Question 1

If the total price of 23 identical toys is Rs. 276, then, assuming the same rate per toy, what will be the total price of 12 such toys?

Accepted Answer

Introduction: This is a direct proportion question in basic arithmetic. We know the total price for 23 toys, and the toys are identical. Therefore the price for any other number of toys can be found by first computing the unit price and then multiplying by the required quantity. Such questions commonly appear under the topic of ratio and proportion or unitary method. Given Data / Assumptions: Total price of 23 toys = Rs. 276. All 23 toys are identical and have the same price. We must find the total price of 12 toys at the same rate per toy. Concept / Approach: When items are sold at a constant price per unit, total cost is: total cost = number of units * cost per unit. We can find cost per unit by: cost per toy = total cost / number of toys. Then multiply this unit cost by 12 to get the price of 12 toys. Step-by-Step Solution: Step 1: Compute the price of one toy. cost per toy = 276 / 23. 23 * 10 = 230 and 23 * 2 = 46, so 23 * 12 = 276. Therefore, 276 / 23 = 12. So each toy costs Rs. 12. Step 2: Find the price of 12 toys. Price of 12 toys = 12 * 12. 12 * 12 = 144. Thus, the price of 12 toys is Rs. 144. Verification / Alternative check: We can also verify by scaling down and then up. If 23 toys cost Rs. 276, then 1 toy costs Rs. 12 as found. If we imagine 24 toys, they would cost 24 * 12 = 288, which is slightly more than 276, confirming that our per toy price is consistent with simple multiples of 23 and 12. The calculation is exact with no rounding. Why Other Options Are Wrong: Rs. 148 and Rs. 156 would correspond to per toy prices slightly greater than Rs. 12, which does not match the original information that 23 toys cost exactly Rs. 276. Rs. 186 would mean each toy costs more than Rs. 15 when multiplied by 12, clearly inconsistent. Rs. 132 would imply a cost per toy of Rs. 11, which would make 23 toys cost 23 * 11 = 253, not Rs. 276. Common Pitfalls: A frequent error is to try to scale directly from 23 to 12 without first finding the unit price, leading to complicated fractions or mistakes. Another mistake is miscalculating 276 / 23 under time pressure. Recognising that 23 * 12 equals 276 allows a quick mental calculation and avoids division errors. Final Answer: The total price of 12 toys at the same rate is Rs. 144.

Question 2

If 20 men can build a wall 56 metres long in 6 days, what length of a similar wall can be built by 35 men in 3 days?

Accepted Answer

Introduction / Context:This is a typical chain rule and work problem. It tests the idea that the amount of work done is directly proportional to the number of workers, the time for which they work and, in this problem, the length of wall constructed. Such questions are common in quantitative aptitude, and a clear understanding of proportional reasoning makes them straightforward to solve.Given Data / Assumptions: Twenty men build a 56 metre wall in 6 days. The work rate of each man is assumed to be constant throughout. The wall built by 35 men in 3 days is of the same type, so work is directly proportional to the length. We must find the length of wall built by 35 men in 3 days. Concept / Approach:Work done is proportional to the product of men, days and length. We can write:men * days * rate is proportional to length.Another neat method is to write that the length is directly proportional to the product men * days when other conditions are identical. Then we can set up a proportion between the known and unknown situations and solve for the missing length.Step-by-Step Solution:Step 1: Compute the effective work input in the first case, which is 20 men working for 6 days.Step 2: In the second case, we have 35 men working for 3 days.Step 3: Since length is directly proportional to men * days, we write: 56 / L = (20 * 6) / (35 * 3).Step 4: Simplify the right hand side: (20 * 6) = 120 and (35 * 3) = 105, so the ratio is 120 / 105 = 8 / 7.Step 5: Thus 56 / L = 8 / 7, which implies L = 56 * 7 / 8 = 7 * 7 = 49 metres.Verification / Alternative check:We can check reasonableness. The product of men and days initially is 20 * 6 = 120. In the new case it is 35 * 3 = 105, which is slightly less. So the new length should be slightly less than 56 metres. However, notice that the proportionality relation was inverted in our ratio, so we ensure the algebra is correct. Directly, we can write L2 = L1 * (men2 * days2) / (men1 * days1) = 56 * (35 * 3) / (20 * 6) = 56 * 105 / 120 = 49. This gives the same result and confirms the calculation.Why Other Options Are Wrong:46 metres would correspond to a different men days ratio and does not satisfy the proportional relation between work inputs.47 metres also fails when substituted back into the men days proportion and is simply a distractor near the correct value.48 metres is closer but still does not satisfy the exact ratio that comes from the product of men and days.Common Pitfalls:A frequent mistake is to invert the proportion incorrectly or to confuse directly proportional with inversely proportional relationships. Another common error is to mix up which product of men and days belongs in the numerator and which in the denominator. Writing the relation clearly as new length = old length * (new men * new days) / (old men * old days) avoids confusion and leads quickly to the correct answer.Final Answer:The length of the wall built by 35 men in 3 days is 49 metres.

Question 3

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

Accepted Answer

Introduction / Context:This is a classic time and work problem using the concept that total work equals the product of the number of workers and the duration when each worker has a constant rate. The question compares two situations with different numbers of men and different durations for the same total work. It is an excellent example of how to set up and solve simple equations based on proportional reasoning.Given Data / Assumptions: A certain number of men, say M, can finish a piece of work in 100 days. If there are 10 fewer men, that is M - 10 men, the work takes 110 days. The nature of work does not change between the two scenarios. Each man works at a constant rate throughout the job. Concept / Approach:The total work can be written as men * days * individual rate. Since the kind of work remains the same, the total amount of work is constant across both scenarios. Thus, we can equate the work expressions for the two cases. The individual rate cancels out, leaving an equation involving M, the original number of men, and the given durations. Solving this equation yields M.Step-by-Step Solution:Step 1: Let the original number of men be M.Step 2: In the first situation, the total work W can be expressed as W = M * 100 * r, where r is the rate of one man.Step 3: In the second situation, the same work W is done by M - 10 men in 110 days, so W = (M - 10) * 110 * r.Step 4: Equate the two expressions for W and cancel r: M * 100 = (M - 10) * 110.Step 5: Expand the right hand side: M * 100 = 110M - 1100. Rearranging gives 110M - 100M = 1100, so 10M = 1100 and M = 110.Verification / Alternative check:We can verify by plugging M = 110 back into both scenarios. With 110 men, the work takes 100 days, corresponding to 110 * 100 = 11000 man days. With 100 men, because 10 have been removed, the work takes 110 days, corresponding to 100 * 110 = 11000 man days again. Both give the same total man days, confirming that the number of men originally was 110.Why Other Options Are Wrong:Seventy five men would give 75 * 100 = 7500 man days in the first case and 65 * 110 = 7150 in the second, which do not match.Eighty two men would produce 82 * 100 = 8200 and 72 * 110 = 7920 man days, which are unequal.One hundred men would give 100 * 100 = 10000 and 90 * 110 = 9900 man days, again inconsistent with a fixed total work.Common Pitfalls:Many students misplace the terms and end up with an incorrect equation such as M * 110 = (M - 10) * 100, which reverses the given condition. Others forget to cancel the common individual rate r, making the algebra seem more complicated than it is. Carefully translating the language into an equation where total work is equal in both cases is the key skill in solving this problem correctly.Final Answer:The original number of men was 110.

Question 4

If 36 men can complete a piece of work in 25 hours, in how many hours will 15 men complete the same work, assuming the work rate is uniform?

Accepted Answer

Introduction / Context:This question is a time and work problem based on inverse proportionality between number of workers and time taken, when total work is fixed and each worker has a constant rate. Such problems commonly appear in aptitude tests to test understanding of how changing the number of workers affects the completion time for the same task.Given Data / Assumptions: Thirty six men can complete the work in 25 hours. We need to find how many hours 15 men will take to complete the same work. All men work at the same constant rate. The total amount of work remains unchanged between the two situations. Concept / Approach:When work is fixed, the relationship between number of workers and time is inversely proportional. That means men * time is constant for a fixed task if work rate per man does not change. Thus, we can relate the two situations by equating the products of men and time. Solving this equation gives the required time for the second group of men.Step-by-Step Solution:Step 1: Let T be the number of hours for which 15 men must work to complete the job.Step 2: In the first case, the total man hours are 36 * 25.Step 3: In the second case, the total man hours are 15 * T.Step 4: Since total work is the same, set the products equal: 36 * 25 = 15 * T.Step 5: Compute 36 * 25 = 900. Then T = 900 / 15 = 60 hours.Verification / Alternative check:We can check proportionality. If the number of men is reduced from 36 to 18, halving the workforce, time should double from 25 to 50 hours. Reducing further from 18 to 9 should again double the time to 100 hours. Since 15 lies between 18 and 9, a time of 60 hours is reasonable and fits the exact algebraic result. A direct calculation of man hours confirms that 36 men for 25 hours and 15 men for 60 hours both contribute 900 man hours, which must correspond to the same total work.Why Other Options Are Wrong:Forty hours would give 15 * 40 = 600 man hours, which is less than the required 900, so the work would remain incomplete.Fifty hours would produce 15 * 50 = 750 man hours, still short of the needed 900.Seventy hours would give 15 * 70 = 1050 man hours, which exceeds the total work and implies extra unused capacity.Common Pitfalls:Some candidates incorrectly treat the relationship as directly proportional and set up equations like 36 / 15 = 25 / T, which reverses the logic. Others forget to check dimensions and end up with non sensible results. Writing the relationship clearly as men1 * time1 = men2 * time2 for the same work is the easiest way to avoid these errors and quickly arrive at the correct answer.Final Answer:Fifteen men will complete the work in 60 hours.

Question 5

Two men and seven boys can do a piece of work in 14 days, while three men and eight boys can do the same work in 11 days. In how many days can eight men and six boys together complete three times this work?

Accepted Answer

Introduction / Context:This is a multi concept time and work problem involving two different types of workers, men and boys, with different daily work rates. The question first presents two conditions that allow us to find the relative efficiencies of a man and a boy. It then asks how long a new combination of men and boys will take to complete three times the original work. Such questions test algebraic translation, proportional reasoning and careful simplification.Given Data / Assumptions: Two men and seven boys can finish one unit of work in 14 days. Three men and eight boys can finish the same unit of work in 11 days. We must find the time taken when eight men and six boys work together to do three times this unit of work. Each man has a constant daily rate of work, and each boy also has a constant daily rate, possibly different from that of a man. Concept / Approach:We introduce variables for daily work rates. Let the daily work done by one man be M units and by one boy be B units. Then the total work in the first case is (2M + 7B) * 14 and in the second case is (3M + 8B) * 11. Since both equal the same amount of work, we can equate them to find the ratio between M and B. After obtaining M and B up to a common multiple, we compute the total amount of work and the daily work of the group of eight men and six boys. Finally, we find the time required for three times the original work.Step-by-Step Solution:Step 1: Express total work W using the first condition: W = (2M + 7B) * 14.Step 2: Express the same work W using the second condition: W = (3M + 8B) * 11.Step 3: Set these equal: (2M + 7B) * 14 = (3M + 8B) * 11.Step 4: Expand both sides: 28M + 98B = 33M + 88B. Rearranging gives 98B - 88B = 33M - 28M, so 10B = 5M and M = 2B.Step 5: Substitute M = 2B into W = (2M + 7B) * 14. Then 2M + 7B = 2 * 2B + 7B = 11B, so W = 11B * 14 = 154B. Three times the work is 3W = 462B. The daily work rate of eight men and six boys is 8M + 6B = 8 * 2B + 6B = 22B. The required time is T = 3W / (8M + 6B) = 462B / 22B = 21 days.Verification / Alternative check:As a check, we can see that if 2 men and 7 boys do the work in 14 days, their combined daily work is W / 14. If 3 men and 8 boys do it in 11 days, their combined daily work is W / 11. The second combination does more work per day, consistent with one extra man and one extra boy. Our algebra identified that a man is equivalent to twice a boy in daily work. When eight men and six boys work together, their rate is 22B per day, and 462B units of work divided by 22B again gives 21 days, reinforcing the earlier result.Why Other Options Are Wrong:Eighteen days would require a higher daily rate than 22B, implying either more workers or more efficient workers than given in the problem.Twenty four days would imply a smaller combined daily rate and would not be consistent with the efficiencies derived from the original conditions.Thirty days would be even slower and clearly contradicts the computed work rate of the group of eight men and six boys.Common Pitfalls:Students sometimes try to apply chain rule directly without setting up variables for the work rates. This often leads to confusion when multiple types of workers are involved. Another common mistake is algebraic: errors when expanding and simplifying the equation that relates M and B. Writing each step carefully, especially when equating total work from two scenarios, prevents these mistakes and leads to a clean solution.Final Answer:Eight men and six boys will take 21 days to complete three times the given work.

Question 6

If 7 spiders make 7 webs in 7 days, in how many days will 1 spider make 1 web, assuming all spiders work at the same constant rate?

Accepted Answer

Introduction / Context:This interesting time and work problem is framed with spiders and webs instead of men and work. The underlying concept is exactly the same: total work is proportional to the number of workers and the time they spend working. Here, we need to interpret the given information correctly to find the time needed for a single spider to produce a single web.Given Data / Assumptions: Seven spiders make seven webs in seven days. All spiders work at the same constant rate and independently. The nature of each web is identical so that the work per web is the same. We must find how many days one spider will take to make one web. Concept / Approach:First, we determine the rate at which webs are produced per spider per day. The information that seven spiders make seven webs in seven days tells us the total output of the group over that period. Using this, we can derive the work rate of one spider and then compute the time required for one spider to complete one web. This is essentially a unitary method problem.Step-by-Step Solution:Step 1: Find the total number of spider days in the given situation. Seven spiders working for seven days produce 7 * 7 = 49 spider days of work.Step 2: The total number of webs produced in that time is given as 7.Step 3: Therefore, the production rate is 7 webs per 49 spider days, which simplifies to 1 web per 7 spider days.Step 4: One spider day is defined as one spider working for one day. Thus, one spider taking 7 spider days means that the same spider works for 7 days to produce one web.Step 5: Hence, one spider will complete one web in 7 days.Verification / Alternative check:We can interpret the original statement in another way. If each spider produces webs independently at the same rate, then in seven days each spider must have produced exactly one web for the total of seven webs to be achieved by seven spiders. Therefore, by simple reasoning, one spider in seven days produces one web, which is consistent with the unitary method result. This double check reinforces the conclusion.Why Other Options Are Wrong:One day would suggest that one spider can produce one web in one day. That would mean seven spiders in seven days would produce 49 webs, which contradicts the given information.Three days would imply a higher individual production rate than is supported by the data. Multiplying back would not yield seven webs in seven days for seven spiders.Fourteen days would imply that a spider works more slowly than indicated by the given scenario, because the group output would be too small if each web took 14 days for one spider.Common Pitfalls:Many students misread the problem and think it describes a situation where 7 spiders together build a single web in 7 days. In fact, the wording clearly states that seven spiders build seven webs. Another source of confusion is mixing up spider days and human intuition about how quickly a spider spins a web. Focusing only on the numerical relationships in the problem avoids these distractions.Final Answer:One spider will make one web in 7 days.

Question 7

A contractor undertakes to complete a piece of work in 40 days. He engages 100 men at the beginning and, after 35 days, engages 100 more men so that the work is finished in time. If he had not engaged the additional men, how many days behind schedul…

Accepted Answer

Introduction / Context:This time and work question explores how adding more workers partway through a project affects completion time. We are asked to compute how much delay would have occurred if these extra workers had not been engaged. Such problems illustrate the concept of total man days and show how to compare different workforce allocations for the same total work.Given Data / Assumptions: The contractor plans to finish the work in 40 days. Initially, he employs 100 men from day 1. After 35 days, he hires 100 additional men, making 200 men in total for the last 5 days. With this arrangement, the work finishes exactly in 40 days. We need to find how many days late the work would have been if the additional 100 men had never been hired. Concept / Approach:The central idea is that total work can be measured in man days, which equal the number of men multiplied by the number of days they work. We first compute the total man days actually used with the extra men. That gives the total amount of work required. Next, we find how many days 100 men alone would need to provide that many man days. The difference between this time and the scheduled 40 days gives the delay.Step-by-Step Solution:Step 1: Calculate man days contributed in the first 35 days: 100 men * 35 days = 3500 man days.Step 2: For the remaining 5 days, 200 men work, so their contribution is 200 * 5 = 1000 man days.Step 3: Total man days required to complete the work are 3500 + 1000 = 4500 man days.Step 4: If only 100 men were employed from the beginning to the end, let T be the total number of days needed. Then 100 * T = 4500, so T = 4500 / 100 = 45 days.Step 5: The planned duration was 40 days. With only 100 men, the work would take 45 days, which is 5 days longer than scheduled.Verification / Alternative check:We can sanity check by comparing average workforce. For the first 35 days, there are 100 men, and for the last 5 days, 200 men. The effective average over 40 days is higher than 100 men, so using only 100 men throughout should naturally take more than 40 days. A difference of 5 days is reasonable. Also, if we think numerically, 100 men working 45 days provide 4500 man days, exactly matching the total work already computed, verifying our algebraic solution.Why Other Options Are Wrong:Three days would correspond to a total of 43 days with 100 men, giving only 4300 man days, which is insufficient to complete the work that actually required 4500 man days.Six days would mean 46 days with 100 men, implying 4600 man days, which is more than required for the project.Nine days would lead to 49 days of work with 100 men, or 4900 man days, which is much greater than the required 4500 man days and clearly not minimal.Common Pitfalls:Many learners forget to add the contributions of both phases correctly or misinterpret the question and try to compute the completion time with 200 men from the start. Another error is failing to convert the question into total man days and instead attempting to apply chain rule mechanically. Using total man days keeps the reasoning organised and helps avoid these mistakes.Final Answer:If no additional men had been engaged, the work would have been finished 5 days behind schedule.

Question 8

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

Accepted Answer

Introduction / Context:This is a chain rule problem involving changes in both the number of workers and daily working hours. The total work is the same in both scenarios. The question checks understanding of how work output scales with both the number of persons and the number of hours per day, and how to combine these changes correctly to determine the required time.Given Data / Assumptions: Thirty nine persons can repair the road in 12 days, working 5 hours each day. We need to find the number of days for 30 persons to complete the same work if they work 6 hours each day. Each person works at a constant rate, and the work done per hour is uniform. The total amount of repair work is unchanged between the two cases. Concept / Approach:The total work can be measured in person hours. In the first scenario, we compute total person hours as persons * days * hours per day. In the second scenario, we set up an expression for person hours using the unknown number of days. Since total work is constant, the two totals must be equal. Solving for the unknown number of days gives the required time.Step-by-Step Solution:Step 1: Compute total person hours in the first case: 39 persons * 12 days * 5 hours per day.Step 2: The total is 39 * 12 * 5 = 39 * 60 = 2340 person hours.Step 3: Let D be the number of days required for 30 persons working 6 hours per day.Step 4: Total person hours in the second case are 30 * D * 6 = 180D.Step 5: Equate total person hours: 180D = 2340. Solving gives D = 2340 / 180 = 13 days.Verification / Alternative check:We can interpret the situation proportionally. Compare the effective daily work. In the first case, daily person hours equal 39 * 5 = 195. In the second case, daily person hours equal 30 * 6 = 180. The new team does slightly less work per day than the first team. The ratio of daily work is 195 / 180 = 13 / 12. Hence time should increase in the same ratio from 12 days to 12 * 13 / 12 = 13 days. This proportional reasoning agrees exactly with the earlier calculation.Why Other Options Are Wrong:Ten days would correspond to 30 * 10 * 6 = 1800 person hours, which is too small compared to the required 2340 person hours.Fourteen days would provide 30 * 14 * 6 = 2520 person hours, which is more than required and implies extra unused capacity.Fifteen days would yield 30 * 15 * 6 = 2700 person hours, even more above the total work and therefore not the minimal correct value.Common Pitfalls:Students often forget to include the hours per day when computing total work and only compare persons and days. Another mistake is flipping ratios incorrectly when applying chain rule formulas. Thinking in terms of total person hours gives a clear physical picture and helps to set up the correct equation easily.Final Answer:Thirty persons working 6 hours a day will complete the work in 13 days.

Question 9

Three pumps working 8 hours a day can empty a tank in 2 days. How many hours per day must 4 pumps work to empty the same tank in 1 day?

Accepted Answer

Introduction / Context:This is a classic chain rule problem involving pumps, time and working hours. It requires understanding that the volume of water pumped is proportional to the number of pumps, the number of hours they run each day and the number of days for which they work. The total volume of the tank remains constant between the two scenarios.Given Data / Assumptions: Three pumps running 8 hours per day can empty the tank in 2 days. We must find the number of hours per day that 4 pumps should run to empty the same tank in 1 day. All pumps have identical constant pumping rates. The capacity of the tank and the rate at which each pump works do not change between the two cases. Concept / Approach:The total volume emptied can be measured in pump hours. In the first case, we know the total pump hours used. In the second case, we express total pump hours in terms of the unknown number of hours per day. Because the tank volume is the same, the total pump hours for both cases must be equal. Solving this equality yields the required daily hours for the second case.Step-by-Step Solution:Step 1: Compute total pump hours in the first case: 3 pumps * 8 hours per day * 2 days.Step 2: This gives 3 * 8 * 2 = 48 pump hours.Step 3: Let H be the number of hours per day that 4 pumps must run to empty the tank in 1 day.Step 4: Total pump hours in the second case are 4 pumps * H hours per day * 1 day = 4H.Step 5: Equate total pump hours: 4H = 48. Solving gives H = 48 / 4 = 12 hours.Verification / Alternative check:We can reason proportionally. Going from 3 pumps to 4 pumps increases pumping capacity by a factor of 4 / 3. To finish in 1 day instead of 2 days, we need to do the same work in half as many days. Combining these adjustments, daily operating hours must change so that overall pump hours remain 48. Running 4 pumps for 12 hours gives 4 * 12 * 1 = 48 pump hours, which matches the first case. Thus the answer is consistent and reasonable.Why Other Options Are Wrong:Nine hours would give 4 * 9 = 36 pump hours in one day, which is not enough to empty the tank that requires 48 pump hours.Ten hours would give 4 * 10 = 40 pump hours, still below the required 48.Eleven hours would give 4 * 11 = 44 pump hours, which remains short of the total work required to empty the tank.Common Pitfalls:Some students confuse days and hours and may mistakenly think that because the number of pumps increased, the hours can be kept unchanged without checking total pump hours. Others may attempt to scale both pumps and days without verifying the net effect. Writing the product pumps * hours per day * days explicitly for each scenario is the most reliable way to avoid confusion and reach the correct result.Final Answer:Four pumps must work 12 hours a day to empty the tank in 1 day.

Question 10

Eighteen binders can bind 900 books in 10 days. How many binders will be required to bind 660 books in 12 days, assuming each binder works at the same rate?

Accepted Answer

Introduction / Context:This work and time problem asks how many workers are required to complete a smaller amount of work over a longer period when the rate of each worker is constant. The situation involves binders and books instead of generic workers and tasks, but the mathematics is the same. Understanding proportional relationships makes such questions quite manageable.Given Data / Assumptions: Eighteen binders can bind 900 books in 10 days. Each binder works at a constant daily rate. We need to find the number of binders required to bind 660 books in 12 days. The nature of the books and binding work does not change between the two scenarios. Concept / Approach:First, we determine the rate at which one binder works, expressed as books per binder per day. Using this rate, we can find how many binder days are needed for 660 books and then divide by the available 12 days to get the required number of binders. This solution uses the unitary method and basic proportional reasoning.Step-by-Step Solution:Step 1: Compute the total binder days in the first situation: 18 binders * 10 days = 180 binder days to bind 900 books.Step 2: Find the binding rate of one binder per day. One binder day therefore corresponds to 900 / 180 = 5 books per binder per day.Step 3: For 660 books, determine total binder days needed. Since each binder day handles 5 books, total binder days required are 660 / 5 = 132 binder days.Step 4: Let N be the number of binders required to finish 660 books in 12 days. Then N * 12 = 132 binder days.Step 5: Solve for N: N = 132 / 12 = 11 binders.Verification / Alternative check:An alternative method is to use direct proportion. The amount of work in the second scenario as a fraction of the first is 660 / 900 = 11 / 15. The time available changes from 10 days to 12 days. Since work is proportional to men * days, we have men2 = men1 * (work2 / work1) * (time1 / time2) = 18 * (11 / 15) * (10 / 12). Simplifying, we get 18 * 11 / 15 * 10 / 12 = 11, which matches the unitary method result.Why Other Options Are Wrong:Twenty two binders would double the workforce, providing 22 * 12 = 264 binder days, which is much more than the 132 binder days needed.Fourteen binders would yield 14 * 12 = 168 binder days, more than necessary and therefore not minimal.Thirteen binders would supply 13 * 12 = 156 binder days, again exceeding the required 132 binder days.Common Pitfalls:Students sometimes mistakenly treat the relationship as direct without accounting for both the change in number of books and the change in days. Others may attempt to plug values into an incorrect chain rule formula without seeing the simple logic behind binder days. Focusing on the basic idea that total work equals rate times time helps structure the reasoning and avoid calculation errors.Final Answer:Eleven binders are required to bind 660 books in 12 days.

Question 11

36 men can complete a piece of work in 18 days. If all men work at the same constant rate, in how many days will 27 men complete the same work?

Accepted Answer

Introduction / Context:This question tests the basic concept of work and time when the amount of work is fixed and the number of workers changes. It is a standard direct and inverse proportion or chain rule problem that appears frequently in aptitude exams, project planning, and productivity based questions. Given Data / Assumptions: 36 men can complete the entire work in 18 days. All men work at the same constant rate. The total work remains the same. We need the time taken by 27 men to complete the same work. Concept / Approach:The total amount of work can be represented in man days. This is the product of the number of men and the number of days. When more men work, the task finishes faster, and when fewer men work, it takes longer. Mathematically, for a fixed work, number of men and number of days are inversely proportional. Step-by-Step Solution:Step 1: Compute total work in man days: Work = 36 * 18.Step 2: 36 * 18 = 648 man days.Step 3: Let the required number of days for 27 men be D.Step 4: Since work is constant, 27 * D = 648.Step 5: Solve for D: D = 648 / 27 = 24 days. Verification / Alternative check:We can also use the inverse proportion shortcut. Time is inversely proportional to number of men, so D2 = D1 * (M1 / M2) = 18 * (36 / 27). The ratio 36 / 27 simplifies to 4 / 3. Thus D2 = 18 * (4 / 3) = 18 * 4 / 3 = 24 days, which matches the detailed calculation. Why Other Options Are Wrong:18 days would mean that reducing the number of men has no effect, which contradicts inverse proportion. 20 and 22 days are less than 24 and would imply that fewer men finish faster, which is impossible for identical workers. 27 days is greater than 24 but does not satisfy the correct proportional relationship obtained from man day calculations. Common Pitfalls: Multiplying instead of dividing when using the ratio of men, leading to incorrect time. Forgetting that work is constant and trying to adjust both men and days independently. Doing mental arithmetic incorrectly when dividing 648 by 27. Final Answer:The 27 men will complete the same work in 24 days.

Question 12

A quarter kilogram of carrots costs 60 paise. At the same rate, how many paise will 200 grams of carrots cost?

Accepted Answer

Introduction / Context:This question is a simple unitary method or chain rule problem involving proportional cost with respect to weight. The idea is that if you know the cost of a certain quantity, you can find the cost of any other quantity of the same item at the same rate. Such problems are very common in shopping, billing, and basic arithmetic aptitude. Given Data / Assumptions: Cost of 1/4 kilogram of carrots (that is, 250 grams) is 60 paise. The price per gram is constant. We need to find the cost of 200 grams of carrots. Concept / Approach:Since cost is directly proportional to weight, we can either find the cost per gram and multiply by 200, or use a simple ratio between 250 grams and 200 grams. The unitary method says that if cost for some quantity is known, cost for 1 unit is total cost divided by quantity, and then we multiply by required quantity. Step-by-Step Solution:Step 1: A quarter kilogram is 250 grams.Step 2: Cost for 250 grams = 60 paise.Step 3: Cost per gram = 60 / 250 paise.Step 4: Required cost for 200 grams = (60 / 250) * 200.Step 5: Compute: 60 * 200 / 250 = 60 * (4 / 5) = 48 paise. Verification / Alternative check:We can also use a direct ratio. If 250 grams cost 60 paise, then 200 grams is 200 / 250 of this weight, which is 4 / 5 of 250 grams. Therefore cost for 200 grams is 4 / 5 of 60 paise, which is again 48 paise. Both approaches confirm the same value. Why Other Options Are Wrong:40 and 42 paise are below the correct proportional value and would correspond to a lower rate than stated. 52 paise is slightly above the correct value and would give a higher price per gram. 60 paise is the cost for 250 grams, not for 200 grams, so it cannot be correct for the smaller quantity. Common Pitfalls: Using 200 / 250 incorrectly as 5 / 4 instead of 4 / 5. Forgetting to convert quarter kilogram into grams and directly comparing 1/4 and 200 without a common unit. Rounding too early instead of keeping exact fractions until the last step. Final Answer:The cost of 200 grams of carrots is 48 paise.

Question 13

In an arithmetic progression, the sum of the 3rd term and the 15th term is equal to the sum of the 6th, 11th and 13th terms. Which term of the progression must be equal to zero?

Accepted Answer

Introduction / Context:This question involves properties of arithmetic progressions and how specific terms can be related using algebra. You are asked to use the definition of general terms of an arithmetic progression and a condition on the sum of several terms, and then deduce which term must be zero. This is a classic reasoning pattern for arithmetic progression questions in aptitude and entrance exams. Given Data / Assumptions: We have an arithmetic progression with first term a and common difference d. Sum of the 3rd and 15th terms equals the sum of the 6th, 11th and 13th terms. The terms are indexed in the standard way: Tn = a + (n - 1) * d. We must find which numbered term of the progression is necessarily equal to zero. Concept / Approach:We express each mentioned term in terms of a and d. Then we use the given equality of sums to obtain an equation relating a and d. From that equation we find a linear relation between a and d. Finally, we determine for which term number k the expression a + (k - 1) * d becomes zero. Step-by-Step Solution:Step 1: Write required terms: T3 = a + 2d, T15 = a + 14d.Step 2: Their sum is T3 + T15 = (a + 2d) + (a + 14d) = 2a + 16d.Step 3: Write other terms: T6 = a + 5d, T11 = a + 10d, T13 = a + 12d.Step 4: Their sum is T6 + T11 + T13 = (a + 5d) + (a + 10d) + (a + 12d) = 3a + 27d.Step 5: Given that the two sums are equal, so 2a + 16d = 3a + 27d.Step 6: Rearrange: 0 = a + 11d, so a = -11d.Step 7: General term Tn = a + (n - 1)d. Substitute a = -11d: Tn = -11d + (n - 1)d = (n - 12) * d.Step 8: For Tn to be zero with d not equal to zero, we require n - 12 = 0, thus n = 12. Verification / Alternative check:If T12 is zero, then Tn = (n - 12) * d. You can quickly check that T3 + T15 equals T6 + T11 + T13 when you express each as multiples of d. This confirms that the 12th term must be zero whenever the condition holds, independent of the actual value of d (except d not equal to zero). Why Other Options Are Wrong:The 7th, 9th, 14th or 15th terms would require different values of n and would make Tn zero at a different index, which would break the derived relation a = -11d. With any of those indices, the equality between the two sums would not hold for a non zero progression. Common Pitfalls: Writing incorrect indices for the terms, for example mixing T3 with a + 3d instead of a + 2d. Forgetting that the common difference d cannot be zero if the problem is to be meaningful. Trying to plug in arbitrary numbers instead of using the general term form and algebra. Final Answer:The term that must be equal to zero is the 12th term of the arithmetic progression.

Question 14

Some persons can complete a piece of work in 12 days. If twice as many persons work at the same rate, in how many days will they complete half of that work?

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Introduction / Context:This problem checks understanding of how work, number of workers, and time are related when the total work changes. It uses the idea of man days and is a typical chain rule situation where you adjust the number of persons and the fraction of work to compute the new time required. Given Data / Assumptions: Some fixed number of persons can finish the entire work in 12 days. Twice that number of persons are now available. They are asked to complete only half of the original work. All persons work at the same constant rate. Concept / Approach:First, represent the total work in terms of man days using the original group. Then compute how many man days are needed for half of that work. Finally, divide the required man days by the new number of persons to get the time. Because work is proportional to the product of persons and days, we can use proportional reasoning to move quickly between scenarios. Step-by-Step Solution:Step 1: Let the original number of persons be P.Step 2: Total work in man days = P * 12.Step 3: Half of this work requires (P * 12) / 2 = 6P man days.Step 4: The new workforce is 2P persons.Step 5: Time required = required man days / number of persons = 6P / (2P) = 3 days. Verification / Alternative check:Observe that you are doubling the workforce but doing only half the work. Doubling workers alone would halve the time from 12 days to 6 days. Doing only half the work again halves the time from 6 days to 3 days. This reasoning also leads directly to 3 days without explicit algebra. Why Other Options Are Wrong:2 days would require more workers or an even smaller fraction of work. 4 or 6 days imply that the effect of more workers and less work has not been fully accounted for. 12 days ignores both changes and is just the original time for full work with the original team. Common Pitfalls: Forgetting that the work is halved and only adjusting for the increased number of persons. Mixing direct and inverse proportion incorrectly when several factors change. Trying to invent an actual number for P and then making arithmetic mistakes instead of using symbols. Final Answer:The doubled group needs 3 days to complete half of the original work.

Question 15

In a queue, Shankar is ninth from the rear end and Althaf is eighth from the front. Nitu is standing somewhere strictly between Shankar and Althaf. What is the minimum possible number of boys standing in the queue?

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Introduction / Context:This is a positions in a queue or ordering question. It tests understanding of how positions counted from the front and from the rear relate to the total number of people, and how to ensure that a third person can stand between two given positions. Questions like this are common in verbal and logical reasoning sections. Given Data / Assumptions: Shankar is 9th from the rear end. Althaf is 8th from the front. Nitu stands somewhere strictly between Shankar and Althaf. We want the minimum possible total number of boys in the queue. Concept / Approach:Let the number of boys in the queue be N. Positions from the front and rear must be consistent: position from front of a person is N minus position from rear plus 1. We use this relationship for Shankar. Then we use the requirement that Nitu must have at least one position strictly between Althaf and Shankar, which gives a minimum distance between their positions and thus a minimum value of N. Step-by-Step Solution:Step 1: Let N be the total number of boys.Step 2: Shankar is 9th from the rear, so his position from the front is N - 9 + 1 = N - 8.Step 3: Althaf is 8th from the front, so his position from the front is 8.Step 4: Nitu is strictly between them, so there must be at least one position between 8 and N - 8.Step 5: For at least one integer between 8 and N - 8, we need (N - 8) - 8 >= 2.Step 6: This simplifies to N - 16 >= 2, so N >= 18.Step 7: For N = 18, Shankar is at position 10 from the front and Althaf at position 8, so Nitu can stand at position 9, which is strictly between them. Verification / Alternative check:If you try N = 17, then Shankar would be at position 17 - 8 = 9 from the front, while Althaf is at 8. There is no integer position strictly between 8 and 9, so Nitu cannot stand between them. Therefore N = 18 is indeed the smallest possible value that allows a position for Nitu in the middle. Why Other Options Are Wrong:Values 16 or 17 are too small and do not leave any gap between Althaf and Shankar for Nitu. Values like 20 or 22 are possible but not minimal; they satisfy the condition but the question specifically asks for the minimum possible number of boys. Common Pitfalls: Forgetting to convert Shankar position from rear to a position from front using the correct formula. Allowing Nitu to be at the same position as one of the boys instead of strictly between them. Assuming that the total is simply the sum of the two positions without considering overlap. Final Answer:The minimum possible number of boys in the queue is 18.

Question 16

A father has Rs. 2000 and wants to distribute this amount among his five children so that each child receives Rs. 30 more than the next younger child. What is the share of the youngest child?

Accepted Answer

Introduction / Context:This question involves distribution of money among children in arithmetic progression. Each older child receives a fixed amount more than the next younger child, and the total is known. Such problems are good practice for arithmetic progressions and basic algebraic equations used in aptitude exams. Given Data / Assumptions: Total money to be distributed = Rs. 2000. There are 5 children. Each child gets Rs. 30 more than the immediately younger child. The youngest child gets the smallest share. Concept / Approach:The shares form an arithmetic progression because the difference between consecutive terms is constant. If we let the youngest child share be x, then the shares of the others are x + 30, x + 60, x + 90 and x + 120. We sum this progression, set equal to the total, and solve for x. Step-by-Step Solution:Step 1: Let the youngest child share be x rupees.Step 2: Then the second youngest gets x + 30, then x + 60, x + 90 and the oldest gets x + 120.Step 3: Total money distributed = x + (x + 30) + (x + 60) + (x + 90) + (x + 120).Step 4: Combine like terms: total = 5x + (30 + 60 + 90 + 120) = 5x + 300.Step 5: Set equal to 2000: 5x + 300 = 2000.Step 6: Subtract 300: 5x = 1700, so x = 1700 / 5 = 340. Verification / Alternative check:Check the individual shares: youngest 340, then 370, 400, 430 and oldest 460. Sum = 340 + 370 + 400 + 430 + 460 = 2000. This matches the total amount available, confirming that the distribution is consistent and that the youngest child share is correct. Why Other Options Are Wrong:Shares like Rs. 280 or Rs. 310 would give totals smaller than 2000 when you build the progression. Rs. 360 or Rs. 400 as shares for the youngest child would produce a sum larger than 2000 for the five children when you add the extra steps of 30 rupees each. Common Pitfalls: Placing x for the middle child instead of the youngest child, which shifts the progression and gives a wrong equation. Adding the constant difference incorrectly and miscomputing the fixed sum 30 + 60 + 90 + 120. Making a calculation error when dividing 1700 by 5. Final Answer:The share of the youngest child is Rs. 340.

Question 17

If 8 men can reap 80 hectares of a field in 24 days, then how many hectares can 36 men reap in 30 days at the same rate of working?

Accepted Answer

Introduction / Context:This is a classic work and chain rule question where both the number of workers and the time change, and you are asked to find the new quantity of work done. The scenario is framed in terms of men reaping hectares, but the underlying logic is about proportionality among men, days and total work. Given Data / Assumptions: 8 men reap 80 hectares in 24 days. All men work at the same constant rate. The number of men increases to 36. The number of days is 30. We need to find how many hectares can be reaped in this new situation. Concept / Approach:Total work done is proportional to the product of number of men, number of days and the per man per day rate. From the first situation, we can compute the rate in hectares per man per day. Then we multiply this rate by the new number of men and new number of days to get the total area that can be reaped. Step-by-Step Solution:Step 1: Compute total man days in the first case: 8 men * 24 days = 192 man days.Step 2: These 192 man days produce 80 hectares.Step 3: Rate of work per man per day = 80 / 192 hectares.Step 4: In the new case, we have 36 men and 30 days, so man days = 36 * 30 = 1080.Step 5: Total area reaped = 1080 * (80 / 192) hectares.Step 6: Simplify: 1080 / 192 = 5.625, and 5.625 * 80 = 450 hectares. Verification / Alternative check:Use proportional reasoning directly. When we change from 8 men to 36 men, the workforce is multiplied by 36 / 8 = 4.5. The time increases from 24 days to 30 days, a factor of 30 / 24 = 1.25. Combined factor = 4.5 * 1.25 = 5.625. Multiply the original 80 hectares by 5.625 to obtain 450 hectares, which matches the detailed calculation. Why Other Options Are Wrong:350, 400 or 425 hectares all assume a smaller combined growth factor than the correct value 5.625. 500 hectares would require a higher factor and thus overestimates the productivity of the new team and duration. Common Pitfalls: Forgetting to change both men and days together, and adjusting for only one of them. Using an incorrect ratio like 8 / 36 instead of 36 / 8, which inverts the effect. Doing arithmetic with large numbers without simplification, leading to calculation errors. Final Answer:The 36 men can reap 450 hectares in 30 days.

Question 18

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

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Introduction / Context:This question deals with meshed gears or toothed wheels, a practical application of inverse proportion. When two cog wheels mesh, the number of teeth and the number of revolutions they make are inversely proportional, because the same number of teeth must pass through the point of contact. This concept is used in mechanical design and appears in aptitude tests under chain rule or ratio topics. Given Data / Assumptions: The small wheel has 6 cogs (teeth). The large wheel has 14 cogs. The wheels are meshed without slipping. The small wheel makes 21 complete revolutions. We need the number of revolutions of the large wheel. Concept / Approach:When gears mesh, teeth passing a given point are the same for both wheels, so total teeth moved = number of revolutions * number of cogs. For the same contact, the product of cogs and revolutions is equal for both wheels. From this we can see that revolutions are inversely proportional to the number of cogs: more teeth mean fewer revolutions for the same tooth travel. Step-by-Step Solution:Step 1: Let R1 be the revolutions of the small wheel and R2 for the large wheel.Step 2: Number of cogs on the small wheel = 6, on the large wheel = 14.Step 3: Teeth moved by small wheel = R1 * 6.Step 4: Teeth moved by large wheel = R2 * 14.Step 5: Since they are meshed, R1 * 6 = R2 * 14.Step 6: Given R1 = 21, so 21 * 6 = R2 * 14, which gives R2 = (21 * 6) / 14.Step 7: Simplify: 21 / 7 = 3, 14 / 7 = 2, so R2 = (3 * 6) / 2 = 18 / 2 = 9 revolutions. Verification / Alternative check:Use direct inverse proportion. Revolutions are in the ratio of the opposite number of cogs: R1 : R2 = 14 : 6. Since R1 is 21, we set 21 : R2 = 14 : 6. Cross multiply to get 21 * 6 = 14 * R2, which again leads to R2 = 9. This quick ratio approach confirms the detailed calculation. Why Other Options Are Wrong:Values like 4 or 6 are too small and do not satisfy the equality of tooth travel. Values like 12 or 14 are too large and would correspond to fewer teeth per wheel than actually given. Only 9 revolutions matches the correct inverse proportion relationship. Common Pitfalls: Assuming revolutions are directly proportional to the number of cogs instead of inversely proportional. Multiplying the number of cogs instead of forming the correct ratio. Simple arithmetic mistakes in simplifying (21 * 6) / 14. Final Answer:The larger wheel makes 9 revolutions when the smaller wheel makes 21 revolutions.

Question 19

A fort has a stock of food that can feed 150 men for 45 days. After 10 days, 25 men leave the fort. For how many more days will the remaining food last for the men who stay?

Accepted Answer

Introduction / Context:This is a typical work and resources problem involving food supply and varying number of people. The total amount of food is fixed, and men consume it at a constant rate. When some men leave, the daily consumption changes, which affects how long the remaining supply lasts. This is a classic chain rule scenario in aptitude tests. Given Data / Assumptions: Food can feed 150 men for 45 days initially. All men consume food at the same constant rate. After 10 days, 25 men leave, so 125 men remain. We must find the additional number of days the remaining food will last. Concept / Approach:The total quantity of food can be expressed in man days, which is the product of the number of men and the number of days they can be fed. First, we calculate the total food in man days. Then we subtract what has already been consumed in the first 10 days. The remaining man days are divided by the new number of men to find how many more days the food will last. Step-by-Step Solution:Step 1: Total food in man days = 150 * 45 = 6750 man days.Step 2: In the first 10 days, 150 men consume 150 * 10 = 1500 man days of food.Step 3: Remaining food in man days = 6750 - 1500 = 5250 man days.Step 4: After 10 days, 25 men leave, so remaining men = 150 - 25 = 125.Step 5: Remaining days the food lasts = 5250 / 125.Step 6: 5250 / 125 = 42 days. Verification / Alternative check:You can check by recombining the whole period. Men consume food for 10 days with 150 men and then for 42 days with 125 men. Total consumption = 150 * 10 + 125 * 42 = 1500 + 5250 = 6750 man days, exactly the initial food supply. So the calculation is consistent. Why Other Options Are Wrong:36 or 38 days would give total consumption below the available 6750 man days. 45 or 54 days would mean the remaining men consume more than the stored food, which is impossible without extra supply. Common Pitfalls: Forgetting to subtract the food already consumed during the first 10 days. Using the original 150 men instead of 125 men when computing the remaining days. Confusing man days with simple days and trying to compare days directly without accounting for number of men. Final Answer:The remaining food will last for 42 days after the 25 men leave.

Question 20

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

Accepted Answer

Introduction / Context:This is a symbolic unitary method question that tests your ability to work with algebraic expressions in place of numerical values. Instead of concrete numbers, the length and the cost are expressed using variables. You must derive the general formula for the cost of a different length of wire at the same rate per metre. Given Data / Assumptions: Cost of x metres of wire = d rupees. The cost per metre is constant. We are asked to find the cost of y metres of wire. All variables x, y and d are positive real numbers. Concept / Approach:Cost is directly proportional to length when the rate per metre is fixed. First we compute the cost per metre by dividing the total cost by the total length. Then we multiply this rate by the new length y to find the new total cost. The process is identical to numerical unitary method, but carried out using algebraic symbols. Step-by-Step Solution:Step 1: Cost of x metres = d rupees.Step 2: Cost per metre = d / x rupees.Step 3: Required length is y metres.Step 4: Cost of y metres = (d / x) * y.Step 5: This simplifies to cost = (y * d) / x = Rs. (yd / x). Verification / Alternative check:You can test the expression with simple numbers. Suppose x = 2 metres and d = 10 rupees, then cost per metre is 5 rupees. For y = 3 metres, cost should be 15 rupees. Plugging into (yd / x) gives (3 * 10) / 2 = 30 / 2 = 15 rupees, which matches. This confirms that Rs. (yd / x) is correct. Why Other Options Are Wrong:Rs. (xy / d) inverts the relationship and has wrong dimensions. Rs. (xd) or Rs. (yd) ignore division by x and therefore do not preserve the unit rate. Rs. (d / xy) is also dimensionally incorrect and would decrease cost for larger y, which is not logical when the rate is fixed. Common Pitfalls: Not computing cost per metre first and trying to guess a direct formula. Mixing up x and y in the numerator and denominator. Forgetting that cost must increase linearly with y when the rate is constant, so any expression where cost decreases with y is incorrect. Final Answer:The cost of y metres of wire at the same rate is Rs. (yd / x).

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