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 …

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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…

Accepted Answer

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 the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

Accepted Answer

Part filled in 4 minutes =4(1/15+1/20) = 7/15 Remaining part =(1-7/15) = 8/15 Part filled by B in 1 minute =1/20 : 8/15 :: 1:x x = (8/15*1*20) = 10 2 3 m i n = 10 m i n 40 s e c The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec

Question 13

A pump can fill a tank with water in 2 hours. Because of a leak, it took 213 hours to fill the tank. The leak can drain all the water of the tank in:

Accepted Answer

If the total area of pump=1 part The pumop take 2 hrs to fill 1 part The pumop take1 hour to fill 1/2 portion Due to lickage The pumop take 7/3 hrs to fill 1 part The pumop take1 hour to fill 3/7 portion Now the difference of area = (1/2-3/7)=1/14 This 1/14 part of water drains in 1 hour Total area=1 part of water drains in (1x14/1)hours= 14 hours So the leak can drain all the water of the tank in 14 hours. Leak will empty the tank in 14 hrs

Question 14

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

Accepted Answer

Time taken by one tap to fill half of the tank = 3 hrs. Part filled by the four taps in 1 hour =4*1/6 =2/3 Remaining part = 1 - 1 2 = 1 2 2 3 : 1 2 : : 1 : x => x = 1 2 * 1 * 3 2 = 3 4 So, total time taken = 3 hrs. 45 mins.

Question 15

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

Accepted Answer

Part filled in 2 hours = 2/6=1/3 Remaining part = 1 - 1 3 = 2 3 (A + B)'s 7 hour's work = 2/3 (A + B)'s 1 hour's work = 2/21 C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work } =1/6-2/21 = 1/14 C alone can fill the tank in 14 hours.

Question 16

Three taps A,B and C can fill a tank in 12,15 and 20 hours respectively. If A is open all the time and B ,C are open for one hour each alternatively, the tank will be full in:

Accepted Answer

A + B ' s 1 h o u r w o r k = 1 12 + 1 15 = 9 60 = 3 20 A + C ' s 1 h o u r w o r k = 1 12 + 1 20 = 8 60 = 2 15 p a r t f i l l e d i n 2 h r s = 3 20 + 2 15 = 17 60 p a r t f i l l e d i n 6 h r s = 3 * 17 60 = 17 20 r e m a i n i n g p a r t = 1 - 17 20 = 3 20 Now, it is the turn of A and B (3/20) part is filled by A and B in 1 hour. Therefore, Total time taken to fill the tank =(6+1)hrs= 7 hrs

Question 17

A cistern is normally filled in 8 hours but takes two hours longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in ?

Accepted Answer

1/8 - 1/x = 1/10 => x = 40 hrs

Question 18

A and B can fill the tank in 60 min and 90 min. There is a leak at 3/4th of the height. If leak is opened alone, it takes 36 min to empty till 3/4th the height. Find the time taken to fill the tank if all of the taps and the leak are opened simultan…

Accepted Answer

39

Question 19

A water tank is two-fifth full.Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes.If both the pipes are open,how long will it take to empty or fill the tank completely?

Accepted Answer

Clearly,pipe B is faster than pipe A and so,the tank will be emptied. part to be emptied = 2/5 part emptied by (A+B) in 1 minute= 1 6 - 1 10 = 1 15 1 15 : 2 5 : : 1 : x 2 5 * 15 = 6 m i n s so, the tank will be emptied in 6 min

Question 20

A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m3 . The emptying capacity of the tank is 10 m3 per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank…

Accepted Answer

Let the filling capacity of the pump be x m 3 /min. Then, emptying capacity of the pump=(x+10) m 3 /min. so, 2400 x - 2400 x + 10 = 8 ⇔ x 2 + 10 x - 3000 = 0 ⇒ x - 50 + x + 60 = 0 ⇔ x = 50

Pipes and Cistern Questions