Question 1

Three pipes A, B, and C together fill a tank in 5 hours. Pipe C is twice as fast as B, and B is twice as fast as A. How long will A alone take to fill the tank?

Accepted Answer

Problem restatement Given relative speeds of three pipes and their combined fill time, find the solo fill time of the slowest pipe A. Given data Combined time = 5 h ⇒ combined rate = 1 ÷ 5 tank/hour. C is twice B; B is twice A. Concept/Approach Let A's rate be r tanks/hour. Then B = 2r and C = 4r. Sum rates, equate to 1/5, solve for r, then take reciprocal for A's time. Step-by-step calculation Total rate = r + 2r + 4r = 7r 7r = 1/5 ⇒ r = 1/35 (tank/hour) Time for A alone = 1 ÷ r = 35 hours Verification/Alternative Individual times: A = 35 h, B = 17.5 h, C = 8.75 h. Harmonic combination 1/35 + 1/17.5 + 1/8.75 = 1/5 tank/hour (checks). Common pitfalls Assuming C is twice A directly (it is actually four times A since C = 2×B and B = 2×A). Final Answer 35 hours

Question 2

Pipes A and B together fill a cistern in 4 hours. If opened separately, B would take 6 hours longer than A. How many hours does A alone need to fill the cistern?

Accepted Answer

Problem restatement Two inlet pipes fill together in 4 hours. Separately, B needs 6 hours more than A. Find A's individual time. Given data Together time = 4 hours. B's time = A's time + 6 hours. Concept/Approach Let A's time be a hours, B's time be a + 6 hours. Use rates: 1/a + 1/(a + 6) = 1/4 and solve the quadratic for a > 0. Step-by-step calculation 1/a + 1/(a + 6) = 1/4(2a + 6) / [a(a + 6)] = 1/44(2a + 6) = a(a + 6) ⇒ 8a + 24 = a2 + 6aa2 − 2a − 24 = 0 ⇒ (a − 6)(a + 4) = 0a = 6 hours (reject negative root) Verification/Alternative A = 6 h ⇒ B = 12 h. Combined rate = 1/6 + 1/12 = 1/4 ⇒ 4 h, matches. Common pitfalls Adding times instead of rates; forgetting to discard the negative quadratic root. Final Answer 6 hours

Question 3

Pipes A, B, and C together fill a tank in 6 hours. After 2 hours, C is closed and A & B fill the rest in 7 hours. How many hours would C alone take to fill the tank?

Accepted Answer

Problem restatement All three fill for 2 hours, then only A and B continue for 7 hours. Find C's solo time. Given data a + b + c = 1/6 (tank per hour) Work done in first 2 hours = 2 / 6 = 1/3 Remaining = 2/3 done by A + B in 7 hours ⇒ a + b = 2/21 Concept/Approach Subtract to get c: c = (1/6) − (2/21). Then invert to get time. Step-by-step calculation c = 1/6 − 2/21 = 7/42 − 4/42 = 3/42 = 1/14So C's time = 14 hours Verification/Alternative Rates: a + b = 2/21; with c = 1/14 ⇒ a + b + c = 2/21 + 1/14 = 4/42 + 3/42 = 1/6. Common pitfalls Using 2 hours of A and B instead of 7 hours when setting a + b; mixing hours and fractions. Final Answer 14 hours

Question 4

Three pipes fill a tank with uniform flow. The first two together take the same time as the third alone. The second pipe is 5 hours faster than the first and 4 hours slower than the third. How long does the first pipe take?

Accepted Answer

Problem restatement Let the individual times be t1, t2, t3 for the first, second, third pipes. Given relational constraints and that the rate of the third equals the combined rate of first and second. Given data t2 = t1 − 5 t2 = t3 + 4 ⇒ t3 = t1 − 9 1/t1 + 1/t2 = 1/t3 Concept/Approach Translate the time relations into rates, substitute t2, t3 in terms of t1, and solve the resulting rational equation. Step-by-step calculation 1/t1 + 1/(t1 − 5) = 1/(t1 − 9)[(t1 − 5) + t1] / [t1(t1 − 5)] = 1/(t1 − 9)(2t1 − 5)(t1 − 9) = t1(t1 − 5)2t12 − 23t1 + 45 = t12 − 5t1t12 − 18t1 + 45 = 0Discriminant = 182 − 4×45 = 144 ⇒ t1 = (18 ± 12)/2 ⇒ 15 or 3Valid time: 15 hours (3 h would make t2 negative) Verification/Alternative t1=15 ⇒ t2=10, t3=6; rates 1/15 + 1/10 = 1/6 (true). Common pitfalls Equating times instead of rates: the statement implies equal rates over the same time interval. Final Answer 15 hours

Question 5

Pipes A and B can fill a tank in 20 min and 30 min respectively. If both are opened together, how long does it take to fill the tank?

Accepted Answer

Problem restatement Combine two constant fill rates to get the time to fill one tank. Given data A's rate = 1/20 tank per min B's rate = 1/30 tank per min Concept/Approach Add rates for parallel work: total rate = 1/20 + 1/30 = 1/12 tank per min ⇒ time = reciprocal. Step-by-step calculation 1/20 + 1/30 = (3 + 2)/60 = 5/60 = 1/12Time = 1 / (1/12) = 12 minutes Common pitfalls Averaging times instead of adding rates. Final Answer 12 minutes

Question 6

Taps A, B, C fill a tank in 12 h, 15 h, and 20 h respectively. A stays open continuously; B and C open alternately for one hour each. In how many hours will the tank be full?

Accepted Answer

Problem restatement Compute total time when A runs continuously and B, C alternate hourly. Given data A = 1/12 per hour B = 1/15 per hour C = 1/20 per hour Concept/Approach Evaluate work done in a 2-hour cycle (A+B then A+C). Multiply full cycles until near completion, then add the next hour as needed. Step-by-step calculation 2-hour work = (A+B) + (A+C) = 2A + B + C= 2/12 + 1/15 + 1/20 = 1/6 + 1/15 + 1/20LCM 60 ⇒ (10 + 4 + 3)/60 = 17/60 per 2 hoursAfter 6 hours (3 cycles): 3 × 17/60 = 51/60Next hour is A+B: 1/12 + 1/15 = 9/60 ⇒ 51/60 + 9/60 = 1Total time = 7 hours Common pitfalls Assuming 6 hours (stopping at cycles) without considering the next alternating hour. Final Answer 7 hours

Question 7

Pipes A and B fill a cistern in 37½ min and 45 min. If both open together but B is closed after some time, the cistern just fills in 30 min. After how many minutes should B be closed?

Accepted Answer

Problem restatement Run A and B together for t minutes, then only A runs, so that total fill time is 30 minutes. Given data A's time = 37.5 min ⇒ rate A = 2/75 per min B's time = 45 min ⇒ rate B = 1/45 per min Total time = 30 min Concept/Approach Work equation: t(A+B) + (30 − t)A = 1 tank ⇒ 30A + tB = 1. Step-by-step calculation 30A = 30 × (2/75) = 4/5 = 0.8tB = 1 − 0.8 = 0.2t = 0.2 / (1/45) = 9Close B after 9 minutes Verification/Alternative First 9 min: work = 9(2/75 + 1/45) = 9( (6 + 5)/225 ) = 9(11/225) = 99/225 = 0.44. Remaining 21 min by A: 21 × 2/75 = 42/75 = 0.56. Total = 1. Common pitfalls Using 37.5 as 37 or 38; keep it as 75/2 to avoid rounding. Final Answer 9 minutes

Question 8

Pipes A, B, C fill a tank in 30, 20, and 10 minutes. All are opened for 3 minutes into an empty tank. A, B, C deliver solutions P, Q, R respectively. What fraction of the tank's liquid is solution R after 3 minutes?

Accepted Answer

Problem restatement Each pipe pours its own solution; find R's proportion after 3 minutes of simultaneous inflow. Given data A rate = 1/30 per min ⇒ 3 min gives 1/10 tank of P B rate = 1/20 per min ⇒ 3 min gives 3/20 tank of Q C rate = 1/10 per min ⇒ 3 min gives 3/10 tank of R Concept/Approach Proportion of R = (volume from C) / (total volume after 3 min). No mixing losses. Step-by-step calculation Total filled = 1/10 + 3/20 + 3/10 = 2/20 + 3/20 + 6/20 = 11/20R contributed = 3/10 = 6/20Proportion R = (6/20) / (11/20) = 6/11 Common pitfalls Dividing C's rate by total rates, instead of using 3-minute volumes. Final Answer 6/11

Question 9

Two pipes fill a tank in 20 min and 24 min; a waste pipe empties at 3 gallons/min. All three together fill the tank in 15 min. What is the tank's capacity (in gallons)?

Accepted Answer

Problem restatement Given two fill rates dependent on capacity and one constant emptying rate, find the capacity from the net fill time. Given data Fillers: V/20 and V/24 gallons per min Waste: 3 gallons per min Net fill time with all open: 15 min ⇒ net rate = V/15 Concept/Approach Set up the net-rate equation: V/20 + V/24 − 3 = V/15 and solve for V. Step-by-step calculation V/20 + V/24 = (6V + 5V)/120 = 11V/12011V/120 − 3 = V/15 = 8V/120(11V − 8V)/120 = 3 ⇒ 3V/120 = 3 ⇒ V/40 = 3V = 120 gallons Verification/Alternative Rates at V = 120: 6 + 5 − 3 = 8 gal/min ⇒ 120/8 = 15 min (checks). Common pitfalls Forgetting that only the waste pipe's rate is capacity-independent. Final Answer 120 gallons

Question 10

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

Accepted Answer

Problem restatement First phase: 1 tap until half-full. Second phase: 4 identical taps for the remaining half. Given data Single-tap time = 6 h ⇒ single rate = 1/6 per h Concept/Approach Compute phase times separately and add. Step-by-step calculation Phase 1: time = (1/2) / (1/6) = 3 hPhase 2: 4 taps ⇒ rate = 4 / 6 = 2/3 per hTime for remaining half = (1/2) / (2/3) = 3/4 h = 45 mTotal time = 3 h + 45 m = 3 h 45 m Common pitfalls Averaging rates across phases instead of computing phase-wise times. Final Answer 3 h 45 m

Question 11

Pipes A and B fill a tank in 5 h and 6 h; pipe C empties it in 12 h. If all three are opened together, how long will the tank take to fill?

Accepted Answer

Problem restatement Combine two inlet rates and one outlet rate to get net fill time. Given data A = 1/5 per h B = 1/6 per h C (outlet) = 1/12 per h Concept/Approach Net rate = A + B − C. Time = 1 / (net rate). Step-by-step calculation Net rate = 1/5 + 1/6 − 1/12= (12 + 10 − 5)/60 = 17/60 per hourTime = 1 / (17/60) = 60/17 hours = 3 9/17 hours ≈ 3 h 31 m 46 s Verification/Alternative Check net positivity: 1/5 + 1/6 > 1/12, so the tank fills. Common pitfalls Adding times instead of rates, or subtracting 1/12 from only one inlet. Final Answer 3 9/17 hours (≈ 3 h 31 m 46 s)

Question 12

A pump can fill a tank in 2 hours. Due to a leak, it takes 2 1/3 hours to fill. In how many hours can the leak alone drain the full tank?

Accepted Answer

Problem restatement A pump fills a tank in 2 hours, but with a leak present, filling takes 2 1/3 hours. Find the time the leak alone would need to empty a full tank. Given data Pump's fill time (alone) = 2 h ⇒ fill rate = 1/2 tank per hour. Fill time with leak = 2 1/3 h = 7/3 h ⇒ net fill rate = 1 ÷ (7/3) = 3/7 tank per hour. Concept/Approach Net rate = pump rate − leak rate. Hence leak rate = pump rate − net rate. Step-by-step calculation Leak rate = 1/2 − 3/7 = (7/14 − 6/14) = 1/14 tank per hour Time for leak alone to drain full tank = 1 ÷ (1/14) = 14 hours Verification/Alternative Check: With leak, net = 1/2 − 1/14 = (7 − 1)/14 = 6/14 = 3/7 tank per hour ⇒ time 7/3 h = 2 1/3 h (matches). Common pitfalls Adding rates instead of subtracting to get the net with a leak (which drains). Final Answer 14 hours

Question 13

Two pipes A and B can fill a tank in 15 min and 20 min. Both are opened; after 4 min, A is closed. What is the total time to fill the tank?

Accepted Answer

Problem restatement Pipes A and B run together for 4 minutes, then only B continues. Find total filling time. Given data A's rate = 1/15 tank/min; B's rate = 1/20 tank/min. Both open for first 4 min, then only B runs. Concept/Approach Compute work done in the first phase, subtract from 1 tank, then compute remaining time with B alone. Step-by-step calculation Rate(A + B) = 1/15 + 1/20 = 7/60 tank/min Work in 4 min = 4 × 7/60 = 28/60 = 7/15 tank Remaining = 1 − 7/15 = 8/15 tank Time with B alone = (8/15) ÷ (1/20) = (8/15) × 20 = 160/15 = 10 2/3 min = 10 min 40 s Total time = 4 min + 10 min 40 s = 14 min 40 s Verification/Alternative Fractional minutes 10 2/3 converts to 10 minutes 40 seconds (since 2/3 of a minute is 40 seconds). Common pitfalls Using A's time after closing instead of B's. Arithmetic slip when converting fractional minutes to seconds. Final Answer 14 minutes 40 seconds

Question 14

Pipes A and B fill a tanker in 60 min and 40 min. B is used for half the total time; A and B together fill for the other half. How many minutes to fill the tanker?

Accepted Answer

Problem restatement Let total time be T minutes. For T/2 minutes only B runs; for T/2 minutes A and B run together. Find T so that 1 full tank is filled. Given data Rate(A) = 1/60 tank/min; Rate(B) = 1/40 tank/min. Rate(A + B) = 1/60 + 1/40 = 1/24 tank/min. Concept/Approach Total work = (time × rate) summed over phases; set equal to 1. Step-by-step calculation Work = (T/2)×(1/40) + (T/2)×(1/24) = T×(1/80 + 1/48) 1/80 + 1/48 = (3/240 + 5/240) = 8/240 = 1/30 Thus T × (1/30) = 1 ⇒ T = 30 minutes Verification/Alternative In 15 min, B alone fills 15/40 = 0.375; in next 15 min, A+B fill 15/24 = 0.625; total = 1.000 (exact). Common pitfalls Assuming sequential halves mean equal work instead of equal time. Final Answer 30 minutes

Question 15

One pipe fills a tank three times as fast as another. Together they fill in 36 minutes. How long will the slower pipe take alone?

Accepted Answer

Problem restatement Two pipes have rates in the ratio 1 : 3 (slower : faster). Together they fill in 36 minutes. Find the slower pipe's solo time. Given data Let slower rate = r tank/min; faster rate = 3r tank/min. Together rate = 1/36 tank/min. Concept/Approach Add rates to match the combined fill rate, then invert to get time. Step-by-step calculation r + 3r = 4r = 1/36 ⇒ r = 1/144 tank/min Slower's time = 1 ÷ r = 144 minutes Verification/Alternative Faster time = 1 ÷ (3r) = 48 min; check: 1/144 + 1/48 = (1 + 3)/144 = 4/144 = 1/36 (ok). Final Answer 144 minutes

Question 16

Pipe A can fill a tank in 45 hours and pipe B can fill it in 36 hours. If both pipes are opened together into an empty tank, in how many hours will the tank be completely full?

Accepted Answer

Introduction / Context: Pipes-and-cistern problems rely on constant rates. When multiple inlets are opened together, their filling rates add to give a combined rate that determines the total time to fill the tank. Given Data / Assumptions: Pipe A fills the tank in 45 hours (rate = 1/45 tank/hour). Pipe B fills the tank in 36 hours (rate = 1/36 tank/hour). The tank is initially empty; both pipes run simultaneously without interruptions or losses. Concept / Approach: For constant rates, net rate = sum of individual rates for inlets. Time to complete one tank = 1 / (net rate). Step-by-Step Solution: Rate(A) = 1/45 tank per hour.Rate(B) = 1/36 tank per hour.Net rate = 1/45 + 1/36.LCM(45, 36) = 180 ⇒ 1/45 = 4/180, 1/36 = 5/180.Net rate = (4 + 5)/180 = 9/180 = 1/20 tank per hour.Time to fill = 1 / (1/20) = 20 hours. Verification / Alternative check: In 20 hours, A contributes 20/45 = 4/9 tank and B contributes 20/36 = 5/9 tank; total 1 tank. Why Other Options Are Wrong: 10 hr and 15 hr underestimate the time by treating the faster pipe as much quicker than it is; 28 hr ignores correct rate addition. Common Pitfalls: Adding times instead of rates or using an incorrect LCM when summing fractions. Final Answer: 20 hr

Question 17

If a single inlet pipe can fill a tank in 6 hours, what fraction of the tank will be filled by this pipe in 1 hour?

Accepted Answer

Introduction / Context: When a job takes T hours, the constant rate per hour is the reciprocal, 1/T of the job per hour. Pipes-and-cistern questions frequently ask for the fraction filled in a given time. Given Data / Assumptions: The pipe fills the entire tank in 6 hours. Filling rate is constant (no leakage or pressure fluctuations). We seek the portion filled in exactly 1 hour. Concept / Approach: Rate = work/time. For one full tank in 6 hours, the per-hour fraction filled is 1/6 of the tank. Step-by-Step Solution: Total work (one full tank) = 1 tank.Time to complete = 6 hours.Rate = 1 / 6 tank per hour.Portion filled in 1 hour = rate * time = (1/6) * 1 = 1/6. Verification / Alternative check: In 6 hours at 1/6 per hour, total filled = 6 * (1/6) = 1 tank, consistent. Why Other Options Are Wrong: 1/3 and 1/4 imply faster filling; 1/5 is arbitrary relative to the given 6-hour completion. Common Pitfalls: Confusing rate with remaining fraction or trying to proportion time and fraction incorrectly. Final Answer: 1/6

Question 18

A cistern is normally filled in 8 hours, but due to a leak at the bottom it takes 10 hours to fill. If the cistern is full, in how many hours will the leak alone empty it?

Accepted Answer

Introduction / Context: When a leak is present, the net filling rate equals the inlet rate minus the leak rate. By comparing normal filling time to delayed filling time, we can deduce the leak’s emptying rate and hence its time to empty a full cistern. Given Data / Assumptions: Normal fill time (no leak) = 8 hours ⇒ inlet rate = 1/8 tank/hour. With leak, fill time = 10 hours ⇒ net rate = 1/10 tank/hour. Leak rate is constant and acts continuously while the inlet runs. Concept / Approach: Net rate = inlet rate − leak rate. Rearranging gives leak rate = inlet rate − net rate. Time to empty (leak alone) is the reciprocal of the leak rate. Step-by-Step Solution: Inlet rate = 1/8.Net rate with leak = 1/10.Leak rate = 1/8 − 1/10 = (5 − 4)/40 = 1/40 tank/hour.Time for leak to empty a full cistern = 1 / (1/40) = 40 hours. Verification / Alternative check: Check: With both acting, 1/8 − 1/40 = (5 − 1)/40 = 4/40 = 1/10 tank/hour, matching the delayed 10-hour fill. Why Other Options Are Wrong: 16, 20, and 25 hours correspond to incorrect differences or inverted fractions. Common Pitfalls: Subtracting times instead of rates, or forgetting to invert to get time from rate. Final Answer: 40 hrs.

Question 19

One pipe can empty a tank in 12 minutes and another pipe can empty it in 16 minutes. If both outlet pipes are opened together, how long will it take to empty a full tank?

Accepted Answer

Introduction / Context: When two outlets drain a tank simultaneously, their emptying rates add. The total time is the reciprocal of the combined rate. Given Data / Assumptions: Outlet 1 empties in 12 minutes ⇒ rate = 1/12 tank/min. Outlet 2 empties in 16 minutes ⇒ rate = 1/16 tank/min. The tank is initially full and both outlets run together. Concept / Approach: Net emptying rate = 1/12 + 1/16. Time to empty = 1 / (net emptying rate). Step-by-Step Solution: LCM(12, 16) = 48.1/12 = 4/48; 1/16 = 3/48.Combined rate = (4 + 3)/48 = 7/48 tank/min.Time to empty = 1 / (7/48) = 48/7 minutes ≈ 6.857 minutes. Verification / Alternative check: Approximate check: 7 minutes at 7/48 per minute would drain 49/48 > 1 tank; slightly less than 7 minutes is consistent with 48/7. Why Other Options Are Wrong: 6 minutes is too short; 7 minutes is a rounded overestimate; 8 minutes is clearly too long. Common Pitfalls: Adding times instead of rates or miscomputing the LCM when adding fractions. Final Answer: 48/7 minutes

Question 20

A cistern is normally filled in 8 hours but takes 2 hours longer (10 hours total) due to a bottom leak. If the cistern is full, how long will the leak alone take to empty it completely?

Accepted Answer

Introduction / Context: Compare normal filling versus delayed filling to deduce the leak’s emptying rate. The leak subtracts from the inlet’s rate, increasing the total fill time. Given Data / Assumptions: No leak fill time = 8 hours ⇒ inlet rate = 1/8 tank/hour. With leak fill time = 10 hours ⇒ net rate = 1/10 tank/hour. Rates remain constant; the leak acts uniformly. Concept / Approach: leak rate = inlet rate − net rate. Emptying time of the leak = reciprocal of leak rate. Step-by-Step Solution: Inlet rate = 1/8.Net rate = 1/10.Leak rate = 1/8 − 1/10 = 1/40 tank/hour.Time to empty = 1 / (1/40) = 40 hours. Verification / Alternative check: With both acting: 1/8 − 1/40 = 4/40 = 1/10 ⇒ matches 10 hours to fill. Why Other Options Are Wrong: 16, 20, and 25 hours arise from wrong fraction differences or inversion mistakes. Common Pitfalls: Subtracting times rather than rates; forgetting to invert rate to get time. Final Answer: 40 hrs.

Pipes and Cistern Questions