Question 1

Two fillers together take 12 h; one is 10 h faster If two pipes running together fill a tank in 12 hours, and one pipe is 10 hours faster than the other, how long does the faster pipe take by itself?

Accepted Answer

Introduction / Context: Difference-of-times problems with a known joint time reduce to a quadratic in the individual times when working in rate form. Choose the physically valid root. Given Data / Assumptions: Together time = 12 h. Let slower = x h; faster = x − 10 h. Rates add: 1/x + 1/(x − 10) = 1/12. Concept / Approach: Clear denominators, solve the quadratic, and discard negative or non-sensible times (x − 10 > 0). Step-by-Step Solution: (1/x) + (1/(x − 10)) = 1/12(2x − 10) / (x(x − 10)) = 1/1212(2x − 10) = x^2 − 10x ⇒ 24x − 120 = x^2 − 10xx^2 − 34x + 120 = 0 ⇒ x = (34 ± 26)/2 ⇒ x = 30 (valid) or 4 (invalid)Slower = 30 h; Faster = 20 h Verification / Alternative check: 1/30 + 1/20 = (2 + 3)/60 = 5/60 = 1/12 as required. Why Other Options Are Wrong: 60, 15, 25 mismatch the quadratic solution. Common Pitfalls: Using time differences directly rather than rates, leading to linear (but wrong) equations. Final Answer: 20 h

Question 2

Early filling, then steady emptying (clarified timing) Pipe P fills in 12 min and pipe R fills in 15 min, while pipe M empties in 6 min. First, only P and R run for 5 min (twice 2.5 min, as intended). Then M is also opened. After how much time <em>f…

Accepted Answer

Introduction / Context: The original stem was ambiguous about whether “emptied” is measured from the start or from adding the outlet. By the Recovery-First Policy, we clarify that the asked time is after opening the outlet, which aligns with the given options and preserves the core intent. Given Data / Assumptions: P: 12 min ⇒ 1/12 per min. R: 15 min ⇒ 1/15 per min. M (outlet): 6 min ⇒ 1/6 per min (negative in net). Phase 1: P+R for 5 min. Phase 2: P+R+M thereafter. Concept / Approach: Compute the volume after Phase 1, then the net emptying rate in Phase 2. Time to empty equals volume / rate. Step-by-Step Solution: Phase 1 fill = 5 * (1/12 + 1/15) = 5 * (9/60) = 3/4 tankPhase 2 net rate = 1/12 + 1/15 − 1/6 = (5 + 4 − 10)/60 = −1/60 per minTime to empty from that state = (3/4) / (1/60) = 45 min Verification / Alternative check: Total elapsed from the very start would be 5 + 45 = 50 min, but the clarified question asks the duration after M is opened. Why Other Options Are Wrong: 50 min is the start-to-empty duration; 25 and 35 are inconsistent with the computed net rate. Common Pitfalls: Trying to “average times” instead of using rates, or forgetting sign when an outlet is added. Final Answer: 45 min

Question 3

Mixture composition under simultaneous inflow Three pipes A, B and C fill a tank in 30 min, 20 min and 10 min respectively. When all are opened together from empty, A brings solution P, B brings Q, and C brings R. What fraction of R is present in th…

Accepted Answer

Introduction / Context: When multiple inlets add different solutions simultaneously, the instantaneous composition is proportional to the inlets’ rates. For a short interval where all run throughout, the mixture’s fraction of each solution equals that solution’s rate divided by the total rate. Given Data / Assumptions: A: 30 min ⇒ 1/30 per min (solution P). B: 20 min ⇒ 1/20 per min (solution Q). C: 10 min ⇒ 1/10 per min (solution R). All three run for the entire 3 min. Concept / Approach: Fraction of R = rate(C) / (rate(A) + rate(B) + rate(C)). Time cancels out because all run for the same duration. Step-by-Step Solution: Total rate = 1/30 + 1/20 + 1/10 = (2 + 3 + 6)/60 = 11/60Rate(C) = 1/10 = 6/60Fraction of R = (6/60) / (11/60) = 6/11 Verification / Alternative check: In 3 min the tank has 3*(11/60) = 11/20 full. Of this, portion from C is 3*(1/10) = 3/10. (3/10) / (11/20) = 6/11. Why Other Options Are Wrong: 8/11, 5/11, 7/11 are not the ratio of C’s rate to the total rate here. Common Pitfalls: Computing fraction using absolute volumes without dividing by the total mixed volume. Final Answer: 6/11

Question 4

Count inlets vs outlets from net fill time Seven pipes connect to a tank; some are inlets (each fills in 10 h) and some are outlets (each empties in 15 h). When all seven are opened together, the tank fills in 2 8/11 hours. How many inlets and how m…

Accepted Answer

Introduction / Context: When identical inlets and outlets are mixed, the net rate is (number of inlets * inlet rate) − (number of outlets * outlet rate). With a known net fill time, we can solve a pair of linear equations for the counts. Given Data / Assumptions: Total pipes = 7. Each inlet: 10 h ⇒ 1/10 per hour. Each outlet: 15 h ⇒ 1/15 per hour (negative). Net fill time = 2 8/11 h = 30/11 h ⇒ net rate = 11/30 per hour. Concept / Approach: Let i = inlets and o = outlets with i + o = 7. Then i/10 − o/15 = 11/30. Solve the linear system. Step-by-Step Solution: i + o = 7i/10 − o/15 = 11/30 ⇒ 3i − 2o = 11 (after multiplying by 30)Substitute o = 7 − i: 3i − 2(7 − i) = 11 ⇒ 5i − 14 = 11 ⇒ 5i = 25 ⇒ i = 5o = 7 − 5 = 2 Verification / Alternative check: Net rate = 5*(1/10) − 2*(1/15) = 1/2 − 2/15 = (15 − 4)/30 = 11/30 ⇒ time = 30/11 h. Why Other Options Are Wrong: Any other split fails the linear system. Common Pitfalls: Using 2 8/11 as 2.8 recurring instead of 30/11; always convert mixed fractions to improper ones. Final Answer: 5, 2

Question 5

Time for partial fill with three inlets Taps A, B and C fill a tank in 20 min, 15 min and 12 min respectively. If all are opened together, how long will they take to fill 40% of the tank?

Accepted Answer

Introduction / Context: Partial-fill questions use the same rate addition as full-fill problems, but the target volume is a fraction (here 40%) of the tank. Given Data / Assumptions: A: 20 ⇒ 1/20 per min; B: 15 ⇒ 1/15 per min; C: 12 ⇒ 1/12 per min. All open from empty; target volume = 0.40 tank. Concept / Approach: Combined rate = sum of individual rates. Time = (target fraction) / (combined rate). Step-by-Step Solution: Combined rate = 1/20 + 1/15 + 1/12 = (3 + 4 + 5) / 60 = 12/60 = 1/5 per minTime for 40% = 0.40 / 0.20 = 2 min Verification / Alternative check: In 2 min, they add 2 * (1/5) = 2/5 = 40%. Why Other Options Are Wrong: 1, 3, 4 min correspond to 20%, 60%, 80% respectively at this rate. Common Pitfalls: Computing each tap’s 40% time separately and averaging—incorrect. Final Answer: 2 min

Question 6

Relative capacity: B is 80% more than A Tap B’s capacity is 80% greater than A’s. If both run together, they fill the tank in 45 hours. How long would B take alone to fill the tank?

Accepted Answer

Introduction / Context: “80% more” means B’s rate is 1.8 times A’s rate. With the joint time known, we solve for the base rate and then get B’s solo time by inversion. Given Data / Assumptions: Together time = 45 h ⇒ combined rate = 1/45 per hour. Let A’s rate = r; then B’s rate = 1.8r = 9r/5. Concept / Approach: r + 9r/5 = 1/45 ⇒ solve for r. Then B-alone time = 1 / (9r/5). Step-by-Step Solution: Combined rate = r + 9r/5 = 14r/5 = 1/45r = (5 / 14) * (1/45) = 1 / 126B’s rate = 9r/5 = 9 / (5*126) = 1 / 70B-alone time = 70 h Verification / Alternative check: 1/126 + 1/70 = (5 + 9)/630 = 14/630 = 1/45. Why Other Options Are Wrong: 72, 66, 48 contradict the derived rate ratio 1 : 1.8. Common Pitfalls: Treating “80% more” as 1.8 more (i.e., 2.8×) instead of 1.8× the base. Final Answer: 70 h

Question 7

Net rate from volumetric velocities with one outlet Taps A, B, C have velocities 42 L/h, 56 L/h and 48 L/h. A and B are inlets; C is an outlet. If all three are opened together from empty and the tank fills in 16 hours, what is the tank’s capacity?

Accepted Answer

Introduction / Context: When volumetric flow rates are given directly, the net inflow is the algebraic sum of the listed L/h values (taking the outlet as negative). Capacity equals net rate multiplied by time to fill. Given Data / Assumptions: A = +42 L/h, B = +56 L/h, C = −48 L/h. Tank fills in 16 h. Flows are steady and additive. Concept / Approach: Net flow = 42 + 56 − 48. Capacity = net flow * time. Step-by-Step Solution: Net = 50 L/hCapacity = 50 * 16 = 800 L Verification / Alternative check: Reverse check: 800 / 16 = 50 L/h, which equals 42 + 56 − 48. Why Other Options Are Wrong: 960, 1600, 2346 are inconsistent with the stated time and flows. Common Pitfalls: Adding the outlet as positive instead of subtracting. Final Answer: 800 L

Question 8

Imperfectly opened taps (reduced discharge) Two taps A and B nominally fill a tank in 25 min and 20 min. Due to partial opening, A allows only 5/6 of its normal flow and B allows only 2/3 of its normal flow. How long will they take together to fill …

Accepted Answer

Introduction / Context: When taps are partially opened, multiply each nominal rate by its effective fraction to get the reduced rate. Then add reduced rates as usual. Given Data / Assumptions: Nominal A: 25 min ⇒ 1/25 per min; effective = (5/6)*(1/25). Nominal B: 20 min ⇒ 1/20 per min; effective = (2/3)*(1/20). Tank initially empty; both run together. Concept / Approach: Compute effective rates and add. Time is the inverse of the total effective rate. Step-by-Step Solution: Rate(A_eff) = (5/6)*(1/25) = 1/30Rate(B_eff) = (2/3)*(1/20) = 1/30Total effective = 1/30 + 1/30 = 1/15 per minTime = 15 min Verification / Alternative check: In 15 min, A contributes 15*(1/30)=1/2 and B contributes 1/2; sum 1 tank. Why Other Options Are Wrong: 14, 13, 12 imply total effective rates greater than 1/15, which are not supported by the given fractions. Common Pitfalls: Reducing the times instead of the rates (that yields wrong arithmetic). Final Answer: 15 min

Question 9

Alternate operation of two inlets and one outlet (A, then B, then C) Taps A and B are inlets (20 min and 30 min), tap C is an outlet (15 min). They are operated alternately for one minute each in the repeating order A → B → C → A → …. How long will …

Accepted Answer

Introduction / Context: Minute-by-minute alternation requires accounting for partial minutes at the end, since the tank may become full during an inlet’s minute before the next outlet minute would start. We therefore compute cycle gains and then finish with a fractional minute if needed. Given Data / Assumptions: A: +1/20 per min; B: +1/30 per min; C: −1/15 per min. Order repeats A → B → C, each exactly 1 min. Stop as soon as the tank becomes full. Concept / Approach: Net gain per 3-minute cycle = 1/20 + 1/30 − 1/15. After many full cycles, a final partial minute by A finishes the remaining fraction. Step-by-Step Solution: Cycle gain = 1/20 + 1/30 − 1/15 = (3 + 2 − 4)/60 = 1/60After 59 cycles (177 min): 59/60 filledRemaining = 1/60; next minute is ATime needed by A = (1/60) / (1/20) = 1/3 minTotal time = 177 + 1/3 = 532/3 min (≈ 177.33 min) Verification / Alternative check: Any longer (e.g., waiting for B or C) would overshoot or reduce level; stopping mid-minute when full is appropriate. Why Other Options Are Wrong: 180, 167, 185 are rounded or mis-sequenced approximations; 532/3 is exact. Common Pitfalls: Forcing whole-minute stops, which is unrealistic for continuous flow; ignoring the outlet’s negative contribution. Final Answer: 532/3 min

Question 10

Leak vs. compensating inflow (capacity from times) A tank has a leak that would empty it in 8 hours. A tap admitting 3 L per minute is turned on, and now the tank empties in 12 hours. What is the capacity of the tank?

Accepted Answer

Introduction / Context: This is a capacity-from-rate-difference problem. Treat the leak as a constant outflow and the tap as a constant inflow. The observed emptying time under compensation identifies the net outflow; solving two linear relations yields the tank volume. Given Data / Assumptions: Leak alone empties in 8 h ⇒ leak rate = V/8 per hour = V/480 per min. With 3 L/min added, emptying time is 12 h ⇒ net outflow = V/12 per hour = V/720 per min. V = tank capacity in liters. Concept / Approach: Net outflow (leak − 3) equals V/720. Set up the equation using leak-alone rate and solve for V. Step-by-Step Solution: Leak rate per min = V/480V/480 − 3 = V/720(V/480 − V/720) = 3 ⇒ V(1/480 − 1/720) = 3V * (1/1440) = 3 ⇒ V = 4320 L Verification / Alternative check: Check: leak 540 L/h; tap 180 L/h in; net 360 L/h ⇒ 4320/360 = 12 h. Why Other Options Are Wrong: 4000, 2250, 4120 do not satisfy both time conditions simultaneously. Common Pitfalls: Mistaking hours for minutes or mixing units inside the same equation. Final Answer: 4320 L

Question 11

Different speeds: A four times faster than B Inlet A is four times as fast as inlet B. If A alone fills the tank in 15 minutes, how long will both together take to fill it?

Accepted Answer

Introduction / Context: Relative-speed statements like “A is four times faster than B” define a rate ratio. Convert the statement into individual rates and then add the rates to get the joint time. Given Data / Assumptions: A alone: 15 min ⇒ rate(A) = 1/15 per min. A is 4 times B ⇒ rate(B) = rate(A) / 4 = 1/60 per min. Tank initially empty; both start together. Concept / Approach: Total rate = rate(A) + rate(B). Time = 1 / total rate. Step-by-Step Solution: rate(A) = 1/15rate(B) = 1/60Total = 1/15 + 1/60 = 5/60 = 1/12 per minTime together = 12 min Verification / Alternative check: In 12 min, A adds 12/15 = 4/5; B adds 12/60 = 1/5; total 1 tank. Why Other Options Are Wrong: 10, 15, 14 contradict the combined rate 1/12. Common Pitfalls: Interpreting “4 times faster” as “A’s time is one quarter,” which is only true for rates, not times unless converted carefully. Final Answer: 12 min

Question 12

Alternate hours starting from A Two inlets A and B can fill the tank in 16 h and 10 h. They are opened alternately for 1 hour each, starting with A. How long will it take to fill the tank (stop immediately when full)?

Accepted Answer

Introduction / Context: Alternate-hour problems are best handled in 2-hour cycles (one hour of A, one hour of B). After a whole number of cycles, a final fractional hour of the next tap may be required to finish the small remainder. Given Data / Assumptions: A: 16 h ⇒ 1/16 per hour; B: 10 h ⇒ 1/10 per hour. Order: A then B, repeating; stop as soon as the tank reaches full. Concept / Approach: Per 2-hour cycle, net fill = 1/16 + 1/10. Accumulate cycles to just below 1 tank, then finish with a partial hour of A. Step-by-Step Solution: 2-hour gain = 1/16 + 1/10 = (5 + 8)/80 = 13/80After 12 hours (6 cycles): 6 * 13/80 = 78/80 = 39/40Remaining = 1/40, next in order is ATime for A = (1/40) / (1/16) = 16/40 = 2/5 h = 24 minTotal time = 12 + 2/5 = 62/5 h (12 h 24 min) Verification / Alternative check: Check contributions: 12 h give 39/40; last 24 min at 1/16 adds 1/40. Why Other Options Are Wrong: 13 h or 14.5 h ignore the exact fractional finish; 12 h 24 min equals 62/5 h (both are correct renderings). Common Pitfalls: Stopping at a whole cycle and forgetting the final fraction with the correct next tap. Final Answer: 62/5 h

Question 13

Alternate minutes starting with the faster pipe Two pipes X and Y fill a cistern in 6 min and 7 min. Starting with X, they are opened alternately for 1 minute each. In what time will the cistern be filled (stop immediately when full)?

Accepted Answer

Introduction / Context: Alternate-minute problems mirror alternate-hour logic but with smaller time quanta. Compute 2-minute cycle gain, then apply a final fractional minute of the next (faster) inlet if necessary. Given Data / Assumptions: X: 6 min ⇒ 1/6 per min; Y: 7 min ⇒ 1/7 per min. Order: X then Y, repeating; stop immediately when full. Concept / Approach: Per 2-min cycle, fill = 1/6 + 1/7 = 13/42. After an integer number of cycles, the remainder is filled by the next minute of X. Step-by-Step Solution: After 3 cycles (6 min): 3 * 13/42 = 39/42 = 13/14Remaining = 1/14; next is X at 1/6 per minTime for X to finish = (1/14) / (1/6) = 6/14 = 3/7 minTotal time = 6 + 3/7 = 45/7 min (≈ 6.4286 min) Verification / Alternative check: In the last partial minute, we stop the instant the tank becomes full; we do not start Y again. Why Other Options Are Wrong: 62/7, 63/7, 65/7 assume whole extra minutes or miscount cycles. Common Pitfalls: Forcing whole-minute boundaries instead of stopping at the exact completion time. Final Answer: 45/7 min

Question 14

Outlet scaling with diameter (area proportional to d^2) A tap of diameter d empties a tank in 40 minutes. Assuming the discharge is proportional to the cross-sectional area, how long will another tap of diameter 2d take to empty the same tank?

Accepted Answer

Introduction / Context: Under similar head/pressure conditions, volumetric discharge of a round tap is proportional to cross-sectional area, which scales with the square of the diameter (d^2). Doubling diameter multiplies area (and hence flow) by 4, reducing time by a factor of 4. Given Data / Assumptions: Tap 1 diameter = d; time = 40 min. Tap 2 diameter = 2d; flow ∝ (2d)^2 = 4 d^2. Same tank, comparable hydraulic conditions; time inversely ∝ area. Concept / Approach: If Q ∝ d^2 and V is fixed, then time t ∝ 1/Q ∝ 1/d^2. Doubling diameter quarters the time. Step-by-Step Solution: t2 = t1 / 4 = 40 / 4 = 10 min Verification / Alternative check: Relative-flow logic: 4× the flow in the larger tap empties the same volume in one-quarter the time. Why Other Options Are Wrong: 20 min halves, not quarters, the time; 5 min would imply 8× flow; 80 min is the opposite trend. Common Pitfalls: Scaling time linearly with diameter instead of with area (d^2). Final Answer: 10 min

Question 15

Two inlet pipes and one outlet pipe act on the same cistern. Pipe A fills the cistern in 12 minutes; Pipe B fills it in 15 minutes; Pipe C (outlet) can empty a full cistern in 6 minutes. First, only A and B are opened for exactly 5 minutes. After th…

Accepted Answer

Introduction / Context: When pipes fill and drain simultaneously, we work with rates (fractions of a tank per minute). We will compute how much water has accumulated after the first phase (only inlets), then use the net rate with the outlet added to determine the time to reach empty from that intermediate level. Given Data / Assumptions: A fills in 12 min ⇒ rate = 1/12 tank/min. B fills in 15 min ⇒ rate = 1/15 tank/min. C empties in 6 min ⇒ rate = −1/6 tank/min. Phase 1: A and B for 5 minutes. Phase 2: A, B, and C together until empty. Concept / Approach: Add inlet rates for Phase 1 to find the filled fraction. Then, for Phase 2, compute the net rate including the outlet (inlet rates minus outlet rate). Time = (remaining volume) / (absolute value of net outflow rate). Step-by-Step Solution: Phase 1 fill in 5 min = 5 * (1/12 + 1/15) = 5 * (5/60 + 4/60) = 5 * (9/60) = 45/60 = 3/4.At the start of Phase 2, tank is 3/4 full.Net rate with all three = 1/12 + 1/15 − 1/6 = (5 + 4 − 10)/60 = −1/60 tank/min.Time to go from 3/4 to empty = (3/4) / (1/60) = 45 minutes. Verification / Alternative check: The negative net indicates draining. Magnitude 1/60 means 1 minute removes 1/60 of capacity; 45 minutes remove 45/60 = 3/4, exactly the amount present after Phase 1. Why Other Options Are Wrong: 38/22/42 min do not match the exact fraction arithmetic; only 45 min empties exactly 3/4 at the given net rate. Common Pitfalls: Forgetting the outlet subtracts from fill rate; mixing “from the start” vs “from when outlet opens.” The question explicitly asks for the emptying time from the moment all three are open. Final Answer: 45 min

Question 16

A tank can be filled by Tap 1 in 20 minutes and by Tap 2 in 25 minutes. Both taps are opened together and run for exactly 5 minutes, after which Tap 2 is turned off. From that moment, how many additional minutes will Tap 1 alone take to completely f…

Accepted Answer

Introduction / Context: This is a two-phase filling problem. We compute the fraction filled in the first phase (both taps on), then determine how long the remaining fraction takes with only Tap 1 working. Given Data / Assumptions: Tap 1 rate = 1/20 tank/min. Tap 2 rate = 1/25 tank/min. Both on for 5 minutes, then Tap 2 is closed. Concept / Approach: Phase 1: fraction filled = time * (rate1 + rate2). Phase 2: remaining fraction divided by rate1 gives the additional minutes required. Step-by-Step Solution: Phase 1 fill = 5 * (1/20 + 1/25) = 5 * ((5 + 4)/100) = 5 * (9/100) = 45/100 = 9/20.Remaining = 1 − 9/20 = 11/20.Tap 1 alone time = (11/20) / (1/20) = 11 minutes. Verification / Alternative check: In 11 min, Tap 1 adds 11/20; total becomes 9/20 + 11/20 = 1, as required. Why Other Options Are Wrong: 17 1/2, 12, 6 min do not match the exact remainder and Tap 1 rate. Common Pitfalls: Computing the 5-minute fill incorrectly or forgetting to subtract to get the remainder. Final Answer: 11 min

Question 17

A cistern has two inlet taps that fill it in 12 minutes and 15 minutes, respectively. There is also a waste pipe (outlet). When all three pipes are opened together on an empty cistern, it becomes full in 20 minutes. How long does the waste pipe alon…

Accepted Answer

Introduction / Context: Combine inlet rates and subtract the unknown outlet rate to match the observed net fill time. Then invert the outlet rate to get its emptying time for a full cistern. Given Data / Assumptions: Inlet A = 1/12 tank/min. Inlet B = 1/15 tank/min. All three together fill in 20 min ⇒ net rate = 1/20 tank/min. Concept / Approach: If w is the outlet rate (tank/min), then 1/12 + 1/15 − w = 1/20. Solve for w, then time = 1/w. Step-by-Step Solution: 1/12 + 1/15 = (5/60 + 4/60) = 9/60 = 3/20.3/20 − w = 1/20 ⇒ w = 2/20 = 1/10 tank/min.Emptying time = 1 / (1/10) = 10 minutes. Verification / Alternative check: Net with all = 3/20 − 1/10 = 3/20 − 2/20 = 1/20, consistent. Why Other Options Are Wrong: 8/12/16 min imply different net rates and would not yield 20 minutes to fill. Common Pitfalls: Adding the outlet by mistake; always subtract outlets. Final Answer: 10 min

Question 18

Two taps can fill a cistern in 10 minutes and 15 minutes, respectively. With the waste pipe (outlet) also open, the cistern is filled in 18 minutes. How long does the waste pipe alone take to empty a full cistern?

Accepted Answer

Introduction / Context: This is similar to standard “pipes and cisterns” problems. We compute the combined inlet rate, subtract the waste rate to get the observed net rate, and then isolate the waste rate to find its emptying time. Given Data / Assumptions: Inlet rates: 1/10 and 1/15 tank/min. Net filling time with waste open = 18 min ⇒ net rate = 1/18 tank/min. Concept / Approach: If w is the waste rate (tank/min), then (1/10 + 1/15) − w = 1/18. Solve for w and invert to get time to empty a full cistern. Step-by-Step Solution: 1/10 + 1/15 = (3 + 2)/30 = 1/6.1/6 − w = 1/18 ⇒ w = 1/6 − 1/18 = 2/18 = 1/9 tank/min.Emptying time = 1 / (1/9) = 9 minutes. Verification / Alternative check: Net = 1/6 − 1/9 = (3 − 2)/18 = 1/18, as given. Why Other Options Are Wrong: 7/13/23 minutes produce wrong net rates and fail the 18-minute condition. Common Pitfalls: Arithmetic with fractions; prefer a common denominator to avoid mistakes. Final Answer: 9 minutes

Question 19

Three pipes act on a tank: Pipe A fills it in 32 minutes, Pipe B fills it in 36 minutes, and Pipe C empties it in 20 minutes. If all three pipes are opened simultaneously, in how many minutes will exactly half of the tank be filled?

Accepted Answer

Introduction / Context: We combine the two inlet rates and subtract the outlet rate to obtain the net rate. Since the target is half the tank, we set (net rate) * time = 1/2 and solve for time. If that value does not match any provided option, “None of these” is appropriate. Given Data / Assumptions: A = 1/32 tank/min. B = 1/36 tank/min. C = −1/20 tank/min. Goal: reach 1/2 tank from empty. Concept / Approach: Net rate r = 1/32 + 1/36 − 1/20. Then time t satisfies r * t = 1/2 ⇒ t = (1/2) / r. Step-by-Step Solution: Compute r with denominator 2880: 1/32 = 90/2880; 1/36 = 80/2880; 1/20 = 144/2880.r = (90 + 80 − 144)/2880 = 26/2880 = 13/1440 tank/min.t = (1/2) / (13/1440) = (1/2) * (1440/13) = 720/13 ≈ 55.38 minutes. Verification / Alternative check: Multiply r by 720/13: (13/1440) * (720/13) = 1/2, confirming correctness. Why Other Options Are Wrong: 16/24/48 min do not match 720/13; hence the correct choice is “None of these.” Common Pitfalls: Dropping the outlet term or using rounded intermediate values. Keep exact fractions until the final step. Final Answer: None of these (exact time = 720/13 minutes ≈ 55.38 min)

Question 20

Pipes A, B, and C together can fill a cistern in 6 hours. After they work together for exactly 2 hours, Pipe C is closed and A and B continue to fill the cistern. With only A and B running, they take 7 more hours to finish filling. How long would Pi…

Accepted Answer

Introduction / Context: Let a, b, c be the hourly fill rates of A, B, and C. The joint rate and the two-phase timeline give two equations, from which we isolate c and invert to get C’s solo time. Given Data / Assumptions: a + b + c = 1/6 job/hour. After 2 hours together, remaining work is filled by A+B in 7 hours ⇒ (a + b) * 7 = remaining. Concept / Approach: Work done in first 2 hours = 2 * (1/6) = 1/3. Remaining = 2/3. Thus (a + b) * 7 = 2/3 ⇒ a + b = 2/21. Then c = 1/6 − 2/21. Step-by-Step Solution: a + b = 2/21.c = 1/6 − 2/21 = (7 − 4)/42 = 3/42 = 1/14 job/hour.C alone time = 1 / (1/14) = 14 hours. Verification / Alternative check: Check totals: First 2 h give 1/3; then 7 h at 2/21 give 14/21 = 2/3; total 1 job. Why Other Options Are Wrong: 10/12/16 hours do not match the derived c = 1/14. Common Pitfalls: Confusing “7 hours to finish” as total time rather than the second phase duration. Final Answer: 14 hours

Pipes and Cistern Questions