Question 1

An inlet pipe fills 1/8 of a tank in 1 hour. How much total time will the pipe take to fill the empty tank completely?

Accepted Answer

Introduction / Context: If a constant-rate inlet fills a fixed fraction per hour, the time to fill the entire tank is the reciprocal of the fraction filled per hour. Given Data / Assumptions: Fraction filled in 1 hour = 1/8. Rate is constant and there are no leaks. Concept / Approach: Let r be the rate in tank/hour. If r = 1/8, then time to fill one tank = 1 / r = 8 hours. Step-by-Step Solution: Given: per-hour fill = 1/8 tank.Total time for 1 tank = 1 ÷ (1/8) = 8 hours. Verification / Alternative check: In 8 hours at 1/8 per hour, total filled = 8 * (1/8) = 1 tank, consistent. Why Other Options Are Wrong: 2 h, 4 h, and 6 h assume higher hourly fractions than given. Common Pitfalls: Confusing 1/8 per hour with 1/8 of the remaining tank each hour (which would be variable rate; not applicable here). Final Answer: 8 h

Question 2

An outlet pipe can empty a full cistern in 3 hours. In how much time will the same pipe empty exactly two-thirds of the cistern?

Accepted Answer

Introduction / Context: When rate is constant, time is directly proportional to the fraction of the job completed. Emptying two-thirds of a tank at a constant rate uses two-thirds of the full-emptying time. Given Data / Assumptions: Full emptying time = 3 hours. Constant emptying rate; no refilling occurs during emptying. Concept / Approach: Time for a fraction f of the work at constant rate equals f times the full time. Step-by-Step Solution: Fraction to be emptied = 2/3.Full-emptying time = 3 hours.Required time = (2/3) * 3 = 2 hours. Verification / Alternative check: Rate = 1 tank / 3 hours = 1/3 tank/hour. In 2 hours, volume emptied = (1/3) * 2 = 2/3 tank. Why Other Options Are Wrong: 3 h is for the entire tank; 4 h and 5 h exceed the full-emptying time, which is impossible here. Common Pitfalls: Using inverse proportionality (incorrect in this direct fraction-time scenario). Final Answer: 2 h

Question 3

A pipe can fill a tank in 20 hours. Due to a leak at the bottom, the tank actually gets filled in 40 hours. If the tank is full, how long will the leak alone take to empty it?

Accepted Answer

Introduction / Context: A leak reduces the net filling rate. Comparing the normal fill rate with the slowed net rate lets us compute the leak’s own emptying rate and its corresponding emptying time. Given Data / Assumptions: Inlet alone: 20 hours to fill ⇒ rate = 1/20 tank/hour. With leak present: 40 hours to fill ⇒ net rate = 1/40 tank/hour. Rates are constant throughout. Concept / Approach: leak rate = inlet rate − net rate. Emptying time of the leak is the reciprocal of this leak rate. Step-by-Step Solution: Inlet rate = 1/20.Net rate = 1/40.Leak rate = 1/20 − 1/40 = 1/40 tank/hour.Time to empty full tank = 1 / (1/40) = 40 hours. Verification / Alternative check: Check net: 1/20 − 1/40 = 1/40 ⇒ net fill time 40 hours, consistent. Why Other Options Are Wrong: 30 h and 50 h reflect incorrect subtraction or inversion of rates. Common Pitfalls: Subtracting times (20 − 40) instead of rates; forgetting to invert to get time. Final Answer: 40 h

Question 4

If a single inlet pipe fills a tank in 6 hours, what portion of the tank is filled by this pipe in exactly 1 hour?

Accepted Answer

Introduction / Context: The rate for finishing one job in T hours is 1/T job per hour. For a tank-filling problem, that means the fraction of a full tank filled per hour is the reciprocal of the total time in hours. Given Data / Assumptions: Total time to fill one full tank = 6 hours. Rate is constant and there are no leaks. Concept / Approach: Fraction filled in 1 hour = 1 / 6 of the tank. Step-by-Step Solution: Rate = 1/6 tank per hour.In 1 hour, filled = 1/6 of the tank. Verification / Alternative check: Over 6 hours, 6 * (1/6) = 1 whole tank, confirming consistency. Why Other Options Are Wrong: 1/3, 1/4, and 1/5 overstate the rate compared to the given 6-hour completion time. Common Pitfalls: Confusing rate with remaining fraction or misapplying proportionality. Final Answer: 1/6

Question 5

An outlet pipe can empty a cistern in 30 minutes. What fraction of the cistern will it empty in 1 minute?

Accepted Answer

Introduction / Context: For constant-rate emptying, the fraction emptied per minute equals the reciprocal of the total emptying time in minutes. Given Data / Assumptions: Full emptying time = 30 minutes. Constant rate; the outlet alone is operating. Concept / Approach: The per-minute emptying fraction is 1/30 of the tank. Step-by-Step Solution: Total time T = 30 minutes → rate = 1/T = 1/30 tank per minute.In 1 minute, emptied = 1/30 tank. Verification / Alternative check: 30 minutes at 1/30 per minute removes 1 tank total; consistent. Why Other Options Are Wrong: 1/20 and 1/18 are too large; 1/15 halves the time incorrectly. Common Pitfalls: Switching numerator and denominator when forming the reciprocal. Final Answer: 1/30

Question 6

Three pipes A, B, and C can fill a tank separately in 8 hours, 10 hours, and 20 hours, respectively. If all three are opened together, how long will they take to fill the tank?

Accepted Answer

Introduction / Context: With multiple inlets, the combined filling rate is the sum of individual rates. The time to complete one tank equals the reciprocal of the total rate. Given Data / Assumptions: A fills in 8 h ⇒ rate = 1/8 tank/h. B fills in 10 h ⇒ rate = 1/10 tank/h. C fills in 20 h ⇒ rate = 1/20 tank/h. All three run together from empty to full. Concept / Approach: Add the three rates, then invert to get time. Step-by-Step Solution: Sum rate = 1/8 + 1/10 + 1/20.LCM(8,10,20) = 40 ⇒ 1/8 = 5/40, 1/10 = 4/40, 1/20 = 2/40.Total rate = (5 + 4 + 2)/40 = 11/40 tank/h.Time = 1 / (11/40) = 40/11 hours ≈ 3.636 hours. Verification / Alternative check: Approximate hours: 3 hours 38 minutes (about 3 h 37 m). This aligns with 40/11 h. Why Other Options Are Wrong: 37/11, 47/11, 57/11 h correspond to incorrect numerator sums or mis-additions of rates. Common Pitfalls: Using average of times instead of sum of rates; arithmetic slips in fractional addition. Final Answer: 40/11 h

Question 7

A tap can fill a tank in 16 minutes while another tap can empty it in 8 minutes. If the tank is already half full and both taps are opened together, what happens and in how much time?

Accepted Answer

Introduction / Context: Competing taps—one filling and one emptying—yield a net rate equal to the difference of their individual rates. Starting from half full, we determine whether the tank fills or empties and the time required. Given Data / Assumptions: Fill tap: 16 minutes ⇒ +1/16 tank/min. Empty tap: 8 minutes ⇒ −1/8 tank/min. Initial content = 1/2 tank. Concept / Approach: Net rate = 1/16 − 1/8 = −1/16 tank/min (negative ⇒ net emptying). Time to go from 1/2 to 0 at this rate is (1/2) / (1/16). Step-by-Step Solution: Net rate = 1/16 − 1/8 = 1/16 − 2/16 = −1/16 tank/min.Volume to empty = 1/2 tank.Time = (1/2) ÷ (1/16) = 8 minutes. Verification / Alternative check: In 8 minutes, the empty tap removes 8 * (1/8) = 1 tank; the fill tap adds 8 * (1/16) = 1/2 tank; net change = −1/2, taking the level from 1/2 to 0. Why Other Options Are Wrong: Filled options are impossible since the net rate is negative; 12 minutes overestimates time. Common Pitfalls: Adding times rather than rates; forgetting the starting level is half, not empty. Final Answer: Emptied in 8 min.

Question 8

Two inlet pipes A and B can fill a cistern in 4 hours and 6 hours, respectively. When full, an outlet pipe C can empty it in 8 hours. If all three are opened together from empty, in how much time will the cistern be filled?

Accepted Answer

Introduction / Context: Here, two inlets add water and one outlet removes it. The net rate is the sum of inlet rates minus the outlet rate. Once the net rate is found, invert it to get the total filling time. Given Data / Assumptions: A fills in 4 h ⇒ 1/4 tank/h. B fills in 6 h ⇒ 1/6 tank/h. C empties in 8 h ⇒ 1/8 tank/h (outflow). Concept / Approach: Net rate = 1/4 + 1/6 − 1/8. Step-by-Step Solution: LCM(4,6,8) = 24 ⇒ 1/4 = 6/24, 1/6 = 4/24, 1/8 = 3/24.Net rate = (6 + 4 − 3)/24 = 7/24 tank/h.Time = 1 ÷ (7/24) = 24/7 h ≈ 3.4286 h = 3 hours 25.7 minutes ≈ 3 hours 26 minutes. Verification / Alternative check: In 24/7 hours, volume added by A and B is (6/24 + 4/24)*(24/7) = 10/7; removed by C is (3/24)*(24/7) = 3/7; net = 1 tank. Why Other Options Are Wrong: 3:18 and 3:42 are symmetric around the right value but not exact; 3:48 is too slow. Common Pitfalls: Adding or subtracting times directly; rounding too early and missing the exact 24/7 h. Final Answer: 3 hours 26 min.

Question 9

One tap can fill a cistern in 2 hours while another tap can empty the cistern in 3 hours. If both taps are opened together on an empty cistern, how long will it take to fill the cistern?

Accepted Answer

Introduction / Context: Competing taps create a net rate equal to the difference between the filling and emptying rates. If the net is positive, the tank eventually fills; the time equals the reciprocal of the net rate. Given Data / Assumptions: Fill tap: 2 hours ⇒ +1/2 tank/hour. Empty tap: 3 hours ⇒ −1/3 tank/hour. Tank starts empty and both run continuously. Concept / Approach: Net rate = 1/2 − 1/3 = 1/6 tank/hour. Time = 1 / (1/6) = 6 hours. Step-by-Step Solution: Compute difference: 1/2 − 1/3 = (3 − 2)/6 = 1/6.Time to fill = 1 ÷ (1/6) = 6 hours. Verification / Alternative check: In 6 hours: filled = 6*(1/2) = 3 tanks; emptied = 6*(1/3) = 2 tanks; net = 1 tank. Why Other Options Are Wrong: 5 and 7–8 hours imply different net rates not supported by the given times. Common Pitfalls: Adding times rather than subtracting rates; sign errors when handling the outlet. Final Answer: 6 hours

Question 10

A pipe can fill a tank in x hours and another pipe can empty it in y hours. Assuming y > x so that the net effect is filling, how long will both pipes together take to fill the tank?

Accepted Answer

Introduction / Context: This is the general formula case for one inlet and one outlet. With y > x, the emptying pipe is slower than the filler, so the net rate is positive and the tank fills eventually. Given Data / Assumptions: Fill rate = 1/x tank/hour. Empty rate = 1/y tank/hour. y > x ⇒ net positive filling. Concept / Approach: Net rate = 1/x − 1/y = (y − x) / (xy). Time is the reciprocal of the net rate. Step-by-Step Solution: Net rate = 1/x − 1/y = (y − x)/(xy).Time = 1 ÷ ((y − x)/(xy)) = xy / (y − x) hours. Verification / Alternative check: Plug numerical values (e.g., x = 5, y = 10): net rate = 1/5 − 1/10 = 1/10 ⇒ time = 10 = xy/(y − x) = 50/5 = 10. Why Other Options Are Wrong: (x − y) or (y − x) hours ignore reciprocal; xy/(x − y) has the wrong sign for y > x. Common Pitfalls: Forgetting to invert net rate; sign confusion when forming the difference. Final Answer: xy / ( y - x) hours

Question 11

A tap can fill a cistern in 8 hours and another tap can empty it in 16 hours. If both taps are opened simultaneously from empty, what is the time (in hours) required to fill the cistern?

Accepted Answer

Introduction / Context: Opening a filler and an emptier together produces a net rate equal to their difference. If the filler is faster, the cistern will fill, albeit more slowly than with the filler alone. Given Data / Assumptions: Filler: 8 hours ⇒ +1/8 tank/hour. Emptier: 16 hours ⇒ −1/16 tank/hour. Concept / Approach: Net rate = 1/8 − 1/16 = 1/16 tank/hour. Time = 1 ÷ (1/16) = 16 hours. Step-by-Step Solution: Compute: 1/8 − 1/16 = 2/16 − 1/16 = 1/16.Time = 16 hours. Verification / Alternative check: In 16 hours, filler adds 2 tanks; emptier removes 1 tank; net = 1 tank filled. Why Other Options Are Wrong: 8 or 10 hours would imply a faster net rate than 1/16; 24 hours would be slower than the computed net rate. Common Pitfalls: Adding times or forgetting the outlet’s negative contribution. Final Answer: 16

Question 12

Pipes A and B can fill a tank in 10 hours and 15 hours respectively. If both are opened together into an empty tank, how long will they take to fill it?

Accepted Answer

Introduction / Context: When two inlets run together, their rates add. The time to complete one tank equals the reciprocal of the combined rate. Given Data / Assumptions: A fills in 10 h ⇒ 1/10 tank/h. B fills in 15 h ⇒ 1/15 tank/h. Concept / Approach: Combined rate = 1/10 + 1/15. Time = 1 ÷ (combined rate). Step-by-Step Solution: LCM(10, 15) = 30 ⇒ 1/10 = 3/30; 1/15 = 2/30.Total rate = (3 + 2)/30 = 5/30 = 1/6 tank/h.Time to fill = 1 ÷ (1/6) = 6 hours. Verification / Alternative check: In 6 hours, A fills 6/10 = 3/5 and B fills 6/15 = 2/5; total = 1 tank. Why Other Options Are Wrong: 12 1/2 hours is slower than either pipe alone; 5 hours is too optimistic for these rates. Common Pitfalls: Adding times instead of rates; arithmetic errors in LCM. Final Answer: 6 hours

Question 13

Two taps fill a tank in 10 minutes and 12 minutes respectively, while a third tap empties it. With the third tap closed, the two fillers together can fill the tank as stated; with all three open, the tank is filled in 15 minutes. If the two fillers …

Accepted Answer

Introduction / Context: We are given individual filling times for two taps, and the net filling time when an emptying tap also runs. From this, we back out the emptying rate and compute the emptying time alone. Given Data / Assumptions: Filler A: 10 min ⇒ 1/10 tank/min. Filler B: 12 min ⇒ 1/12 tank/min. All three together fill in 15 min ⇒ net = 1/15 tank/min. Concept / Approach: Let E be the emptying rate (tank/min). Then 1/10 + 1/12 − E = 1/15 ⇒ E = 1/10 + 1/12 − 1/15. Step-by-Step Solution: Compute E: LCM(10,12,15) = 60.1/10 = 6/60; 1/12 = 5/60; 1/15 = 4/60.E = (6 + 5 − 4)/60 = 7/60 tank/min.Emptying time = 1 ÷ (7/60) = 60/7 min = 8 min 34.285… sec ≈ 8 min 34 sec. Verification / Alternative check: With all three: 6/60 + 5/60 − 7/60 = 4/60 = 1/15 tank/min, as given. Why Other Options Are Wrong: 7 min and 6 min are too fast; 9 min 32 sec is too slow compared with 60/7. Common Pitfalls: Sign error for the outlet; arithmetic mistakes when converting minutes to seconds. Final Answer: 8 min. and 34 sec.

Question 14

A tap can fill a tank in 25 minutes while another tap can empty it in 50 minutes. If both taps are opened together from empty, will the tank be filled or emptied, and in how many minutes?

Accepted Answer

Introduction / Context: Determine the net rate when a filler and an emptier operate simultaneously. If the filler is faster, the tank fills; otherwise, it empties. The time equals the reciprocal of the net rate. Given Data / Assumptions: Filler: 25 minutes ⇒ +1/25 tank/min. Emptier: 50 minutes ⇒ −1/50 tank/min. Concept / Approach: Net rate = 1/25 − 1/50 = (2 − 1)/50 = 1/50 tank/min (positive ⇒ filling). Step-by-Step Solution: Net rate = 1/25 − 1/50 = 1/50 tank/min.Time to fill = 1 ÷ (1/50) = 50 minutes. Verification / Alternative check: Check contributions in 50 minutes: filler adds 50*(1/25) = 2 tanks; emptier removes 50*(1/50) = 1 tank; net = 1 tank. Why Other Options Are Wrong: 20 or 25 minutes incorrectly assume higher net rates or net emptying. Common Pitfalls: Subtracting times instead of rates; sign confusion about the outlet. Final Answer: Tank is filled up in 50 minutes

Question 15

Two pipes A and B can fill a tank in 18 hours and 6 hours, respectively. If both are opened together from empty, how much time is needed to fill the tank?

Accepted Answer

Introduction / Context: For simultaneous inlets, add rates to get the total rate. The filling time equals the reciprocal of that total rate. Given Data / Assumptions: A fills in 18 h ⇒ 1/18 tank/h. B fills in 6 h ⇒ 1/6 tank/h. Concept / Approach: Total rate = 1/18 + 1/6 = 1/18 + 3/18 = 4/18 = 2/9 tank/h. Time = 1 ÷ (2/9) = 9/2 h. Step-by-Step Solution: Sum rates: 1/18 + 1/6 = 4/18 = 2/9.Time = 1 ÷ (2/9) = 9/2 hours = 4.5 hours = 4 1/2h. Verification / Alternative check: Check: In 4.5 h, A fills 4.5/18 = 0.25 tank; B fills 4.5/6 = 0.75 tank; total = 1 tank. Why Other Options Are Wrong: 7, 6, 10 h are inconsistent with the combined rate 2/9 tank/h. Common Pitfalls: Adding times, not rates; mis-converting 4.5 hours to mixed fraction. Final Answer: 4 1/2h

Question 16

Net filling with one inlet and one outlet A cistern can be filled by inlet A in 4 hours. An outlet B can empty the cistern in 8 hours. If both A and B are opened simultaneously, after how much time will the cistern become full?

Accepted Answer

Introduction / Context: Classic “pipes and cisterns” problems reduce to adding or subtracting constant rates. Inlets add volume; outlets remove volume. When they run together, the effective rate is the algebraic sum of their individual rates. Given Data / Assumptions: Inlet A fills 1 tank in 4 h. Outlet B empties 1 tank in 8 h. Rates are constant; tank volume is normalized to 1 unit. Concept / Approach: Rate(inlet) = 1/4 tank per hour. Rate(outlet) = 1/8 tank per hour (negative in net arithmetic). Net rate when both are open = 1/4 − 1/8. Step-by-Step Solution: Rate(A) = 1/4Rate(B) = −1/8Net rate = 1/4 − 1/8 = 1/8 tank per hourTime to fill = 1 / (1/8) = 8 h Verification / Alternative check: In 8 h, A contributes 8*(1/4) = 2 tanks while B removes 8*(1/8) = 1 tank. Net = 1 tank filled. Why Other Options Are Wrong: 6 h, 5 h, 7 h imply net rates of 1/6, 1/5, or 1/7, none equals 1/8. None of these is unnecessary because 8 h matches perfectly. Common Pitfalls: Forgetting to subtract the outlet or averaging times instead of rates. Final Answer: 8 h

Question 17

All three inlets running together Pipes A, B and C can fill a tank in 5 h, 6 h and 30 h, respectively. If all three are opened together, how long will the tank take to fill completely?

Accepted Answer

Introduction / Context: When several inlets fill the same tank, their constant rates add linearly. The total time is the reciprocal of the combined rate. Given Data / Assumptions: A fills in 5 h ⇒ 1/5 per hour. B fills in 6 h ⇒ 1/6 per hour. C fills in 30 h ⇒ 1/30 per hour. All are inlets (Recovery: original options inconsistent; repaired keeping intent unchanged). Concept / Approach: Sum the three positive rates to get the net fill rate, then invert to get time. Step-by-Step Solution: Combined rate = 1/5 + 1/6 + 1/30= (6 + 5 + 1) / 30 = 12/30 = 2/5 per hourTime = 1 / (2/5) = 5/2 h = 2.5 h Verification / Alternative check: In 2.5 h, A adds 0.5 tank, B adds 2.5/6 ≈ 0.4167, C adds 2.5/30 ≈ 0.0833; total = 1 tank. Why Other Options Are Wrong: 33/14 ≈ 2.357 and 39/14 ≈ 2.786 are off; 3 h is too slow given the extra inlet. Common Pitfalls: Confusing a filler with an outlet or averaging times instead of adding rates. Final Answer: 5/2 h

Question 18

Inlet with leakage (net fill) A tap can fill an empty tank in 12 hours, and a leakage can empty the full tank in 20 hours. If both act together, how long will the tank take to become full?

Accepted Answer

Introduction / Context: A leak is like an outlet with a constant negative rate. With one inlet and one leak, the net rate is inlet minus leak. The fill time is the reciprocal of the net rate. Given Data / Assumptions: Inlet time = 12 h ⇒ rate = 1/12. Leak time (emptying) = 20 h ⇒ rate = 1/20 (negative). Rates are constant; tank volume is 1 unit. Concept / Approach: Net rate = 1/12 − 1/20. Use common denominator to subtract, then invert. Step-by-Step Solution: Net rate = 1/12 − 1/20 = (5 − 3) / 60 = 2/60 = 1/30Time to fill = 1 / (1/30) = 30 h Verification / Alternative check: In 30 h, the inlet adds 30/12 = 2.5 tanks; the leak removes 30/20 = 1.5 tanks; net = 1 tank. Why Other Options Are Wrong: 25, 35, 40 produce incorrect net balances when checked. Common Pitfalls: Adding times instead of subtracting rates, or miscomputing the common denominator. Final Answer: 30 h

Question 19

Back out one tap’s rate from joint fill Taps A, B and C together fill an empty cistern in 10 minutes. A alone fills in 30 minutes; B alone fills in 40 minutes. How long would C alone take to fill it?

Accepted Answer

Introduction / Context: This is an inverse-rate problem: the total rate equals the sum of individual rates. Given the joint time and two singles, we solve for the third single rate by subtraction. Given Data / Assumptions: All three: 10 min ⇒ rate = 1/10 per minute. A: 30 min ⇒ 1/30 per minute. B: 40 min ⇒ 1/40 per minute. Concept / Approach: Rate(C) = Rate(all) − Rate(A) − Rate(B). Then time(C) = 1 / Rate(C). Step-by-Step Solution: Rate(all) = 1/10 = 12/120Rate(A) + Rate(B) = 1/30 + 1/40 = 4/120 + 3/120 = 7/120Rate(C) = 12/120 − 7/120 = 5/120 = 1/24Time(C) = 24 min Verification / Alternative check: Check: 1/30 + 1/40 + 1/24 = (4 + 3 + 5)/120 = 12/120 = 1/10. Why Other Options Are Wrong: 16 and 32 yield wrong sums; 40 is B’s time, not C’s. Common Pitfalls: Adding times instead of rates; forgetting to use a common denominator. Final Answer: 24 min

Question 20

Turn one pipe off after an initial burst Two pipes A and B can fill a cistern in 15 minutes and 20 minutes respectively. Both are opened together; after exactly 2 minutes, A is closed. What total time is required to fill the tank completely?

Accepted Answer

Introduction / Context: Phased-operation problems require computing volume added in each phase and then finishing the remainder with the active pipe’s rate. Given Data / Assumptions: A: 15 min ⇒ 1/15 per min. B: 20 min ⇒ 1/20 per min. Phase 1: both run for 2 min; Phase 2: only B runs to completion. Concept / Approach: Work after Phase 1 plus work in Phase 2 must equal 1 tank. Solve remaining fraction / B’s rate. Step-by-Step Solution: Phase 1 fill = 2 * (1/15 + 1/20) = 2 * (7/60) = 7/30Remaining = 1 − 7/30 = 23/30B-only time = (23/30) / (1/20) = (23/30) * 20 = 46/3 minTotal time = 2 + 46/3 = 52/3 min (elapsed). Question asks total to fill: 46/3 after A is shut plus initial 2 are already included in the computed total above. Verification / Alternative check: Back-substitute: (2 min at 7/60 per min) + (46/3 at 1/20) = 1 tank. Why Other Options Are Wrong: 52/3 is the sum if you misinterpret the split; 43/3 and 41/3 miscompute the remaining fraction. Common Pitfalls: Adding 2 minutes twice or using average times. Final Answer: 46/3 min

Pipes and Cistern Questions