Question 1

Engine consumption comparison: 9 engines of type A consume 24 tonnes of coal when each runs 8 h/day. If 3 engines of type A consume as much as 4 engines of type B, how much coal is required for 8 engines of type B each running 13 h/day?

Accepted Answer

Introduction / Context: This question involves two engine types with a stated consumption equivalence. By first finding consumption per engine-hour for type A and then using the given equivalence to infer type B’s consumption, we can compute the required coal for type B in the new schedule. Given Data / Assumptions: Type A: 9 engines * 8 h/day consume 24 tonnes ⇒ per engine-hour for A is constant Equivalence: 3 A-engines consume as much as 4 B-engines (same time) Target: 8 B-engines running 13 h/day; find total coal Concept / Approach: Let a be tonnes per engine-hour for A; then 9*8*a = 24 ⇒ a = 1/3. From 3A ≡ 4B, per engine-hour for B is b = (3/4)*a = 1/4. Then compute 8*13*b for the required total consumption. Step-by-Step Solution: Type A per engine-hour a = 24 / (9*8) = 24/72 = 1/3 tonneType B per engine-hour b = (3/4) * a = (3/4) * (1/3) = 1/4 tonneTotal for 8 B-engines, 13 h: 8 * 13 * (1/4) = 26 tonnes Verification / Alternative check: Equivalent A-engines for 8 B-engines: 4B ≡ 3A ⇒ 8B ≡ 6A. Consumption via A-rate: 6 * 13 * (1/3) = 26 tonnes—same result, confirming consistency. Why Other Options Are Wrong: 20, 23: underestimate consumption given the derived 1/4 tonne per engine-hour for B. 39: overestimates; requires higher per-hour usage than implied by the equivalence. 24: still under the computed requirement of 26. Common Pitfalls: Applying 3A = 4B backwards (treating B as more consumptive), or forgetting to convert to per engine-hour before scaling. Final Answer: 26 metric tonnes

Question 2

Scaling workforce to hit a target: If 40 men make 30 boxes in 20 days, how many more men are needed to make 60 boxes in 25 days (same efficiency)?

Accepted Answer

Introduction / Context: This is a direct work-rate proportionality problem. Boxes made are proportional to men * days (assuming identical hours and efficiency). We first compute output per man-day from the first scenario and then determine required men for the new target, finally reporting “how many more” than 40 are needed. Given Data / Assumptions: Base: 40 men make 30 boxes in 20 days Target: 60 boxes in 25 days Same daily hours and same efficiency Concept / Approach: Boxes per man-day = 30 / (40*20). For target, let M men be needed so that M * 25 * (boxes per man-day) = 60. Solve for M, then subtract 40 to get “additional men”. Step-by-Step Solution: Boxes per man-day = 30 / 800 = 0.0375Let M be required men: M * 25 * 0.0375 = 60M = 60 / (25 * 0.0375) = 60 / 0.9375 = 64Additional men = 64 − 40 = 24 Verification / Alternative check: Proportion method: Men ∝ (Boxes)/(Days) with same productivity ⇒ M = 40 * (60/30) * (20/25) = 40 * 2 * 0.8 = 64. Additional = 24. Same result by a quicker ratio approach. Why Other Options Are Wrong: 20, 27, 28, 18: Do not match the computed increment from 40 to the required 64 men. Common Pitfalls: Forgetting it asks for “more men” (M − 40), or inverting a ratio (using 25/20 instead of 20/25). Keep ratio direction consistent. Final Answer: 24

Question 3

Combined men-and-boys work rates: 2 men and 7 boys finish a job in 14 days; 3 men and 8 boys finish it in 11 days. In how many days will 8 men and 6 boys finish three times this job?

Accepted Answer

Introduction / Context: This problem involves two worker types (men and boys) with different efficiencies. By forming two equations from the given completion times, we can solve for the individual daily rates, combine them for a new team, and then scale the duration for three times the original amount of work. Given Data / Assumptions: (2m + 7b) completes 1 job in 14 days ⇒ daily rate = 1/14 (3m + 8b) completes 1 job in 11 days ⇒ daily rate = 1/11 Find time for 8m + 6b to complete 3 jobs Concept / Approach: Let m = work/day by one man and b = work/day by one boy. Solve the linear system: 2m + 7b = 1/14 and 3m + 8b = 1/11. Then compute the combined rate of 8m + 6b and invert to get time for 1 job; finally multiply by 3 for three jobs. Step-by-Step Solution: From 2m + 7b = 1/14 and 3m + 8b = 1/11Subtracting (after scaling): 5b = 3/14 − 2/11 = 5/154 ⇒ b = 1/154Then 2m + 7*(1/154) = 1/14 ⇒ 2m + 1/22 = 1/14 ⇒ 2m = 4/154 ⇒ m = 1/77Rate of 8m + 6b = 8/77 + 6/154 = 16/154 + 6/154 = 22/154 = 11/77 per dayTime for 1 job = 1 / (11/77) = 7 days ⇒ Time for 3 jobs = 3 * 7 = 21 days Verification / Alternative check: Sanity: 8 men contribute 8/77 and 6 boys 6/154 per day; sum is 11/77, a clean fraction yielding 7 days per job. Scaling to triple work preserves linearity: 3 * 7 = 21 days. Why Other Options Are Wrong: 18, 34, 39, 24: Do not match the computed rate and scaled duration for triple work. Common Pitfalls: Arithmetic slips in fraction subtraction or forgetting to take the reciprocal when converting rate to time. Keep fractions aligned over a common denominator. Final Answer: 21

Question 4

Work–men–days (direct proportion): If 20 men can build a similar wall 56 meters long in 6 days, what length of such a wall can 35 men build in 3 days?

Accepted Answer

Introduction / Context: This is a classic unitary-method (work–time) question. For similar work, total output is directly proportional to the product men * days, assuming all men work at the same constant rate. We use this proportionality to scale from one scenario to another. Given Data / Assumptions: 20 men in 6 days build 56 meters of wall. We need the length built by 35 men in 3 days (same wall type and productivity per man). Rates are constant and additive across workers. Concept / Approach: Total work (length of wall) ∝ men * days. Hence, length2 = length1 * (men2 * days2) / (men1 * days1). Compute once and select the matching option. Step-by-Step Solution: Known output: 56 m for (20 men * 6 days) = 120 man-days.Required scenario: (35 men * 3 days) = 105 man-days.Per man-day output = 56 / 120 = 0.4667 m.Length for 105 man-days = 105 * 0.4667 ≈ 49 m. Verification / Alternative check: Use direct ratio: length2 = 56 * (35*3)/(20*6) = 56 * 105 / 120 = 56 * 0.875 = 49 m. Same result via either method. Why Other Options Are Wrong: 40, 43, 48: All underestimate the proportional increase from 120 to 105 man-days relative to the 56 m benchmark; only 49 fits the precise scaling. Common Pitfalls: Dividing by the wrong base (e.g., scaling by men only and forgetting days) or assuming inverse proportion. Here, output grows with both men and days. Final Answer: 49

Question 5

Meshing gears (inverse revolutions): A 6-cog wheel meshes with a larger 14-cog wheel. When the smaller wheel makes 21 revolutions, how many revolutions does the larger wheel make?

Accepted Answer

Introduction / Context: For meshed gears, revolutions are inversely proportional to the number of cogs (teeth). Fewer cogs imply more turns to pass the same number of teeth, and vice versa. We apply inverse proportion to map revolutions between the two wheels. Given Data / Assumptions: Smaller wheel: 6 cogs; Larger wheel: 14 cogs. Smaller wheel makes 21 revolutions. No slipping; perfect meshing; same tooth engagement count. Concept / Approach: rev_small * cogs_small = rev_large * cogs_large. Solve for rev_large using the product equality of engaged teeth counts. Step-by-Step Solution: rev_large = rev_small * cogs_small / cogs_large= 21 * 6 / 14 = 126 / 14 = 9 Verification / Alternative check: Tooth counts: small engages 21 * 6 = 126 teeth; for the big gear to see 126 teeth with 14 per revolution, it must make 126/14 = 9 revolutions. Why Other Options Are Wrong: 4 or 7: Undercount the necessary turns to match 126 engaged teeth. 14: Would over-rotate the larger gear compared to meshing constraints. Common Pitfalls: Using direct proportion (wrong here) instead of inverse proportion. Always remember: more cogs ⇒ fewer revolutions for a fixed tooth transfer. Final Answer: 9

Question 6

Manpower adjustment on a fixed deadline: A contractor plans to finish a job in 40 days. He employs 100 men initially and adds 100 more after 35 days, finishing on time. If he had not added extra men, how many days late would the work be?

Accepted Answer

Introduction / Context: This is a work–time problem converting varying workforce over time into total man-days. Once the total man-days for completion are known, we can compute the time if only the original crew worked throughout and compare with the target to find the delay. Given Data / Assumptions: Target duration = 40 days. First 35 days: 100 men. Last 5 days: 200 men (100 original + 100 additional). Uniform productivity; man-days add linearly. Concept / Approach: Total work W equals the sum of man-days used in the actual schedule. Then compute how long 100 men alone would take: time_alone = W / 100. The delay = time_alone − 40, if positive. Step-by-Step Solution: Total work W = (100 men * 35 days) + (200 men * 5 days)= 3500 + 1000 = 4500 man-days.If only 100 men worked: time = 4500 / 100 = 45 days.Delay relative to 40-day target = 45 − 40 = 5 days late. Verification / Alternative check: Check proportionality: replacing the final 5 days of 100 men with 200 men adds 500 extra man-days, exactly the shortfall that would have caused a 5-day delay at 100 man-days/day. Why Other Options Are Wrong: 7, 12, 18: Do not match the precise calculation showing a 5-day delay. None: Correct here because 5 is not listed among the numerical choices. Common Pitfalls: Forgetting that “additional 100 men” means total 200 in the last 5 days, or miscomputing total man-days. Always aggregate man-days and then divide by the constant crew size for alternative schedules. Final Answer: None

Question 7

Inverse men–days relation: 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work (same rate)?

Accepted Answer

Introduction / Context: With constant productivity, work done equals men * days. For a fixed job, men and days are inversely proportional. Reducing workers increases the required days proportionally. Given Data / Assumptions: 36 men finish in 18 days. Find days for 27 men at same rate. Concept / Approach: Use men1 * days1 = men2 * days2. Solve for days2 with the given numbers. Step-by-Step Solution: 36 * 18 = 27 * days2days2 = (36 * 18) / 27 = (36/27) * 18 = (4/3) * 18 = 24 Verification / Alternative check: Total man-days = 36 * 18 = 648. With 27 men, days = 648 / 27 = 24. Same result. Why Other Options Are Wrong: 23, 34, 39: Do not preserve the constant man-days 648 for the fixed job. Common Pitfalls: Using direct instead of inverse proportion between men and days for fixed work. Keep the product men * days constant for the given job. Final Answer: 24

Question 8

Scaling workforce and scope: Some persons can complete a job in 12 days. If the number of persons is doubled, how long will it take to complete half of that job?

Accepted Answer

Introduction / Context: For fixed productivity, time varies directly with work and inversely with manpower. Here both the team size and the amount of work change, so we combine both effects carefully. Given Data / Assumptions: Original: N persons finish the full job in 12 days. New plan: 2N persons, and only half of the original job. Concept / Approach: Total work W = N * 12 (person-days). Half work is W/2. With 2N persons, time t = (W/2) / (2N) = W / (4N). Substitute W = 12N to solve. Step-by-Step Solution: W = N * 12.t = (W/2) / (2N) = (12N/2) / (2N) = 6 / 2 = 3 days. Verification / Alternative check: Check proportionality: doubling persons halves the time; halving the work halves the time again. Starting from 12 days: 12/2/2 = 3 days. Why Other Options Are Wrong: 16, 12, 8: Ignore the compounded effect of both changes (work halved and manpower doubled). Common Pitfalls: Applying only one of the two changes or adding/dividing in the wrong order. Always express in person-days to avoid confusion. Final Answer: 3

Question 9

Same job, different crew size: If 36 men can do a piece of work in 25 hours, in how many hours will 15 men do the same work (same rate)?

Accepted Answer

Introduction / Context: Work done is men * hours for a fixed job. With fewer men, the required hours increase so that the product remains constant. We compute total man-hours and then divide by the new crew size. Given Data / Assumptions: 36 men complete in 25 hours. New crew: 15 men. Concept / Approach: Total man-hours = men * hours = constant for the job. So hours2 = (36 * 25) / 15. Step-by-Step Solution: Total man-hours = 36 * 25 = 900.Hours for 15 men = 900 / 15 = 60 hours. Verification / Alternative check: Proportionality: reducing men by a factor of 36/15 = 2.4 increases time by 2.4; 25 * 2.4 = 60. Why Other Options Are Wrong: 45, 55, 72: Do not preserve the constant 900 man-hours for completion. Common Pitfalls: Multiplying instead of dividing by the new crew size or mixing units. Keep the men * hours product equal to the known job total. Final Answer: 60

Unitary Method Questions