Question 1

Two pipes A and B can fill a cistern in 10 minutes and 15 minutes respectively. A person opens both pipes together but accidentally leaves a waste pipe open as well. He expects the cistern to be full when only A and B would have filled it, but at th…

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Introduction / Context:This question involves two inlet pipes and one waste pipe. The person expects the cistern to be full at a certain time based on the filling rates of the two inlets, but discovers a waste pipe is also open. After closing the waste pipe, the cistern becomes full 3 minutes later. From this information we must deduce the emptying rate of the waste pipe and hence how long it would take to empty a full cistern on its own. Given Data / Assumptions: Pipe A alone fills the cistern in 10 minutes. Pipe B alone fills the cistern in 15 minutes. The person initially opens both A and B together, but the waste pipe is also open. If only A and B were open, the cistern would be full in 6 minutes. At that 6 minute mark, he discovers the waste pipe is open, closes it, and the cistern becomes full 3 minutes later. Concept / Approach:We first find the combined filling rate of A and B. Since the person discovers the problem at the time when A and B alone should have filled the cistern (6 minutes), we know that up to that time all three pipes have been operating. We compute the fraction filled after 6 minutes with all three pipes, then deduce the waste pipe rate from the fact that the remaining volume is filled in 3 minutes by A and B only. Step-by-Step Solution:Step 1: Rate of A = 1 / 10 cistern per minute.Step 2: Rate of B = 1 / 15 cistern per minute.Step 3: Combined rate of A and B = 1 / 10 + 1 / 15 = (3 + 2) / 30 = 1 / 6 cistern per minute.Step 4: If only A and B were working, they would fill the cistern in 6 minutes.Step 5: Let waste pipe rate be W cistern per minute (emptying), so net rate with all three open is 1 / 6 - W.Step 6: In the first 6 minutes, volume filled = 6 * (1 / 6 - W) = 1 - 6W.Step 7: After the waste pipe is closed, only A and B fill the cistern. In the next 3 minutes, volume added = 3 * (1 / 6) = 1 / 2.Step 8: This 1 / 2 must be exactly the remaining fraction of the cistern, so remaining fraction after 6 minutes = 1 / 2.Step 9: Therefore, 1 - 6W = 1 / 2, so 6W = 1 / 2 and W = 1 / 12 cistern per minute.Step 10: Time taken by waste pipe alone to empty a full cistern = 1 / W = 12 minutes. Verification / Alternative check:Check the partial filling: with W = 1 / 12, net rate during the first 6 minutes = 1 / 6 - 1 / 12 = 1 / 12 cistern per minute. In 6 minutes that fills 6 * (1 / 12) = 1 / 2 of the cistern. Then A and B alone fill the remaining half in 3 minutes at rate 1 / 6, since 3 * (1 / 6) = 1 / 2. The total time is 9 minutes, consistent with the story, and the calculation of W is confirmed. Why Other Options Are Wrong:10 minutes, 8 minutes and 15 minutes correspond to waste pipe rates that do not result in exactly half the tank remaining after 6 minutes and thus do not satisfy the information about the last 3 minutes of filling. Common Pitfalls:Some students incorrectly assume the first phase lasts an unknown time rather than the 6 minutes determined by the A and B filling time. Others mix up net rates, adding instead of subtracting the waste pipe rate. Keeping track of volumes and writing equations carefully for each phase prevents such errors. Final Answer:The waste pipe alone can empty a full cistern in 12 minutes.

Question 2

Two inlet pipes can fill a tank in 20 minutes and 24 minutes respectively, and a waste pipe can empty water from the tank at the rate of 3 gallons per minute. When all three pipes are opened together, the tank is filled in 15 minutes. What is the ca…

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Introduction / Context:This problem mixes fractional filling rates with an absolute emptying rate in gallons per minute. Two pipes fill the tank, while a waste pipe empties it at a fixed volume rate. We must find the total capacity of the tank given the net effect when all three operate together. Given Data / Assumptions: Pipe 1 fills the tank in 20 minutes. Pipe 2 fills the tank in 24 minutes. The waste pipe empties water at 3 gallons per minute. All three pipes operating together fill the tank in 15 minutes. Concept / Approach:Let the capacity of the tank be V gallons. The two inlets fill at rates V / 20 and V / 24 gallons per minute. The waste pipe empties at 3 gallons per minute. The net filling rate is then V / 20 + V / 24 - 3 gallons per minute. Since the tank is filled in 15 minutes, the net rate must also equal V / 15 gallons per minute. Setting these expressions equal allows us to solve for V. Step-by-Step Solution:Step 1: Let tank capacity be V gallons.Step 2: Rate of pipe 1 = V / 20 gallons per minute.Step 3: Rate of pipe 2 = V / 24 gallons per minute.Step 4: Rate of waste pipe = 3 gallons per minute (emptying).Step 5: Net rate with all three pipes = V / 20 + V / 24 - 3 gallons per minute.Step 6: Because the tank is filled in 15 minutes, net rate must also equal V / 15 gallons per minute.Step 7: Set up the equation: V / 20 + V / 24 - 3 = V / 15.Step 8: Compute V / 20 + V / 24 = V * (1 / 20 + 1 / 24) = V * (6 / 120 + 5 / 120) = 11V / 120.Step 9: So we have 11V / 120 - 3 = V / 15.Step 10: Multiply through by 120 to clear denominators: 11V - 360 = 8V.Step 11: Rearrange: 11V - 8V = 360, so 3V = 360 and V = 120.Step 12: Therefore, the capacity of the tank is 120 gallons. Verification / Alternative check:Check the rates with V = 120. Pipe 1 fills at 120 / 20 = 6 gallons per minute. Pipe 2 fills at 120 / 24 = 5 gallons per minute. Combined inlets give 11 gallons per minute. Subtracting the waste rate of 3 gallons per minute, net rate is 8 gallons per minute. In 15 minutes, total volume filled = 8 * 15 = 120 gallons, which matches the tank capacity. Why Other Options Are Wrong:60 gallons and 100 gallons produce net rates that do not match the given 15 minute fill time when substituted into the equation. 180 gallons would imply larger inlet rates and would not satisfy the relation that all three together fill the tank exactly in 15 minutes. Common Pitfalls:Students sometimes forget that the waste pipe rate is in gallons per minute, not a fraction of the tank per minute, and treat it as 1 / t. Others add instead of subtract the waste rate, which reverses its effect. Careful unit tracking and consistent use of the tank capacity symbol V avoid these issues. Final Answer:The capacity of the tank is 120 gallons.

Question 3

An empty reservoir has a capacity of 30 litres. An inlet pipe fills water at 5 litres per minute, and an outlet pipe empties water at 4 litres per minute. The pipes operate alternately for one-minute intervals, with the inlet pipe operating first. A…

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Introduction / Context:This problem involves alternating inflow and outflow. One pipe fills the reservoir for a minute, then the other pipe empties for a minute, and this pattern repeats. We must track the volume after each minute to find the first time the reservoir reaches exactly 30 litres. This is a good example of a discrete step process rather than a constant-rate process. Given Data / Assumptions: Reservoir capacity = 30 litres. Inlet pipe rate = 5 litres per minute. Outlet pipe rate = 4 litres per minute. Pipes operate in 1-minute intervals, starting with the inlet pipe. The water level cannot exceed 30 litres; we stop once it hits 30 litres for the first time. Concept / Approach:Since the pipes act in alternate 1-minute blocks, the volume in the reservoir changes in discrete steps. In each 2-minute cycle, the net effect is an increase of 1 litre (fill 5, then empty 4). We can model the process cycle by cycle and then determine at which minute within a cycle the reservoir first reaches 30 litres. Step-by-Step Solution:Step 1: Start with 0 litres at time 0.Step 2: Minute 1 (inlet): volume increases by 5 litres to 5 litres.Step 3: Minute 2 (outlet): volume decreases by 4 litres to 1 litre.Step 4: Thus, every 2-minute cycle increases the volume by 1 litre overall.Step 5: After n complete 2-minute cycles, volume = n litres. Time elapsed = 2n minutes.Step 6: We also have an inlet action at the start of each new cycle. After n cycles and one more inlet minute, volume = n + 5 litres.Step 7: We need to find the smallest integer minute when the volume first reaches 30 litres.Step 8: From the stepwise pattern, just before the final filling minute, volume should be 25 litres so that a +5 litre fill takes it to exactly 30.Step 9: Volume equals 1 litre after the first 2-minute cycle, 2 litres after the second, and so on. So volume equals 25 litres after 25 cycles.Step 10: 25 cycles correspond to 25 * 2 = 50 minutes.Step 11: At minute 51, the inlet pipe runs and adds 5 litres, taking the volume from 25 litres to 30 litres for the first time. Verification / Alternative check:If we simulate the process, we see the volume sequence at the end of inlet minutes as 5, 6, 7, and so on, and at the end of outlet minutes as 1, 2, 3, and so on. Following this pattern shows that we reach 25 litres after 50 minutes and then hit 30 litres after the inlet runs at minute 51, which corroborates the analytical reasoning. Why Other Options Are Wrong:49.5 minutes is not an integer and does not match the discrete 1-minute switching pattern. 50 minutes leaves the reservoir at 25 litres, not yet full. 52 minutes would mean one extra outlet minute after the reservoir has already become full, which is beyond the first time it hits full capacity. Common Pitfalls:Students often treat the net rate as 1 litre per 2 minutes and divide 30 by 0.5 to get 60 minutes, ignoring that the reservoir may become full during the filling minute within a cycle. It is important to track the sequence minute by minute, especially for discrete alternating processes, rather than relying solely on an average net rate per cycle. Final Answer:The reservoir will become completely full for the first time after 51 minutes.

Question 4

Water flows into a rectangular tank that is 200 m long and 150 m wide through a pipe of cross section 0.3 m x 0.2 m at a speed of 20 km/h. In how many hours will the water level in the tank rise by 12 m?

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Introduction / Context: This question tests the standard pipes and cistern idea that time taken to fill a container equals volume divided by flow rate. Here the tank is large and water is flowing through a pipe with a given cross section and speed, so we must convert the speed into consistent units and then compute the time required for the water level to rise by a fixed height of 12 m. Given Data / Assumptions: - Length of tank = 200 m - Width of tank = 150 m - Rise in water level required = 12 m - Pipe cross section = 0.3 m x 0.2 m - Flow speed in pipe = 20 km per hour = 20000 m per hour - Assume uniform flow and no leakage or spillage Concept / Approach: The key ideas are: - Volume of water required = base area of tank * rise in level - Discharge (flow rate) of pipe = cross sectional area of pipe * velocity of water in the pipe - Time = required volume / flow rate, with all quantities in consistent units (here in cubic metres and hours). Step-by-Step Solution: Step 1: Base area of tank = length * width = 200 * 150 = 30000 m^2. Step 2: Volume needed for 12 m rise = 30000 * 12 = 360000 m^3. Step 3: Cross sectional area of pipe = 0.3 * 0.2 = 0.06 m^2. Step 4: Convert speed: 20 km/h = 20000 m/h. Step 5: Flow rate of pipe = 0.06 * 20000 = 1200 m^3 per hour. Step 6: Time required = volume / rate = 360000 / 1200 = 300 hours. Verification / Alternative check: We can simplify the fraction 360000 / 1200 by dividing numerator and denominator by 100 to get 3600 / 12 = 300. Units also check out as hours since we used m^3 and m^3 per hour. The magnitude is large because the tank is extremely big compared with the relatively small pipe. Why Other Options Are Wrong: 200 hours: This is too small and would correspond to a higher flow rate than given by the pipe dimensions and speed. 240 hours: Also underestimates the required time because it assumes either a smaller volume or a larger discharge than the actual values. 270 hours: Slightly closer but still not equal to the exact ratio 360000 / 1200, so it is not correct. Common Pitfalls: A frequent mistake is to forget to convert 20 km/h into metres per hour, or to mix hours and minutes. Another error is to treat 12 m as the total depth instead of the rise in level, although here both interpretations match because the initial level is not specified. Always ensure that area, velocity and volume are in compatible units before computing time. Final Answer: The required time for the water level in the tank to rise by 12 m is 300 hours.

Question 5

Twelve buckets of water, each of capacity 13.5 litres, are just enough to fill a tank. How many buckets will be required to fill the same tank completely if each bucket now has a capacity of 9 litres?

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Introduction / Context: This is a direct capacity and volume comparison problem. The tank capacity remains fixed, but the size of each bucket changes. The idea is to compute the volume of the tank from the first situation and then find how many smaller buckets are needed to hold the same volume of water. Given Data / Assumptions: - Number of buckets in the first case = 12 - Capacity of each bucket in the first case = 13.5 litres - Capacity of each bucket in the second case = 9 litres - The tank is completely filled in both situations - No water is spilled or wasted in either case Concept / Approach: We use the relationship: - Tank capacity = number of buckets * capacity per bucket. The tank capacity found from the first situation must equal total water from the second situation. Then the required number of buckets in the second case is tank capacity divided by capacity of each smaller bucket. Step-by-Step Solution: Step 1: Compute tank capacity from the first case: capacity = 12 * 13.5 litres. Step 2: 12 * 13.5 = 162, so the tank holds 162 litres. Step 3: In the second case, capacity of each bucket = 9 litres. Step 4: Required number of buckets = total volume / capacity of one new bucket. Step 5: Number of buckets = 162 / 9 = 18. Verification / Alternative check: We can verify by multiplying back: 18 buckets * 9 litres per bucket = 162 litres, which matches the tank capacity found earlier. Therefore the count of 18 buckets is consistent and correct. Why Other Options Are Wrong: 8 buckets: 8 * 9 = 72 litres, which is much less than the required 162 litres. 15 buckets: 15 * 9 = 135 litres, still short of the full tank capacity. 16 buckets: 16 * 9 = 144 litres, also less than 162 litres, so the tank would not be full. Common Pitfalls: A common mistake is to assume that if the capacity reduces by a certain factor, the number of buckets changes by the same numeric difference rather than by a ratio. Students may also confuse 13.5 with 13 or 14 while doing mental multiplication. Writing the intermediate step for the tank capacity avoids such careless errors. Final Answer: The tank will be filled completely by 18 buckets of capacity 9 litres each.

Question 6

One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then in how many minutes will the slower pipe alone fill the tank?

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Introduction / Context: This is a classic pipes and cistern time and work question. Two pipes working together fill a tank in a known time, and one pipe is three times as fast as the other. We must express their rates in terms of a single variable, use the combined rate, and then find the time for the slower pipe. Given Data / Assumptions: - Let the slower pipe be pipe S and the faster pipe be pipe F - Pipe F is three times as fast as pipe S - Together S and F fill the tank in 36 minutes - Tank is assumed empty at the start and there is no leakage Concept / Approach: If a pipe fills a tank in T minutes, its rate is 1/T tank per minute. We set up rates in terms of a single variable for the slower pipe, add the rates for the combined work, and equate this to the given overall filling rate. Then we solve for the time of the slower pipe. Step-by-Step Solution: Step 1: Let time taken by the slower pipe alone be x minutes, so its rate is 1/x tank per minute. Step 2: Faster pipe is three times as fast, so its rate is 3/x tank per minute and its time alone would be x/3 minutes. Step 3: Combined rate = 1/x + 3/x = 4/x tank per minute. Step 4: Together they fill the tank in 36 minutes, so combined rate is 1/36 tank per minute. Step 5: Set 4/x = 1/36, so x = 4 * 36 = 144 minutes. Verification / Alternative check: If the slower pipe takes 144 minutes, the faster pipe takes 144/3 = 48 minutes. Rates are 1/144 and 1/48. Their sum equals 1/144 + 1/48 = (1 + 3) / 144 = 4/144 = 1/36, confirming that together they indeed fill the tank in 36 minutes. Why Other Options Are Wrong: 81 minutes: This would make the faster pipe time 27 minutes, giving a combined time different from 36 minutes. 108 minutes: Gives a combined rate higher than required, leading to a filling time less than 36 minutes. 192 minutes: Too slow, leading to a combined time greater than 36 minutes. Common Pitfalls: A common mistake is to assume that if one pipe is three times as fast, then its time is simply 36 / 3 without setting up the equation, which leads to wrong results. Another error is to mix up time and rate, or to add times directly instead of adding rates. Always work with rates when dealing with combined work problems. Final Answer: The slower pipe alone will fill the tank in 144 minutes.

Question 7

Two pipes A and B can fill a tank in 16 hours and 12 hours respectively. The capacity of the tank is 240 litres. If both pipes are opened together and then closed after 2 hours, how many more litres of water are still needed to fill the tank complet…

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Introduction / Context: This problem combines the idea of fractional tank filling with an actual volume in litres. Two pipes fill the tank together for a limited time, and we must determine the remaining volume needed to fill the tank completely. It is an application of rates, fractions of work done, and unit conversion to litres. Given Data / Assumptions: - Pipe A fills the entire tank in 16 hours - Pipe B fills the entire tank in 12 hours - Tank capacity = 240 litres - Both pipes run together for 2 hours and then are closed - No leakage or overflow is present Concept / Approach: We first compute the combined fraction of the tank filled in the 2 hours using their rates in terms of the fraction of the tank per hour. Then we convert this fraction to litres to get water already stored and subtract from the total capacity to find the remaining volume. Step-by-Step Solution: Step 1: Rate of pipe A = 1/16 tank per hour. Step 2: Rate of pipe B = 1/12 tank per hour. Step 3: Combined rate = 1/16 + 1/12 = (3 + 4) / 48 = 7/48 tank per hour. Step 4: In 2 hours they fill fraction = 2 * 7/48 = 14/48 = 7/24 of the tank. Step 5: Fraction of tank still empty = 1 - 7/24 = 17/24. Step 6: Remaining volume in litres = 17/24 * 240 = 17 * 10 = 170 litres. Verification / Alternative check: We can also compute the volume already filled: 7/24 of 240 = 7 * 10 = 70 litres. The tank capacity is 240 litres, so remaining = 240 - 70 = 170 litres. Both methods match, which confirms the calculation. Why Other Options Are Wrong: 70 litres: This is the amount already filled, not the amount remaining. 90 litres: Corresponds to an incorrect fraction of the tank and does not match either method. 190 litres: Would imply that only 50 litres have been filled, which conflicts with the calculated combined rate of the two pipes. Common Pitfalls: Students often mix up the filled part with the remaining part. Another frequent error is to add the times directly instead of the rates. Also, some may forget to convert the fractional result of the tank into litres using the given total capacity, which is essential when the answer is expected in litres. Final Answer: After closing both pipes, an additional 170 litres of water are needed to fill the tank completely.

Question 8

A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. If pipe B is used alone for the first half of the total time and then both pipes A and B are used together for the second half, how many minutes in total ar…

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Introduction / Context: This question checks understanding of variable rates over different time intervals. A slower and a faster pipe are operated in different combinations during two equal halves of the total filling time. We must correctly express the filled fraction in each half and ensure that the sum equals one full tank. Given Data / Assumptions: - Pipe A alone fills the tank in 60 minutes - Pipe B alone fills the tank in 40 minutes - For the first half of the total time only pipe B is used - For the second half of the total time both A and B are used together - Tank is initially empty and there is no leakage Concept / Approach: Let the total filling time be t minutes. Then each half period lasts t/2 minutes. In the first half we use only the rate of B, and in the second half we use the combined rates of A and B. The sum of the fractions of the tank filled in these two halves must equal 1. We then solve for t. Step-by-Step Solution: Step 1: Rate of A = 1/60 tank per minute. Step 2: Rate of B = 1/40 tank per minute. Step 3: In the first half (t/2 minutes), only B works, so fraction filled = (t/2) * 1/40 = t/80. Step 4: In the second half (t/2 minutes), A and B together work, combined rate = 1/60 + 1/40 = (2 + 3) / 120 = 5/120 = 1/24 tank per minute. Step 5: Fraction filled in second half = (t/2) * 1/24 = t/48. Step 6: Total fraction filled = t/80 + t/48 = 1. Take LCM 240: (3t + 5t) / 240 = 8t / 240 = t/30 = 1, so t = 30 minutes. Verification / Alternative check: In 15 minutes (half of 30 minutes), B alone fills 15 * 1/40 = 15/40 = 3/8 of the tank. In the next 15 minutes, A and B together fill 15 * 1/24 = 15/24 = 5/8 of the tank. Total = 3/8 + 5/8 = 1 full tank, which confirms the result. Why Other Options Are Wrong: 31 minutes or 29 minutes: These do not satisfy the equation formed from the rates and lead to a total fraction slightly more or less than 1. 28 minutes: Also fails the rate equation and underestimates the time needed to fill the tanker. Common Pitfalls: One common mistake is to assume that the tanker is worked on by both pipes for 30 minutes and then adjusted, rather than defining t properly and splitting it into equal halves. Another pitfall is to add the times instead of adding the fractions of the tank filled. Always work with rates and fractions of work when intervals change. Final Answer: The tanker will be filled completely in a total of 30 minutes under the given operating pattern.

Question 9

Pipe A is an inlet pipe that can fill an empty cistern in 69 hours, while pipe B is an outlet pipe that can empty a full cistern in 46 hours. Starting with the cistern full, the two pipes are operated one at a time for one hour each, beginning with …

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Introduction / Context: This problem involves alternating usage of a filling pipe and an emptying pipe on an initially full cistern. The net effect over a repeated two hour cycle must be studied to find how long it takes until the cistern becomes empty. It is a good example of combining rates and dealing with long time spans expressed in days and hours. Given Data / Assumptions: - Pipe A fills a completely empty cistern in 69 hours - Pipe B empties a completely full cistern in 46 hours - The cistern is initially full - First hour: only pipe B is open - Second hour: only pipe A is open - This two hour pattern repeats (B then A, one hour each) - No other leaks or sources of water are present Concept / Approach: We compute the fraction of the cistern emptied or filled by each pipe in one hour. Over a two hour cycle (B then A) we find the net change in water level. Since the pattern repeats identically, the volume after k cycles is a linear sequence. We determine the number of cycles needed until the cistern is exactly empty and then convert the total time into days and hours. Step-by-Step Solution: Step 1: Rate of pipe B (emptying) = 1/46 cistern per hour. Step 2: Rate of pipe A (filling) = 1/69 cistern per hour. Step 3: First hour (pipe B only): change in level = -1/46. Step 4: Second hour (pipe A only): change in level = +1/69. Step 5: Net change per 2 hour cycle = -1/46 + 1/69. Step 6: Take LCM of 46 and 69 which is 138, so net change = (-3 + 2) / 138 = -1/138 of the cistern per 2 hour cycle. Step 7: Let the cistern be full at time zero. After k cycles, volume fraction = 1 - k/138. Step 8: For the cistern to be just empty, set 1 - k/138 = 0 which gives k = 138 cycles. Step 9: Each cycle lasts 2 hours, so total time = 138 * 2 = 276 hours. Step 10: Convert 276 hours into days and hours: 1 day = 24 hours; 276 / 24 = 11 days with remainder 276 - 11 * 24 = 12 hours. Verification / Alternative check: We can confirm by noting that in 138 cycles the net fraction removed is 138 * (1/138) = 1 cistern. Since the net effect per cycle is always negative, the cistern will be empty exactly at the end of a full number of cycles. The conversion from hours to days is also straightforward: 11 * 24 = 264 hours, plus 12 hours equals 276 hours, which matches the computed total. Why Other Options Are Wrong: 11 days 10 hours: Corresponds to 274 hours, which is slightly less than the required 276 hours, so the cistern would not yet be empty. 11 days 7 hours: Even smaller total time and therefore definitely insufficient to empty the cistern. 1 day 13 hours: Much too small compared with the slow net emptying rate of 1/138 per 2 hours. Common Pitfalls: A common mistake is to average the individual times rather than the rates, or to forget that B empties and A fills, so their effects have opposite signs. Another pitfall is to try to track hour by hour instead of grouping into cycles, which makes the problem tedious and error prone. Always compute net change over one full pattern cycle and then scale up. Careless conversion between hours and days can also lead to wrong final options. Final Answer: The cistern will become empty after a total of 11 days 12 hours of alternating the two pipes as described.

Question 10

Taps X and Y can fill a tank in 30 minutes and 40 minutes respectively, while tap Z can empty a full tank in 60 minutes. If all three taps are opened together and kept running continuously, in how many minutes will the tank be completely filled?

Accepted Answer

Introduction / Context: This question involves two inlet taps and one outlet tap operating at the same time. We need to find the net filling rate and then the time required to fill the tank completely. It is a typical pipes and cistern situation with both filling and emptying effects acting simultaneously. Given Data / Assumptions: - Tap X fills the tank in 30 minutes - Tap Y fills the tank in 40 minutes - Tap Z empties a full tank in 60 minutes - All three taps are opened together and kept open - The tank is initially empty and has no other leakage Concept / Approach: If a tap fills a tank in T minutes, its rate is 1/T of the tank per minute. If a tap empties the tank in T minutes, its rate is -1/T of the tank per minute. When taps operate together, their rates add algebraically. After we compute the net rate, time is simply the reciprocal of this rate. Step-by-Step Solution: Step 1: Rate of X = 1/30 tank per minute. Step 2: Rate of Y = 1/40 tank per minute. Step 3: Rate of Z (emptying) = -1/60 tank per minute. Step 4: Net rate when all three are open = 1/30 + 1/40 - 1/60. Step 5: Take LCM of 30, 40 and 60 which is 120. Then rates are 4/120, 3/120 and -2/120 respectively. Step 6: Net rate = (4 + 3 - 2) / 120 = 5/120 = 1/24 tank per minute. Step 7: Time required to fill the tank = 1 divided by 1/24 = 24 minutes. Verification / Alternative check: We can also reason proportionally. In 24 minutes, X would fill 24/30 = 4/5 of the tank, Y would fill 24/40 = 3/5 of the tank, and Z would empty 24/60 = 2/5 of the tank. Net filled fraction = 4/5 + 3/5 - 2/5 = 5/5 = 1 whole tank, which confirms the time of 24 minutes. Why Other Options Are Wrong: 48 minutes and 72 minutes: These assume much slower net filling rates; if used, the net fraction filled would exceed 1 for the given inflows and outflow. None of these: This is incorrect because 24 minutes is an exact solution and matches the given options list. Common Pitfalls: The main mistake is to add or subtract the times instead of adding the rates. Some may also forget that the emptying tap has a negative contribution. Miscalculating the LCM or the fraction operations can also lead to wrong net rates. Writing each rate explicitly and verifying the arithmetic helps avoid these errors. Final Answer: When all three taps run together, the tank will be completely filled in 24 minutes.

Question 11

The capacity of a tank is 2400 cubic metres. A pump can be used both for filling and for emptying the tank. Its emptying capacity is 10 m^3 per minute higher than its filling capacity, and it takes 8 minutes less to empty the tank than to fill it. W…

Accepted Answer

Introduction / Context: Here a single pump can either fill or empty a tank, with different capacities in each mode. The emptying capacity is known to be larger by a fixed amount, and the time difference for filling and emptying is given. We use the basic relationship time = volume / rate to set up an equation in one variable and then solve for the filling capacity. Given Data / Assumptions: - Tank capacity = 2400 m^3 - Let filling capacity of the pump = f m^3 per minute - Then emptying capacity of the pump = f + 10 m^3 per minute - Time to fill the tank = 2400 / f minutes - Time to empty the tank = 2400 / (f + 10) minutes - Emptying takes 8 minutes less than filling Concept / Approach: We use the equation based on the given time difference: time to empty = time to fill minus 8 minutes. Substituting time expressions in terms of f gives a rational equation. Solving this equation gives a positive value of f that matches one of the given options. Step-by-Step Solution: Step 1: Time to fill = 2400 / f minutes. Step 2: Time to empty = 2400 / (f + 10) minutes. Step 3: Given that emptying is 8 minutes faster: 2400 / (f + 10) = 2400 / f - 8. Step 4: Rearrange: 2400 / (f + 10) - 2400 / f = -8. Step 5: Combine fractions: 2400[f - (f + 10)] / [f(f + 10)] = -8 which simplifies to 2400 * (-10) / [f(f + 10)] = -8. Step 6: This gives 24000 / [f(f + 10)] = 8, so f(f + 10) = 24000 / 8 = 3000. Step 7: Solve f^2 + 10f - 3000 = 0. The positive root is f = 50 m^3 per minute. Verification / Alternative check: If f = 50, then emptying capacity is 60 m^3 per minute. Time to fill = 2400 / 50 = 48 minutes. Time to empty = 2400 / 60 = 40 minutes. The difference in times is indeed 48 - 40 = 8 minutes, so f = 50 m^3 per minute is consistent with all the conditions. Why Other Options Are Wrong: 46, 44 or 48 cubic metres per minute: These values do not satisfy the quadratic equation f^2 + 10f - 3000 = 0 and fail to produce a time difference of exactly 8 minutes between filling and emptying when substituted into the time formulas. Common Pitfalls: Students may incorrectly write the time relation as filling time minus emptying time equals 8, reversing the sign. Another common error is algebraic manipulation when combining fractions or solving the quadratic equation. Careful stepwise substitution and simplification prevents such mistakes. Final Answer: The filling capacity of the pump is 50 cubic metres per minute.

Question 12

Pipe A can fill a tank in 16 minutes and pipe B can empty the same tank in 24 minutes. If both pipes are opened together, after how many minutes should pipe B be closed so that the tank is exactly full in a total time of 30 minutes?

Accepted Answer

Introduction / Context: This problem mixes a filling pipe and a draining pipe that initially work together, but one pipe is later closed. We must choose the correct moment to close the outlet pipe so that the tank becomes exactly full after a specified total time. This needs careful handling of rates and time intervals. Given Data / Assumptions: - Pipe A fills the tank in 16 minutes - Pipe B empties the tank in 24 minutes - Both start together at time zero - Total time for the tank to become full = 30 minutes - Pipe B will be closed after some time t minutes, while A continues Concept / Approach: We divide the process into two phases: an initial phase where both A and B are open, and a second phase where only A runs. We then compute the fraction of the tank filled in each phase in terms of t. The sum of these fractions must equal 1. Solving this equation gives the required time t after which pipe B should be closed. Step-by-Step Solution: Step 1: Rate of pipe A = 1/16 tank per minute. Step 2: Rate of pipe B = -1/24 tank per minute (since it empties). Step 3: Combined rate with both open = 1/16 - 1/24. Step 4: Take LCM of 16 and 24 which is 48. So combined rate = (3 - 2) / 48 = 1/48 tank per minute. Step 5: Let pipe B be open for t minutes. Fraction filled in this period = t * 1/48 = t/48. Step 6: Remaining time with only pipe A = 30 - t minutes. Fraction filled during this time = (30 - t) * 1/16. Step 7: Total fraction filled = t/48 + (30 - t)/16 = 1. Step 8: Convert to common denominator 48: t/48 + 3(30 - t)/48 = 1, so t + 90 - 3t = 48. Step 9: Simplify: 90 - 2t = 48, so 2t = 42 and t = 21 minutes. Verification / Alternative check: For the first 21 minutes, both pipes run, filling 21 * 1/48 = 21/48 = 7/16 of the tank. Remaining fraction = 1 - 7/16 = 9/16. Only A runs for the last 9 minutes, and A fills 9 * 1/16 = 9/16 of the tank. Total = 7/16 + 9/16 = 1 full tank, matching the requirement. Why Other Options Are Wrong: 24, 20 or 22 minutes: Each of these values for t fails to satisfy the equation t/48 + (30 - t)/16 = 1. They would either overfill or underfill the tank by the end of 30 minutes. Common Pitfalls: Learners sometimes forget to treat the outlet as a negative rate or incorrectly assume that the tank is half full at half time. Another frequent error is incorrect algebra when solving the linear equation for t. Writing out the equation clearly and checking the arithmetic avoids such mistakes. Final Answer: Pipe B should be closed after 21 minutes so that the tank is full in exactly 30 minutes.

Question 13

The efficiencies of two filling pipes are in the ratio 4 : 5. A third pipe is an emptying pipe whose efficiency is two thirds of the average efficiency of the first two pipes, and it can empty a completely full tank in 36 minutes. In how many minute…

Accepted Answer

Introduction / Context: This problem relates efficiencies of three pipes: two inlets and one outlet. The outlet pipe rate is linked to the average rate of the first two pipes, and its emptying time is given. From this information we can derive the absolute rates of the filling pipes and then find the time they need to fill the tank together. Given Data / Assumptions: - Efficiency ratio of two filling pipes A and B = 4 : 5 - Average efficiency of A and B = (4 + 5) / 2 = 4.5 parts - Emptying pipe C has efficiency = two thirds of this average = 2/3 * 4.5 = 3 parts - Pipe C alone empties a full tank in 36 minutes - All efficiencies are measured in the same units of tank per minute Concept / Approach: We treat the given ratio numbers as proportional to actual rates. Since C empties the tank in 36 minutes, we can equate its proportional efficiency to the real rate 1/36 and find the scale factor. Using this factor we find actual rates for A and B. Then we add the rates of A and B to get their combined rate and find the time to fill one tank. Step-by-Step Solution: Step 1: Let the common efficiency unit be k. Then rates are: A = 4k, B = 5k, C = 3k. Step 2: Pipe C empties a full tank in 36 minutes, so 3k = 1/36 tank per minute. Step 3: Therefore k = 1 / (36 * 3) = 1/108. Step 4: Actual rate of A = 4k = 4/108 = 1/27 tank per minute. Step 5: Actual rate of B = 5k = 5/108 tank per minute. Step 6: Combined rate of A and B = 1/27 + 5/108 = 4/108 + 5/108 = 9/108 = 1/12 tank per minute. Step 7: Time required by A and B together = 1 divided by 1/12 = 12 minutes. Verification / Alternative check: We can check quickly: if A and B together fill the tank in 12 minutes, they fill 1/12 of the tank per minute. Using k = 1/108, the sum of their proportional rates 4k + 5k equals 9k which is 9/108 = 1/12, exactly matching the calculated combined rate. Why Other Options Are Wrong: 16, 14 and 20 minutes: These times correspond to smaller combined rates than 1/12 tank per minute. When back substituted, they are inconsistent with the emptying information that C empties the tank in 36 minutes based on efficiency two thirds of the average of A and B. Common Pitfalls: Many students forget that the given efficiencies are in ratio form and must be linked to an actual rate using the emptying condition. Another mistake is computing the average of 4 and 5 incorrectly, or using one third instead of two thirds. Careful reading and consistent use of proportional reasoning are essential. Final Answer: Both filling pipes together can fill the tank in 12 minutes.

Question 14

A single tap can fill a tank in 6 hours. After this tap alone has filled half of the tank, three more identical taps are opened. What is the total time taken to fill the tank completely from empty?

Accepted Answer

Introduction / Context: This is a typical staged filling problem where initially one tap works alone and later more identical taps are opened. Because the taps are identical, their combined rate is a simple multiple of the single tap rate. We split the process into two phases and sum the times for each phase. Given Data / Assumptions: - One tap alone fills the tank in 6 hours - This tap is used alone until half the tank is full - After half the tank is filled, three more similar taps are opened, making four taps in total - All taps are identical and there are no leaks Concept / Approach: For a single tap with filling time 6 hours, its rate is 1/6 of the tank per hour. In the second phase, four identical taps work, so their combined rate is 4 times that of the single tap, that is 4/6 = 2/3 of the tank per hour. The first phase fills half the tank, and the second phase fills the remaining half. We compute time for each phase separately and then add them. Step-by-Step Solution: Step 1: Rate of one tap = 1/6 tank per hour. Step 2: Time to fill half the tank with one tap = (1/2) / (1/6) = 1/2 * 6 = 3 hours. Step 3: After this, 1/2 of the tank remains. Step 4: In the second phase, there are 4 identical taps, so combined rate = 4 * 1/6 = 4/6 = 2/3 tank per hour. Step 5: Time to fill the remaining half = (1/2) / (2/3) = 1/2 * 3/2 = 3/4 hour. Step 6: 3/4 hour equals 45 minutes. Step 7: Total time = 3 hours + 45 minutes = 3 hours 45 minutes. Verification / Alternative check: We can verify by computing total work. Phase one: 3 hours * 1/6 tank per hour = 1/2 tank. Phase two: 45 minutes is 0.75 hours and the rate is 2/3 tank per hour, so filled amount = 0.75 * 2/3 = 1/2 tank. Total = 1/2 + 1/2 = 1 tank, confirming that the times are correct. Why Other Options Are Wrong: 4 hours 15 minutes, 3 hours 24 minutes, 4 hours 51 minutes: Each of these totals would imply either too much or too little work done when multiplied by the appropriate rates. They do not satisfy both phase requirements simultaneously. Common Pitfalls: A common error is to think that adding more taps simply divides the initial time by the number of taps without considering that this only applies if they all operate for the entire duration. Another mistake is miscalculating the time for the second phase when converting between fractions of an hour and minutes. Final Answer: The tank will be completely filled in a total of 3 hours 45 minutes.

Question 15

One filling pipe A is three times as fast as a second filling pipe B and therefore takes 32 minutes less than B to fill a cistern. If both pipes A and B are opened together when the cistern is empty, in how many minutes will the cistern be completel…

Accepted Answer

Introduction / Context: This problem uses a relationship between the speeds of two filling pipes. Pipe A is three times as fast as pipe B, and we know the difference between their individual filling times. By expressing both times in terms of one variable, we can find each individual time and then their combined time to fill the cistern. Given Data / Assumptions: - Pipe A fills the cistern three times as fast as pipe B - Time taken by pipe A is 32 minutes less than time taken by pipe B - The cistern is initially empty - Both pipes are opened together, and we want the time until the cistern is full Concept / Approach: Let the time taken by the slower pipe B be x minutes. Since A is three times as fast, A will take x/3 minutes. The difference between these times is given as 32 minutes. Once we find x and x/3, we convert them into rates and add the rates to find the combined filling time. Step-by-Step Solution: Step 1: Let time taken by pipe B alone be x minutes. Step 2: Time taken by pipe A alone = x/3 minutes because it is three times as fast. Step 3: Given time difference: x - x/3 = 32. Step 4: Simplify: (2/3)x = 32, so x = 32 * 3 / 2 = 48 minutes. Step 5: Therefore, time for A alone = 48 / 3 = 16 minutes. Step 6: Rate of A = 1/16 tank per minute; rate of B = 1/48 tank per minute. Step 7: Combined rate = 1/16 + 1/48 = (3 + 1) / 48 = 4/48 = 1/12 tank per minute. Step 8: Time taken together = 1 divided by 1/12 = 12 minutes. Verification / Alternative check: Check the difference in individual times: B takes 48 minutes, A takes 16 minutes, and 48 - 16 = 32 minutes which matches the problem statement. Also, in 12 minutes A alone would fill 12/16 = 3/4 of the tank and B alone would fill 12/48 = 1/4 of the tank, so together they indeed fill one full tank in 12 minutes. Why Other Options Are Wrong: 6, 8 or 10 minutes: These times imply combined rates that are inconsistent with the derived individual rates and the given 32 minute difference. They would result in either more or less than one full tank being filled. Common Pitfalls: One common mistake is to assume that the faster pipe time is three times longer instead of three times shorter, which reverses the relation. Another error is to mis-handle the fraction when solving x - x/3 = 32. Always write out the equation clearly and simplify step by step. Final Answer: Together, pipes A and B will fill the cistern in 12 minutes.

Question 16

A water tank in a village is normally filled in 8 hours by an inlet pipe, but due to a leak at the bottom it now takes 2 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 question explores the effect of a leak on the filling process. The inlet pipe alone can fill the tank in a known time, but with the leak present the filling time increases. From the difference in effective rates, we deduce the rate of the leak and then compute the time taken by the leak alone to empty a full tank. Given Data / Assumptions: - Inlet alone fills the tank in 8 hours - With the leak present, the tank fills in 10 hours - The tank is initially full when we later consider the leak acting alone - There are no other inlets or outlets Concept / Approach: We treat the inlet rate as positive and the leak rate as negative. The effective net rate when both are acting equals 1/10 of the tank per hour. The inlet rate is 1/8 per hour. The leak rate is then inlet rate minus net rate. Once we have the leak rate, its time to empty a full tank is the reciprocal of its magnitude. Step-by-Step Solution: Step 1: Rate of inlet = 1/8 tank per hour. Step 2: Net rate with leak present (longer filling time) = 1/10 tank per hour. Step 3: Let leak rate be L tanks per hour (negative number). Then 1/8 + L = 1/10. Step 4: Solve for L: L = 1/10 - 1/8 = (4 - 5) / 40 = -1/40. Step 5: Magnitude of leak rate = 1/40 tank per hour. Step 6: Time taken by leak alone to empty the full tank = 1 / (1/40) = 40 hours. Verification / Alternative check: Check the net rate: Inlet rate 1/8 minus leak rate 1/40 equals (5 - 1) / 40 = 4/40 = 1/10 tank per hour, which indeed corresponds to a filling time of 10 hours. This confirms that the leak empties at 1/40 tank per hour and needs 40 hours to empty the full tank. Why Other Options Are Wrong: 34, 36 and 38 hours: These times would correspond to leak rates that do not satisfy both the original 8 hour filling time without a leak and the 10 hour filling time with the leak. Substituting them back into the rate equation would not yield a consistent net rate of 1/10 tank per hour. Common Pitfalls: Students sometimes subtract the times instead of the rates, or misinterpret the direction of the leak rate. It is also easy to confuse which time corresponds to which scenario. Always write separate rate equations for the normal case and the leak case, and then subtract to isolate the leak rate. Final Answer: The leak alone will empty a full tank in 40 hours.

Question 17

Three taps A, B and C can fill a tank in 20 minutes, 30 minutes and 40 minutes respectively. All three taps are opened together. After 5 minutes tap A is closed, and 6 minutes later tap B is also closed. At that moment a leak develops which, acting …

Accepted Answer

Introduction / Context: This is a multi stage pipes and cistern problem with changing conditions. Initially three filling taps are open, then one tap is closed, then another tap is closed, and finally a leak develops while the last tap continues to fill. We must carefully track the fraction of the tank filled in each phase and then compute the remaining part under the combined action of the last tap and the leak. Given Data / Assumptions: - Tap A fills the tank in 20 minutes - Tap B fills the tank in 30 minutes - Tap C fills the tank in 40 minutes - All taps start together at time zero - After 5 minutes, tap A is closed - After a further 6 minutes (that is, at 11 minutes from the start), tap B is closed - At the moment tap B is closed, a leak starts that can empty a full tank in 70 minutes - From 11 minutes onward, tap C and the leak act together until the tank is full Concept / Approach: We divide the filling into three phases: phase one with A, B and C; phase two with B and C only; and phase three with C and the leak. We compute the filled fraction in each phase using the corresponding net rates and times. The remaining fraction at the start of phase three is then filled by the net rate of C and the leak, from which we find the additional time required. Step-by-Step Solution: Step 1: Rates: A = 1/20, B = 1/30, C = 1/40 tank per minute. Step 2: Phase one (0 to 5 minutes): all three taps are open. Combined rate = 1/20 + 1/30 + 1/40. Step 3: LCM of 20, 30 and 40 is 120. So rates are 6/120, 4/120 and 3/120. Combined = 13/120 tank per minute. Step 4: Fraction filled in first 5 minutes = 5 * 13/120 = 65/120 = 13/24. Step 5: Phase two (5 to 11 minutes): A is closed, B and C remain open. Combined rate = 1/30 + 1/40 = (4 + 3) / 120 = 7/120 tank per minute. Step 6: Duration of phase two = 6 minutes, so additional filled fraction = 6 * 7/120 = 42/120 = 7/20. Step 7: Fraction filled by time 11 minutes = 13/24 + 7/20. LCM of 24 and 20 is 120, so 13/24 = 65/120 and 7/20 = 42/120, giving total 107/120. Step 8: Remaining fraction at 11 minutes = 1 - 107/120 = 13/120. Step 9: Phase three: only C and the leak work. Rate of leak = -1/70 tank per minute. Step 10: Net rate in phase three = 1/40 - 1/70. LCM of 40 and 70 is 280, so net rate = (7 - 4) / 280 = 3/280 tank per minute. Step 11: Let t be the time in minutes for phase three. Then t * 3/280 = 13/120. Step 12: Solve for t: t = (13/120) * (280/3) = (13 * 280) / 360 = 364 / 36 = 91/9 ≈ 10.11 minutes. Step 13: Total time from the start = 11 + 91/9 = (99/9 + 91/9) = 190/9 ≈ 21.11 minutes. Verification / Alternative check: We can check by approximate reasoning: about 89 percent of the tank is filled in the first 11 minutes, leaving roughly 11 percent. The net rate in phase three is slightly more than 0.01 tank per minute (3/280 ≈ 0.0107). Filling 0.1083 of the tank at this rate requires just over 10 minutes, which matches 91/9 ≈ 10.11 minutes. Adding this to the initial 11 minutes gives about 21.11 minutes overall. Why Other Options Are Wrong: 24.315, 26.166 and 22.154 minutes: These times are all larger than the correct 21.11 minutes and would imply that the tank is more than full in the third phase when combined with the phases one and two calculations. Common Pitfalls: The main challenges are keeping track of the different phases and remembering when the leak starts. Some students mistakenly assume the leak acts from the beginning or confuse which taps are open during each interval. Careful phase separation, clear rate computations and consistent fractions are vital to avoid errors. Final Answer: The tank becomes completely full after a total of approximately 21.11 minutes from the start.

Question 18

A cistern has a leak that would empty the cistern in 20 minutes. When a tap is turned on, admitting water at 4 litres per minute into the cistern, the cistern is emptied in 24 minutes. What is the capacity of the cistern in litres?

Accepted Answer

Introduction / Context: This question combines an outflow due to a leak and an inflow due to a tap. Even though a tap is adding water, the leak is strong enough that the cistern still becomes empty, but in a different time. By comparing the two scenarios, we can compute the capacity of the cistern in litres. Given Data / Assumptions: - Leak alone empties the cistern in 20 minutes - Tap admits water at 4 litres per minute - With both leak and tap acting, the cistern is emptied in 24 minutes - Let capacity of cistern be V litres - Flow rates are assumed constant Concept / Approach: When the leak acts alone, the outflow rate is V / 20 litres per minute. When both leak and tap act together, the net outflow is V / 24 litres per minute, because the cistern still empties. Thus, net outflow = leak outflow minus inflow from the tap. This gives an equation in V which we solve to get the capacity. Step-by-Step Solution: Step 1: Let capacity of cistern = V litres. Step 2: Leak rate alone (outflow) = V / 20 litres per minute. Step 3: Tap inflow rate = 4 litres per minute. Step 4: With both tap and leak, cistern empties in 24 minutes, so net outflow = V / 24 litres per minute. Step 5: Net outflow = leak outflow minus tap inflow: V / 20 - 4 = V / 24. Step 6: Rearrange: V / 20 - V / 24 = 4. Step 7: Take LCM of 20 and 24 which is 120, so (6V - 5V) / 120 = 4, giving V / 120 = 4. Step 8: Therefore V = 4 * 120 = 480 litres. Verification / Alternative check: With V = 480, leak rate = 480 / 20 = 24 litres per minute. Net outflow with tap running is 480 / 24 = 20 litres per minute. This equals 24 - 4, so the equation is satisfied. The numbers are consistent with the times and the tap inflow rate. Why Other Options Are Wrong: 360, 320 and 420 litres: Substituting any of these values into the rate equations would not yield a net outflow equal to V / 24 while maintaining a leak rate corresponding to emptying the tank in 20 minutes and a tap inflow of 4 litres per minute. Common Pitfalls: Some learners mistakenly add the tap inflow to the leak rate instead of subtracting, forgetting that the leak removes water. Others may attempt to work with times directly rather than establishing clear rate equations. Always relate time and volume through rate = volume / time and then combine rates with correct signs. Final Answer: The cistern can hold 480 litres of water.

Question 19

Two filling taps P and Q have constant rates of 40 litres per minute and 60 litres per minute respectively, and together they can fill a tank in 8 minutes. If a waste tap can empty the completely filled tank in 32 minutes, what is the rate of the wa…

Accepted Answer

Introduction / Context: This question involves two filling taps with known individual rates and a waste tap that empties the full tank in a given time. From the filling rates, we can deduce the capacity of the tank. Using the emptying time, we can then find the emptying rate in litres per minute for the waste tap. Given Data / Assumptions: - Tap P inflow rate = 40 litres per minute - Tap Q inflow rate = 60 litres per minute - Together P and Q fill the tank in 8 minutes - Waste tap empties a completely full tank in 32 minutes - All rates are constant, and there are no other leaks or taps Concept / Approach: First we find the tank capacity by using the combined inflow when P and Q work together. Then, knowing that the waste tap empties this capacity in 32 minutes, we divide capacity by time to get its emptying rate. This is a straightforward application of capacity = rate * time. Step-by-Step Solution: Step 1: Combined inflow rate of P and Q = 40 + 60 = 100 litres per minute. Step 2: Time to fill the tank with P and Q together = 8 minutes. Step 3: Capacity of tank = combined rate * time = 100 * 8 = 800 litres. Step 4: Waste tap empties the full tank in 32 minutes. Step 5: Emptying rate of waste tap = tank capacity / emptying time = 800 / 32 litres per minute. Step 6: 800 / 32 = 25 litres per minute. Verification / Alternative check: If the waste tap runs alone at 25 litres per minute, in 32 minutes it removes 32 * 25 = 800 litres, which exactly matches the capacity found from the filling taps. This consistency confirms the calculation. Why Other Options Are Wrong: 34, 22 and 18 litres per minute: These rates, when multiplied by 32 minutes, do not match the 800 litre capacity implied by the filling information. Therefore they cannot satisfy both the filling and emptying conditions of the problem. Common Pitfalls: A typical mistake is misinterpreting the initial line and thinking the combined rate is 40 litres in 8 minutes or similar. Another error is computing the wrong capacity due to arithmetic mistakes in multiplication. Writing the formula capacity = rate * time explicitly helps avoid confusion. Final Answer: The waste tap empties water at a rate of 25 litres per minute.

Question 20

Pipe K can fill a tank in 30 minutes. Pipe L can fill the same tank five times as fast as pipe K. If both pipes K and L are opened together when the tank is empty, after how many minutes will the tank overflow?

Accepted Answer

Introduction / Context: This is a straightforward pipes and cistern question where one pipe is several times faster than another. From the individual times, we find their rates, sum them to get the combined rate, and then compute the time taken to fill the tank starting from empty. The overflow will occur at the same time as the tank first becomes full. Given Data / Assumptions: - Pipe K fills the tank in 30 minutes - Pipe L fills the tank five times as fast as K - Both pipes are opened simultaneously when the tank is empty - No leakage or outlet is present Concept / Approach: If K fills the tank in 30 minutes, its rate is 1/30 tank per minute. If L is five times as fast, its rate is 5/30 tank per minute and its individual filling time is 30 / 5 = 6 minutes. The combined rate is the sum of the individual rates, and then time to fill is simply the reciprocal of the combined rate. Step-by-Step Solution: Step 1: Rate of K = 1/30 tank per minute. Step 2: L is five times as fast, so rate of L = 5 * 1/30 = 5/30 = 1/6 tank per minute. Step 3: Combined rate of K and L working together = 1/30 + 1/6. Step 4: Convert 1/6 to a denominator of 30: 1/6 = 5/30, so combined rate = 1/30 + 5/30 = 6/30 = 1/5 tank per minute. Step 5: Time taken to fill the tank from empty = 1 divided by 1/5 = 5 minutes. Verification / Alternative check: In 5 minutes, K alone would have filled 5 * 1/30 = 1/6 of the tank, and L alone would have filled 5 * 1/6 = 5/6 of the tank. Together that is 1/6 + 5/6 = 1 full tank, which matches the requirement that the tank is full in 5 minutes. Why Other Options Are Wrong: 3, 2 and 4 minutes: These times correspond to higher combined rates that do not match the given individual times. For example, in 4 minutes the total filled fraction would be 4 * (1/5) = 4/5, which is not a full tank. Common Pitfalls: Students sometimes misread “five times as fast” as adding 5 minutes or dividing the time wrongly. Remember that being multiple times as fast refers to rate, not to time directly. The correct way is to multiply or divide the time inversely and then derive the rate accurately. Final Answer: The tank will overflow after 5 minutes when both pipes K and L are open together.

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