Question 1

Pipes and Cisterns – Closing one pipe after a short interval: Two pipes P and Q can fill a cistern in 12 minutes and 15 minutes, respectively. Both are opened together; after 3 minutes, pipe P is closed. How much additional time will be required to …

Accepted Answer

Introduction / Context: When one pipe is shut after a while, split the process into phases: (1) both pipes operate; (2) the remaining pipe continues alone. Add the volumes (fractions) done in each phase to reach the full tank. Given Data / Assumptions: P alone: 12 min ⇒ rate = 1/12 per min. Q alone: 15 min ⇒ rate = 1/15 per min. Both open for first 3 min; then only Q continues. Concept / Approach: Compute work done in 3 minutes by both pipes, subtract from 1 to get the remainder, then divide by Q’s rate to obtain the extra time. Step-by-Step Solution: In 3 min, both pipes do: 3*(1/12 + 1/15) = 3*( (5 + 4)/60 ) = 3*(9/60) = 27/60 = 9/20.Remaining = 1 − 9/20 = 11/20.Q’s rate = 1/15 per min ⇒ extra time = (11/20) / (1/15) = (11/20)*15 = 165/20 = 8.25 min = 8 min 15 sec. Verification / Alternative check: Convert to seconds if preferred: 0.25 min = 15 sec; total is 8 min 15 sec. Why Other Options Are Wrong: 5, 7:30, 9 minutes fail to match the computed remainder at Q’s solo rate. Common Pitfalls: Multiplying times directly or forgetting to split into phases properly. Final Answer: 8 min 15 sec

Question 2

Pipes and Cisterns – Individual times exceed the joint time by fixed minutes: Two pipes A and B fill a tank. When both are opened together, they fill it in T minutes. Pipe A alone takes 16 minutes more than T, and pipe B alone takes 9 minutes more t…

Accepted Answer

Introduction / Context: This is a parameterized pipes problem: each solo time is an additive offset from the unknown joint time T. The physics is the same: work rates add. We set up an equation in T and solve. Given Data / Assumptions: Together: T min ⇒ rate(A + B) = 1/T per min. A alone: T + 16 ⇒ rate A = 1/(T + 16). B alone: T + 9 ⇒ rate B = 1/(T + 9). Concept / Approach: Because rates add, 1/(T + 16) + 1/(T + 9) = 1/T. Clear denominators and solve the resulting quadratic for the positive, realistic T. Step-by-Step Solution: 1/(T + 16) + 1/(T + 9) = 1/T.Cross-multiply: T(T + 9 + T + 16) = (T + 16)(T + 9).2T^2 + 25T = T^2 + 25T + 144 ⇒ T^2 = 144 ⇒ T = 12 (take positive root). Verification / Alternative check: Check rates: 1/28 + 1/21 = (3 + 4)/84 = 7/84 = 1/12, which equals 1/T. Why Other Options Are Wrong: 8, 10, 14, 15 minutes fail to satisfy the rate equation when substituted into 1/(T + 16) + 1/(T + 9) = 1/T. Common Pitfalls: Adding times rather than rates; sign mistakes when clearing denominators. Final Answer: 12 min

Question 3

Pipes and Cisterns – Two inlets and one outlet with net partial fill in one hour: Three pipes are connected to a tank. In one hour, pipe 1 can fill 1/2 of the tank, pipe 2 can fill 1/3 of the tank, and pipe 3 is an outlet (empties). When all three a…

Accepted Answer

Introduction / Context: Net effect over one hour is known. Express the outlet’s unknown rate as x (per hour) and solve using the fact that the resulting net equals 7/12 tank per hour. Given Data / Assumptions: Pipe 1 rate = 1/2 per h. Pipe 2 rate = 1/3 per h. Pipe 3 is an outlet with rate = x per h (to be subtracted). Net with all open = 7/12 per h. Concept / Approach: Set up the rate equation: 1/2 + 1/3 − x = 7/12 and solve for x, then invert to get the emptying time. Step-by-Step Solution: 1/2 + 1/3 = 5/6 = 10/12.10/12 − x = 7/12 ⇒ x = 3/12 = 1/4 per h.Therefore, outlet alone empties in 1 / (1/4) = 4 h. Verification / Alternative check: Recombine: 1/2 + 1/3 − 1/4 = 6/12 + 4/12 − 3/12 = 7/12 per h, matches the statement. Why Other Options Are Wrong: 2, 3, 5, and 6 h would not produce 7/12 per h net. Common Pitfalls: Using 1/2 + 1/3 = 2/5 (incorrect) or forgetting to subtract the outlet. Final Answer: 4 h

Question 4

Pipes and Cisterns – Two inlets plus a constant waste pipe (capacity asked): Two inlet pipes can fill a tank in 20 minutes and 24 minutes, respectively. A waste pipe empties water at a constant 6 gallons per minute. With all three operating together…

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Introduction / Context: Here, two variable rates depend on tank capacity (inlet pipes), while the waste pipe has a fixed volumetric rate (6 gallons per minute). Form a linear equation in the unknown capacity and solve. Given Data / Assumptions: Pipe 1: capacity/20 gallons per min. Pipe 2: capacity/24 gallons per min. Waste: 6 gallons per min (subtract). Net fill time with all pipes: 15 min ⇒ net rate = capacity/15 gallons per min. Concept / Approach: Set (V/20) + (V/24) − 6 = V/15 and solve for V, the tank capacity. Step-by-Step Solution: (V/20) + (V/24) − 6 = V/15.LCM 120 ⇒ 6V + 5V − 720 = 8V ⇒ 11V − 720 = 8V ⇒ 3V = 720 ⇒ V = 240. Verification / Alternative check: Check numerically: Inlets sum = 12 + 10 = 22 gpm; net must be V/15 = 16 gpm ⇒ waste = 22 − 16 = 6 gpm, consistent. Why Other Options Are Wrong: 210, 150, 180, and 50 gallons do not satisfy the equation with a 6 gpm waste and 15-minute completion. Common Pitfalls: Forgetting to keep units consistent (minutes vs hours) or omitting the constant waste term. Final Answer: 240 gallon

Question 5

Pipes and Cisterns – Open both taps briefly, then continue with the slower tap: A tank can be filled by tap 1 in 20 minutes and by tap 2 in 60 minutes. Both taps are opened for 5 minutes, after which tap 1 is closed. How much additional time will it…

Accepted Answer

Introduction / Context: We handle this in two phases: the first 5 minutes with both taps, then the remaining fraction with just the slower tap. Using rates ensures exact accounting of the filled fraction. Given Data / Assumptions: Tap 1: 20 min ⇒ 1/20 per min. Tap 2: 60 min ⇒ 1/60 per min. Phase 1 duration = 5 min with both; Phase 2 is only tap 2. Concept / Approach: Compute the fraction filled in the first 5 minutes, subtract from 1, then divide the remainder by tap 2’s rate. Step-by-Step Solution: In 5 min: fraction = 5*(1/20 + 1/60) = 5*( (3 + 1)/60 ) = 5*(4/60) = 1/3.Remaining = 1 − 1/3 = 2/3.Tap 2 alone rate = 1/60 ⇒ remaining time = (2/3) / (1/60) = 40 min. Verification / Alternative check: Equivalent total time = 5 + 40 = 45 min overall, but the question asks for the time after closing tap 1, i.e., 40 min. Why Other Options Are Wrong: 20, 30, 35, 45 minutes correspond to misreading the question or miscomputing the first 5-minute contribution. Common Pitfalls: Answering with 45 min (total) rather than the requested additional time after closing tap 1. Final Answer: 40 min

Question 6

Pipes and Cisterns – Staggered opening times, including a strong outlet later: A cistern has three pipes A, B, and C. Pipe A fills the cistern in 3 hours and pipe B in 4 hours. Pipe C empties a full cistern in 1 hour. The pipes are opened in order a…

Accepted Answer

Introduction / Context: When pipes open at different times, break the timeline into intervals with constant rates. First, only A runs; then A and B; finally A, B, and the strong outlet C act together, potentially reversing the net flow. Given Data / Assumptions: A fills in 3 h ⇒ 1/3 per h. B fills in 4 h ⇒ 1/4 per h. C empties in 1 h ⇒ 1 per h (negative). Intervals: [3–4 pm] A only; [4–5 pm] A + B; [after 5 pm] A + B − C. Concept / Approach: Accumulate the filled fraction at each stage, then after C opens the net rate becomes negative and the cistern empties from its then-current level. Step-by-Step Solution: 3–4 pm: A adds 1/3 ⇒ level = 1/3.4–5 pm: A + B add 1/3 + 1/4 = 7/12 ⇒ new level = 1/3 + 7/12 = 11/12.After 5 pm: net rate = 1/3 + 1/4 − 1 = 7/12 − 1 = −5/12 per h.Time to empty from 11/12 at rate 5/12 per h = (11/12) / (5/12) = 11/5 = 2.2 h = 2 h 12 min.Empty time = 5:00 pm + 2 h 12 min = 7:12 pm. Verification / Alternative check: Reasonable since a strong outlet opens late, quickly draining near-full cistern. Why Other Options Are Wrong: 6:15 pm is too early; 8:12 pm and 8:35 pm are too late given the high emptying rate after 5 pm. Common Pitfalls: Forgetting that after 5 pm the net flow is outward (negative rate). Final Answer: 7:12 pm

Question 7

Pipes and Cisterns – Three taps with flow proportional to square of diameter: Three inlet taps have diameters 1 cm, 4/3 cm, and 2 cm. The volume flow through each is proportional to the square of its diameter. The largest (2 cm) tap alone fills the …

Accepted Answer

Introduction / Context: When flow is proportional to the square of diameter (v ∝ d^2), relative rates can be set in those ratios. One tap’s absolute time fixes the unit scaling, allowing us to compute the combined rate and time. Given Data / Assumptions: Diameters: 1 cm, 4/3 cm, 2 cm ⇒ squares 1, 16/9, 4. Largest (2 cm) alone fills in 61 min. Tank capacity normalized to W; proportionality is linear in rates. Concept / Approach: Let “1 unit” of rate correspond to the 1 cm tap. Since 4 units correspond to the 2 cm tap and that equals W/61 per min, we get the unit rate and can sum all three taps’ rates. Step-by-Step Solution: Let 4 units = W/61 per min ⇒ 1 unit = W/244 per min.Total units = 1 + 16/9 + 4 = (9 + 16 + 36)/9 = 61/9.Combined rate = (61/9) * (W/244) = W * 61 / 2196 per min.Time = W / (W * 61 / 2196) = 2196 / 61 = 36 min. Verification / Alternative check: Because 61 × 36 = 2196, the arithmetic cancels exactly, giving an integer number of minutes. Why Other Options Are Wrong: 44, 45, 155/18, and 40 min do not align with the squared-diameter proportionality anchored by the 61-minute reference. Common Pitfalls: Using diameters (d) instead of d^2 for proportionality, or mixing units. Final Answer: 36 min

Question 8

Pipes and Cisterns – Two inlets with a bottom leak causing delay: Two pipes can fill a cistern in 14 hours and 16 hours, respectively. Owing to a leak at the bottom, the cistern actually takes 92 minutes longer to fill than it would without the leak…

Accepted Answer

Introduction / Context: The leak reduces the net filling rate, increasing the total time by a known amount. By comparing the “no-leak” combined rate with the “with-leak” rate, we can isolate the leak’s emptying rate and invert it. Given Data / Assumptions: Pipe rates: 1/14 and 1/16 per h. No-leak time = (14*16)/(14+16) = 224/30 = 112/15 h = 7 h 28 min. Observed time with leak = 7 h 28 min + 92 min = 9 h. Concept / Approach: Let r = 1/14 + 1/16. With leak, net = 1/9. Leak rate l = r − 1/9. The leak’s emptying time is 1/l. Step-by-Step Solution: r = 1/14 + 1/16 = (8 + 7)/112 = 15/112 per h.l = r − 1/9 = 15/112 − 1/9 = (135 − 112)/1008 = 23/1008 per h.Leak emptying time = 1 / (23/1008) = 1008/23 h ≈ 43.83 h. Verification / Alternative check: Check: 1/14 + 1/16 − 1/(1008/23) = 15/112 − 23/1008 = (135 − 23)/1008 = 112/1008 = 1/9 ⇒ 9 hours with leak. Why Other Options Are Wrong: Other choices do not reproduce the net 9-hour fill time when combined with the two inlet rates. Common Pitfalls: Adding times instead of rates, or converting 92 minutes incorrectly. Final Answer: 1008/23 h

Question 9

Pipes and Cisterns – Turn off one tap after a while, find the remaining minutes: Three taps A, B, and C can fill a cistern in 10, 15, and 20 minutes respectively. All are opened together, but C is turned off after 3 minutes. How many additional minu…

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Introduction / Context: We again separate the process into phases. First, all three taps contribute for 3 minutes. Then only A and B continue to finish the remaining fraction. Given Data / Assumptions: A: 10 min ⇒ 1/10 per min. B: 15 min ⇒ 1/15 per min. C: 20 min ⇒ 1/20 per min. Phase 1 duration: 3 minutes with all three; Phase 2: A + B only. Concept / Approach: Compute the fraction completed in the first 3 minutes and subtract from 1 to get the remaining fraction. Then divide by the A + B combined rate. Step-by-Step Solution: In 3 minutes: (1/10 + 1/15 + 1/20)*3 = (6/60 + 4/60 + 3/60)*3 = (13/60)*3 = 13/20.Remaining = 1 − 13/20 = 7/20.A + B rate = 1/10 + 1/15 = (3 + 2)/30 = 1/6 per min.Time = (7/20) / (1/6) = (7/20)*6 = 42/20 = 2.1 min = 2 min 6 sec. Verification / Alternative check: 2.1 min × 60 sec/min = 126 sec ⇒ 2 min 6 sec; sums correctly to a full tank. Why Other Options Are Wrong: 1:06, 2:00, 2:30, 3:08 do not match the computed remainder at 1/6 per minute. Common Pitfalls: Forgetting to convert the decimal 0.1 min into seconds or miscomputing the first 3 minutes’ contribution. Final Answer: 2 min 6 sec

Question 10

Pipes and Cisterns – Reduced initial flow, then full flow completes in a known time: Two pipes can fill a cistern in 30 minutes and 36 minutes, respectively. Initially both are partially clogged so that only 5/6 of the normal water flows through the…

Accepted Answer

Introduction / Context: The process has two phases: a reduced-flow phase and a full-flow phase. The total work (1 tank) equals the sum of work done in each phase. The second phase duration is given (15.5 minutes) at full rate; we use that to infer how much was already completed, and hence how long the reduced phase lasted. Given Data / Assumptions: Normal rates: 1/30 and 1/36 per min. Reduced rates: (5/6)*(1/30) = 1/36; (9/10)*(1/36) = 1/40. Full-flow (normal) total rate: 1/30 + 1/36 = 11/180 per min. From removal onward: time to finish = 15.5 min = 31/2 min. Concept / Approach: Remaining portion at the moment of removal = (full rate) * 15.5 = (11/180)*(31/2) = 341/360. Therefore, the portion completed during reduced flow = 1 − 341/360 = 19/360. Divide this by the reduced combined rate to find the reduced-phase duration. Step-by-Step Solution: Reduced combined rate = 1/36 + 1/40 = 19/360 per min.Work done in reduced phase = 19/360 of tank.Time in reduced phase = (19/360) / (19/360) = 1 minute. Verification / Alternative check: Check totals: reduced 1 min + full 15.5 min = 16.5 min overall; full-phase work equals 341/360, reduced equals 19/360; sum is 1. Why Other Options Are Wrong: Any duration other than 1 minute would not make the remaining fraction equal to 341/360 at the switch. Common Pitfalls: Treating 15.5 minutes as the time to fill from empty at full rate; it is only for the remaining portion after obstructions are removed. Final Answer: 1 minute

Question 11

Pipes and Cisterns – A leak slows one inlet; deduce leak strength and B’s new time: Pipe A fills the cistern in 30 minutes and pipe B in 40 minutes. Because of a crack (leak) at the bottom, pipe A now needs 40 minutes to fill the cistern alone. With…

Accepted Answer

Introduction / Context: The leak reduces each inlet’s effective rate by the same leak rate. Using the observed slowdown for A (from 30 to 40 minutes), we can compute the leak rate and then apply it to B to find B’s new filling time. The leak’s emptying time is the reciprocal of its rate. Given Data / Assumptions: A normal = 1/30 per min; with leak = 1/40 per min. B normal = 1/40 per min; leak rate is the same for both. Concept / Approach: Leak rate l = (A normal) − (A with leak) = 1/30 − 1/40. Then B with leak = (B normal) − l. Step-by-Step Solution: l = 1/30 − 1/40 = (4 − 3)/120 = 1/120 per min.B with leak = 1/40 − 1/120 = (3 − 1)/120 = 1/60 per min ⇒ B now takes 60 min (1 hour).Leak emptying time = 1 / (1/120) = 120 min = 2 hours. Verification / Alternative check: For A, 1/30 − 1/120 = 1/40, which matches the stated slowdown, confirming leak computation. Why Other Options Are Wrong: Options implying a 1-hour leak or a 2-hour B time contradict the rate arithmetic derived from A’s slowdown. Common Pitfalls: Subtracting times rather than rates; mixing minutes and hours. Final Answer: B fills in an hour and the leak empties in 2 hours

Question 12

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 classical pipes and cisterns question involving changing conditions during the filling process. Initially, two pipes work together, and then one pipe is closed. To solve such problems, we translate times into rates of work, compute how much of the tank is filled in each phase, and then determine the total time taken. Given Data / Assumptions: Pipe A fills the tank in 15 minutes, so its rate is 1/15 of the tank per minute. Pipe B fills the tank in 20 minutes, so its rate is 1/20 of the tank per minute. Both pipes are opened together for the first 4 minutes. After 4 minutes, pipe A is closed and only pipe B continues to fill the tank. Concept / Approach: We use the idea that work done equals rate multiplied by time. We first find the combined filling rate of A and B, then compute the fraction of the tank filled in 4 minutes. From that, we find the remaining fraction and calculate how long pipe B alone needs to fill the rest. Finally we add the two time segments. 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 = (4 + 3) / 60 = 7 / 60 tank per minute.Step 4: In the first 4 minutes, amount filled = 4 * (7 / 60) = 28 / 60 = 7 / 15 of the tank.Step 5: Remaining fraction of the tank = 1 - 7 / 15 = 8 / 15.Step 6: Now only B works, with rate 1/20 tank per minute.Step 7: Time taken by B to fill the remaining 8 / 15 = (8 / 15) / (1 / 20) = (8 / 15) * 20 = 160 / 15 = 32 / 3 minutes.Step 8: 32 / 3 minutes equals 10 minutes 40 seconds.Step 9: Total time = initial 4 minutes + 10 minutes 40 seconds = 14 minutes 40 seconds. Verification / Alternative check: We can convert everything into seconds to double check. The total time is 14 minutes 40 seconds, which is 880 seconds. In 880 seconds, pipe B works for 880 seconds, and pipe A works only for the first 240 seconds. Converting to minutes and applying rates will again give exactly one full tank, confirming our answer. Why Other Options Are Wrong: 10 minutes 20 seconds and 11 minutes 45 seconds come from mixing up the remaining fraction or miscomputing combined rates. 12 minutes 30 seconds underestimates the total time and usually indicates that the 4 minutes of joint work and the later solo work of B were not separated correctly. Common Pitfalls: Students often forget to compute the remaining fraction after the first phase and instead try to average the times. Another common error is to mis-handle fractions when converting into minutes and seconds, leading to small but important numerical mistakes. Always compute rates carefully and break the problem into clear stages. Final Answer: The tank will be completely filled in 14 minutes 40 seconds.

Question 13

A pump can fill a tank with water in 2 hours when there is no leak. Because of a leak in the tank, it now takes 2 hours 20 minutes to fill the tank. In how much time will the leak alone empty a full tank?

Accepted Answer

Introduction / Context: This is a pipes and cisterns problem with one inlet (the pump) and one outlet (the leak). The presence of the leak reduces the effective filling rate, causing the tank to take longer to fill. From the difference in times, we can calculate the leak rate and then find how long the leak alone would take to empty the tank. Given Data / Assumptions: The pump alone fills the tank in 2 hours. With the leak active, the net filling time is 2 hours 20 minutes. 2 hours 20 minutes equals 7 / 3 hours. We assume the tank capacity is 1 full tank unit for easier calculation. Concept / Approach: Let the capacity of the tank be 1 unit. Then the pump rate and the net rate with the leak can be written as fractions of the tank per hour. The leak rate is the difference between the pump rate and the net rate. Once we know the leak rate, we can invert it to get the time taken by the leak alone to empty the full tank. Step-by-Step Solution: Step 1: Pump rate alone = 1 / 2 tank per hour.Step 2: Net rate with leak = 1 / (7 / 3) = 3 / 7 tank per hour.Step 3: Let leak rate (emptying) be L tank per hour. Then pump rate - leak rate = net rate.Step 4: So 1 / 2 - L = 3 / 7.Step 5: Solve for L: L = 1 / 2 - 3 / 7 = (7 / 14) - (6 / 14) = 1 / 14 tank per hour.Step 6: If the leak alone is open, it empties 1 / 14 of the tank per hour.Step 7: Therefore, time taken by the leak alone to empty the full tank = 1 / (1 / 14) = 14 hours. Verification / Alternative check: Check the combined rate: pump rate minus leak rate = 1 / 2 - 1 / 14 = (7 / 14) - (1 / 14) = 6 / 14 = 3 / 7, which matches the earlier net rate. Filling at 3 / 7 tank per hour, the tank will indeed take 7 / 3 hours or 2 hours 20 minutes to fill, confirming our calculations. Why Other Options Are Wrong: 7 hours and 8 hours are too small and correspond to larger leak rates, which would cause much more delay than 20 extra minutes. 12 hours is closer but still incorrect and results from arithmetic mistakes when subtracting fractions or converting mixed times to fractional hours. Common Pitfalls: Students often forget to convert 2 hours 20 minutes into a single fraction (7 / 3 hours) and instead use inconsistent units. Others mistakenly add the leak rate instead of subtracting it, as if the leak also filled the tank. Remember that outlet rates reduce net filling speed. Carefully convert all times into hours or minutes before proceeding with algebra. Final Answer: The leak alone will empty a full tank in 14 hours.

Question 14

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

Accepted Answer

Introduction / Context: This is a two-phase filling problem. Initially, one tap is running, and later additional identical taps are opened, increasing the total filling rate. We need to carefully handle the fraction of the tank filled in each phase and then add the times. Given Data / Assumptions: One tap alone fills the tank in 6 hours. After half the tank is filled, three more taps identical to the first are opened. All taps are assumed to have constant flow rates. We consider the tank capacity as 1 full unit for easier calculation. Concept / Approach: We treat the filling process in two parts. In the first part, a single tap fills half the tank. In the second part, four identical taps (the original one plus three more) fill the remaining half. Using rate times time equals work done, we compute the time for each phase and then sum them. Step-by-Step Solution: Step 1: Let the tank capacity be 1 unit. One tap fills it in 6 hours, so its rate is 1 / 6 tank per hour.Step 2: Time to fill half the tank with one tap = (1 / 2) / (1 / 6) = 3 hours.Step 3: After this, three more identical taps are opened. There are now 4 taps in total.Step 4: Combined rate of 4 taps = 4 * (1 / 6) = 4 / 6 = 2 / 3 tank per hour.Step 5: Remaining volume to be filled = 1 - 1 / 2 = 1 / 2 tank.Step 6: Time to fill the remaining half with 4 taps = (1 / 2) / (2 / 3) = (1 / 2) * (3 / 2) = 3 / 4 hours.Step 7: 3 / 4 hours equals 45 minutes.Step 8: Total filling time = 3 hours + 45 minutes = 3 hours 45 minutes. Verification / Alternative check: We can compute how much work is done per quarter hour in the second phase. In 15 minutes (1 / 4 hour) with 4 taps, the amount filled is (2 / 3) * (1 / 4) = 1 / 6. Therefore, in 45 minutes (3 such intervals) the taps fill 3 * (1 / 6) = 1 / 2 of the tank, which confirms our earlier calculation that 45 minutes are enough to complete the remaining half. Why Other Options Are Wrong: 3 hours 15 minutes assumes an incorrect combined rate or mistakenly uses only three taps in the second phase. 4 hours and 4 hours 15 minutes both overestimate the time, indicating that the contribution of multiple taps in the second phase was not properly accounted for. Common Pitfalls: Some students forget to distinguish between the two phases and incorrectly assume four taps work from the start. Others mis-handle the fractional calculations, such as adding rates incorrectly or forgetting that only half the tank remains when extra taps are opened. Always compute how much of the tank is filled in each phase and use the correct number of taps and rates for each period. Final Answer: The tank will be completely filled in 3 hours 45 minutes.

Question 15

Three pipes A, B and C together can fill a tank in 6 hours. After they work together for 2 hours, pipe C is closed and pipes A and B together fill the remaining part of the tank in 7 hours. How many hours would pipe C alone take to fill the tank?

Accepted Answer

Introduction / Context: This is a standard pipes and cisterns problem involving three pipes, where one pipe is later closed. We must use the information about their combined work and the work done after closing one pipe to deduce the individual rate of pipe C and then find how long C alone would need to fill the tank. Given Data / Assumptions: Pipes A, B and C together fill the tank in 6 hours. All three pipes run together for 2 hours. After 2 hours, pipe C is closed. Pipes A and B together fill the remaining part of the tank in 7 hours. Tank capacity is taken as 1 unit. Concept / Approach: We convert the given times into rates. First we obtain the combined rate of A + B + C. Then using the information about how long A and B take to fill the remaining fraction, we find the combined rate of A + B alone. Subtracting these gives the rate of C alone. Finally, we invert that rate to find the time required by C working by itself. Step-by-Step Solution: Step 1: Combined rate of A + B + C = 1 / 6 tank per hour.Step 2: In the first 2 hours, amount filled by all three pipes = 2 * (1 / 6) = 1 / 3 tank.Step 3: Remaining fraction of the tank after 2 hours = 1 - 1 / 3 = 2 / 3.Step 4: A and B together fill this remaining 2 / 3 in 7 hours, so rate of A + B = (2 / 3) / 7 = 2 / 21 tank per hour.Step 5: Therefore, rate of C alone = (A + B + C rate) - (A + B rate) = 1 / 6 - 2 / 21.Step 6: Compute 1 / 6 - 2 / 21 = (7 / 42) - (4 / 42) = 3 / 42 = 1 / 14 tank per hour.Step 7: Time taken by C alone to fill the full tank = 1 / (1 / 14) = 14 hours. Verification / Alternative check: We can check the arithmetic by recomputing combined rates. If C fills at 1 / 14, then A + B rate must be 1 / 6 - 1 / 14 = (7 - 3) / 42 = 4 / 42 = 2 / 21, which matches our earlier value. With A + B working at 2 / 21, they need 7 hours to fill 2 / 3 of the tank, since 7 * (2 / 21) = 14 / 21 = 2 / 3. This confirms the consistency of our solution. Why Other Options Are Wrong: 10 hours and 12 hours correspond to faster rates than allowed by the given conditions and do not satisfy the combined time relationships. 16 hours is too long, implying a slower C that would require a different pattern of filling times for A and B than the one described. Common Pitfalls: Many students wrongly assume that the shares of work of A, B and C can be read directly from the times, without computing actual rates. Others try to combine fractions in their heads and miscalculate 1 / 6 - 2 / 21. It is safer to write fractions with a common denominator step by step and then simplify carefully. Final Answer: Pipe C alone would fill the tank in 14 hours.

Question 16

Three taps A, B and C can fill a tank in 12 hours, 15 hours and 20 hours respectively. Tap A is kept open all the time, while taps B and C are opened one at a time in alternate hours. In how many hours will the tank be completely filled?

Accepted Answer

Introduction / Context: This problem involves varying filling rates over time. Tap A runs continuously, while taps B and C take turns for one hour each. To handle this, we consider a full 2-hour cycle and compute how much of the tank is filled in each cycle. Then we see how many such cycles are needed and whether a partial cycle is required to reach exactly one full tank. Given Data / Assumptions: Tap A alone fills the tank in 12 hours. Tap B alone fills the tank in 15 hours. Tap C alone fills the tank in 20 hours. Tap A is open continuously. Taps B and C are opened alternately, each for one hour at a time. Concept / Approach: We compute the rates of A, B and C in terms of tank per hour. Then we calculate how much of the tank is filled in the first hour (A with B) and in the second hour (A with C). Adding these gives the work done in one 2-hour cycle. We then determine how many such cycles and possibly an extra hour are needed to fill the tank. Step-by-Step Solution: Step 1: Rate of A = 1 / 12 tank per hour.Step 2: Rate of B = 1 / 15 tank per hour.Step 3: Rate of C = 1 / 20 tank per hour.Step 4: In hour 1, A and B together work. Combined rate = 1 / 12 + 1 / 15 = (5 + 4) / 60 = 9 / 60 = 3 / 20.Step 5: In hour 2, A and C together work. Combined rate = 1 / 12 + 1 / 20 = (5 + 3) / 60 = 8 / 60 = 2 / 15.Step 6: Work done in a 2-hour cycle = 3 / 20 + 2 / 15 = (9 + 8) / 60 = 17 / 60 of the tank.Step 7: After 3 full cycles (6 hours), work done = 3 * (17 / 60) = 51 / 60 of the tank.Step 8: Remaining work = 1 - 51 / 60 = 9 / 60 = 3 / 20.Step 9: The next hour is like hour 1, with A and B working together at 3 / 20 tank per hour.Step 10: In that one extra hour, they exactly fill the remaining 3 / 20, so total time = 6 + 1 = 7 hours. Verification / Alternative check: We can verify by computing total work after 7 hours directly: in the first, third, fifth and seventh hours, taps A and B work; in the second, fourth and sixth hours, taps A and C work. Summing 4 times the work of A+B (4 * 3 / 20) and 3 times the work of A+C (3 * 2 / 15) gives 12 / 20 + 6 / 15 = 3 / 5 + 2 / 5 = 1 full tank. Why Other Options Are Wrong: 6 hours would give only 51 / 60 of the tank, which is not completely full. 20/3 hours and 15/2 hours do not correspond to any integer number of cycles and do not yield exactly one full tank when we sum each hour’s contribution. Common Pitfalls: Some candidates try to average the rates of B and C instead of handling the alternating pattern correctly. Others forget that we must consider whole cycles and may stop at 6 hours without checking whether the tank is completely full. The safest approach is to compute work done over a full 2-hour cycle and then proceed carefully. Final Answer: The tank will be completely filled in 7 hours.

Question 17

A cistern is normally filled by an inlet pipe in 8 hours, but because of a leak at the bottom it now takes 10 hours to fill. If the cistern is full and only the leak is left open, in how many hours will the cistern be emptied?

Accepted Answer

Introduction / Context: This is another inlet outlet problem where a leak slows down the filling of a cistern. We know how long the inlet takes alone and how long the cistern takes to fill when both inlet and leak are active. From that information, we can determine the leak rate and then find the time for the leak alone to empty a full cistern. Given Data / Assumptions: Inlet alone fills the cistern in 8 hours. With the leak also open, the cistern fills in 10 hours. We take the cistern capacity as 1 unit. The leak continuously drains water at a constant rate. Concept / Approach: We represent the inlet rate and the net filling rate as fractions of the cistern per hour. The leak rate is simply the difference between the inlet rate and the net rate. Once we know the leak rate, the time to empty the cistern is the reciprocal of that rate. Step-by-Step Solution: Step 1: Inlet rate alone = 1 / 8 cistern per hour.Step 2: Net rate with leak = 1 / 10 cistern per hour.Step 3: Let leak rate be L cistern per hour (this is an emptying rate).Step 4: Inlet rate minus leak rate equals net rate, so 1 / 8 - L = 1 / 10.Step 5: Solve for L: L = 1 / 8 - 1 / 10 = (5 / 40) - (4 / 40) = 1 / 40 cistern per hour.Step 6: Leak alone empties 1 / 40 of the cistern per hour.Step 7: Therefore, time taken by the leak alone to empty a full cistern = 1 / (1 / 40) = 40 hours. Verification / Alternative check: If the leak rate is 1 / 40, then the net filling rate with inlet and leak together is 1 / 8 - 1 / 40 = (5 - 1) / 40 = 4 / 40 = 1 / 10 cistern per hour, which means the cistern is filled in 10 hours. This matches the given information, so our leak rate and emptying time are consistent. Why Other Options Are Wrong: 20 hours and 28 hours are too small and correspond to leak rates of 1 / 20 or 1 / 28, which would slow down filling more than observed. 36 hours is closer but still incorrect and comes from miscalculating the difference between 1 / 8 and 1 / 10. Common Pitfalls: Some students incorrectly add the rates instead of subtracting, as if both pipe and leak were filling. Others subtract the times instead of subtracting the rates, which is not valid. Always convert times into rates (1 / time) and work with those when combining effects of multiple pipes or leaks. Final Answer: The leak alone will empty a full cistern in 40 hours.

Question 18

Pipes A and B can fill a tank in 60 minutes and 90 minutes respectively. There is a leak at three-fourths of the height of the tank. If the leak alone is opened when the tank is full, it takes 36 minutes to lower the water level from full down to th…

Accepted Answer

Introduction / Context: This problem combines multiple ideas: two inlet pipes, one leak, and the fact that the leak only operates when the water level is above three-fourths of the tank height. We must carefully separate the stages of filling and use the information about how fast the leak can lower the water level from full to three-fourths of the height. Given Data / Assumptions: Pipe A fills the full tank in 60 minutes. Pipe B fills the full tank in 90 minutes. The leak is located at three-fourths of the tank height. When the tank is full and only the leak is open, it takes 36 minutes to lower the level to three-fourths of the height. The tank has uniform cross section, so volume is proportional to height. Concept / Approach: Because the tank cross section is uniform, lowering the level from full to three-fourths of the height corresponds to removing one-fourth of the tank volume. Thus the leak alone empties one-fourth of the tank in 36 minutes. With two inlets and one leak, we consider two stages: from empty to three-fourths of the tank (when the leak is not yet active), and from three-fourths to full (when the leak is active along with the inlets). Step-by-Step Solution: Step 1: Let the full tank volume be 1 unit.Step 2: Rate of pipe A = 1 / 60 tank per minute.Step 3: Rate of pipe B = 1 / 90 tank per minute.Step 4: Leak alone removes 1 / 4 of the tank in 36 minutes, so leak rate = (1 / 4) / 36 = 1 / 144 tank per minute.Step 5: Combined filling rate of A and B = 1 / 60 + 1 / 90 = (3 + 2) / 180 = 5 / 180 = 1 / 36 tank per minute.Step 6: First stage: from empty to three-fourths of the tank. In this stage the leak does not operate, so the rate is 1 / 36.Step 7: Time for stage one = (3 / 4) / (1 / 36) = (3 / 4) * 36 = 27 minutes.Step 8: Second stage: from three-fourths to full. Now both inlets and the leak are active.Step 9: Net rate in stage two = (1 / 36) (A and B) minus 1 / 144 (leak) = (4 / 144 - 1 / 144) = 3 / 144 = 1 / 48 tank per minute.Step 10: Volume to fill in stage two = 1 - 3 / 4 = 1 / 4 tank.Step 11: Time for stage two = (1 / 4) / (1 / 48) = (1 / 4) * 48 = 12 minutes.Step 12: Total time to fill the tank = 27 + 12 = 39 minutes. Verification / Alternative check: We can confirm the leak rate: if it empties at 1 / 144 tank per minute, in 36 minutes it empties 36 / 144 = 1 / 4 of the tank, which matches the given drop from full to three-fourths level. Using this rate in the net filling computation gives a reasonable total time less than one hour, consistent with the speeds of the pipes. Why Other Options Are Wrong: 27 minutes corresponds only to the first stage and ignores the slower second stage where the leak counteracts the inlets. 33 minutes and 37 minutes both underestimate the contribution of the leak. Any value smaller than 39 minutes would imply a faster net filling rate during the second stage than is actually possible given the leak rate. Common Pitfalls: One common error is to assume the leak operates from the beginning, rather than only after the water reaches its height. Another is to misinterpret “three-fourths of the height” as something other than three-fourths of the volume in a uniform tank. Also, some students forget to compute the net rate correctly when the leak is active. Properly dividing the process into two phases avoids these mistakes. Final Answer: The tank will be completely filled in 39 minutes.

Question 19

A water tank is two-fifths full. Pipe A can fill the whole tank in 10 minutes and pipe B can empty the whole tank in 6 minutes. If both pipes are opened together, will the tank be emptied or filled, and how long will it take for the process to compl…

Accepted Answer

Introduction / Context: This problem combines an inlet pipe and an outlet pipe starting from a partially filled tank. We must determine whether the net effect is filling or emptying and then compute how long it takes to either reach full or become empty. Such questions are solved by comparing the rates of filling and emptying and then applying them to the current volume. Given Data / Assumptions: The tank is initially two-fifths full. Pipe A fills the entire tank in 10 minutes. Pipe B empties the entire tank in 6 minutes. Both pipes are opened at the same time. Tank capacity is taken as 1 unit. Concept / Approach: We convert the times for A and B into rates. Then we compute the net rate when both A and B are open together. If the net rate is negative, the tank is being emptied. Knowing the initial content, we divide that volume by the magnitude of the net rate to find the time required for the tank to become empty. Step-by-Step Solution: Step 1: Let tank capacity be 1 unit. Initially, volume of water = 2 / 5 unit.Step 2: Rate of pipe A (filling) = 1 / 10 tank per minute.Step 3: Rate of pipe B (emptying) = 1 / 6 tank per minute.Step 4: Net rate when both are open = 1 / 10 - 1 / 6.Step 5: Compute 1 / 10 - 1 / 6 = (3 / 30 - 5 / 30) = -2 / 30 = -1 / 15 tank per minute.Step 6: The negative sign shows the tank is being emptied at a net rate of 1 / 15 tank per minute.Step 7: Time to empty the tank from 2 / 5 full to empty = (2 / 5) / (1 / 15) = (2 / 5) * 15 = 6 minutes. Verification / Alternative check: We can check the net effect after 6 minutes: net emptying rate is 1 / 15 per minute, so in 6 minutes, total emptied volume is 6 * (1 / 15) = 6 / 15 = 2 / 5, which equals the initial volume of water. That means the tank just becomes empty after 6 minutes, matching our calculation. Why Other Options Are Wrong: 6 minutes to fill would require a positive net rate and an initial content lower than final, which is not the case. 9 minutes to empty or 9 minutes to fill both correspond to using incorrect net rates or miscalculating the initial fraction of water. Common Pitfalls: Some students mistakenly add the rates, as if both pipes were filling, which gives the wrong net effect. Others forget to account for the initial partial filling and instead compute the time to empty a full tank. Always first determine whether the net flow is into or out of the tank, and then apply that net rate to the actual starting volume. Final Answer: The tank will be emptied, and it will take 6 minutes to empty completely.

Question 20

A booster pump can be used both for filling and for emptying a tank of capacity 2400 cubic metres. The emptying rate of the pump is 10 cubic metres per minute higher than its filling rate, and the pump needs 8 minutes less time to empty the full tan…

Accepted Answer

Introduction / Context: This problem involves a single pump that can operate in two modes: filling and emptying. We are given the total capacity of the tank, the difference between the two rates, and the difference in times taken in the two modes. The goal is to find the filling rate of the pump. This is solved by setting up an algebraic equation based on time equals volume divided by rate. Given Data / Assumptions: Tank capacity = 2400 cubic metres. Emptying rate is 10 cubic metres per minute greater than filling rate. Time taken to empty the tank is 8 minutes less than the time taken to fill it. Concept / Approach: Let the filling rate be F cubic metres per minute. Then the emptying rate is F + 10 cubic metres per minute. The time to fill the tank is 2400 / F minutes. The time to empty it is 2400 / (F + 10) minutes. We are told that emptying is 8 minutes faster, so we can set up the equation 2400 / F - 2400 / (F + 10) = 8 and solve for F. Step-by-Step Solution: Step 1: Let filling rate = F m^3 per minute.Step 2: Then emptying rate = F + 10 m^3 per minute.Step 3: Time to fill the tank = 2400 / F minutes.Step 4: Time to empty the tank = 2400 / (F + 10) minutes.Step 5: Given that emptying is 8 minutes faster, we write 2400 / F - 2400 / (F + 10) = 8.Step 6: Divide both sides by 8 to simplify: 300 / F - 300 / (F + 10) = 1.Step 7: Multiply through by F(F + 10): 300(F + 10) - 300F = F(F + 10).Step 8: Left side simplifies to 300F + 3000 - 300F = 3000.Step 9: So F(F + 10) = 3000, giving the quadratic equation F^2 + 10F - 3000 = 0.Step 10: Solve this equation: F^2 + 10F - 3000 = 0 has a positive root F = 50.Step 11: Therefore, filling rate of the pump = 50 cubic metres per minute. Verification / Alternative check: If F = 50, then filling time = 2400 / 50 = 48 minutes. Emptying rate = 60 m^3 per minute, so emptying time = 2400 / 60 = 40 minutes. The difference is 48 - 40 = 8 minutes, which matches the problem statement. Hence our value for F is correct. Why Other Options Are Wrong: 60 m^3 per minute would make emptying rate 70, and the time difference would not equal 8 minutes. 72 m^3 per minute or 40 m^3 per minute similarly fail to satisfy the time difference when used in the equation and thus do not fit all the constraints. Common Pitfalls: One common mistake is to take 2400 / (F + 10) - 2400 / F = 8, reversing the sign of the time difference. Another is to cancel 2400 incorrectly or to forget that rates and times are reciprocals. Always express each time as volume divided by rate and then use the given information about the difference in times to set up a precise algebraic equation. Final Answer: The filling capacity of the pump is 50 m^3 per minute.

Pipes and Cistern Questions