Question 1

In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days will one cow eat one bag?

Accepted Answer

Given data 40 cows consume 40 bags in 40 days. Concept / Approach Use the work-rate idea: consumption is proportional to (number of cows) × (days). Step-by-step calculationLet r = bags consumed per cow per day.Total bags = (cows) × (days) × r ⇒ 40 = 40 × 40 × rr = 40 / 1600 = 1/40 bag per cow per dayTime for 1 cow to consume 1 bag = 1 ÷ (1/40) = 40 days Final Answer40 days

Question 2

36 men can finish a job in 18 days. In how many days will 27 men complete the same work (same efficiency)?

Accepted Answer

Concept / Approach Assuming equal efficiency, Work = Men × Days (constant for the same job). Step-by-step calculation36 × 18 = 27 × DD = (36 × 18) ÷ 27 = (4/3) × 18 = 24 Quick checkFewer men (27 < 36) ⇒ more days; 24 > 18, consistent. Final Answer24 days

Question 3

Three pumps, each working 8 hours per day, can empty a tank in 2 days. How many hours per day must four pumps work to empty the tank in 1 day?

Accepted Answer

Given data 3 pumps × 8 h/day × 2 days empty the tank Find hours/day for 4 pumps to finish in 1 day Concept / Approach Total work (emptying the tank) is proportional to the total pump-hours. So, (number of pumps) × (hours per day) × (days) = constant for the same tank. Step-by-step calculation Total pump-hours needed = 3 × 8 × 2 = 48 pump-hoursLet required hours/day for 4 pumps be H. Then 4 × H × 1 = 48H = 48 ÷ 4 = 12 hours/day Verification 4 pumps × 12 h = 48 pump-hours, same as original requirement. Common pitfalls Adding or averaging times; this is a direct work (rate × time) problem. Final Answer 12 hours/day

Question 4

Four mat-weavers weave 4 mats in 4 days. At the same rate, how many mats will 8 mat-weavers weave in 8 days?

Accepted Answer

Given data 4 weavers → 4 mats in 4 days Find mats made by 8 weavers in 8 days at the same rate Concept / Approach Mats are proportional to (number of weavers) × (days), assuming identical constant rates. Step-by-step calculation Baseline output per (weaver·day): k = 4 mats ÷ (4 weavers × 4 days) = 4/16 = 0.25For 8 weavers in 8 days: mats = 0.25 × 8 × 8 = 16 Alternative check Scaling factor = (8×8)/(4×4) = 64/16 = 4 ⇒ 4 mats × 4 = 16 mats. Common pitfalls Adding weavers and days instead of multiplying. Final Answer 16 mats

Question 5

A flagstaff 17.5 m high casts a 40.25 m shadow. Under similar conditions, what is the height of a building that casts a 28.75 m shadow?

Accepted Answer

Given data Flagstaff: height = 17.5 m, shadow = 40.25 m Building: shadow = 28.75 m; find its height Concept / Approach With the same sun elevation, triangles are similar ⇒ height is proportional to shadow length. Step-by-step calculation Height / Shadow = constantBuilding height = 17.5 × (28.75 / 40.25)= 17.5 × 0.713422… = 12.5 m Verification 17.5/40.25 ≈ 0.4348, 12.5/28.75 ≈ 0.4348 (matches). Common pitfalls Using inverse ratio (40.25/28.75) erroneously. Final Answer 12.5 m

Question 6

An industrial loom weaves 0.128 m of cloth per second. Approximately how many seconds are needed to weave 25 m of cloth?

Accepted Answer

Given data Speed = 0.128 m/s Distance (cloth length) = 25 m Concept / Approach Time = Distance / Speed, keeping units consistent (metres, seconds). Step-by-step calculation t = 25 / 0.128 = 195.3125 sApproximately = 195 s (nearest second) Verification 0.128 × 195 ≈ 24.96 m (very close to 25 m); rounding is appropriate for 'approximately'. Common pitfalls Inverting the fraction (using 0.128/25). Converting to minutes inadvertently (units must remain seconds). Final Answer 195 s

Question 7

If 7 spiders make 7 webs in 7 days, then one spider will make one web in how many days?

Accepted Answer

Given data 7 spiders → 7 webs in 7 days. Concept / Approach Work is proportional to spider-days. Compute rate per spider-day. Step-by-step calculation Total spider-days = 7 spiders × 7 days = 49 spider-daysWebs made = 7 ⇒ rate = 7 / 49 = 1/7 web per spider-dayFor 1 spider to make 1 web: days = 1 ÷ (1/7) = 7 days Verification 1 spider in 7 days produces 1 web at 1/7 per day; scales consistently. Common pitfalls Assuming 1 day by symmetry; must consider rate via spider-days. Final Answer 7 days

Question 8

A camp has food for 120 men or 200 children. If 150 children have eaten, how many men can be fed with the remaining food?

Accepted Answer

Given data Food equivalent: 120 men ≡ 200 children 150 children have consumed their share Concept / Approach Use equivalence: 1 child = (120/200) man = 3/5 man. Convert remaining children-equivalents to men. Step-by-step calculation Remaining children-equivalents = 200 − 150 = 50 childrenConvert to men: 50 × (3/5) = 30 men Verification Alternatively, per child consumption = 1/200 of stock; leftover = 50/200 = 1/4 stock; 1/4 of 120 men = 30. Common pitfalls Using inverse factor 5/3 instead of 3/5 when converting children to men. Final Answer 30 men

Question 9

A wheel with 6 cogs meshes with a larger wheel with 14 cogs. When the smaller wheel makes 21 revolutions, how many revolutions does the larger wheel make?

Accepted Answer

Given data Small wheel: 6 cogs Large wheel: 14 cogs Small wheel revolutions = 21 Concept / Approach Teeth engagements are equal at the mesh. Total teeth passed by small wheel = total teeth passed on large wheel's rim. Revolutions are inversely proportional to the number of cogs. Step-by-step calculation Teeth passed by small wheel = 21 × 6 = 126 teethRevolutions of large wheel = 126 ÷ 14 = 9 Verification Inverse ratio check: revL / revS = 6/14 ⇒ revL = 21 × 6/14 = 9. Common pitfalls Using direct ratio 14/6 instead of inverse. Final Answer 9

Question 10

If the cost of x metres of wire is d rupees, what is the cost of y metres at the same rate?

Accepted Answer

Given data Cost for x m = d rupees Find cost for y m at same unit rate Concept / Approach Unit rate = (cost per metre) = d/x rupees per metre. Step-by-step calculation Cost for y m = (d/x) × y = (dy)/x rupees Verification For y = x, formula gives d rupees, consistent. Common pitfalls Using dx/y (that would be cost for x/y metres). Final Answer dy/x rupees

Question 11

Six identical machines produce 270 bottles per minute at a constant rate. At this rate, how many bottles can ten such machines produce in four minutes?

Accepted Answer

Given data 6 machines → 270 bottles/min Find output of 10 machines in 4 minutes Concept / Approach Rate per machine is constant; total rate scales linearly with number of machines. Step-by-step calculation Per-machine rate = 270 / 6 = 45 bottles/min10 machines rate = 10 × 45 = 450 bottles/minIn 4 minutes: 450 × 4 = 1800 bottles Verification Proportionality check: factor (10/6) in machines and ×4 minutes ⇒ 270 × (10/6) × 4 = 270 × (5/3) × 4 = 270 × (20/3) = 90 × 20 = 1800. Common pitfalls Multiplying by 6 again or forgetting the 4-minute duration. Final Answer 1800 bottles

Question 12

If a quarter kilogram of potatoes costs 60 paise, how many paise will 200 grams cost?

Accepted Answer

Given data 0.25 kg costs 60 paise Find cost for 200 g (= 0.2 kg) Concept / Approach Use unitary method (cost is proportional to weight). Step-by-step calculation Cost per kg = 60 ÷ 0.25 = 240 paise/kgCost for 0.2 kg = 240 × 0.2 = 48 paise Verification Proportion check: 200 g is 4/5 of 250 g; 4/5 × 60 = 48 paise. Common pitfalls Converting grams to kg incorrectly or mixing paise and rupees. Final Answer 48 paise

Question 13

A fort has food for 150 men for 45 days. After 10 days, 25 men leave. For how many days will the remaining food last?

Accepted Answer

Given data Initial capacity: food for 150 men for 45 days After 10 days at 150 men, 25 men leave ⇒ remaining men = 125 Concept / Approach Use man-days: total food = (men) × (days). Food consumed is proportional to men × time. Step-by-step calculationTotal food (man-days) = 150 × 45 = 6750Food used in first 10 days = 150 × 10 = 1500Remaining food (man-days) = 6750 − 1500 = 5250With 125 men, days left = 5250 ÷ 125 = 42 Verification42 × 125 = 5250 man-days, matching the remainder. Common pitfalls Subtracting days instead of using man-days. Forgetting that the consumption rate changes when men leave. Final Answer42 days

Question 14

A man completes 5⁄8 of a job in 10 days. At this rate, how many more days are needed to finish the job?

Accepted Answer

Given data Work done in 10 days = 5/8 of job Concept / Approach Rate = (work)/(time). Remaining work = 1 − 5/8 = 3/8. Step-by-step calculationDaily rate = (5/8) ÷ 10 = 1/16 of the job per dayTime for remaining 3/8 = (3/8) ÷ (1/16) = (3/8) × 16 = 6 Quick checkTotal time = 10 + 6 = 16 days → 16 × (1/16) = 1 job. Final Answer6 days

Question 15

39 persons can repair a road in 12 days, working 5 hours per day. In how many days will 30 persons, working 6 hours per day, complete the work?

Accepted Answer

Concept / Approach Work is proportional to (persons) × (hours per day) × (days) when efficiency is constant. Step-by-step calculationTotal work (person-hours) = 39 × 5 × 12 = 2340Daily capacity with 30 persons at 6 h/day = 30 × 6 = 180 person-hours/dayDays required = 2340 ÷ 180 = 13 Verification13 × 180 = 2340 person-hours, equal to the original work. Final Answer13 days

Question 16

In a work and time (chain rule) problem, 20 men can build a straight wall 56 meters long in 6 days. Under the same working conditions and efficiency, what length of a similar wall can be built by 35 men in 3 days?

Accepted Answer

Introduction / Context:This is a classic work and time question based on the chain rule. It checks your understanding of how the amount of work done (here, length of wall built) varies directly with the number of workers and the number of days, assuming all men work at the same constant rate. Given Data / Assumptions:- 20 men can build a wall 56 meters long in 6 days.- All men work at the same constant rate.- We want the length of a similar wall built by 35 men in 3 days under the same conditions.- Work (wall length) is directly proportional to (number of men) * (number of days). Concept / Approach:The total work done is proportional to the product of men and days. If r is the length of wall built by one man in one day, then:- Work = men * days * r.We first use the initial situation to determine the effective combined rate, and then apply that rate to the new combination of men and days to find the new length of wall. Step-by-Step Solution:Step 1: Let r be the length (in meters) built by one man in one day.Step 2: From the first condition: 20 men * 6 days * r = 56 meters.Step 3: So, 120r = 56, giving r = 56 / 120 meters per man-day.Step 4: For 35 men working 3 days, total man-days = 35 * 3 = 105.Step 5: Length built L = 105 * r = 105 * (56 / 120).Step 6: Simplify: 105 / 120 = 7 / 8, so L = 56 * (7 / 8) = 49 meters. Verification / Alternative check:We can use direct proportion: length is proportional to men and days. Multiply by factor (35 / 20) for men and (3 / 6) for days: 56 * (35 / 20) * (3 / 6) = 56 * (7 / 4) * (1 / 2) = 56 * (7 / 8) = 49 meters, confirming the same result. Why Other Options Are Wrong:46, 47 and 48 meters are all less than the correct proportional result and arise from incorrect partial scaling or arithmetic errors. 50 meters is slightly more than the true value and would require a higher work rate than given. None of these satisfy the proportional relationship exactly. Common Pitfalls:Students often forget that both men and days affect the result, or they scale only by men and ignore the change in days, or vice versa. Another mistake is to invert the ratios when applying the chain rule. Always remember that work ∝ men * days, and keep fractions in the correct orientation while simplifying. Final Answer:The required length of the similar wall is 49 meters.

Question 17

A certain number of men can finish a piece of work in 100 days. If there were 10 fewer men available, it would take 10 days more to finish the same work. How many men were there originally at the start of the work?

Accepted Answer

Introduction / Context:This work and time question tests your ability to model the relationship between number of workers, duration, and total work using simple algebra. By comparing two scenarios with different numbers of men and days, you can set up an equation to find the unknown original workforce strength. Given Data / Assumptions:- Let the original number of men be x.- With x men, the work is finished in 100 days.- With (x - 10) men, the same work takes 110 days (10 days more).- Work done is directly proportional to (number of men) * (number of days).- All men work at the same constant rate. Concept / Approach:Total work is fixed. If W is the total work and each man works at the same rate, then:- W = x * 100 (man-days) in the first case.- W = (x - 10) * 110 (man-days) in the second case.Since both expressions represent the same total work, we can equate them and solve for x, the original number of men. Step-by-Step Solution:Step 1: Express total work using the first situation: W = x * 100.Step 2: Express total work using the second situation: W = (x - 10) * 110.Step 3: Set the two expressions equal: x * 100 = (x - 10) * 110.Step 4: Expand the right-hand side: 100x = 110x - 1100.Step 5: Rearrange: 1100 = 110x - 100x = 10x.Step 6: Solve for x: x = 1100 / 10 = 110. Verification / Alternative check:With 110 men, total work W = 110 * 100 = 11,000 man-days. If there are 10 fewer men, i.e. 100 men, to do the same work, days needed = 11,000 / 100 = 110 days. This matches the condition given in the problem, confirming that 110 is correct. Why Other Options Are Wrong:If you choose 100 or 120, the resulting man-days in the second scenario will not equal the first scenario's total, breaking the equality of work. Values like 75 or 82 arise from algebraic or arithmetic mistakes, such as mis-distributing 110 or misplacing terms when equating the two expressions. Common Pitfalls:Common errors include forgetting that total work must be equal in both cases, mixing up which scenario has more days, or incorrectly simplifying the equation (for example, subtracting 110 from both sides instead of handling the x terms correctly). Writing the basic equation of man-days carefully and solving stepwise avoids these mistakes. Final Answer:The original number of men was 110.

Question 18

If 36 men can complete a piece of work in 25 hours of continuous work, then in how many hours will 15 men complete the same work, assuming all men work at the same constant rate?

Accepted Answer

Introduction / Context:This question is a straightforward application of the inverse proportionality between number of workers and time taken when the total work is fixed. It checks whether you can calculate total man-hours and then adjust the time when the workforce changes. Given Data / Assumptions:- 36 men can complete the work in 25 hours.- We need the time required if only 15 men are working.- Total work remains the same in both cases.- All men work at identical and constant efficiency. Concept / Approach:Total work can be measured in man-hours. If the work requires a certain number of man-hours, then:- Work W = men * hours.First, we compute the man-hours needed using the initial data (36 men, 25 hours). Then we divide that same number of man-hours by the new number of men (15 men) to get the new time. Step-by-Step Solution:Step 1: Calculate the total man-hours required using the first situation: W = 36 * 25.Step 2: Compute: 36 * 25 = 900 man-hours.Step 3: For 15 men, let the required time be T hours.Step 4: Then 15 * T = 900 man-hours (since total work is unchanged).Step 5: Solve for T: T = 900 / 15.Step 6: T = 60 hours. Verification / Alternative check:The time is inversely proportional to the number of men when total work is fixed. So, T2 / T1 = M1 / M2 = 36 / 15 = 12 / 5 = 2.4. Since the original time T1 is 25 hours, T2 = 25 * 2.4 = 60 hours. This matches our earlier calculation. Why Other Options Are Wrong:Values like 40 or 50 hours are too small and would imply that fewer men somehow finish the work faster, which contradicts the inverse relationship. 70 hours is larger than necessary and would represent more than the required man-hours. Only 60 hours preserves the original total of 900 man-hours. Common Pitfalls:Students might mistakenly assume a direct proportion (more men, more time) instead of the correct inverse proportion, or they may forget to hold total work constant. Miscalculations like 36 * 25 or dividing 900 incorrectly are also common. Always compute total man-hours first and then divide by the new workforce to stay consistent. Final Answer:15 men will complete the work in 60 hours.

Question 19

Two men and seven boys can complete a certain piece of work in 14 days, while three men and eight boys can complete the same work in 11 days. At the same rates of work, in how many days will eight men and six boys together complete three times this …

Accepted Answer

Introduction / Context:This is a classic men-and-boys work problem using the chain rule and simultaneous equations. The question tests your ability to convert verbal information into algebraic equations, find individual or combined work rates, and then scale the result to a multiple of the original work. Given Data / Assumptions:- 2 men and 7 boys complete 1 unit of work in 14 days.- 3 men and 8 boys complete the same 1 unit of work in 11 days.- All men have the same efficiency; all boys have the same efficiency.- We must find the time for 8 men and 6 boys to complete 3 times the original work. Concept / Approach:Let the daily work rates be:- m = work done per man per day.- b = work done per boy per day.Then the total daily work of a group is the sum of individual rates. Using the two given scenarios, we set up two equations in m and b, solve them, and then use the resulting combined rate for 8 men and 6 boys to calculate the time for three units of work. Step-by-Step Solution:Step 1: Total work W is taken as 1 unit.Step 2: From the first group: (2m + 7b) * 14 = W → 2m + 7b = W / 14.Step 3: From the second group: (3m + 8b) * 11 = W → 3m + 8b = W / 11.Step 4: Equate W from both forms: (2m + 7b) * 14 = (3m + 8b) * 11.Step 5: Expand: 28m + 98b = 33m + 88b → -5m + 10b = 0 → m = 2b.Step 6: Substitute m = 2b into W expression using the first case: 2m + 7b = 2(2b) + 7b = 11b, so W = 11b * 14 = 154b.Step 7: For 8 men and 6 boys, daily work rate = 8m + 6b = 8(2b) + 6b = 22b.Step 8: For three times the work, total work = 3W = 3 * 154b = 462b.Step 9: Time T = total work / rate = 462b / 22b = 462 / 22 = 21 days. Verification / Alternative check:You can verify that W = 154b is consistent in both original groups by checking (3m + 8b) * 11 = (6b + 8b) * 11 = 14b * 11 = 154b. This confirms the intermediate calculations and makes the final time of 21 days reliable. Why Other Options Are Wrong:Values like 18, 24, 28 or 30 days come from arithmetic slips such as using 1W instead of 3W, forgetting to multiply the original work by 3, or mis-simplifying 462 / 22. None of them keep the ratio of work to rate correct for the threefold work case. Common Pitfalls:Many students forget that the last part requires three times the original work, not just one unit. Others incorrectly solve the simultaneous equations or mis-handle the proportionality between m and b. Carefully solving for m and b and recomputing the combined rate avoids these mistakes. Final Answer:Eight men and six boys will complete three times the work in 21 days.

Question 20

Seven spiders can weave seven complete webs in seven days. Assuming all spiders work at the same constant rate, in how many days will one spider weave one web?

Accepted Answer

Introduction / Context:This question is a neat illustration of unitary method and work-rate reasoning. It appears simple at first glance but can be confusing if you only look at the numbers superficially. The key is to think in terms of total spider-days needed per web rather than trying to guess directly. Given Data / Assumptions:- 7 spiders weave 7 webs in 7 days.- All spiders are equally efficient and work continuously at the same rate.- We want the time required for 1 spider to weave 1 web. Concept / Approach:The best way to handle this is to calculate the total work and express it in spider-days. Then you can find how many spider-days are required to make a single web. Finally, you translate spider-days into days for one spider. Total work is measured in “webs”, while effort is measured in “spider-days”. Step-by-Step Solution:Step 1: Compute total spider-days used in the given situation: 7 spiders * 7 days = 49 spider-days.Step 2: The result of this effort is 7 webs. So, 7 webs require 49 spider-days.Step 3: Therefore, 1 web requires 49 / 7 = 7 spider-days.Step 4: If 1 spider is working alone, each day contributes exactly 1 spider-day.Step 5: To accumulate 7 spider-days, one spider will need 7 days.Step 6: Hence, one spider will weave one web in 7 days. Verification / Alternative check:You can also reason proportionally: the situation “7 spiders make 7 webs in 7 days” is symmetric. Intuitively, that means each spider effectively makes 1 web in 7 days, since there are 7 spiders and 7 webs. This matches the more formal spider-day calculation. Why Other Options Are Wrong:1 day and 3 days are too short and would imply each spider makes multiple webs in 7 days, contradicting the original data. 14 or 21 days are too long and would give fewer than 7 webs in 7 days if all spiders were working at those speeds. Only 7 days is consistent with the total of 7 webs from 7 spiders in 7 days. Common Pitfalls:Some learners misinterpret the statement and think “7 spiders, 7 webs, 7 days” automatically means 1 day for 1 web, forgetting that the work is shared. Others mix up proportionality, assuming work per spider per day is 1 web, which clearly contradicts the total. Using spider-days as a unit of effort helps avoid these errors. Final Answer:One spider will make one web in 7 days.

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