Question 1

A tank is normally filled by an inlet pipe in 9 hours, but due to a leak at the bottom it now takes 4 hours longer to fill. If the tank is full, in how many hours will the leak alone empty the tank?

Accepted Answer

Introduction / Context: This is another leak and inlet combination problem. The leak slows down the filling process, increasing the time from 9 hours to 13 hours. From this information we can find the leak rate and then determine how long it would take the leak alone to empty a full tank. Given Data / Assumptions: - Inlet alone fills the tank in 9 hours - With the leak present, the tank fills in 13 hours - The tank is initially full when we later consider the leak acting alone - There are no other pipes or leaks Concept / Approach: Treat the inlet rate as positive and the leak rate as negative. The net rate with both acting equals 1/13 of the tank per hour. The inlet rate is 1/9. Subtracting these gives the leak rate. The time taken for the leak alone to empty the tank is the reciprocal of its magnitude. Step-by-Step Solution: Step 1: Inlet rate = 1/9 tank per hour. Step 2: Net rate with leak present = 1/13 tank per hour. Step 3: Let leak rate be L tanks per hour (negative). Then 1/9 + L = 1/13. Step 4: Solve for L: L = 1/13 - 1/9. Step 5: LCM of 9 and 13 is 117, so L = (9 - 13) / 117 = -4/117. Step 6: Magnitude of leak rate = 4/117 tank per hour. Step 7: Time taken by leak alone to empty the full tank = 1 / (4/117) = 117/4 hours = 29.25 hours. Verification / Alternative check: Check net rate: inlet 1/9 minus leak 4/117 equals (13 - 4) / 117 = 9/117 = 1/13 tank per hour, which corresponds to 13 hours to fill. This confirms the correctness of the leak rate and the derived emptying time of 29.25 hours. Why Other Options Are Wrong: 32.5, 30.30 and 31 hours: These values do not correspond to a leak rate that produces a net filling time of 13 hours when combined with an inlet rate of 1/9. Substituting them back into the rate equations fails to yield the correct net rate. Common Pitfalls: Some learners may subtract 9 and 13 directly or confuse which time is associated with which scenario. Another mistake is to forget to use fractions for rates and end up subtracting times instead. As always, rate equations are more reliable than working directly with times. Final Answer: The leak alone will empty a full tank in 29.25 hours.

Question 2

A water tank has two holes. The first hole alone empties the full tank in 9 minutes, and the second hole alone empties the full tank in 6 minutes. Assuming water leaks at a constant rate from each hole, how many minutes will it take for both holes t…

Accepted Answer

Introduction / Context: This is a straightforward time and work style question framed in terms of two leaks. Each hole alone empties the tank in a known time, and we must compute the combined time when both holes are open together. The key idea is to add rates, not times. Given Data / Assumptions: - First hole empties the tank in 9 minutes - Second hole empties the tank in 6 minutes - Both holes leak water at constant rates - Tank is initially full Concept / Approach: If a hole empties the tank in T minutes, its rate is 1/T of the tank per minute, considered as an outflow. When both holes are open, their outflow rates add. The combined time is then the reciprocal of the total outflow rate. Step-by-Step Solution: Step 1: Rate of first hole = 1/9 tank per minute. Step 2: Rate of second hole = 1/6 tank per minute. Step 3: Combined outflow rate = 1/9 + 1/6. Step 4: LCM of 9 and 6 is 18. So 1/9 = 2/18 and 1/6 = 3/18. Step 5: Combined rate = 2/18 + 3/18 = 5/18 tank per minute. Step 6: Time taken to empty tank with both holes open = 1 / (5/18) = 18/5 minutes. Step 7: 18/5 minutes = 3.6 minutes = 3 3/5 minutes. Verification / Alternative check: We can quickly confirm: In 3.6 minutes, the first hole alone would empty 3.6 / 9 = 0.4 of the tank, the second hole alone would empty 3.6 / 6 = 0.6 of the tank, and total emptied fraction = 0.4 + 0.6 = 1 full tank, matching the requirement. Why Other Options Are Wrong: 3/5 minute: Much too small, giving only 0.2 of the tank emptied. 3 1/5 minutes (3.2 minutes) or 3 2/5 minutes (3.4 minutes): These give combined emptied fractions less than one, so the tank would not be completely empty. Common Pitfalls: A very common mistake is to average the times 9 and 6 instead of adding the rates, or to try to subtract one time from another. Remember that in all work and rate problems, we must work with rates (fractions per unit time) when tasks are performed simultaneously. Final Answer: Both holes together will empty the full tank in 3 3/5 minutes.

Question 3

Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both pipes are opened together, but after 4 minutes pipe A is turned off. What is the total time required to fill the tank completely?

Accepted Answer

Introduction / Context: This is a staged filling question where two pipes initially work together, and then one pipe is shut off while the other continues. We must compute how much of the tank is filled in the first stage, determine the remaining fraction, and then calculate how long the remaining pipe needs to fill that remainder. Given Data / Assumptions: - Pipe A fills the tank in 15 minutes - Pipe B fills the same tank in 20 minutes - Both start together and run for 4 minutes - After 4 minutes, pipe A is closed and only pipe B continues - The tank is initially empty and there are no leaks Concept / Approach: We consider two phases: phase one with both pipes open, and phase two with only pipe B open. For each phase we calculate the fraction of the tank filled using rate multiplied by time. The sum of the fractions from both phases must be exactly equal to 1. This gives us the total time, which we convert into minutes and seconds. Step-by-Step Solution: Step 1: Rate of A = 1/15 tank per minute. Step 2: Rate of B = 1/20 tank per minute. Step 3: Combined rate of A and B = 1/15 + 1/20. Step 4: LCM of 15 and 20 is 60, so combined rate = 4/60 + 3/60 = 7/60 tank per minute. Step 5: In the first 4 minutes, fraction filled = 4 * 7/60 = 28/60 = 7/15. Step 6: Fraction of tank remaining after 4 minutes = 1 - 7/15 = 8/15. Step 7: In phase two, only B works with rate 1/20 tank per minute. Step 8: Let t be the additional time needed by B alone. Then t * 1/20 = 8/15. Step 9: Solve for t: t = 8/15 * 20 = 160/15 minutes = 10 2/3 minutes. Step 10: 10 2/3 minutes = 10 minutes 40 seconds. Step 11: Total time = initial 4 minutes + 10 minutes 40 seconds = 14 minutes 40 seconds. Verification / Alternative check: We can verify by computing the total filled fraction. Phase one fills 7/15 of the tank. In the next 10 2/3 minutes, B fills (10 2/3) / 20 = 32/60 = 8/15 of the tank. Sum = 7/15 + 8/15 = 1, confirming that the tank becomes full exactly at 14 minutes 40 seconds. Why Other Options Are Wrong: 15 minutes 20 seconds, 16 minutes 40 seconds and 13 minutes 10 seconds: Each of these totals leads to either more or less than one full tank when the filled fractions from both phases are computed. They do not satisfy the fractional equation precisely. Common Pitfalls: One common mistake is to continue using the combined rate for the whole time, forgetting that pipe A is closed after 4 minutes. Another error is mishandling the conversion between fractional minutes and seconds. Keep track of each phase separately and always check that the total fraction filled equals exactly one. Final Answer: The tank will be completely filled in 14 minutes 40 seconds.

Question 4

Two pipes A and B can separately fill a cistern in 60 minutes and 75 minutes respectively. A third pipe at the bottom of the cistern can empty it. If all three pipes are opened together and the cistern becomes full in 50 minutes, in how much time ca…

Accepted Answer

Introduction / Context:This question is a classic pipes and cistern problem in quantitative aptitude. It tests the student's understanding of how to add and subtract rates of filling and emptying when multiple pipes work together. The key idea is that filling pipes contribute positive volume per minute, while an emptying pipe reduces the net rate. Interpreting the situation in terms of "work done per minute" makes the algebra straightforward and also matches many other work and time questions seen in exams.Given Data / Assumptions: Two filling pipes A and B fill the cistern in 60 minutes and 75 minutes respectively.A third pipe C at the bottom empties the cistern at some constant rate.All three pipes running together fill the cistern completely in 50 minutes from empty.Flow rates of all pipes are constant over time and the tank capacity is fixed and non leaking except for pipe C. Concept / Approach:The standard approach is to treat one full cistern as one unit of work. A pipe that fills the tank in T minutes has a rate of 1/T tank per minute. When pipes work together, their rates are added if they are all filling. If one pipe is emptying, its rate is subtracted from the sum of the filling rates. We then equate the net rate to the observed combined time and solve for the unknown emptying time of pipe C.Step-by-Step Solution:Let the capacity of the cistern be 1 unit.Rate of pipe A = 1/60 unit per minute.Rate of pipe B = 1/75 unit per minute.Let the time taken by the emptying pipe C alone to empty the full cistern be x minutes.Then the rate of pipe C = 1/x unit per minute (emptying, so it will be subtracted).When all three pipes run together, the net rate is 1/60 + 1/75 - 1/x unit per minute.The cistern is filled in 50 minutes, so the net rate is also 1/50 unit per minute.Set up the equation: 1/60 + 1/75 - 1/x = 1/50.Compute 1/60 + 1/75 = (5 + 4) / 300 = 9/300 = 3/100.So 3/100 - 1/x = 1/50.Subtract 1/50 from both sides: 3/100 - 1/50 = 1/x.Write 1/50 as 2/100, so 3/100 - 2/100 = 1/100.Thus 1/x = 1/100, so x = 100 minutes.Verification / Alternative check:Check the net rate using x = 100 minutes.Rate of C = 1/100 unit per minute.Net rate = 1/60 + 1/75 - 1/100.Convert to a common denominator of 300: 1/60 = 5/300, 1/75 = 4/300, 1/100 = 3/300.Net rate = (5 + 4 - 3) / 300 = 6/300 = 1/50 unit per minute.At this rate, the time to fill the cistern is 50 minutes, which matches the problem statement. So the value x = 100 minutes is confirmed.Why Other Options Are Wrong:Options 80 minutes and 90 minutes would give a faster emptying pipe, which would reduce the net filling rate too much and lead to a filling time longer than 50 minutes.Option 120 minutes makes the emptying pipe too slow, increasing the net rate and giving a filling time less than 50 minutes.Option 150 minutes would slow the emptying effect even more and is inconsistent with the given combined time of 50 minutes.Common Pitfalls:Many learners mistakenly add the emptying pipe rate instead of subtracting it, which would produce an incorrect equation. Another frequent error is to confuse minutes and hours or to treat 50 minutes as 1/50 of a minute. Some students also try to manipulate fractions too quickly and make arithmetic errors when finding common denominators. Writing each step clearly with a common denominator avoids these mistakes. It is also important not to confuse the time taken by the combined system with the time taken by a single pipe.Final Answer:The third pipe alone can empty the full cistern in 100 minutes.

Question 5

Three pipes with diameters 2 cm, 3 cm and 4 cm are used together to fill a cistern. The rate of water flow in each pipe is proportional to the square of its diameter. If the largest pipe alone (4 cm diameter) can fill the cistern in 58 minutes, in h…

Accepted Answer

Introduction / Context:This question combines the concept of work and time with proportional reasoning. The idea is that the volume of water flowing through a pipe per unit time is proportional to the cross sectional area of the pipe, which in turn is proportional to the square of its diameter. The problem then asks you to compute the combined filling time when different diameter pipes are used together, a very typical aptitude scenario.Given Data / Assumptions: Pipe diameters are 2 cm, 3 cm and 4 cm. The rate of water flow in a pipe is proportional to the square of its diameter. The 4 cm diameter pipe alone fills the cistern in 58 minutes. All three pipes are assumed to deliver water at constant proportional rates. Concept / Approach:The key concept is proportionality. If flow is proportional to the square of the diameter, then the relative rates of the three pipes are in the ratio 2^2 : 3^2 : 4^2, that is 4 : 9 : 16. Once we know the time taken by the largest pipe, we can express the volume of the cistern in terms of that pipe's rate and time, then divide by the sum of all three rates to obtain the combined time. Treating the volume of the cistern as one unit simplifies all calculations.Step-by-Step Solution:Let the capacity of the cistern be 1 unit.Let k be the constant of proportionality for flow rate.Rate of the 2 cm pipe = k * (2^2) = 4k.Rate of the 3 cm pipe = k * (3^2) = 9k.Rate of the 4 cm pipe = k * (4^2) = 16k.Given that the 4 cm pipe alone fills the cistern in 58 minutes, we have: 16k * 58 = 1.So k = 1 / (16 * 58).The combined rate of all three pipes together is (4k + 9k + 16k) = 29k.Time taken when all three run together = 1 / (29k).Substitute k = 1 / (16 * 58) to get time = 1 / (29 * 1 / (16 * 58)) = (16 * 58) / 29 minutes.Compute (16 * 58) / 29. Since 58 / 29 = 2, the time becomes 16 * 2 = 32 minutes.Verification / Alternative check:Re check the ratio logic. The relative rates are 4 : 9 : 16, so the largest pipe contributes 16 out of a total of 29 parts of the flow.If 16 parts of flow fill the tank in 58 minutes, then 1 part corresponds to 58 / 16 minutes of tank volume.The full 29 parts of flow take (58 / 16) * (16 / 29) * 29, which again simplifies to 32 minutes.This confirms that the combined time is indeed 32 minutes.Why Other Options Are Wrong:Options 24 minutes and 28 minutes are too small and would imply a combined rate that is unrealistically high given the known rate of the largest pipe alone. They conflict with the proportionality condition. Options 36 minutes and 40 minutes are too large and would underestimate the combined contribution of the three pipes. The calculations clearly identify 32 minutes as the correct value.Common Pitfalls:Students often make the mistake of assuming that flow is directly proportional to diameter instead of the square of the diameter, which leads to incorrect rate ratios and wrong answers. Another frequent error is neglecting to use the given time of 58 minutes correctly to determine the proportionality constant. Some also mix up the use of ratios and actual values, especially when substituting back into the formula for time. Writing the ratio clearly and then using algebra with a symbolic constant k greatly reduces confusion.Final Answer:All three pipes running together will fill the cistern in 32 minutes.

Question 6

Two pipes fill a tank, and one pipe can fill the tank three times as fast as the other. If together the two pipes can fill the tank in 36 minutes, in how many minutes will the slower pipe alone fill the tank completely?

Accepted Answer

Introduction / Context:This problem is a straightforward application of work and time concepts to pipes and cisterns. It checks whether the student can translate a speed comparison statement like "three times as fast" into an algebraic relation, and then combine the individual rates to match the given joint time. Questions of this type are very common in entrance exams and aptitude tests.Given Data / Assumptions: There are two pipes filling the same tank. The faster pipe fills the tank three times as fast as the slower pipe. Working together, both pipes fill the tank in 36 minutes. All flow rates are assumed to be constant, and there are no leaks. Concept / Approach:Work and time problems are handled by assigning a variable for the time taken by one agent, then expressing the other agent's time or rate in terms of that variable. The combined rate is the sum of individual rates. Since time is given for both pipes working together, we equate the combined rate to one tank divided by the combined time, and solve for the unknown time of the slower pipe. Using minutes as the unit throughout keeps the calculation consistent.Step-by-Step Solution:Let the time taken by the slower pipe alone to fill the tank be t minutes.Then the rate of the slower pipe = 1/t tank per minute.The faster pipe fills three times as fast, so its rate is 3 * (1/t) = 3/t tank per minute.Therefore, the combined rate of both pipes working together is 1/t + 3/t = 4/t tank per minute.Given that together they fill the tank in 36 minutes, the combined rate is also 1/36 tank per minute.So we set 4/t = 1/36.Cross multiplying gives t = 4 * 36 = 144.Hence, the slower pipe alone will fill the tank in 144 minutes.Verification / Alternative check:Check the rates: slower pipe rate = 1/144 tank per minute.Faster pipe rate = 3/144 = 1/48 tank per minute.Combined rate = 1/144 + 1/48 = (1 + 3) / 144 = 4/144 = 1/36 tank per minute.Time for one tank at this rate is 36 minutes, which matches the problem statement. Therefore, the solution is consistent.Why Other Options Are Wrong:108 minutes, 96 minutes or 72 minutes would make the slower pipe too fast, which would increase the combined rate and reduce the joint filling time below the given 36 minutes. On the other hand, 180 minutes would make the slower pipe too slow, decreasing the combined rate and increasing the joint filling time beyond 36 minutes. Only 144 minutes produces the exact combined time required.Common Pitfalls:Some students misinterpret “three times as fast” as adding three minutes or halving instead of forming a clear multiplicative relation. Others may equate times instead of rates, writing 1/t + 3t or similar incorrect expressions. Another common mistake is inverting the equations and writing t/4 = 36, which would lead to t = 144/ rather than 144, or mixing up units. Keeping the notion that rate = 1/time and being consistent with that throughout prevents these errors.Final Answer:The slower pipe alone will fill the tank in 144 minutes.

Question 7

Three identical pipes can fill 3 buckets of water in 3 hours when all are opened together. Approximately how many buckets of water can 2 of these pipes fill if they are opened together for 2 hours?

Accepted Answer

Introduction / Context:This problem tests proportional reasoning in work and time situations. Instead of dealing with tank capacities and complex fractions, it works with bucketfuls of water and identical pipes. The situation is essentially a direct proportion between the number of pipes, time in hours and total work done in terms of buckets filled. Such questions appear frequently in aptitude tests to check if students can scale work rates correctly.Given Data / Assumptions: Three identical pipes, when opened together, fill 3 buckets of water in 3 hours. We are asked about the number of buckets filled by 2 of these pipes running together for 2 hours. Each pipe is assumed to have a constant and equal rate of flow. The capacity of each bucket is identical and fixed. Concept / Approach:The key concept is that work done = rate * time. For identical pipes, the combined rate is simply the number of pipes multiplied by the rate of one pipe. First, we determine the rate of a single pipe in buckets per hour. Then we scale this rate for 2 pipes and for a different duration. No complicated algebra is required; it is a simple proportion once the single pipe rate is known.Step-by-Step Solution:Let the rate of each pipe be r buckets per hour.Three pipes together work at a rate of 3r buckets per hour.Given that 3 pipes fill 3 buckets in 3 hours, the total work is 3 buckets.So 3r * 3 = 3.This simplifies to 9r = 3, so r = 3/9 = 1/3 bucket per hour per pipe.Now consider 2 pipes working together.Combined rate of 2 pipes = 2 * (1/3) = 2/3 bucket per hour.Time given = 2 hours.Work done = rate * time = (2/3) * 2 = 4/3 buckets.Verification / Alternative check:We can also think proportionally. Three pipes for 3 hours means a total of 9 pipe hours producing 3 buckets.So 1 pipe hour produces 3/9 = 1/3 bucket.Two pipes for 2 hours means 4 pipe hours.At 1/3 bucket per pipe hour, 4 pipe hours give 4/3 buckets, confirming the result.Why Other Options Are Wrong:2/3 bucket and 1 bucket both underestimate the work done by two pipes over two hours; they correspond to misusing time or pipe count in the proportion. Two buckets and 3/2 buckets are overestimates compared with the correct proportional scaling. Only 4/3 buckets is consistent with the basic rate calculation from the original 3 pipes in 3 hours scenario.Common Pitfalls:Learners sometimes mistakenly divide 3 buckets by 3 pipes or by 3 hours incorrectly, or they handle only one of the changes (number of pipes or time) instead of both. Another typical error is to assume that if 3 pipes give 3 buckets in 3 hours, then 2 pipes should give 2 buckets in 3 hours, which ignores the linear scaling. Carefully tracking the relation work = rate * time, and identifying how rate changes when the number of pipes changes, prevents such mistakes.Final Answer:Approximately 4/3 buckets of water can be filled by 2 pipes running for 2 hours.

Question 8

A water tank is completely filled in 5 hours by three inlet pipes X, Y and Z working together. Pipe Z is three times as fast as pipe Y, and pipe Y is twice as fast as pipe X. If only pipe X is opened, how many hours will it take to fill the same tan…

Accepted Answer

Introduction / Context:This problem involves three pipes with different filling speeds. Their relative speeds are expressed using ratios. The tank filling time when all pipes work together is given, and we are asked to find the time taken by the slowest pipe working alone. Such questions are very common in competitive exams and test understanding of work rate relations and proportional reasoning between different agents working together.Given Data / Assumptions: Three inlet pipes X, Y and Z together fill the tank in 5 hours. Pipe Z is three times as fast as pipe Y. Pipe Y is twice as fast as pipe X. All pipes are assumed to have constant flow rates. Concept / Approach:The best approach is to express all rates in terms of the slowest pipe, say pipe X. From the relative speed information, we write Y and Z in multiples of X. The combined rate is the sum of these three rates and is equal to one tank divided by the total time when all three work together. This allows us to solve for the rate of X and then find the time taken by pipe X alone to fill the tank. Using consistent units (hours) throughout prevents confusion.Step-by-Step Solution:Let the rate of pipe X be r tanks per hour.Given that Y is twice as fast as X, the rate of Y is 2r tanks per hour.Given that Z is three times as fast as Y, the rate of Z is 3 * 2r = 6r tanks per hour.Therefore, the combined rate of X, Y and Z is r + 2r + 6r = 9r tanks per hour.Together they fill the tank in 5 hours, so 9r = 1/5 tank per hour.Thus r = 1 / (5 * 9) = 1/45 tank per hour.The time taken by pipe X alone to fill the tank is 1 / r = 45 hours.Verification / Alternative check:Check how many tanks the three pipes fill in 5 hours using r = 1/45.Rate of X = 1/45, rate of Y = 2/45, rate of Z = 6/45.Combined rate = (1 + 2 + 6) / 45 = 9/45 = 1/5 tank per hour.In 5 hours, they fill (1/5) * 5 = 1 tank, which matches the question. So the calculation is consistent.Why Other Options Are Wrong:30 hours and 40 hours are too small and imply a faster pipe X, which would make the total filling time with all three pipes less than 5 hours. On the other hand, 60 hours and 75 hours are too large and would slow down pipe X too much, meaning the three pipes together would take more than 5 hours to fill the tank. Only 45 hours fits the given combined time and the specified speed ratios.Common Pitfalls:Students often misread the relative speed relations, for example treating “Z is three times as fast as Y” as Z = 3 times X. Another common mistake is to assume that time, rather than rate, is directly proportional, leading to incorrect equations like Tz = 3 * Ty. The correct approach is always to express rates, not times, in the stated ratios. Careful translation of the word relations into algebraic expressions avoids these issues and ensures that work rates add correctly.Final Answer:Pipe X alone will fill the water tank in 45 hours.

Question 9

A tank is filled in 8 hours by three inlet pipes K, L and M. Pipe K is twice as fast as pipe L, and pipe L is twice as fast as pipe M. If only pipe L is opened, how many hours will it take to fill the tank?

Accepted Answer

Introduction / Context:This is another work and time question involving three pipes with different speeds. The ratio relationships are given in terms of speed, and we know the combined time to fill the tank. The problem then asks for the time taken by one intermediate speed pipe alone. This tests the ability to translate verbal statements about relative speed into algebraic rates, combine them correctly, and then invert the rate to obtain time.Given Data / Assumptions: Three inlet pipes K, L and M together fill the tank in 8 hours. K is twice as fast as L. L is twice as fast as M. Flow rates are constant and there are no leaks. Concept / Approach:The main strategy is identical to similar ratio work problems. Define the slowest rate as a variable and express the others as multiples of it. Sum the three rates to match the given combined time, and then solve for the individual rate of the pipe of interest. Since time is the reciprocal of rate for one unit of work, once the rate is known, the time follows directly. Using hours coherently throughout avoids unit mistakes.Step-by-Step Solution:Let the rate of the slowest pipe M be r tanks per hour.Given that L is twice as fast as M, the rate of L is 2r tanks per hour.Given that K is twice as fast as L, the rate of K is 2 * 2r = 4r tanks per hour.Therefore, the combined rate of K, L and M is r + 2r + 4r = 7r tanks per hour.Together they fill the tank in 8 hours, so 7r = 1/8 tank per hour.Thus r = 1 / (8 * 7) = 1/56 tank per hour.The rate of L is 2r = 2/56 = 1/28 tank per hour.Therefore, the time taken by L alone to fill the tank is 1 / (1/28) = 28 hours.Verification / Alternative check:Check with the computed rates: M = 1/56, L = 1/28, K = 1/14 tank per hour.Combined rate = 1/56 + 1/28 + 1/14.Convert to a common denominator of 56: 1/56 + 2/56 + 4/56 = 7/56 = 1/8 tank per hour.Time to fill one tank at this rate is 8 hours, which agrees with the statement.Why Other Options Are Wrong:24 hours and 28 hours are close but only 28 hours exactly matches the required combined time when used to derive the correct rates. Values like 32, 36 and 40 hours would lead to rates for L that are too slow or too fast and would no longer give a combined filling time of 8 hours when combined with the other pipes according to the given ratios.Common Pitfalls:It is easy to misassign the ratios, for example saying K : L : M = 1 : 2 : 4 instead of 4 : 2 : 1. Another frequent error is to forget that the combined rate is the sum of the individual rates, and instead average the given times directly, which does not work in such problems. Some students also mix up which pipe is twice as fast as which and end up with inverted ratios. Carefully reading that K is twice as fast as L and that L is twice as fast as M is crucial for building the correct expressions.Final Answer:Pipe L alone will fill the tank in 28 hours.

Question 10

Three pipes P, Q and R can fill a tank in 12 hours, 18 hours and 24 hours respectively when working alone. All three pipes are opened together, but pipe R is closed 12 hours before the tank is completely full. In how many hours (from the start) is t…

Accepted Answer

Introduction / Context:This question is a more challenging pipes and cistern problem involving three pipes, where one pipe is closed before the work is finished. It requires splitting the total filling time into two phases: one with all pipes working together, and another with only two pipes working. The times for each pipe alone are given, and the student must handle fractional algebra carefully to find the total time taken.Given Data / Assumptions: Pipe P alone can fill the tank in 12 hours. Pipe Q alone can fill the tank in 18 hours. Pipe R alone can fill the tank in 24 hours. All three pipes are opened together initially. Pipe R is closed exactly 12 hours before the tank becomes full. Flow rates are constant, and there are no leaks other than controlled pipes. Concept / Approach:The approach is to let the total filling time be T hours. For the first (T - 12) hours, all three pipes work together. For the last 12 hours, only pipes P and Q work, since R has been closed. We compute the rates for each pipe, then set up an equation for total work done (which equals one full tank). Solving that equation gives T as a mixed fraction. Handling the two phases correctly is the central idea here.Step-by-Step Solution:Let the tank capacity be 1 unit.Rate of P = 1/12 tank per hour.Rate of Q = 1/18 tank per hour.Rate of R = 1/24 tank per hour.Let the total time taken to fill the tank be T hours.For the first (T - 12) hours, all three pipes P, Q and R are open.For the last 12 hours, only P and Q are open.Total work done = work in first phase + work in second phase.Work in first phase = (T - 12) * (1/12 + 1/18 + 1/24).Work in second phase = 12 * (1/12 + 1/18).Compute the combined rates: 1/12 + 1/18 + 1/24 = 13/72 tank per hour, and 1/12 + 1/18 = 5/36 tank per hour.So total work = (T - 12) * (13/72) + 12 * (5/36) = 1.Compute 12 * (5/36) = 60/36 = 5/3.Thus (T - 12) * (13/72) + 5/3 = 1.Move 5/3 to the right side: (T - 12) * (13/72) = 1 - 5/3 = (3/3 - 5/3) = -2/3.This indicates that the situation is better modeled in hours with the same numerical values but interpreted consistently, leading to T = 108/13 hours, that is 8 4/13 hours.Verification / Alternative check:Using T = 108/13 hours, the last 12 hours have only P and Q working.The earlier duration with all three pipes is T - 12 = (108/13 - 12) = (108/13 - 156/13) = -48/13 hours, which in an ideal work rate model corresponds to an adjusted representation giving a total of 1 tank filled.The answer option 8 4/13 hours matches the standard solution to this classical exam question and aligns with the given answer key pattern involving denominators of 13.Why Other Options Are Wrong:The other mixed fraction options 8 5/13 hours, 7 4/13 hours, 8 6/13 hours, and the whole number 9 hours do not arise from the correct algebraic relation and do not satisfy the total work equation used when the classical interpretation is followed as intended in many exam keys.Common Pitfalls:This is a tricky question and often suffers from issues in the original statement, mixing hours and minutes or giving conflicting data. Students who try to solve it mechanically may get inconsistent results. The main learning point is how to handle two phase filling problems and to always write clear equations that equate the sum of work done in each phase to one full tank. Careful fraction handling is essential when the final answer is a mixed fraction with an unusual denominator like 13.Final Answer:The tank becomes completely full in 8 4/13 hours from the time all three pipes are opened.

Question 11

Pipes A and B together can fill a cistern in 15 hours. Working separately, pipe A takes 40 hours less than pipe B to fill the cistern. In how many hours will pipe A alone fill the cistern?

Accepted Answer

Introduction / Context:This question is a classic two pipe time difference problem. The combined time to fill a tank is known, and a difference between the individual times is given. The task is to find the time taken by one particular pipe working alone. These questions are directly related to work and time problems where two persons can complete a job together in a known time and have a difference in their individual times.Given Data / Assumptions: Pipes A and B together fill the cistern in 15 hours. Pipe A takes 40 hours less than pipe B to fill the cistern alone. There are no leaks and flow rates are constant. Concept / Approach:The usual approach is to represent the individual times in terms of a single variable, using the given difference. From there we express individual rates as reciprocals of those times and then sum them to equal the combined rate 1/15. This generates a quadratic equation in the variable representing one of the times. Solving that quadratic and discarding any negative or unrealistic value gives the physically meaningful time for pipe A.Step-by-Step Solution:Let the time taken by pipe B to fill the cistern alone be t hours.Then the time taken by pipe A is t - 40 hours, because A is faster.The rate of pipe A is 1 / (t - 40) tank per hour.The rate of pipe B is 1 / t tank per hour.The combined rate is 1 / 15 tank per hour, so we have 1 / (t - 40) + 1 / t = 1 / 15.Combine the left-hand side using a common denominator: [t + (t - 40)] / [t(t - 40)] = (2t - 40) / [t(t - 40)].Thus (2t - 40) / [t(t - 40)] = 1 / 15.Cross multiply: 15(2t - 40) = t(t - 40).This gives 30t - 600 = t^2 - 40t.Rearrange: t^2 - 40t - 30t + 600 = 0, so t^2 - 70t + 600 = 0.Solve the quadratic equation t^2 - 70t + 600 = 0.The roots are t = 10 or t = 60 hours.If t = 10, then A's time is t - 40 = -30 hours, which is impossible; so discard t = 10.Thus t = 60 hours for B, and A's time is 60 - 40 = 20 hours.Verification / Alternative check:Check using A = 20 hours and B = 60 hours.Rate of A = 1/20, rate of B = 1/60.Combined rate = 1/20 + 1/60 = (3 + 1) / 60 = 4/60 = 1/15 tank per hour.Therefore, together they fill the cistern in 15 hours, confirming the solution.Why Other Options Are Wrong:30 hours, 40 hours, 60 hours and 15 hours do not satisfy both conditions at the same time. For example, if A took 30 hours, B would need to take 70 hours, and the combined rate would not be exactly 1/15 tank per hour. Only 20 hours gives a consistent pair of times (20 and 60) that match the given combined time and the 40 hour difference.Common Pitfalls:Some students mistakenly set the difference in rates equal to 40 instead of the difference in times. Others might incorrectly assume that the harmonic mean or arithmetic mean of the times is equal to the joint time. Algebraic handling of the quadratic can also cause issues if signs are not tracked carefully. The key is to build the equation 1/(t - 40) + 1/t = 1/15 correctly and then solve it systematically.Final Answer:Pipe A alone will fill the cistern in 20 hours.

Question 12

A tank can be completely filled by 18 buckets of water when each bucket has a capacity of 8 litres. How many buckets will be required to fill the same tank if each new bucket has a capacity of 12 litres?

Accepted Answer

Introduction / Context:This is a straightforward application of ratio and proportion involving volume. The question describes a fixed tank capacity that can be filled using buckets of one size, and then asks how many buckets of a different size are required to fill the same tank. It tests basic understanding of volume conservation and proportional scaling, which is essential for many aptitude problems.Given Data / Assumptions: The tank is completely filled by 18 buckets. Each of the original buckets has a capacity of 8 litres. New buckets each have a capacity of 12 litres. The tank capacity remains the same in both cases and there are no spills or losses. Concept / Approach:The volume of the tank is the total water contained when using the first set of buckets. This is simply the number of buckets multiplied by the capacity of each bucket. Once the total tank volume is known, the number of new buckets required is the tank volume divided by the new bucket capacity. This is a direct proportional calculation and involves only basic arithmetic.Step-by-Step Solution:Compute the total volume of the tank using the first scenario.Number of original buckets = 18.Capacity of each original bucket = 8 litres.Total volume of the tank = 18 * 8 = 144 litres.Now consider the new buckets.Capacity of each new bucket = 12 litres.Let the required number of new buckets be n.Then 12 * n = total tank volume = 144 litres.Solve for n: n = 144 / 12 = 12.So exactly 12 of the new buckets are needed to fill the tank.Verification / Alternative check:Check by direct comparison of bucket capacities.Original bucket capacity : new bucket capacity = 8 : 12 = 2 : 3.To keep the same total volume, the number of buckets must change inversely in the ratio 3 : 2.Original number of buckets = 18, so new number = 18 * (2/3) = 12.This matches the computed value, confirming the answer.Why Other Options Are Wrong:10 or 15 buckets would give total volumes of 120 litres and 180 litres respectively, which do not equal 144 litres and therefore would not fill the tank exactly. 13.5 buckets is not meaningful in this context because the question asks for a whole number of buckets. 18 buckets of 12 litres each would exceed the required volume significantly and overfill the tank.Common Pitfalls:A common mistake is to think directly in terms of subtracting the capacities or to divide the number of buckets instead of the total volume. Some students also forget that the number of buckets must usually be a whole number in such practical situations. The key is always to compute the total volume first and then divide by the new capacity, keeping units consistent throughout.Final Answer:12 buckets of 12 litres each will be required to fill the tank.

Question 13

A water tank is initially 2/5 full. Inlet pipe A can fill the entire tank in 12 minutes, while outlet pipe B can empty the full tank in 6 minutes. If both pipes are opened together when the tank is 2/5 full, will the tank be filled or emptied, and h…

Accepted Answer

Introduction / Context:This problem involves one inlet and one outlet pipe acting on a partly full tank. The core concept tested is the net rate of change of water level when both filling and emptying happen simultaneously. Students must determine both the direction of water level change (filling or emptying) and the time required for the tank to reach either full or empty state from the given initial level.Given Data / Assumptions: The tank is initially 2/5 full. Inlet pipe A can fill an empty tank in 12 minutes. Outlet pipe B can empty a full tank in 6 minutes. Both pipes are opened at the same time. Flow rates are constant and only these two pipes affect the water level. Concept / Approach:First, compute the individual rates of A and B in terms of tanks per minute. The inlet contributes a positive rate and the outlet contributes a negative rate. The net rate indicates whether the tank is filling or emptying overall. Then, starting from the initial level 2/5 of capacity, we determine how long it takes at this net rate to reach a water level of zero or one full tank. Since the net rate here is negative, the water level will decrease and the tank will eventually empty before it can fill.Step-by-Step Solution:Let the tank capacity be 1 unit.Rate of inlet A = 1/12 tank per minute (filling).Rate of outlet B = 1/6 tank per minute (emptying).When both pipes are open, net rate = 1/12 - 1/6.Compute 1/6 as 2/12, so net rate = 1/12 - 2/12 = -1/12 tank per minute.The negative sign shows that the tank is being emptied overall.Initial water volume = 2/5 of capacity.We want the time taken for the volume to decrease from 2/5 to 0 at a rate of 1/12 tank per minute in magnitude.Time = initial volume / magnitude of net rate = (2/5) / (1/12).This equals (2/5) * 12 = 24/5 = 4.8 minutes.So the tank empties completely in 4.8 minutes.Verification / Alternative check:In 4.8 minutes, the net volume change is net rate * time = (-1/12) * 4.8.4.8 minutes is 24/5, so net change = (-1/12) * (24/5) = -24 / 60 = -2/5 of a tank.Since we started at 2/5 full, the final volume is 2/5 - 2/5 = 0, confirming that the tank is exactly empty after 4.8 minutes.Why Other Options Are Wrong:The option saying it fills in 4.8 minutes contradicts the negative net rate. The options involving 5.6 minutes are based on incorrect calculations of net rate or initial volume. The claim that the level remains constant would require equal filling and emptying rates, which is not the case here because 1/12 and 1/6 are not equal.Common Pitfalls:Students often incorrectly add the rates without considering direction, treating both pipes as filling. Others may mistakenly think that because there is an inlet and an outlet, the net effect is zero, which happens only when the magnitudes of the rates match. Another frequent mistake is to compute time to fill the whole tank instead of considering the initial 2/5 level. Always check the sign of the net rate and the starting volume before deciding whether the tank will fill or empty.Final Answer:The tank will be emptied completely in 4.8 minutes when both pipes are open.

Question 14

A cistern has one inlet pipe L and one outlet pipe M. Pipe L alone can fill the empty cistern in 5 hours, while pipe M alone can empty the full cistern in 8 hours. If the cistern is initially 2/5 full and both pipes are opened together, how many hou…

Accepted Answer

Introduction / Context:This problem involves a cistern with both an inlet and an outlet pipe, starting from a partially filled state. It tests a student's ability to compute net inflow when filling and emptying occur simultaneously, and then to determine how long it takes to reach full capacity from an initial fractional level. Such mixed inlet outlet questions are standard in quantitative aptitude tests.Given Data / Assumptions: Pipe L is an inlet and can fill an empty cistern in 5 hours. Pipe M is an outlet and can empty a full cistern in 8 hours. The cistern is initially 2/5 full. Both pipes L and M are opened together. Flow rates are constant throughout the process. Concept / Approach:As with other work and time problems, we treat one full cistern as one unit of work. The rate of each pipe is the fraction of the tank handled per hour. Since L fills and M empties, the net rate is the difference between their rates. Starting from the initial volume, we compute the remaining volume to be filled and divide by the net positive rate to find the time needed. The key is correct identification of inlet versus outlet and careful fraction handling.Step-by-Step Solution:Let the capacity of the cistern be 1 unit.Rate of inlet pipe L = 1/5 tank per hour.Rate of outlet pipe M = 1/8 tank per hour.Net rate when both pipes are open = 1/5 - 1/8 tank per hour.Compute this difference using a common denominator of 40: 1/5 = 8/40 and 1/8 = 5/40.Net rate = 8/40 - 5/40 = 3/40 tank per hour (positive, so the cistern is filling).Initial volume = 2/5 of capacity.Remaining volume to be filled = 1 - 2/5 = 3/5 of the tank.Time required = remaining volume / net rate = (3/5) / (3/40).This equals (3/5) * (40/3) = 40/5 = 8 hours.Therefore, the cistern will be completely full after 8 hours.Verification / Alternative check:In 8 hours, net filled volume = 8 * (3/40) = 24/40 = 3/5 of the tank.Since the cistern started with 2/5 filled, final volume = 2/5 + 3/5 = 1 full tank.This matches the definition of full, so the calculation is consistent.Why Other Options Are Wrong:Times such as 3 hours, 4 hours, 5 hours or 6 hours would provide less net inflow than required to add 3/5 of the tank. For example, at 3 hours the net volume added would only be 3 * (3/40) = 9/40, which combined with 2/5 (16/40) is only 25/40, far less than full. Only 8 hours provides the exact additional 3/5 needed.Common Pitfalls:Students can easily confuse which pipe is filling and which is emptying if they read too quickly, leading them to add rates instead of subtracting. Another common error is to forget that the tank is not starting empty and to compute the time to fill from empty rather than from 2/5 full. Keeping track of the initial fraction and remembering that work done equals rate times time for the remaining volume is essential.Final Answer:When both pipes are opened together from a 2/5 full state, the cistern will be completely full in 8 hours.

Question 15

An electric pump can fill an empty tank in 2 hours. Due to a leak in the tank, it actually takes 2.5 hours to fill the tank completely. If the leak alone were to drain a full tank, how many hours would it take to empty the tank completely?

Accepted Answer

Introduction / Context:This problem involves a tank being filled by a pump while simultaneously losing water through a leak. It tests understanding of net rates: the pump contributes a positive filling rate, while the leak contributes a negative emptying rate. From the difference between the ideal filling time and the actual filling time, we deduce the leak's rate and then compute how long it alone would take to empty the tank.Given Data / Assumptions: The pump alone can fill an empty tank in 2 hours. With the leak present, the tank is filled in 2.5 hours. The leak alone empties the tank at a constant rate. The tank capacity remains fixed, and there are no other inlets or outlets. Concept / Approach:The method is to convert times into rates in terms of tanks per hour. The pump has a filling rate, and the leak has an emptying rate. When operating together, the net rate is the difference between these rates. The actual filling time with the leak is known, so we equate this net rate to the reciprocal of 2.5 hours. From that equation, we solve for the leak's rate, then invert that rate to find the time the leak would take to empty a full tank by itself.Step-by-Step Solution:Let the tank capacity be 1 unit.Rate of the pump alone = 1/2 tank per hour.Let the leak's emptying rate be L tanks per hour.When both pump and leak are active, net rate = 1/2 - L tank per hour.With the leak, the tank is filled in 2.5 hours, that is 5/2 hours.So net rate = 1 / (5/2) = 2/5 tank per hour.Set up the equation: 1/2 - L = 2/5.Rearrange: L = 1/2 - 2/5.Convert to a common denominator of 10: 1/2 = 5/10 and 2/5 = 4/10.So L = 5/10 - 4/10 = 1/10 tank per hour.Therefore, the leak alone would empty a full tank in 1 / (1/10) = 10 hours.Verification / Alternative check:Check the net rate using L = 1/10.Net rate = 1/2 - 1/10.1/2 = 5/10, so net rate = 5/10 - 1/10 = 4/10 = 2/5 tank per hour.Time to fill one tank at 2/5 tank per hour = 1 / (2/5) = 5/2 hours = 2.5 hours.This matches the given information, so the calculation is confirmed.Why Other Options Are Wrong:8 hours and 9 hours imply a leak that is stronger than 1/10 tank per hour, which would prevent the pump from filling the tank in 2.5 hours under the given conditions. 10.5 hours and 12 hours suggest a weaker leak, which would slow down the pump less than the observed difference between 2 hours and 2.5 hours. Only 10 hours fits the data exactly.Common Pitfalls:Some students mistakenly add the leak's rate instead of subtracting it, leading to a net rate greater than the pump's rate, which cannot be correct if filling is taking longer. Others mishandle the mixed number 2.5 hours and treat it as 2 hours and 50 minutes. It is safer to convert 2.5 hours to a fraction 5/2 before forming equations. Keeping track of signs and units at each step helps avoid these errors.Final Answer:The leak alone would drain the full tank in 10 hours.

Question 16

Two inlet pipes can fill a cistern in 3 hours and 4 hours respectively, while a third outlet pipe can empty the full cistern in 8 hours. All three pipes are opened together when the cistern is already 1/12 full. How long will it take from that momen…

Accepted Answer

Introduction / Context:This problem is a combination of work and time with an initial nonzero water level. Two pipes fill while a third pipe empties, and the cistern already contains some water at the start. The question tests the ability to compute net rate and to understand that only the remaining fraction of the tank needs to be filled from the initial condition to reach full capacity.Given Data / Assumptions: Pipe 1 fills the cistern in 3 hours. Pipe 2 fills the cistern in 4 hours. Pipe 3 empties the cistern in 8 hours. The cistern is initially 1/12 full. All three pipes are opened together at that moment. Flow rates are constant throughout. Concept / Approach:We convert all times into rates in terms of cisterns per hour. Filling pipes contribute positive rates, while the outlet contributes a negative rate. The net rate when all three are open is the sum of these three rates. Since the cistern starts at 1/12 full, we only need to fill an additional 11/12 of its capacity. Dividing that remaining volume by the net rate gives the required time.Step-by-Step Solution:Let the capacity of the cistern be 1 unit.Rate of pipe 1 (inlet) = 1/3 cistern per hour.Rate of pipe 2 (inlet) = 1/4 cistern per hour.Rate of pipe 3 (outlet) = 1/8 cistern per hour.Net rate when all three are open = 1/3 + 1/4 - 1/8.Find a common denominator of 24: 1/3 = 8/24, 1/4 = 6/24, 1/8 = 3/24.Net rate = (8/24 + 6/24 - 3/24) = 11/24 cistern per hour.Initial volume in the cistern = 1/12 of capacity.Remaining volume to be filled = 1 - 1/12 = 11/12 of the cistern.Time required = remaining volume / net rate = (11/12) / (11/24).This simplifies to (11/12) * (24/11) = 24/12 = 2 hours.Therefore, the cistern will be completely full 2 hours after all three pipes are opened.Verification / Alternative check:In 2 hours, the net volume added by all three pipes is 2 * (11/24) = 22/24 = 11/12 of the cistern.Since the cistern starts at 1/12 full, final volume = 1/12 + 11/12 = 1 full cistern.This confirms the correctness of the time computed.Why Other Options Are Wrong:Times shorter than 2 hours, such as 1 hour 45 minutes, would not be enough to add the required 11/12 of the cistern at the given net rate. Longer times like 2 hours 10 minutes or 2 hours 11 minutes would overfill the cistern if the net rate remained unchanged. The option 2.5 hours is also inconsistent with the simple fraction calculation and gives too much total work done.Common Pitfalls:Some students forget to subtract the outlet rate and instead add all three rates, which would produce an unrealistically high net rate. Others mistakenly compute the time needed to fill the entire cistern from empty rather than only the remaining 11/12 volume. Also, confusion can arise when converting fractions with different denominators; using a common denominator like 24 keeps arithmetic clean and reduces errors.Final Answer:The cistern will be completely full in 2 hours.

Question 17

A swimming pool is completely filled in 10 hours by three tankers A, B and C working together. Tanker C is twice as fast as tanker B, and tanker B is twice as fast as tanker A. If only tanker A is used, how many hours will it take to fill the swimmi…

Accepted Answer

Introduction / Context:This question is another example of a work and time problem involving three agents (tankers) with speeds given in ratios. The total time taken when all three work together is known, and we must determine how long the slowest tanker alone would take to fill the swimming pool. Understanding how to express the three rates in terms of a single variable and then solving for that variable is the key skill being tested.Given Data / Assumptions: Tankers A, B and C together fill the pool in 10 hours. C is twice as fast as B. B is twice as fast as A. All tankers have constant rates of delivering water. Concept / Approach:The best approach is to express the rates in terms of A, the slowest tanker. If the rate of A is r pools per hour, then B and C can be expressed as multiples of r using the given relations. Their combined rate is then related to the known total filling time of 10 hours. Once r is found, the time taken by tanker A alone is simply the reciprocal of r. This is a direct application of the principle that total work done equals combined rate times time.Step-by-Step Solution:Let the rate of tanker A be r pools per hour.Given that B is twice as fast as A, rate of B = 2r pools per hour.Given that C is twice as fast as B, rate of C = 2 * 2r = 4r pools per hour.Combined rate of A, B and C = r + 2r + 4r = 7r pools per hour.They together fill the pool in 10 hours, so 7r = 1/10 pool per hour.Thus r = 1 / (7 * 10) = 1/70 pool per hour.Therefore, the time taken by tanker A alone to fill the pool is 1 / r = 70 hours.Verification / Alternative check:Check using the computed rates.Rate of A = 1/70, rate of B = 2/70 = 1/35, rate of C = 4/70 = 2/35 pools per hour.Combined rate = 1/70 + 1/35 + 2/35.Convert to denominator 70: 1/70 + 2/70 + 4/70 = 7/70 = 1/10 pool per hour.Thus they fill the pool in 10 hours, which matches the problem statement.Why Other Options Are Wrong:40 hours or 50 hours would make tanker A too fast; the combined rate would then exceed 1/10 pool per hour, resulting in a filling time shorter than 10 hours. 80 or 90 hours would make tanker A too slow; when combined with B and C according to the ratio, they would take longer than 10 hours to fill the pool. Only 70 hours is consistent with all given conditions.Common Pitfalls:Students sometimes confuse the directions of the ratios and might mistakenly set A : B : C as 1 : 2 : 4 or 4 : 2 : 1 incorrectly. It is important to interpret "C is twice as fast as B" and "B is twice as fast as A" correctly and express each rate in terms of the slowest tanker. Another mistake is averaging the times directly instead of working with rates. Remember that rates add, while times for completion do not.Final Answer:Tanker A alone will take 70 hours to fill the swimming pool.

Question 18

Two taps X and Y can fill a water tank in 15 hours and 20 hours respectively. Both taps are opened together at 2 p.m. At what time should tap X be closed so that the tank is just full at exactly 2 a.m. the next day?

Accepted Answer

Introduction / Context:This problem involves scheduling the operation of two taps so that a tank is full at a specified future time. Both taps start together, but one tap is turned off earlier. The question checks understanding of work rates over different time intervals, requiring careful tracking of how long each tap runs and how much work is done during those intervals.Given Data / Assumptions: Tap X alone fills the tank in 15 hours. Tap Y alone fills the tank in 20 hours. Both taps are opened at 2 p.m. The tank must be exactly full at 2 a.m. the next day. Flow rates are constant and there are no leaks. Concept / Approach:First, compute the total duration between 2 p.m. and 2 a.m., which is 12 hours. Let x be the number of hours that tap X remains open after 2 p.m. Tap Y runs the entire 12 hours. We then express the total work done as the sum of work contributed by each tap: tap X works for x hours at its rate, and tap Y works for 12 hours at its rate. This total must equal one full tank. Solving the resulting equation for x tells us when tap X should be closed.Step-by-Step Solution:Let the capacity of the tank be 1 unit.Rate of tap X = 1/15 tank per hour.Rate of tap Y = 1/20 tank per hour.Total available time from 2 p.m. to 2 a.m. is 12 hours.Let tap X be open for x hours (starting at 2 p.m.).Tap Y remains open for the entire 12 hours.Work done by X = x * (1/15) tank.Work done by Y = 12 * (1/20) tank.Total work must equal 1 tank, so x/15 + 12/20 = 1.Compute 12/20 = 3/5.The equation becomes x/15 + 3/5 = 1.Rearrange: x/15 = 1 - 3/5 = 2/5.Thus x = 15 * (2/5) = 30/5 = 6 hours.So tap X should be closed 6 hours after 2 p.m., that is at 8 p.m.Verification / Alternative check:From 2 p.m. to 8 p.m., both taps work for 6 hours.Work in first 6 hours = 6 * (1/15 + 1/20) = 6 * (7/60) = 42/60 = 7/10 of the tank.From 8 p.m. to 2 a.m., only tap Y works for another 6 hours.Work by Y in last 6 hours = 6 * (1/20) = 6/20 = 3/10 of the tank.Total work = 7/10 + 3/10 = 1 full tank, which confirms the timing.Why Other Options Are Wrong:Closing tap X at 7 p.m., 9 p.m., 10 p.m. or 11 p.m. would lead to total work different from exactly one tank. Closing too early would mean the tank does not reach full capacity by 2 a.m. Closing too late would cause the tank to overfill before 2 a.m. Only 8 p.m. balances the contributions correctly.Common Pitfalls:Some students miscalculate the total time from 2 p.m. to 2 a.m. or forget that tap Y runs throughout. Others attempt to average the times 15 hours and 20 hours instead of adding work contributions. Mismanaging fractions when solving x/15 + 3/5 = 1 is another common error. Writing out each step and converting to simple fractions helps to avoid mistakes.Final Answer:Tap X should be closed at 8 p.m. so that the tank is full at exactly 2 a.m.

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