forgetting that increasing hours/day allows fewer days for the same total work. Final Answer:32

Question 12

Machine-rate scaling: 6 identical machines produce 270 bottles per minute. At the same constant rate, how many bottles will 10 such machines produce in 4 minutes?

Accepted Answer

Introduction / Context:When identical machines work at a constant rate, output scales directly with both the number of machines and the time. Convert to per-machine output and then multiply up to the new count and duration. Given Data / Assumptions: 6 machines ⇒ 270 bottles/minute. All machines are identical; rates are constant. Find output for 10 machines in 4 minutes. Concept / Approach:Per-machine rate = total rate / number of machines. New total = per-machine rate * new machines * new time. Ensure units (minutes) are consistent. Step-by-Step Solution: Per-machine rate = 270 / 6 = 45 bottles/minute 10 machines ⇒ 10 * 45 = 450 bottles/minute In 4 minutes ⇒ 450 * 4 = 1800 bottles Verification / Alternative check:Proportionality: scaling machines by 10/6 and time by 4 multiplies output by (10/6)*4 = 6.666… ⇒ 270 * 6.666… = 1800, same result. Why Other Options Are Wrong:1600, 1700, 2000, 2500 do not match the linear scaling from 270 bottles/min for 6 machines. Common Pitfalls:Forgetting to multiply by time, or dividing by the wrong machine count when computing per-machine rate. Final Answer:1800

Question 13

Price scaling (dozens): If 357 apples cost ₹1517.25, estimate the price of 49 dozens of similar apples (nearest sensible option).

Accepted Answer

Introduction / Context:Use the unitary method: find the unit price (per apple) from the given bundle, then multiply by the requested quantity (in dozens). Rounding to the nearest logical option is appropriate when the computed value sits between choices. Given Data / Assumptions: 357 apples cost ₹1517.25. Required quantity = 49 dozens = 49 * 12 = 588 apples. Concept / Approach:Unit price p = total cost / number of apples. Then target cost = p * 588. Exact arithmetic shows a value extremely close to ₹2500, matching the best option. Step-by-Step Solution: Unit price = 1517.25 / 357 = 4.25 per apple (since 357 * 4.25 = 1517.25) Cost for 588 apples = 588 * 4.25 = (600 − 12) * 4.25 = 2550 − 51 = ₹2499 Nearest sensible option ≈ ₹2500 Verification / Alternative check:Direct multiplication: 588 * 4 = 2352; plus 0.25 * 588 = 147 ⇒ 2352 + 147 = ₹2499; confirms the result. Why Other Options Are Wrong:₹2800 and ₹2900 overshoot by 300–400; ₹2400 undershoots by ~99; ₹2650 is still high vs exact ₹2499. Common Pitfalls:Converting dozens incorrectly or rounding the unit price before scaling to 588 apples, which can create larger errors. Final Answer:Rs. 2500

Question 14

Map scale conversion: On a map, 0.6 cm represents 6.6 km. If two points are 80.5 cm apart on the map, what is their actual separation?

Accepted Answer

Introduction / Context:Scale problems convert a drawing length to the real-world length via proportionality. First obtain the km per cm, then multiply by the measured cm to get the actual distance in km. Keep units clearly labeled throughout. Given Data / Assumptions: Scale: 0.6 cm ↔ 6.6 km. Map separation = 80.5 cm. Concept / Approach:Compute 1 cm equivalent = 6.6 / 0.6 = 11 km/cm. Then multiply the map length by 11 to get the real distance. This collapses the proportionality to simple multiplication. Step-by-Step Solution: 1 cm on map corresponds to 6.6 / 0.6 = 11 km Actual distance = 80.5 cm * 11 km/cm = 885.5 km Verification / Alternative check:Check: 0.6 cm → 6.6 km means each 0.6 cm gives 6.6 km. 80.5 / 0.6 ≈ 134.167 scale units; 134.167 * 6.6 ≈ 885.5 km. Same result. Why Other Options Are Wrong:870, 872.5, 895, 880 km are deviations from the exact proportional conversion of 11 km per cm. Common Pitfalls:Inverting the ratio (using 0.6/6.6) or forgetting to scale all the way from cm to km consistently. Final Answer:885.5 km

Question 15

Work proportionality (similar wall): 8 persons build a 140 m wall in 42 days. In how many days can 30 persons build a similar wall of length 100 m?

Accepted Answer

Introduction / Context:When work is directly proportional to length, and productivity is constant per person per day, the required days scale with wall length and inversely with workforce size. Convert the first scenario into person-days per meter to answer the second scenario quickly. Given Data / Assumptions: 8 persons build 140 m in 42 days. Find days for 30 persons to build 100 m of a similar wall. Concept / Approach:Person-days per meter = (persons * days) / length. Multiply by target length to get total person-days needed, then divide by the new number of persons to get days. Step-by-Step Solution: Person-days per meter = (8*42) / 140 = 336 / 140 = 12/5 = 2.4 Total person-days for 100 m = 2.4 * 100 = 240 With 30 persons, days = 240 / 30 = 8 days Verification / Alternative check:Proportionality: Time ∝ length / persons ⇒ T2 = T1 * (100/140) * (8/30) * (persons/?) after re-arranging yields the same numeric result when applied carefully. Why Other Options Are Wrong:11, 13, 19, 10 days do not match the calculated 240 person-days divided by 30 persons. Common Pitfalls:Forgetting to scale by both length and workforce, or mixing days and persons in the proportion incorrectly. Final Answer:8 days

Question 16

Headcount from two schedules: A group can finish a job in 100 days. With 10 fewer people, the job would take 110 days. How many people were in the original group?

Accepted Answer

Introduction / Context:This is a classic two-schedule unitary-method problem. Express total work in person-days using one schedule, then equate it to the other schedule with a different headcount to solve for the original number of people directly and exactly. Given Data / Assumptions: Original team finishes in 100 days. With 10 fewer people, it takes 110 days. Per-person productivity is constant. Concept / Approach:Let the original headcount be n and total work be W. Then W = 100n (person-days). With 10 fewer, W = 110(n − 10). Equate and solve for n algebraically. Step-by-Step Solution: 100n = 110(n − 10) 100n = 110n − 1100 ⇒ 10n = 1100 ⇒ n = 110 Verification / Alternative check:Total work W = 110 * 100 = 11000 person-days. With 100 people (110 − 10), W / 100 = 110 days, as stated. Why Other Options Are Wrong:180, 190, 196, 120 do not satisfy 100n = 110(n − 10). Common Pitfalls:Setting up 100(n − 10) = 110n (incorrect inversion) or cancelling 100s prematurely and losing track of the equation's balance. Final Answer:110

Question 17

Rationing with reinforcements: A garrison of 500 persons had provisions for 27 days. After 3 days, 300 more persons arrive. For how many more days will the remaining food last?

Accepted Answer

Introduction / Context:Food supply questions are really person-day accounting problems. Compute the total person-days available, subtract what has been consumed, then divide the remainder by the new headcount to get remaining days at the same ration level. Given Data / Assumptions: Initial stock covers 500 persons for 27 days ⇒ 500*27 person-days of food. After 3 days at 500 persons, 300 more persons arrive. Ration per person per day unchanged. Concept / Approach:Total person-days are conserved until consumed. Remaining days = (remaining person-days) / (current headcount). Track consumption during the first 3 days, then recalculate with the new total headcount (800 persons). Step-by-Step Solution: Total stock = 500 * 27 = 13500 person-days Consumed in first 3 days = 500 * 3 = 1500 Remaining stock = 13500 − 1500 = 12000 person-days New headcount = 500 + 300 = 800 persons Remaining days = 12000 / 800 = 15 days Verification / Alternative check:Sanity check: With more people, remaining days must be fewer than 27 − 3 = 24; 15 is reasonable. Why Other Options Are Wrong:26, 27, 28 ignore consumption and added headcount; 18 is not supported by the exact person-day computation. Common Pitfalls:Forgetting to subtract the 3-day consumption or dividing the original stock by the new headcount directly without accounting for the initial usage. Final Answer:15 days

Question 18

Hostel rationing: Provisions for 250 men last 40 days. If 50 men leave, how long will the food last at the same daily ration?

Accepted Answer

Introduction / Context:Again, think in person-days. A fixed stock measured in person-days divided by the new headcount gives the extended duration once some people leave while ration per person per day remains unchanged. Given Data / Assumptions: Initial coverage: 250 men for 40 days. 50 men leave ⇒ new headcount = 200. Same daily ration per person. Concept / Approach:Total stock = 250*40 person-days. With 200 men, duration = stock / 200. No initial consumption needs subtraction because the change is immediate in the problem statement. Step-by-Step Solution: Stock = 250 * 40 = 10000 person-days New duration = 10000 / 200 = 50 days Verification / Alternative check:Proportionality: duration scales inversely with headcount; reducing from 250 to 200 increases days by factor 250/200 = 1.25 ⇒ 40 * 1.25 = 50 days. Why Other Options Are Wrong:58, 67 overshoot; 45 undershoots; only 50 matches exact person-day math. Common Pitfalls:Applying partial consumption incorrectly or changing ration rate, which the question does not state. Final Answer:50 days

Question 19

Camp rationing with arrivals: Food was sufficient for 2000 people for 54 days. After 15 days, more people arrived and the food lasted only 20 more days. How many people arrived?

Accepted Answer

Introduction / Context:Track stock in person-days, subtract the initial consumption, then divide the remainder by the new total headcount to match the shortened remaining duration. Solve for the unknown additional headcount algebraically and exactly. Given Data / Assumptions: Initial stock: 2000 people for 54 days ⇒ 108000 person-days. First 15 days at 2000 people. After that, food lasted 20 more days with the enlarged population. Concept / Approach:Remaining stock after 15 days = 108000 − (2000*15). Let new headcount be 2000 + x. Then (2000 + x)*20 must equal the remaining stock. Solve for x (arrivals). Step-by-Step Solution: Remaining stock = 108000 − 30000 = 78000 person-days (2000 + x) * 20 = 78000 ⇒ 2000 + x = 3900 ⇒ x = 1900 Verification / Alternative check:New headcount = 3900; 3900 * 20 = 78000 matches remaining stock. Numbers balance exactly. Why Other Options Are Wrong:2400, 1940, 2220, 2100 do not satisfy the exact person-day equality for 20-day residual life. Common Pitfalls:Forgetting to subtract the first 15 days of consumption or mixing people added with total headcount after arrival. Final Answer:1900

Question 20

Pump-hours scaling: Three pumps working 8 hours/day empty a tank in 2 days. How many hours/day must four pumps work to empty the tank in 1 day?

Accepted Answer

Introduction / Context:Total work for the tank equals total pump-hours required. Compute the pump-hours from the first schedule, then distribute that requirement over the new number of pumps and the new duration to find the needed hours/day. Given Data / Assumptions: 3 pumps * 8 hours/day * 2 days empties the tank. Find daily hours for 4 pumps to finish in 1 day (same pump type and rate). Concept / Approach:Total pump-hours required = pumps * hours/day * days. This remains constant across equivalent schedules. Solve for the unknown daily hours in the second schedule by equating pump-hours. Step-by-Step Solution: Pump-hours required = 3 * 8 * 2 = 48 pump-hours With 4 pumps in 1 day, hours/day = 48 / 4 = 12 Verification / Alternative check:Check: 4 pumps * 12 hours = 48 pump-hours, same requirement as the original schedule; hence the tank is emptied in 1 day. Why Other Options Are Wrong:11 or 14 hours/day do not hit the constant 48 pump-hours target; 19 and 16 are also inconsistent. Common Pitfalls:Confusing total hours with hours/day or forgetting to multiply by the number of pumps in both scenarios. Final Answer:12

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