Question 1

Garrison rationing with arrivals: Food is sufficient for 1000 soldiers for 1 month (assume 30 days). After 10 days, 1000 more soldiers join. For how many additional days will the remaining food last?

Accepted Answer

Introduction / Context: Measure the stock in person-days, subtract the initial 10-day consumption, then divide the remainder by the new total headcount (after reinforcements) to compute how many more days the rations will last at the same rate. Given Data / Assumptions: Total stock initially: 1000 soldiers * 30 days = 30000 person-days. After 10 days at 1000 soldiers, remaining stock = 30000 − 10000 = 20000 person-days. Then 1000 more soldiers join ⇒ total 2000 soldiers. Concept / Approach: Remaining duration = remaining person-days / new headcount. Keep the month as 30 days unless otherwise specified; this does not affect the proportional result because we use person-days directly. Step-by-Step Solution: Remaining stock = 20000 person-days New headcount = 2000 soldiers Remaining days = 20000 / 2000 = 10 days Verification / Alternative check: With double headcount, remaining days should be halved from 20 (if still 1000) to 10; this matches the arithmetic. Why Other Options Are Wrong: 25, 20, 15, 12 are inconsistent with exact person-day balance after reinforcements. Common Pitfalls: Using calendar month length inconsistently or forgetting to subtract the first 10 days of consumption before reinforcements arrive. Final Answer: 10 days

Question 2

Two-person team timing: A man and a boy together finish a job in 24 days. If, for the last 6 days, the man works alone and the job finishes in 26 days total, how long would the boy take alone?

Accepted Answer

Introduction / Context: Express daily work as rates. The pair’s joint rate is known from the 24-day completion. The altered schedule provides a second equation involving the man’s solo work in the last 6 days. Solve to find each individual rate, then invert to get solo times. Given Data / Assumptions: (Man + Boy) complete in 24 days ⇒ m + b = 1/24 job/day. In the 26-day schedule, first 20 days both work; last 6 days man works alone. Concept / Approach: Total work equals work in first 20 days plus work in last 6 days. Use m + b and the 26-day completion to find m, then compute b = (1/24 − m). Boy’s solo time = 1 / b days. Step-by-Step Solution: 20(m + b) + 6m = 1 20*(1/24) + 6m = 1 ⇒ 20/24 + 6m = 1 ⇒ 5/6 + 6m = 1 6m = 1 − 5/6 = 1/6 ⇒ m = 1/36 b = 1/24 − 1/36 = (3 − 2)/72 = 1/72 ⇒ Boy alone takes 72 days Verification / Alternative check: Check rates: Together 1/24; man 1/36; boy 1/72. First 20 days: 20*(1/24) = 5/6. Last 6 days (man only): 6*(1/36) = 1/6. Total = 5/6 + 1/6 = 1 job. Valid. Why Other Options Are Wrong: 73, 49, 62, 84 contradict the computed b = 1/72 job/day. Common Pitfalls: Assuming both worked for all 26 days or misallocating the last 6 days to both workers. Also, mixing times directly instead of using rates leads to errors. Final Answer: 72 days

Question 3

Staggered workforce: 25 men can reap a field in 20 days. After working some days with all 25, 15 men leave and the remaining 10 must finish in 37.5 days. After how many days should 15 men leave?

Accepted Answer

Introduction / Context: Break total work into two phases with different workforce sizes. Compute total job in man-days from the baseline, then equate it to the sum of man-days delivered in phase 1 (25 men for x days) and phase 2 (10 men for 37.5 days) to solve for x. Given Data / Assumptions: 25 men finish in 20 days ⇒ total work = 25 * 20 = 500 man-days. Phase 1: 25 men for x days. Phase 2: 10 men for 37.5 days (after 15 leave). Concept / Approach: Total man-days supplied must equal the job’s 500 man-days. Hence 25x + 10*37.5 = 500. Solve for x to find when the reduction should occur to still finish on time. Step-by-Step Solution: 25x + 10*37.5 = 500 25x + 375 = 500 ⇒ 25x = 125 ⇒ x = 5 days Verification / Alternative check: Man-days supplied = 25*5 + 10*37.5 = 125 + 375 = 500; equals the job size, confirming the timing. Why Other Options Are Wrong: 10, 9, 7, 6 days give totals other than 500 man-days and therefore miss the required completion in 37.5 days overall. Common Pitfalls: Trying to average days or workers instead of using man-day equality across phases. Final Answer: After 5 days

Question 4

Equivalent work rates: If 12 men or 18 women can finish a job in 14 days, how long will 8 men and 16 women together take to finish it?

Accepted Answer

Introduction / Context: From the two equivalences, deduce individual rates for a man and a woman relative to the whole job. Add the rates for the mixed team (8 men + 16 women) and invert to get the time to finish the complete job under constant productivity assumptions. Given Data / Assumptions: 12 men finish in 14 days ⇒ 12*14 man-days = job ⇒ 1 man = 1/168 job/day. 18 women finish in 14 days ⇒ 18*14 woman-days = job ⇒ 1 woman = 1/252 job/day. Team: 8 men + 16 women. Concept / Approach: Team rate = 8*(1/168) + 16*(1/252). Compute carefully using common denominators to avoid arithmetic slips, then time = 1 / (team rate). Step-by-Step Solution: Men's part: 8/168 = 1/21 Women's part: 16/252 = 4/63 Team rate = 1/21 + 4/63 = 3/63 + 4/63 = 7/63 = 1/9 job/day Time = 1 / (1/9) = 9 days Verification / Alternative check: Relative rate: 1 man = 1.5 women (since 12 men ≡ 18 women). Then 8 men + 16 women ≡ 8 men + (16/1.5) men ≈ 8 + 10.666… = 18.666… men. Since 12 men take 14 days, 1 man/day rate = 1/168; thus time ≈ (job)/(18.666…/168) = 168/18.666… = 9 days. Confirms result. Why Other Options Are Wrong: 10, 12, 14, 8 days contradict the exact combined rate 1/9 job/day. Common Pitfalls: Adding times instead of rates, or mixing up the man-to-woman equivalence (1 man = 1.5 women). Final Answer: 9 days

Question 5

Cutting into equal pieces: A rod 192 m long is cut into small pieces, each 3.2 m in length. How many pieces are obtained?

Accepted Answer

Introduction / Context: For uniform cutting, the number of pieces equals total length divided by piece length, provided the total length is an exact multiple. This is a straightforward unitary method with careful decimal handling (3.2 m = 32/10 m). Given Data / Assumptions: Total length = 192 m. Each piece length = 3.2 m. No wastage; exact division assumed. Concept / Approach: Compute N = total length / piece length. Use fraction conversion to avoid decimal errors: 3.2 = 32/10. Then N = 192 / (32/10) = 192 * (10/32). Reduce to find an integer if divisible. Step-by-Step Solution: 3.2 m = 32/10 m N = 192 / (32/10) = 192 * (10/32) = (192/32) * 10 = 6 * 10 = 60 pieces Verification / Alternative check: Multiply back: 60 pieces * 3.2 m = 192 m, matching the original length with no remainder. Why Other Options Are Wrong: 52, 62, 64, 68 are not equal to 192 / 3.2. Only 60 satisfies the exact division. Common Pitfalls: Dividing by 32 instead of 3.2, or moving the decimal the wrong way when converting 3.2 to a fraction. Final Answer: 60

Question 6

Tables and chairs price relation: The cost of 2 tables is equal to the cost of 5 chairs. If the price difference between one table and one chair is Rs 1200, find the cost (price) of one chair.

Accepted Answer

Introduction / Context: This unitary-method question uses two linear relations between the price of a table and the price of a chair. By translating the statement “cost of 2 tables equals cost of 5 chairs” and the given price difference into equations, we can determine the price of one chair directly. Given Data / Assumptions: 2 * (price of one table) = 5 * (price of one chair) (price of one table) − (price of one chair) = Rs 1200 All prices are in rupees and refer to identical items. Concept / Approach: Let T be the price of a table and C be the price of a chair. From 2T = 5C, we get T = 2.5C. Using the difference T − C = 1200, substitute T = 2.5C and solve for C. This is a straightforward substitution problem leading to one unknown. Step-by-Step Solution: Given 2T = 5C ⇒ T = 2.5CT − C = 1200 ⇒ 2.5C − C = 12001.5C = 1200 ⇒ C = 1200 / 1.5 = 800 Verification / Alternative check: Compute T: T = 2.5 * 800 = 2000. Check the difference: 2000 − 800 = 1200 (matches). Check the ratio: 2T = 2 * 2000 = 4000 and 5C = 5 * 800 = 4000 (matches). Why Other Options Are Wrong: Rs. 500, Rs. 600, Rs. 700, Rs. 400: None satisfy both 2T = 5C and (T − C) = 1200 when solved consistently. Common Pitfalls: Using 2T = 5C as T = 5C/2 and then forgetting to substitute carefully; or mixing up the difference (C − T instead of T − C). Always keep the unknowns labeled clearly to avoid sign errors. Final Answer: Rs. 800

Question 7

Unit conversion and unitary method: If the cost of 1/4 kg of a commodity is Rs 0.60, what is the cost of 200 g of the same commodity?

Accepted Answer

Introduction / Context: This is a classic unitary-method and unit-conversion question. We convert kilograms to grams, find a per-gram cost (or scale directly), and then compute the cost for the requested weight. Getting the conversion right avoids the most common mistakes. Given Data / Assumptions: 1/4 kg costs Rs 0.60 We need the cost of 200 g 1 kg = 1000 g Concept / Approach: First, compute the cost per kilogram from the given quarter kilogram price. Then multiply by the fraction of a kilogram represented by 200 g (which is 0.2 kg). Alternatively, use direct proportionality from 250 g to 200 g since 1/4 kg = 250 g. Step-by-Step Solution: 1/4 kg = 0.25 kg costs Rs 0.60 ⇒ 1 kg costs 0.60 / 0.25 = Rs 2.40200 g = 0.2 kg ⇒ Cost = 2.40 * 0.2 = Rs 0.48Rs 0.48 = 48 paise Verification / Alternative check: Proportion from grams: 250 g costs 60 paise ⇒ 1 g costs 60/250 = 0.24 paise ⇒ 200 g costs 200 * 0.24 = 48 paise. Same result confirms correctness. Why Other Options Are Wrong: 42, 40, 50, 60 paise: These arise from rounding or wrong gram-to-kilogram scaling. Only 48 paise matches exact proportionality. Common Pitfalls: Using 1 kg = 100 g (wrong) or converting 200 g incorrectly as 0.02 kg. Also, mixing paise and rupees (e.g., treating 0.48 as 48 rupees) leads to huge errors. Final Answer: 48 paise

Question 8

Approximate cost via unitary method: If 3 dozen mangoes cost Rs 245, estimate (approximately) the cost of 363 mangoes.

Accepted Answer

Introduction / Context: We are given a price for 3 dozen (36) mangoes and asked to find the approximate cost for 363 mangoes. The unitary method (finding cost per item) then scaling up to the target quantity is the standard approach. Because the answer choices are round, a small rounding is acceptable. Given Data / Assumptions: 36 mangoes cost Rs 245 Target quantity = 363 mangoes Approximate answer acceptable; choose the nearest among options Concept / Approach: Compute cost per mango as 245/36 and then multiply by 363. Keep a few decimals to avoid drift. Finally, compare with the options and choose the closest figure. Step-by-Step Solution: Cost per mango ≈ 245 / 36 ≈ 6.8056Estimated cost for 363 ≈ 6.8056 * 363Breakdown: 6.8056*(300 + 60 + 3) ≈ 2041.68 + 408.336 + 20.417 ≈ 2470.433 Verification / Alternative check: Rounding per mango to 6.81 gives 6.81*363 ≈ 2473. Best nearby options are Rs 2400 (difference ≈ 70) and Rs 2600 (difference ≈ 130). Rs 2400 is closer. Why Other Options Are Wrong: Rs. 2000, Rs. 2200: underestimate the computed total by over 200. Rs. 2600: overshoots by ~130. Rs. 2300: still further than Rs. 2400 from the estimate. Common Pitfalls: Forgetting that 3 dozen = 36 (not 24) or mixing up the per-dozen price with per-piece price. Keep consistent units throughout to avoid large percentage errors. Final Answer: Rs. 2400

Question 9

Weight-length proportionality (uniform rod): A 13 m long rod weighs 23.4 kg. Assuming uniform density, what is the weight of a 6 m long rod of the same material and cross-section?

Accepted Answer

Introduction / Context: For a uniform rod (same material and cross-section), weight is directly proportional to length. The unitary method gives the per-metre weight, which we then multiply by the required length. This is a standard proportionality application in aptitude and basic physics contexts. Given Data / Assumptions: Length L1 = 13 m, Weight W1 = 23.4 kg Find weight W2 for length L2 = 6 m Uniform rod: weight ∝ length Concept / Approach: Compute weight per metre as W1 / L1, then multiply by L2. Because both numbers divide neatly, the final result is exact to one decimal place without complex rounding. Step-by-Step Solution: Weight per metre = 23.4 / 13 = 1.8 kg/mRequired weight = 1.8 * 6 = 10.8 kg Verification / Alternative check: Proportion check: W2/W1 = L2/L1 = 6/13. Thus W2 = 23.4 * (6/13) = 23.4 * 0.461538… = 10.8 kg (matches the direct unitary calculation). Why Other Options Are Wrong: 7.2, 9.0: underestimate the proportional weight. 12.4, 18.0: overshoot the proportional weight for 6 m. Common Pitfalls: Dividing 23.4 by 6 instead of by 13, or mixing metres and kilograms. Always compute per-unit length correctly before scaling to the target length. Final Answer: 10.8 kg

Question 10

Scaling up from a given batch price: If 357 mangoes cost Rs 1517.25, what is the approximate cost of 49 dozen (588) mangoes?

Accepted Answer

Introduction / Context: We are given a total price for a specific count of mangoes and asked to estimate the price for another, larger count. The key is to find a clean per-item price and then scale. Conveniently, the numbers here divide exactly, minimizing rounding error. Given Data / Assumptions: 357 mangoes cost Rs 1517.25 Target = 49 dozen = 49 * 12 = 588 mangoes Choose the closest option to the calculated amount Concept / Approach: Unitary method: find price per mango, then multiply by 588. Since 1517.25/357 simplifies nicely, we get an exact per-item price and, hence, an exact scaled total before rounding to the nearest option. Step-by-Step Solution: Cost per mango = 1517.25 / 357 = 4.25Cost for 588 mangoes = 588 * 4.25 = 588 * (4 + 0.25) = 2352 + 147 = 2499 Verification / Alternative check: Since the computed amount is Rs 2499 exactly, the nearest option is Rs 2500 (difference Rs 1). Other options are substantially further away, so rounding sensibly selects Rs 2500. Why Other Options Are Wrong: Rs. 2600, Rs. 3000: overestimates by 101 and 501 respectively. Rs. 2200, Rs. 2400: underestimate by 299 and 99 respectively. Common Pitfalls: Forgetting that 49 dozen = 588 or attempting to compute per dozen directly without first confirming the per-unit price. The clean 4.25 per mango makes scaling straightforward. Final Answer: Rs. 2500

Question 11

All-in cost per thousand bricks: Cost of 7000 bricks is Rs 5740. Luggage/transport to the site costs Rs 805. What is the total cost per 1000 bricks (including transport)?

Accepted Answer

Introduction / Context: This is an average cost question: we combine purchase cost and transport cost to get a total, then convert to a per-1000 basis using the unitary method. Such problems appear frequently in procurement and construction cost estimation contexts. Given Data / Assumptions: Material cost for 7000 bricks = Rs 5740 Transport cost (luggage) = Rs 805 Total bricks = 7000 Concept / Approach: Total cost = purchase + transport. Then, cost per thousand = (total cost / total bricks) * 1000. Because 7000 is a multiple of 1000, division is simple. Step-by-Step Solution: Total cost = 5740 + 805 = Rs 6545Per 1000 bricks = 6545 / 7 = Rs 935 Verification / Alternative check: Reverse check: Rs 935 per thousand * 7 = Rs 6545 (matches total). The arithmetic is exact—no rounding needed. Why Other Options Are Wrong: Rs. 900, Rs. 800, Rs. 750, Rs. 980: Do not equal 6545/7 exactly and reflect either under- or over-estimates. Common Pitfalls: Forgetting to include the transport cost, or computing per 100 instead of per 1000. Always convert using the correct scaling factor matching the asked unit basis. Final Answer: Rs. 935

Question 12

Packaging and transport capacity: Each box holds 5 dozen toys and 98 boxes fit in one tempo. How many tempos are needed to lift 29,400 toys in one round?

Accepted Answer

Introduction / Context: This is a capacity and batching problem. We convert “dozens” to units, find the number of toys per box, then compute toys per tempo and finally the number of tempos required for the total. Careful stepwise conversion prevents miscounts. Given Data / Assumptions: 1 box = 5 dozen toys = 60 toys 1 tempo holds 98 boxes Total toys to move = 29,400 Concept / Approach: First, compute toys per tempo: (toys/box) * (boxes/tempo). Then, divide the overall requirement by per-tempo capacity. Since the numbers divide cleanly, no rounding ambiguity arises. Step-by-Step Solution: Toys per box = 5 * 12 = 60Toys per tempo = 60 * 98 = 5880Tempos needed = 29400 / 5880 = 5 Verification / Alternative check: Back-check: 5 tempos * 5880 toys/tempo = 29,400 toys, matching the target exactly, confirming the division was exact and no extra tempo is necessary. Why Other Options Are Wrong: 4, 6, 7, 8: These would transport fewer or more toys than required in one round. Only 5 meets the exact total without deficit or surplus. Common Pitfalls: Using 5 toys = 1 dozen (reversing the definition) or forgetting to multiply by 98 boxes per tempo. Always convert dozens properly (1 dozen = 12). Final Answer: 5

Question 13

Bulk furniture costing: Each chair costs Rs 214 and each table costs Rs 937. Find the approximate total cost of 6 dozen chairs and 4 dozen tables.

Accepted Answer

Introduction / Context: This problem scales unit prices to bulk quantities expressed in dozens. We compute the totals for chairs and tables separately and then add them. Because the exact value lies close to a round number, an “approximate” choice among thousands is appropriate. Given Data / Assumptions: Chair price = Rs 214; quantity = 6 dozen = 72 chairs Table price = Rs 937; quantity = 4 dozen = 48 tables Choose the nearest thousand-level option Concept / Approach: Compute each subtotal via unitary method, sum them, and round to the nearest option. The arithmetic is straightforward and exact before rounding to the provided coarse choices. Step-by-Step Solution: Chairs total = 72 * 214 = 15408Tables total = 48 * 937 = 44976Grand total = 15408 + 44976 = 60384 Verification / Alternative check: 60384 is within 384 of 60000 and 1616 below 62000; it is closest to Rs 60000 among the provided options, so 60000 is the appropriate approximate pick. Why Other Options Are Wrong: Rs. 58000, Rs. 55000: underestimate the total. Rs. 62000: farther from 60384 than 60000 is. Rs. 57000: still further away than Rs. 60000. Common Pitfalls: Confusing dozen counts or mis-multiplying 937 * 48. Breaking 48 as 50 − 2 helps: 937*50 − 937*2 = 46850 − 1874 = 44976. Final Answer: Rs. 60000

Question 14

Work-rate scaling (pumps and time): Three pumps working 8 hours/day empty a tank in 2 days. How many hours/day should four pumps work to empty the same tank in 1 day (same efficiency)?

Accepted Answer

Introduction / Context: Work completed is proportional to (number of pumps) * (hours per day) * (days), assuming constant efficiency. We match total pump-hours between scenarios to keep the completed work equal and solve for the unknown daily hours. Given Data / Assumptions: Scenario 1: 3 pumps * 8 h/day * 2 days Scenario 2: 4 pumps * H h/day * 1 day Same tank, same efficiency Concept / Approach: Equate total pump-hours in both scenarios. This linear direct-proportion model is the foundation of most pump/work/time aptitude questions and is identical to man-hour logic in work-rate problems. Step-by-Step Solution: Scenario 1 pump-hours = 3 * 8 * 2 = 48Scenario 2: 4 * H * 1 = 4HEquate: 4H = 48 ⇒ H = 12 Verification / Alternative check: With 4 pumps for 12 hours, total pump-hours = 48, matching the first scenario. Therefore, the same amount of work (the tank) is completed in 1 day. Why Other Options Are Wrong: 10, 8, 9, 15: Do not yield 48 pump-hours when multiplied by 4, so the total work would be underdone or overdone relative to the tank’s requirement. Common Pitfalls: Forgetting to multiply by the number of days, or trying to average hours without accounting for the change in pump count. Always use total pump-hours equivalence. Final Answer: 12

Question 15

Equivalent man-hours across scenarios: 39 persons repair a road in 12 days at 5 h/day. In how many days will 30 persons at 6 h/day complete the same work?

Accepted Answer

Introduction / Context: The quantity of work is proportional to total man-hours = persons * hours/day * days. If the work is the same, equate the man-hours of the two scenarios and solve for the unknown number of days in the second scenario. This is a direct application of the unitary method for work-rate problems. Given Data / Assumptions: Scenario A: 39 persons, 12 days, 5 h/day Scenario B: 30 persons, D days, 6 h/day Same work; constant efficiency Concept / Approach: Man-hours A = Man-hours B ⇒ 39*12*5 = 30*D*6. Solve for D algebraically. Avoid canceling prematurely to prevent arithmetic slips. Step-by-Step Solution: 39*12*5 = 2340 man-hours30*6*D = 180DEquate: 2340 = 180D ⇒ D = 2340 / 180 = 13 Verification / Alternative check: Back-check: 30*6*13 = 2340 man-hours, which equals the original requirement, so the same work is completed. Why Other Options Are Wrong: 9, 12, 14, 10: These values do not balance man-hours to 2340 under 30 workers at 6 h/day. Common Pitfalls: Confusing proportionality (fewer workers at more hours) or dropping a factor of hours/day. Always compute total man-hours explicitly. Final Answer: 13

Question 16

Generalized work formula with parameters: “P men” working “P hours/day” complete “P units” in “P days”. How much work is done by “Q men” at “Q hours/day” in “Q days” (in terms of P and Q)?

Accepted Answer

Introduction / Context: This algebraic work-rate problem asks you to deduce per man-hour productivity from one scenario and then apply it to another. It tests understanding that total work equals (men) * (hours/day) * (days) * (productivity per man-hour). Given Data / Assumptions: P men at P h/day for P days produce P units Find work by Q men at Q h/day for Q days Productivity is constant across scenarios Concept / Approach: From the P-scenario, compute productivity k (units per man-hour). Then compute work for the Q-scenario using the same k. Finally, simplify the expression purely in terms of P and Q to match an option. Step-by-Step Solution: P men * P h/day * P days * k = P units⇒ P^3 * k = P ⇒ k = 1 / P^2For Q men: Work = Q men * Q h/day * Q days * k = Q^3 * (1/P^2) = Q^3 / P^2 Verification / Alternative check: Dimensional check: if P = Q, expression gives P^3/P^2 = P, consistent with the original statement; hence the form is correct. Why Other Options Are Wrong: P2Q2, P2Q, Q2P2, PQ: These combine powers improperly and fail the sanity check when P = Q. Common Pitfalls: Assuming work is simply proportional to men*hours*days without factoring in k from the first scenario, or mis-simplifying exponents. Keep the per man-hour productivity explicit. Final Answer: Q3/P2 units

Question 17

Food equivalence (men vs children): A camp has food for 120 men or 200 children (same duration). If 150 children have taken the meal, how many men can be catered with the remaining food?

Accepted Answer

Introduction / Context: This problem uses equivalence of consumption between men and children. If total food is sufficient for 120 men (or 200 children) for the same period, we can convert children-consumed food into the “men” equivalent and then subtract from the total men-capacity to find how many men can still be served. Given Data / Assumptions: Total food = capacity for 120 men = capacity for 200 children 150 children have eaten Consumption rates are constant over the considered period Concept / Approach: Equate consumption rates: 120 men ≡ 200 children ⇒ 1 man ≡ 200/120 = 5/3 children ⇒ 1 child ≡ 3/5 man. Convert 150 children to men-equivalents and subtract from 120 to get remaining men-capacity. Step-by-Step Solution: Men-equivalent of 1 child = 3/5 manMen-equivalent of 150 children = 150 * 3/5 = 90 menRemaining men-capacity = 120 − 90 = 30 men Verification / Alternative check: Children remaining capacity: 200 − 150 = 50 children. Convert 50 children back to men: 50 * 3/5 = 30 men—consistent from either direction. Why Other Options Are Wrong: 56, 45, 36, 27: These do not match the consumption equivalence computed from the given ratio. Common Pitfalls: Inverting the equivalence (treating 1 child ≡ 5/3 men) or forgetting that the total food corresponds to the same time period for both groups. Final Answer: 30

Question 18

Man-hours balancing: 10 men working 6 h/day complete a job in 18 days. How many hours/day must 15 men work to finish the job in 12 days?

Accepted Answer

Introduction / Context: For constant efficiency, work done equals total man-hours. We use the first scenario to get the total man-hours required, and then solve for the daily hours in the second scenario so that the man-hours match the same total work. Given Data / Assumptions: Scenario A: 10 men, 6 h/day, 18 days Scenario B: 15 men, H h/day, 12 days Same job; efficiencies unchanged Concept / Approach: Equate man-hours required: 10*6*18 = 15*H*12. Solve for H (hours/day) algebraically. This is the standard unitary-method application in work-rate problems. Step-by-Step Solution: Total man-hours = 10*6*18 = 1080Set 15*H*12 = 1080 ⇒ 180H = 1080H = 1080 / 180 = 6 hours/day Verification / Alternative check: Back-check: 15 * 6 * 12 = 1080, same as Scenario A, so the job can be finished in the new schedule with 6 hours/day. Why Other Options Are Wrong: 4, 3, 5, 8: These produce total man-hours different from 1080, so the job would be under- or over-scheduled. Common Pitfalls: Forgetting to multiply by days, or mis-canceling factors when solving the proportion. Keep all factors visible until the final step. Final Answer: 6

Question 19

Scaling examiners’ effort: 9 examiners can examine a certain number of answer books in 12 days by working 5 h/day. How many hours/day should 4 examiners work to examine twice that number in 30 days?

Accepted Answer

Introduction / Context: This is a proportional scaling of man-hours with a changed team size and target workload (twice the original). We equate required man-hours using the same per hour productivity and solve for the daily hours needed in the new scenario. Given Data / Assumptions: Original: 9 examiners, 12 days, 5 h/day ⇒ base workload W New: 4 examiners, 30 days, H h/day ⇒ target workload 2W Efficiency per examiner-hour is constant Concept / Approach: Man-hours scale linearly with workload. Therefore, 9*12*5 corresponds to W; for 2W, we require twice that many examiner-hours distributed as 4*30*H. Solve for H. Step-by-Step Solution: Base man-hours = 9*12*5 = 540Needed for double work = 2*540 = 1080Equation: 4*30*H = 1080 ⇒ 120H = 1080 ⇒ H = 9 Verification / Alternative check: Back-check: 4 * 30 * 9 = 1080 man-hours, exactly double the original 540, so twice the workload is feasible in 30 days. Why Other Options Are Wrong: 11, 13, 17, 7: These values do not yield 1080 man-hours with 4 examiners over 30 days. Common Pitfalls: Forgetting to double the original workload; or equating 9*12*5 directly to 4*30*H without the factor of 2, which would understate hours/day. Final Answer: 9

Question 20

Food stock after departures (man-days logic): A fort had food for 150 men for 45 days. After 10 days, 25 men left. For how many more days will the remaining food last?

Accepted Answer

Introduction / Context: This problem uses the concept of man-days: total food stock can be represented as “men × days”. After some days of consumption, the remaining stock can be reallocated among a new number of men to find how long it will last. This is a staple of work-and-time style aptitude problems. Given Data / Assumptions: Total initial capacity: 150 men for 45 days ⇒ 150*45 man-days Consumption for first 10 days by 150 men After 10 days, remaining men = 150 − 25 = 125 Concept / Approach: Compute total man-days initially, subtract man-days already consumed, and then divide what remains by the new daily consumption (125 men per day) to get remaining days of supply. Step-by-Step Solution: Initial stock = 150 * 45 = 6750 man-daysConsumed in 10 days = 150 * 10 = 1500 man-daysRemaining stock = 6750 − 1500 = 5250 man-daysRemaining days = 5250 / 125 = 42 days Verification / Alternative check: Had nobody left, days left after 10 days would be (6750 − 1500)/150 = 35 days. Fewer men remaining (125) increases the duration: 5250/125 = 42, consistent with intuition. Why Other Options Are Wrong: 39, 37, 32, 56: Do not match the computed man-day balance. Common Pitfalls: Subtracting 10 from 45 directly without accounting for the departure of 25 men, or forgetting to convert everything into man-days before recomputing duration. Final Answer: 42

Unitary Method Questions