Question 1

In $1$ minute, $\frac{3}{7}$ of a bucket is filled. The rest of the bucket can be filled in

Accepted Answer

Concept & Unitary Method: The problem can be solved using the unitary method. If a certain fraction of work is completed in a given time, we can find the time taken for the remaining fraction by first calculating the time for the whole work, or by setting up a direct proportion. Step-by-Step Solution: • Given: Part of the bucket filled in 1 minute = (3/7). • Calculate remaining part: The whole bucket is 1. The remaining part to be filled is 1 - (3/7) = (4/7). • Set up proportion: Time taken to fill (3/7) of the bucket = 1 minute. Time taken to fill 1 (whole) bucket = 1 × (7/3) = (7/3) minutes. • Find time for remaining part: Time taken to fill (4/7) of the bucket = (Time for whole bucket) × (Remaining part) = (7/3) × (4/7) = (4/3) minutes. Exam Strategy & Shortcut: Use direct ratios. The parts are 3 and 4 (since 3/7 is filled, 4/7 remains). If 3 parts take 1 minute, then 1 part takes (1/3) minute. Therefore, 4 parts will take 4 × (1/3) = (4/3) minutes. Common Pitfall: A common mistake is finding the time to fill the *entire* bucket ((7/3) minutes) and mistakenly selecting that as the answer, forgetting the question asks for the time to fill the *rest* of the bucket. Final Answer: Therefore, the correct answer is (4/3) minutes.

Question 2

Two pipes A and B can fill a tank in $20$ and $30$ minutes respectively. If both the pipes are used together, how long will it take to fill the tank?

Accepted Answer

Concept & Combined Work Rate: When two pipes work together, their combined rate of filling is the sum of their individual rates. If Pipe A takes x time and Pipe B takes y time, their combined 1-unit work is (1/x) + (1/y). Combined Time = (xy/x + y) Step-by-Step Solution: • Given: Time taken by Pipe A = 20 minutes. Time taken by Pipe B = 30 minutes. • Calculate 1-minute work: Part filled by A in 1 minute = (1/20) Part filled by B in 1 minute = (1/30) • Calculate combined 1-minute work: Part filled by (A + B) in 1 minute = (1/20) + (1/30) = (3 + 2/60) = (5/60) = (1/12) • Find total time: Since (1/12) of the tank is filled in 1 minute, the entire tank is filled in 12 minutes. Exam Strategy & Shortcut: Use the direct formula for two pipes filling: (a × b/a + b). Calculation: (20 × 30/20 + 30) = (600/50) = 12 minutes. This is much faster than adding fractions manually. Common Pitfall: A common blunder is taking the average of the times ((20 + 30/2) = 25 minutes). Two pipes working together will always fill the tank faster than the fastest individual pipe. Final Answer: Therefore, the correct answer is 12 minutes.

Question 3

A pipe can fill a tank in $x$ hours and another pipe can empty it in $y$ ($y > x$) hours. If both the pipes are open, in how many hours will the tank be filled?

Accepted Answer

Concept & Net Filling Rate: When one pipe fills and another empties a tank, their net work is the difference between their individual rates. Because y > x, the filling pipe is faster, meaning the tank will eventually fill. Net Work Rate = Filling Rate - Emptying Rate Step-by-Step Solution: • Identify Rates: Part filled in 1 hour by the first pipe = (1/x) Part emptied in 1 hour by the second pipe = (1/y) • Calculate Net Rate: Net part filled in 1 hour when both are open = (1/x) - (1/y) • Simplify the Expression: (1/x) - (1/y) = (y - x/xy) • Find Total Time: The total time taken to fill the tank is the reciprocal of the net hourly rate. Time taken = (1/\frac{y - x){xy}} = (xy/y - x) hours. Exam Strategy & Shortcut: Memorize the generic algebraic formula for one inlet and one outlet pipe: \frac{Product of times}{Difference of times}. Since the tank is being filled, the denominator must be positive, which is (y - x) because y > x. This instantly points to (xy/y - x). Common Pitfall: A common mistake is choosing (xy/x - y). Since we are given y > x, (x - y) would result in a negative number of hours, which is physically impossible. Always subtract the smaller time value from the larger one in the denominator for filling scenarios. Final Answer: Therefore, the correct answer is (xy/y - x) hours.

Question 4

A tap can completely fill a water tank in $8$ hours. The water tank has a hole in it through which the water leaks out. The leakage will cause the full water tank to get empty in $12$ hours. How much time will it take for the tap to fill the tank co…

Accepted Answer

Concept & Net Filling Rate with Leakage: When a tank is being filled by a tap but simultaneously drained by a leak, the effective filling rate is the difference between the tap's filling rate and the leak's emptying rate. Net Rate = Filling Rate - Emptying Rate Step-by-Step Solution: • Given Rates: Time taken by the tap to fill the tank = 8 hours. Part filled by the tap in 1 hour = (1/8). Time taken by the leak to empty the tank = 12 hours. Part emptied by the leak in 1 hour = (1/12). • Calculate Net Rate: Net part filled in 1 hour = (1/8) - (1/12) • Solve the Fraction: Find the LCM of 8 and 12, which is 24. Net part filled = (3/24) - (2/24) = (1/24) • Find Total Time: Since (1/24) of the tank is filled in 1 hour, it will take exactly 24 hours to fill the entire tank. Exam Strategy & Shortcut: Use the direct formula for one inlet and one outlet: (xy/y - x). Here, x = 8 and y = 12. Calculation: (8 × 12/12 - 8) = (96/4) = 24 hours. This bypasses fraction arithmetic entirely. Common Pitfall: A common mistake is adding the two rates ((1/8) + (1/12)) instead of subtracting them, which assumes both are filling the tank, leading to an incorrect answer of 4.8 hours. Final Answer: Therefore, the correct answer is 24 hours.

Question 5

A tap can fill a tank in $48$ minutes whereas another tap can empty it in $2$ hours. If both the taps are opened at $11 	ext{ : } 40 	ext{ A.M.}$, then the tank will be filled at

Accepted Answer

Concept & Unit Conversion with Net Rates: Before calculating the combined rate of filling and emptying pipes, all time values must be converted to the same unit (e.g., minutes). The net work is then found by subtracting the emptying rate from the filling rate. Step-by-Step Solution: • Unit Conversion: Filling pipe time = 48 minutes. Emptying pipe time = 2 hours = 2 × 60 = 120 minutes. • Calculate Individual Rates: Part filled in 1 minute = (1/48) Part emptied in 1 minute = (1/120) • Calculate Net Rate: Net part filled in 1 minute = (1/48) - (1/120) LCM of 48 and 120 is 240. Net part filled = (5/240) - (2/240) = (3/240) = (1/80) • Calculate Total Time & Final Clock Time: The tank will be full in 80 minutes, which is equal to 1 hour and 20 minutes. Start time: 11:40 A.M. Add 1 hour 20 minutes: 11:40 A.M. + 1 hour = 12:40 P.M. 12:40 P.M. + 20 minutes = 1:00 P.M. Exam Strategy & Shortcut: Using the (ab/b - a) shortcut: (48 × 120/120 - 48) = (48 × 120/72). Cancel 24 from 48 and 72 to get (2 × 120/3) = 2 × 40 = 80 minutes. Adding 80 mins to 11:40 gives 1:00 P.M. quickly. Common Pitfall: A frequent error is forgetting to convert 2 hours into 120 minutes, leading to an incorrect setup like (1/48) - (1/2), which produces negative (impossible) results. Final Answer: Therefore, the correct answer is 1 : 00 P.M..

Question 6

A cistern has two pipes. One can fill it with water in $8$ hours and other can empty it in $5$ hours. In how many hours will the cistern be emptied if both the pipes are opened together when $\frac{3}{4}$ of the cistern is already full of water?

Accepted Answer

Concept & Net Emptying Rate: When an emptying pipe acts faster than a filling pipe, the tank will experience a net loss of volume over time. The net emptying rate is the emptying rate minus the filling rate. Net Emptying Rate = (1/Empty Time) - (1/Fill Time) Step-by-Step Solution: • Identify Rates: Part emptied by the outlet pipe in 1 hour = (1/5) Part filled by the inlet pipe in 1 hour = (1/8) • Calculate Net Rate: Since (1/5) > (1/8), the cistern is being emptied. Net part emptied in 1 hour = (1/5) - (1/8) = (8 - 5/40) = (3/40) • Determine Remaining Work: The cistern is not full; it is only (3/4) full. We only need to empty this fraction. • Calculate Total Time: Time required to empty (3/4) of the cistern = \frac{Volume to empty}{Net hourly rate} Time = (\frac{3/4)}{(3/40)} = (3/4) × (40/3) = (40/4) = 10 hours. Exam Strategy & Shortcut: Use the LCM total work method. Let the total capacity be 40 units (LCM of 8 and 5). Fill rate = +5 units/hr. Empty rate = -8 units/hr. Net rate = -3 units/hr (emptying). Volume to empty = (3/4) of 40 = 30 units. Time to empty = (30/3) = 10 hours. Common Pitfall: A common mistake is calculating the time to empty a *full* cistern ((40/3) = 13(1/3) hours) and selecting option (d), completely ignoring the critical detail that the cistern is only (3/4) full. Final Answer: Therefore, the correct answer is 10 hours.

Question 7

Two pipes A and B can separately fill a cistern in 60 minutes and 75 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened then the cistern is full in 50 minutes. In ho…

Accepted Answer

Concept & Efficiency Method: Instead of working with fractions, we can assume the total capacity of the cistern to be the Least Common Multiple (LCM) of the given times. This gives us whole-number efficiencies (work done per minute) for each pipe, making calculations much simpler. Efficiency = \frac{Total Work (Capacity)}{Time} Step-by-Step Solution: 1. Find Total Capacity: Let the total capacity of the cistern be the LCM of 60, 75, and 50. LCM(60, 75, 50) = 300 units. 2. Calculate Individual Efficiencies: • Efficiency of Pipe A = 300 / 60 = +5 units/min (filling). • Efficiency of Pipe B = 300 / 75 = +4 units/min (filling). 3. Calculate Combined Efficiency: When all three pipes (A, B, and C) are open, the tank fills in 50 minutes. • Net Efficiency (A + B + C) = 300 / 50 = +6 units/min. 4. Find Efficiency of Pipe C: Let the efficiency of the emptying pipe C be x. • A + B + C = 6 • 5 + 4 + x = 6 • 9 + x = 6 \Rightarrow x = -3 units/min. (The negative sign indicates it is emptying the tank). 5. Calculate Time for Pipe C: Time taken by pipe C to empty the full tank alone = \frac{Total Capacity}{Efficiency of C} • Time = 300 / 3 = 100 minutes. Exam Strategy & Shortcut: Use the LCM approach whenever dealing with multiple pipes or workers. It avoids error-prone fraction additions. Once you see 60, 75, and 50, mentally fix 300 as the volume. Then it's just basic arithmetic: 5 + 4 - Leak = 6. Leak is 3. 300 / 3 = 100. Common Pitfall: A common mistake is assuming the third pipe fills the tank and adding its rate, or setting up the fraction equation incorrectly as (1/60) + (1/75) + (1/x) = (1/50) without paying attention to the negative sign required for an emptying pipe. Final Answer: Therefore, the correct answer is 100 min.

Question 8

A pump can fill a tank with water in 2 hours. Because of a leak, it took $2 \frac{1}{3}$ hours to fill the tank.  The leak can drain all the water of the tank in

Accepted Answer

Concept & Net Work Rate: The net rate at which a tank is filled when there is a leak is the difference between the filling rate of the pump and the emptying rate of the leak. Net Work Rate = Rate of Pump - Rate of Leak Step-by-Step Solution: 1. Determine Individual Rates: • Time taken by the pump alone = 2 hours. • Work done by the pump in 1 hour = (1/2) of the tank. 2. Determine Combined Rate: • Time taken by the pump and leak together = 2 (1/3) = (7/3) hours. • Work done by the (pump + leak) in 1 hour = (1/7/3) = (3/7) of the tank. 3. Calculate Leak Rate: Let the leak empty the tank in x hours. Its work in 1 hour is (1/x). • Pump Rate - Leak Rate = Net Rate • (1/2) - (1/x) = (3/7) • (1/x) = (1/2) - (3/7) • (1/x) = (7 - 6/14) = (1/14) 4. Find Total Time for Leak: • Since the leak does (1/14) of the work in 1 hour, it will empty the full tank in 14 hours. Exam Strategy & Shortcut: Using the LCM method: Let total capacity = LCM(2, 7/3). A multiple of 14 works well. Let Capacity = 14 units. Efficiency of Pump = 14 / 2 = 7 units/hr. Efficiency of Pump + Leak = 14 / (7/3) = 6 units/hr. Leak Efficiency = (Pump + Leak) - Pump = 6 - 7 = -1 unit/hr. Time for leak to empty tank = 14 / 1 = 14 hours. Common Pitfall: A frequent error is subtracting the times directly (e.g., 7/3 - 2 = 1/3) rather than subtracting their reciprocal rates (work done per hour). Always operate on rates, never on total times directly. Final Answer: Therefore, the correct answer is 14 hours.

Question 9

One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in

Accepted Answer

Concept & Relative Efficiency: If a pipe is n times faster than another, its efficiency is n times that of the slower pipe. Time and efficiency are inversely proportional. Total work can be assumed as Total Efficiency × Total Time. Total Work = (Efficiency_1 + Efficiency_2) × Time Together Step-by-Step Solution: 1. Define Efficiencies: • Let the efficiency of the slower pipe = 1 unit/min. • The faster pipe is 3 times as fast, so its efficiency = 3 units/min. 2. Calculate Total Efficiency: • Combined efficiency of both pipes = 1 + 3 = 4 units/min. 3. Determine Total Work (Tank Capacity): • They fill the tank together in 36 minutes. • Total Capacity = Combined Efficiency × Time = 4 × 36 = 144 units. 4. Calculate Time for Slower Pipe: • Time taken by slower pipe = \frac{Total Capacity}{Efficiency of Slower Pipe} • Time = (144/1) = 144 minutes. Exam Strategy & Shortcut: Think purely in ratios. Speed ratio = 3:1. Total speed = 4. The slower pipe's speed is 1 out of the total 4. Therefore, it will take 4 times as long as the combined time. 36 mins × 4 = 144 mins. Common Pitfall: Students often set up a time equation as x + 3x = 36, completely misinterpreting the relationship between time and speed. The correct time relationship is (1/x) + (1/3x) = (1/36). The efficiency method avoids fractions entirely. Final Answer: Therefore, the correct answer is 144 min.

Question 10

Container X can hold three times as much liquid as Container Y. If Container X is used alone, it requires 60 full pours to completely fill an empty tank. If both containers are used simultaneously for each pour, how many total pours will be needed t…

Accepted Answer

Concept & Capacity Ratio: When the capacities of two containers are given as a ratio, we can assign arbitrary unit values to their capacities. Total volume is the product of capacity and the number of iterations required to fill it. Total Volume = Capacity of Container × Number of Pours Step-by-Step Solution: 1. Assign Relative Capacities: • Let the capacity of Container Y = 1 unit per pour. • Container X holds three times as much, so its capacity = 3 units per pour. 2. Calculate Total Tank Volume: • Container X takes 60 pours to fill the tank. • Total Tank Volume = Capacity of X × 60 = 3 × 60 = 180 units. 3. Calculate Combined Capacity: • If both are used together, the combined volume per pour = 3 + 1 = 4 units. 4. Determine Total Pours Required: • Total pours = \frac{Total Tank Volume}{Combined Capacity} • Total pours = (180/4) = 45 pours. Exam Strategy & Shortcut: Think purely in efficiency ratios. X : Y = 3 : 1. Total work = 3 × 60 = 180. Together, they pour 4 parts per turn. Total turns = 180 / 4 = 45. This avoids any complex fractions or algebra. Common Pitfall: A common mistake is trying to calculate the number of pours Container Y would take individually (180 pours) and then using standard pipe-filling formulas ((1/60) + (1/180)). While this works, it is overcomplicated. Volumes per pour can be added directly. Final Answer: Therefore, the correct answer is 45.

Question 11

Two supply tubes, M and N, can fill a large container in 60 minutes and 40 minutes, respectively. How long will it take to fill the empty container if tube N is active for the first half of the total time, and both tubes M and N are active together …

Accepted Answer

Concept & Time-Splitting Equations: When the problem splits the *total time* into fractions (e.g., halves) where different combinations of workers operate, set up an algebraic equation based on the total time rather than the volume. Total Work = (Rate_1 × Time_1) + (Rate_2 × Time_2) Step-by-Step Solution: 1. Determine Capacity and Rates: • Let total capacity = LCM(60, 40) = 120 units. • Rate of Tube M = 120 / 60 = 2 units/min. • Rate of Tube N = 120 / 40 = 3 units/min. 2. Set Up Time Variables: • Let the total time taken to fill the container be 2t minutes. • First half of the time = t minutes. • Second half of the time = t minutes. 3. Formulate the Equation: • During the first half (t mins), only N operates. Work done = 3 × t. • During the second half (t mins), both M and N operate. Work done = (2 + 3) × t = 5 × t. • Total work equals the full capacity: 3t + 5t = 120. 4. Solve for Total Time: • 8t = 120 • t = 120 / 8 = 15 minutes. • The total time is 2t = 2 × 15 = 30 minutes. Exam Strategy & Shortcut: Understand that the average efficiency across the whole period is just the average of the two halves. First half efficiency = 3. Second half efficiency = 5. Average efficiency = (3 + 5) / 2 = 4 units/min. Total time = \frac{Total Capacity}{Average Efficiency} = (120/4) = 30 minutes. This shortcut completely bypasses algebra. Common Pitfall: Solving for t and hastily selecting 15 minutes as the final answer. Remember that t only represents *half* of the time. You must multiply by 2 to get the total time. Final Answer: Therefore, the correct answer is 30 min.

Question 12

Two pipes can fill a tank with water in 15 and 12 hours respectively and a third pipe can empty it in 4 hours. If the pipes be opened in order at 8, 9 and 11 a.m. respectively, the tank will be emptied at (S.S.C., 2005)

Accepted Answer

Concept & Net Emptying Rate: When fill pipes and an emptying pipe operate simultaneously, assign positive efficiencies to filling and negative to emptying. If the net efficiency is negative, the tank will empty. Net Efficiency = \sum Fill Rates - \sum Empty Rates Step-by-Step Solution: 1. Given: Pipe A fills in 15 hrs, Pipe B fills in 12 hrs, Pipe C empties in 4 hrs. 2. Calculation: - Total capacity = LCM(15, 12, 4) = 60 units. - Efficiency of A = +60 / 15 = +4 units/hr. - Efficiency of B = +60 / 12 = +5 units/hr. - Efficiency of C = -60 / 4 = -15 units/hr. 3. Calculate the volume filled before the emptying pipe opens at 11 a.m.: - 8 a.m. to 11 a.m. (3 hrs): Pipe A is open. Work = 3 * 4 = 12 units. - 9 a.m. to 11 a.m. (2 hrs): Pipe B is open. Work = 2 * 5 = 10 units. - Total water in the tank at 11 a.m. = 12 + 10 = 22 units. 4. Calculate emptying time after 11 a.m.: - After 11 a.m., all pipes are open. - Net efficiency = +4 + 5 - 15 = -6 units/hr (tank is emptying at 6 units/hr). - Time required to empty the 22 units = 22 / 6 = 11/3 hours = 3 hours and 40 minutes. 5. Determine final time: - 11:00 a.m. + 3 hours 40 minutes = 2:40 p.m. Exam Strategy & Shortcut: Find the accumulated volume quickly: A works 3 hours, B works 2 hours \rightarrow 3(4) + 2(5) = 22 units. Net rate is -6. Time is 22/6 = 3(2/3) hours. Convert 2/3 hour to 40 mins and add to 11 a.m. Common Pitfall: A common mistake is trying to empty the entire capacity (60 units) rather than just the amount of water that actually collected in the tank by 11 a.m. (22 units). Final Answer: Therefore, the correct answer is 2 : 40 p.m.

Question 13

Tap A fills a tank in 4 hours whereas tap B empties the full tank in 24 hours. A and B are opened alternately for 1 hour each. Every 2 hours the level of water is found to increase by 0.5 m. The depth of the tank is

Accepted Answer

Concept & Proportionality of Volume and Depth: In uniform tanks, the fraction of the tank filled is directly proportional to the fraction of the total depth filled. We analyze the net work done in one complete alternate cycle. Fraction Filled × Total Depth = Observed Depth Increase Step-by-Step Solution: 1. Given: Tap A fills in 4 hours, Tap B empties in 24 hours. They run alternately. 2. Calculation: - In 1 hour, Tap A fills 1/4 of the tank. - In 1 hour, Tap B empties 1/24 of the tank. - Since they operate alternately, a full cycle takes 2 hours. - Net part of the tank filled in 2 hours = (1/4) - (1/24) - Make denominators equal: 1/4 = 6/24. - Net filled = 6/24 - 1/24 = 5/24 of the tank. 3. Relate volume to depth: - We are given that in 2 hours, the water level increases by 0.5 m. - Therefore, 5/24 of the total depth (D) corresponds to 0.5 m. - (5/24) × D = 0.5 4. Solve for Depth (D): - D = 0.5 × (24 / 5) - D = 1/2 × 24 / 5 = 12 / 5 = 2.4 meters. Exam Strategy & Shortcut: Recognize that 2 hours is exactly one cycle. Net fill per cycle is 1/4 - 1/24 = 5/24. If 5/24 of the tank is 0.5 meters, then 1/24 of the tank is 0.1 meters. The full tank (24/24) is 24 × 0.1 = 2.4 meters. Common Pitfall: A common error is confusing the alternate hour setup and accidentally calculating the net rate assuming they run simultaneously for a full 2 hours, which leads to incorrect capacity ratios. Final Answer: Therefore, the correct answer is 2.4 m.

Question 14

Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full? (Campus Recruitment, 2003)

Accepted Answer

Concept & Alternate Working: When pipes are opened on alternate hours, calculate the work done in a full 2-hour cycle. Determine how many complete cycles fit into the total work, and then evaluate the remaining work sequentially. Work per Cycle = Efficiency_A + Efficiency_B Step-by-Step Solution: 1. Given: Time for A = 6 hours, Time for B = 4 hours. A opens first. 2. Calculation: - Total capacity = LCM(6, 4) = 12 units. - Efficiency of A = 12 / 6 = 2 units/hr. - Efficiency of B = 12 / 4 = 3 units/hr. 3. Analyze the 2-hour cycle: - 1st hour (A works): fills 2 units. - 2nd hour (B works): fills 3 units. - Total work in 1 cycle (2 hours) = 2 + 3 = 5 units. 4. Calculate maximum full cycles: - Max cycles fitting in 12 units = 12 // 5 = 2 cycles. - Work done in 2 cycles (4 hours) = 2 * 5 = 10 units. 5. Process remaining work: - Remaining work = 12 - 10 = 2 units. - The 5th hour is Pipe A's turn. - Pipe A fills 2 units per hour. So it will take exactly 2 / 2 = 1 hour to finish the remaining 2 units. 6. Total time = 4 hours (from cycles) + 1 hour (from A) = 5 hours. Exam Strategy & Shortcut: Mental math with LCM: Cap = 12. Cycle = 5 units/2hrs. Two cycles give 10 units in 4 hours. Remaining is 2 units. Next is A's turn, and A's capacity is exactly 2 units/hr, taking exactly 1 more hour. Total = 5. Common Pitfall: A common mistake in alternate hour problems is dividing the total capacity directly by the combined efficiency without separating it into cycles. Doing 12 / 5 = 2.4 × 2 = 4.8 hours is incorrect because it ignores the discrete hourly turns. Final Answer: Therefore, the correct answer is 5.

Question 15

Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is :

Accepted Answer

Concept & Determining Waste Pipe Efficiency: The capacity of a tank can be determined by finding the time it takes for a specific pipe (with a known volumetric rate) to complete its work, and then multiplying that time by the rate. Capacity = Time to Empty × Volumetric Empty Rate Step-by-Step Solution: 1. Given: Pipe A fills in 20 min. Pipe B fills in 24 min. Waste Pipe C empties at 3 gal/min. A + B + C fill in 15 min. 2. Calculation: - Let total capacity = LCM(20, 24, 15) = 120 units. - Efficiency of A = 120 / 20 = +6 units/min. - Efficiency of B = 120 / 24 = +5 units/min. - Net Efficiency (A + B + C) = 120 / 15 = +8 units/min. 3. Find Efficiency of Pipe C: - E_A + E_B + E_C = 8 - 6 + 5 + E_C = 8 - 11 + E_C = 8 \Rightarrow E_C = -3 units/min. 4. Relate conceptual units to actual volume: - The waste pipe empties 3 units per minute. - We are given that it empties exactly 3 gallons per minute. - Therefore, 1 unit = 1 gallon. 5. Determine Capacity: - Total capacity is 120 units. - Since 1 unit = 1 gallon, capacity = 120 gallons. Exam Strategy & Shortcut: Fractions approach: Waste pipe time 1/C = 1/20 + 1/24 - 1/15 = 6/120 + 5/120 - 8/120 = 3/120 = 1/40. So it takes 40 minutes to empty. Capacity = 40 min × 3 gal/min = 120 gallons. Common Pitfall: A common error is confusing the net fill rate (15 minutes) as an empty rate, leading to an incorrect algebraic setup. Always note that they "can fill the tank in 15 minutes". Final Answer: Therefore, the correct answer is 120 gallons.

Question 16

A cistern can be filled by two pipes filling separately in 12 and 16 minutes separately. Both the pipes are opened together for a certain time but being clogged, only $\frac{7}{8}$ of the full quantity of water flows through the former and only $\fr…

Accepted Answer

Concept & Variable Efficiency: When pipes operate at reduced capacity, calculate their new working efficiencies based on the given fractions. Determine the work done during the full capacity phase to find the remaining work that was completed during the clogged phase. Time = \frac{Work Completed}{Working Efficiency} Step-by-Step Solution: 1. Given: Pipe 1 fills in 12 min, Pipe 2 fills in 16 min. Last 3 minutes are at full capacity. 2. Calculation: - Let total capacity = LCM(12, 16) = 48 units. - Normal efficiency of Pipe 1 = 48 / 12 = 4 units/min. - Normal efficiency of Pipe 2 = 48 / 16 = 3 units/min. - Normal combined efficiency = 4 + 3 = 7 units/min. 3. Calculate clogged efficiencies: - Clogged Pipe 1 = 4 × (7/8) = 3.5 units/min. - Clogged Pipe 2 = 3 × (5/6) = 2.5 units/min. - Clogged combined efficiency = 3.5 + 2.5 = 6 units/min. 4. Work backward from the end: - In the last 3 minutes at full flow, work done = 3 × 7 = 21 units. - Remaining work that was done while clogged = 48 - 21 = 27 units. 5. Calculate time spent clogged: - Time clogged = Work done clogged / Clogged efficiency - Time clogged = 27 / 6 = 4.5 minutes, or 4(1/2) minutes. Exam Strategy & Shortcut: Using the LCM of 48 gives clean numbers. Normal rate is 7. Clogged rate is 4(7/8) + 3(5/6) = 3.5 + 2.5 = 6. The final 3 mins at rate 7 fill 21 units. Thus, 48 - 21 = 27 units were filled at the clogged rate of 6. 27 / 6 = 4.5 mins. Common Pitfall: A common mistake is trying to set up an algebraic equation for the total time T, taking T-3 as the clogged time, which often leads to algebraic errors compared to the straightforward reverse work method. Final Answer: Therefore, the correct answer is 4(1/2) min.

Question 17

Three pipes can fill a reservoir in 10, 15 and 20 hours respectively. If the three taps are opened one after another in the given order, with a certain fixed time gap between them, the reservoir fills in 5 hours. The time gap is

Accepted Answer

Concept & Staggered Openings with Uniform Gaps: When pipes are opened at fixed intervals, the total time each pipe operates can be expressed algebraically relative to the first pipe. The sum of the work done by all pipes must equal the total capacity. Work = (Rate_A × Time_A) + (Rate_B × Time_B) + (Rate_C × Time_C) Step-by-Step Solution: 1. Given: Times are 10, 15, and 20 hours. Total fill time is 5 hours. 2. Calculation: - Total capacity = LCM(10, 15, 20) = 60 units. - Efficiency A = 60 / 10 = 6 units/hr. - Efficiency B = 60 / 15 = 4 units/hr. - Efficiency C = 60 / 20 = 3 units/hr. 3. Define time active for each pipe: - Let the fixed time gap be t hours. - Pipe A is open for the full 5 hours. Work = 5 × 6 = 30 units. - Pipe B is open for (5 - t) hours. Work = 4 × (5 - t) units. - Pipe C is open for (5 - 2t) hours. Work = 3 × (5 - 2t) units. 4. Set up the total work equation: - 30 + 4(5 - t) + 3(5 - 2t) = 60 - 30 + 20 - 4t + 15 - 6t = 60 - 65 - 10t = 60 - 10t = 5 - t = 0.5 hours. 5. Convert to minutes: - 0.5 hours = 30 minutes. Exam Strategy & Shortcut: Pipe A does half the work (5/10) instantly, leaving 30 units (out of 60). If there were no gap, B and C would work 5 hours too, doing 20 + 15 = 35 units. The "extra" 5 units of work (35 - 30) comes from the time they were *closed*. B loses 4t and C loses 6t (2t × 3). So 10t = 5, giving t = 0.5 hours. Common Pitfall: A standard error is assigning the time gap as t for B and t for C instead of accumulating the gap (i.e., making C wait 2t). Final Answer: Therefore, the correct answer is 30 min.

Question 18

A bath can be filled by the cold water pipe in 10 minutes and by the hot water pipe in 15 minutes. A person leaves the bathroom after turning on both the pipes. He returns just when the bath should have been full. Finding however, the waste pipe was…

Accepted Answer

Concept & Equivalent Work Volumes: If an expected outcome isn't met due to a leak, the "extra" work required to fix the deficit equals exactly the amount of work the leak undid during the time it was active. Volume Emptied by Leak = Volume Refilled to Catch Up Step-by-Step Solution: 1. Given: Cold pipe (A) fills in 10 min. Hot pipe (B) fills in 15 min. Waste pipe (C) is open initially. 2. Calculation: - Total capacity = LCM(10, 15) = 30 units. - Efficiency A = 30 / 10 = 3 units/min. - Efficiency B = 30 / 15 = 2 units/min. - Normal combined efficiency (A + B) = 3 + 2 = 5 units/min. 3. Determine expected time of return: - Expected time to fill 30 units = 30 / 5 = 6 minutes. - The person returns after 6 minutes. During this time, the waste pipe was open. 4. Analyze the catch-up phase: - He closes the waste pipe, and A + B fill the tank in 4 more minutes. - Volume filled in these 4 minutes = 4 min × 5 units/min = 20 units. - This means exactly 20 units were missing at the 6-minute mark. 5. Calculate waste pipe efficiency: - The waste pipe emptied these 20 units during the initial 6 minutes it was open. - Efficiency of waste pipe = 20 units / 6 min = 10/3 units/min. 6. Determine time to empty full tank: - Time = Total Capacity / Waste Efficiency = 30 / (10/3) = (30 × 3) / 10 = 9 minutes. Exam Strategy & Shortcut: Equate the works directly: The work done by the waste pipe in 6 minutes is equal to the work done by the filling pipes in 4 minutes. So, 6 × E_{waste} = 4 × E_{fill}. Since E_{fill} takes 6 minutes for the whole tank, the whole tank is 6 × E_{fill}. We need time T = Total / E_{waste}. Substitute E_{waste} = (4/6) E_{fill}. Time = 6 / (4/6) = 36/4 = 9 minutes. Common Pitfall: A common pitfall is calculating the net efficiency with all three pipes and creating a complex equation, when simply equating the leaked volume to the make-up volume is much faster and clearer. Final Answer: Therefore, the correct answer is 9 minutes.

Question 19

A large fresh water reservoir is fitted with two types of feeder pipes – hot water pipes and cold water pipes. Six cold water pipes alone can fill the reservoir in 12 hours. 3 cold water pipes and 9 hot water pipes together can fill the reservoir in…

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Concept & Logic: To find the time taken by the hot water pipes, we need to determine the filling rate (work done per hour) for both a single hot water pipe and a single cold water pipe. We can set up a system of linear equations based on their combined rates. Step-by-Step Solution: 1. Let the rate of 1 cold water pipe be C and the rate of 1 hot water pipe be H. 2. 6 cold water pipes can fill the reservoir in 12 hours. 6C = (1/12) C = (1/72) (Reservoir per hour) 3. 3 cold water pipes and 9 hot water pipes can fill the reservoir in 8 hours. 3C + 9H = (1/8) 4. Substitute the value of C into the second equation: 3 ((1/72)) + 9H = (1/8) (1/24) + 9H = (1/8) 5. Solve for H: 9H = (1/8) - (1/24) 9H = (3 - 1/24) = (2/24) = (1/12) H = (1/108) (Reservoir per hour) 6. Now, find the rate of 5 hot water pipes: 5H = 5 × (1/108) = (5/108) 7. The time taken by 5 hot water pipes alone is the reciprocal of their combined rate: Time = (108/5) hours = 21.6 hours 8. Convert the decimal part into minutes: 0.6 hours = 0.6 × 60 minutes = 36 minutes Total time = 21 hrs 36 min. Exam Strategy & Shortcut: Instead of finding individual rates immediately, you can substitute the group of cold pipes directly. Since 6 cold pipes take 12 hours, 3 cold pipes take 24 hours. Their rate is 1/24. Subtracting this from the combined rate (1/8) gives the rate of 9 hot pipes. Common Pitfall: Students often forget to convert the decimal fraction of an hour (0.6) into minutes correctly, sometimes mistakenly writing it as 60 minutes or leaving it as a decimal. Final Answer: Therefore, the correct answer is 21 hrs 36 min.

Question 20

A town is supplied with water from a big overhead tank which is fed with a constant volume of water regularly. When the tank is full, if 32000 gallons are used daily, the supply fails in 50 days. However, if 37000 gallons are used daily, the supply …

Accepted Answer

Concept & Logic: This problem models a tank with a constant initial volume and a continuous, constant inflow. To ensure the supply never fails, the daily usage must exactly equal the daily inflow rate. If usage exceeds inflow, the initial reserve depletes over time. Let initial volume = V and daily inflow = r. Total volume used over d days = V + d × r. Step-by-Step Solution: 1. Let V be the initial capacity of the tank and r be the rate of constant inflow per day. 2. Case 1: 32000 gallons used daily for 50 days. V + 50r = 32000 × 50 V + 50r = 1600000 --- (Equation 1) 3. Case 2: 37000 gallons used daily for 40 days. V + 40r = 37000 × 40 V + 40r = 1480000 --- (Equation 2) 4. Subtract Equation 2 from Equation 1 to eliminate V: (V + 50r) - (V + 40r) = 1600000 - 1480000 10r = 120000 r = 12000 gallons/day 5. For the supply to never fail, the daily usage must not exceed the daily inflow. Thus, the maximum sustainable usage is equal to the inflow rate. Maximum usage = r = 12000 gallons Exam Strategy & Shortcut: Focus on the differences. The difference in total water consumed is due to the difference in inflow over those extra days. Total consumed in 50 days = 1600k Total consumed in 40 days = 1480k Difference = 120k gallons. This extra 120k came from 10 extra days of inflow. Inflow rate = 120k / 10 = 12k gallons/day. This is the sustainable rate. Common Pitfall: A common mistake is trying to solve for V and then getting confused about how to apply it. The key is realizing that "never failing" means operating strictly on the renewable inflow. Final Answer: Therefore, the correct answer is 12000 gallons.

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