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Question 1 / 20
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1
Person A can finish a particular task in 12 days. Given that Person B is 60% more efficient than A, how many days will it take A and B to complete the same task if they work together?
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Solution: Step 1: Determine the efficiency ratio between A and B. If A's efficiency is 100%, B's efficiency is 100% + 60% = 160%. Efficiency A : B = 100 : 160 = 5 : 8. Step 2: Determine the time ratio. Time taken is inversely proportional to efficiency. So, Time A : Time B = 8 : 5. Step 3: Calculate the time B takes alone. Given A takes 12 days. If 8 units of time correspond to 12 days, then 1 unit = 12/8 = 1.5 days. B takes 5 units of time, so B takes 5 * 1.5 = 7.5 days (or 15/2 days). Step 4: Calculate their combined 1-day work rate. A's 1-day work = 1/12. B's 1-day work = 1/(15/2) = 2/15. Combined (A+B)'s 1-day work = 1/12 + 2/15. To add fractions, find the Least Common Multiple (LCM) of 12 and 15, which is 60. (A+B)'s 1-day work = (5/60) + (8/60) = 13/60. Step 5: Determine the total time taken to complete the job together. If they complete 13/60 of the work in 1 day, they will finish the entire job together in 60/13 days.
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20 women can finish a task in 16 days, and 16 men can complete the same task in 15 days. Determine the ratio of the work capacity of a man to that of a woman.
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Solution: Step 1: Calculate the total 'woman-days' required for the work: 20 women * 16 days = 320 woman-days. Step 2: This means 1 woman's 1 day's work is 1/320 of the total work. Step 3: Calculate the total 'man-days' required for the work: 16 men * 15 days = 240 man-days. Step 4: This means 1 man's 1 day's work is 1/240 of the total work. Step 5: The required ratio is (1 man's 1 day's work) : (1 woman's 1 day's work). Step 6: Set up the ratio: (1/240) : (1/320). Step 7: To simplify, multiply both sides by the LCM of 240 and 320 (which is 960): (1/240)*960 : (1/320)*960 = 4 : 3. Step 8: The ratio between the capacity of a man and a woman is 4:3.
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Two pipes can fill a tank in 40 minutes and 48 minutes, respectively. A waste pipe can empty water at a rate of 3 gallons per minute. When all three pipes are operated together, the tank is filled in 30 minutes. What is the total capacity of the tank in gallons?
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Solution: Step 1: Pipe 1's filling rate = 1/40 tank per minute. Step 2: Pipe 2's filling rate = 1/48 tank per minute. Step 3: When all three pipes work together (two filling, one emptying), the tank fills in 30 minutes. Step 4: So, the net combined filling rate of all three pipes = 1/30 tank per minute. Step 5: Let the waste pipe's emptying rate be 1/W (part of tank per minute). Step 6: The equation for the combined rate is: (1/40) + (1/48) - (1/W) = 1/30. Step 7: Rearrange to find the waste pipe's emptying rate: 1/W = (1/40) + (1/48) - (1/30). Step 8: Find a common denominator for 40, 48, and 30 (LCM is 240). Step 9: 1/W = (6/240) + (5/240) - (8/240). Step 10: Calculate the emptying rate: (6 + 5 - 8) / 240 = 3/240 = 1/80 tank per minute. Step 11: So, the waste pipe alone can empty the entire tank in 80 minutes. Step 12: The waste pipe empties at a rate of 3 gallons per minute. Step 13: Total capacity of the tank = (Emptying rate in gallons/minute) * (Time taken by waste pipe to empty the tank alone). Step 14: Capacity = 3 gallons/minute * 80 minutes = 240 gallons.
4
Three pipes, A, B, and C, can fill a tank. Pipes A and B together fill it in 6 hours. When pipe C is added, the tank fills in 5 hours. Pipe A fills at twice the rate of pipe B. How long do pipes B and C take to fill the tank together?
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Solution: Step 1: Let pipe B's rate = x tanks/hour Step 2: Pipe A's rate = 2x tanks/hour Step 3: Combined rate of A and B = x + 2x = 3x tanks/hour Step 4: Time for A and B to fill = 6 hours → 3x * 6 = 1 tank → x = 1/18 tanks/hour Step 5: Combined rate with C = 1 tank / 5 hours = 1/5 tanks/hour Step 6: Rate of C = (1/5) - 3x = (1/5) - (3/18) = (1/5) - (1/6) = (6 - 5)/30 = 1/30 tanks/hour Step 7: Combined rate of B and C = x + (1/30) = (1/18) + (1/30) = (5 + 3)/90 = 8/90 = 4/45 tanks/hour Step 8: Time for B and C = 1 / (4/45) = 45/4 = 11.25 hours
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Pipe X can fill a tank in 12 minutes, while pipes Y and Z together can fill it in 6 minutes. When pipe Y fills for the first minute, followed by pipe X for the next two minutes, and this pattern repeats, the tank is filled in 16 minutes. How many minutes does the fastest pipe take to fill the tank alone?
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Solution: Step 1: Let the tank capacity = 1 unit Step 2: Pipe X rate = 1/12 unit/min Step 3: Pipes Y and Z combined rate = 1/6 unit/min Step 4: In 1 cycle (3 min): Y fills 1/6 unit, X fills 2*(1/12) = 1/6 unit Step 5: Total per cycle = 1/6 + 1/6 = 1/3 unit Step 6: In 15 min (5 cycles): 5*(1/3) = 5/3 units Step 7: Remaining = 1 - 5/3 = 1/3 unit Step 8: Y fills remaining 1/3 unit in 1/3 * 6 = 2 min Step 9: Total time = 15 + 1 = 16 min (verification) Step 10: Fastest pipe (Y) rate = 1/6 unit/min → 6 min to fill tank alone
6
If 20 men can complete a work in C/2 days and 30 women can complete the same work in C/3 days, how many days will it take for 20 men and 30 women to complete the work together?
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Solution: Step 1: 20 men's 1 day work = 2/C. Step 2: 30 women's 1 day work = 3/C. Step 3: Combined 1 day work of 20 men and 30 women = 2/C + 3/C = 5/C. Step 4: Time taken to complete the work together = C / 5 days.
7
A project requires 18 workers to complete in 28 days, working 9 hours daily. How many hours per day must 21 workers work to finish the same project in 36 days?
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Solution: Step 1: Calculate total man-hours for original scenario = 18 workers * 28 days * 9 hours/day = 4536 man-hours Step 2: Let x = hours/day for 21 workers in 36 days Step 3: Set up equation: 21 workers * 36 days * x hours/day = 4536 Step 4: Solve for x: 756x = 4536 Step 5: x = 4536 / 756 = 6 hours/day
8
Two individuals, Person A and Person B, can complete a painting job in 9 hours and 18 hours, respectively. How long will it take for them to complete twice the work together?
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Solution: Step 1: Find individual work rates. Person A's rate = 1/9 of the work per hour, Person B's rate = 1/18 of the work per hour. Step 2: Calculate combined work rate = 1/9 + 1/18 = 3/18 = 1/6 of the work per hour. Step 3: Time to complete one work = 6 hours. Step 4: Time to complete twice the work = 2 * 6 = 12 hours.
9
A and B can finish a task together in 3 days. They begin working together, but B leaves after 2 days. If A then completes the remaining work alone in two additional days, how many days would B require to complete the entire work by himself?
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Solution: Step 1: Calculate the combined 1-day work rate of A and B. - If A and B together complete the work in 3 days, their combined 1-day work rate is 1/3. Step 2: Calculate the amount of work done by A and B together in the first 2 days. - Work done in 2 days = (1/3) × 2 = 2/3 of the total work. Step 3: Calculate the remaining work. - Remaining work = 1 - (2/3) = 1/3 of the total work. Step 4: Determine A's individual 1-day work rate. - A completes the remaining 1/3 of the work in 2 additional days. - A's 1-day work rate = (1/3) / 2 = 1/6 of the total work. Step 5: Calculate B's individual 1-day work rate. - B's 1-day work rate = (A + B)'s 1-day work rate - A's 1-day work rate - B's 1-day work rate = (1/3) - (1/6) - Find a common denominator (6): B's 1-day work rate = (2/6) - (1/6) = 1/6 of the total work. Step 6: Determine the number of days B alone would take to complete the work. - If B completes 1/6 of the work in one day, then B will take 6 days to complete the entire work.
10
If 28 men can complete 7/8 of a particular piece of work in one week, then how many men must be employed to finish the remaining portion of the work in an additional week?
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Solution: Step 1: Use the Man-Days-Work formula (M1 * D1 / W1 = M2 * D2 / W2). Here, M = Number of men, D = Number of days (or weeks in this case), W = Amount of work. Step 2: Identify the given values for the first scenario: M1 = 28 men, D1 = 1 week, W1 = 7/8 of the work. Step 3: Determine the parameters for the second scenario: D2 = 1 week (another week). Remaining work W2 = 1 - (7/8) = 1/8 of the work. M2 (number of men needed) is unknown. Step 4: Substitute the values into the formula: (28 * 1) / (7/8) = (M2 * 1) / (1/8). Step 5: Simplify both sides of the equation: Left side: 28 / (7/8) = 28 * (8/7) = 4 * 8 = 32. Right side: M2 / (1/8) = M2 * 8. Step 6: Now the equation is: 32 = 8 * M2. Step 7: Solve for M2: M2 = 32 / 8 = 4 men. Step 8: Therefore, 4 men must be engaged to complete the remaining work in another week.
11
A 6 cm long cigarette burns completely in 15 minutes if no puff is taken. For every puff, it burns three times as fast during the duration of that puff. If the cigarette burns itself out in a total of 13 minutes, and each average puff lasted 3 seconds, how many puffs did the smoker take?
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Solution: Step 1: Calculate the normal burning rate of the cigarette. * Normal length = 6 cm, Normal time = 15 minutes = 15 * 60 = 900 seconds. * Normal Rate = 6 cm / 900 seconds = 1/150 cm/second. Step 2: Calculate the enhanced burning rate during a puff. * Puff Rate = 3 * Normal Rate = 3 * (1/150) cm/second = 1/50 cm/second. Step 3: Let 'n' be the number of puffs taken. * Each puff lasts 3 seconds, so total time spent puffing = 3n seconds. Step 4: The total burning time is 13 minutes = 13 * 60 = 780 seconds. Step 5: Calculate the time spent burning without puffs. * Time_normal_burn = Total burning time - Total time spent puffing = (780 - 3n) seconds. Step 6: Set up an equation for the total length burned: * Length burned during puffs + Length burned normally = Total length * (Puff Rate * Time_puffing) + (Normal Rate * Time_normal_burn) = 6 cm * (1/50 * 3n) + (1/150 * (780 - 3n)) = 6. Step 7: Solve the equation for 'n': * Multiply the entire equation by the LCM of 50 and 150, which is 150: * (3n/50) * 150 + ((780 - 3n)/150) * 150 = 6 * 150 * 3 * (3n) + (780 - 3n) = 900 * 9n + 780 - 3n = 900 * 6n + 780 = 900 * 6n = 900 - 780 * 6n = 120 * n = 120 / 6 = 20. Step 8: The smoker took 20 puffs.
12
A can complete a work in 25 days and B can complete the same work in 20 days. They start working together, but B leaves after 4 days. A continues to work alone. In how many days will the entire work be completed?
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Solution: Step 1: Determine the total work using the LCM method. Total work = LCM(25, 20) = 100 units. Step 2: Calculate the efficiency (work per day) of A and B. Efficiency of A = Total work / Days taken by A = 100 / 25 = 4 units/day. Efficiency of B = Total work / Days taken by B = 100 / 20 = 5 units/day. Step 3: Calculate the work done by A and B together in the first 4 days. Combined efficiency of A + B = 4 + 5 = 9 units/day. Work done by A and B in 4 days = 9 units/day * 4 days = 36 units. Step 4: Calculate the remaining work after B leaves. Remaining work = Total work - Work done by A and B = 100 - 36 = 64 units. Step 5: Calculate the time A alone takes to complete the remaining work. Time taken by A = Remaining work / Efficiency of A = 64 units / 4 units/day = 16 days. Step 6: Calculate the total number of days to complete the entire work. Total days = Days A and B worked together + Days A worked alone = 4 days + 16 days = 20 days. Therefore, the entire work will be completed in 20 days.
13
A cistern can be filled by two pipes in 20 minutes and 30 minutes, respectively. Both pipes are opened. The first pipe is turned off after some time, and the cistern is completely filled by the second pipe alone in an additional 10 minutes. After how many minutes was the first pipe turned off?
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Solution: Step 1: Let 'x' be the number of minutes after which the first pipe is turned off. Step 2: Pipe 1 fills 1/20 of the cistern per minute. Pipe 2 fills 1/30 of the cistern per minute. Step 3: During the first 'x' minutes, both pipes are open. Their combined fill rate = 1/20 + 1/30. Step 4: The combined rate (LCM of 20 and 30 is 60): (3/60) + (2/60) = 5/60 = 1/12 of the cistern per minute. Step 5: Work done by both pipes in 'x' minutes = x * (1/12) = x/12. Step 6: For the last 10 minutes, only Pipe 2 is open. Work done by Pipe 2 in 10 minutes = 10 * (1/30) = 1/3 of the cistern. Step 7: The total work done (filling the cistern) is 1. So, x/12 + 1/3 = 1. Step 8: Subtract 1/3 from both sides: x/12 = 1 - 1/3. Step 9: Calculate: x/12 = 2/3. Step 10: Solve for x: x = (2/3) * 12 = 8. Step 11: The first pipe must be turned off after 8 minutes.
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A device fills a container in 24 hours. Due to a defect, it now takes 36 hours to fill the same container. If the container is full, how long will the defect take to empty it completely?
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Solution: Step 1: Let the work rate of the device be 1/24 (container/hour) Step 2: Let the work rate of the defect be -x (container/hour) Step 3: Net work rate with defect = 1/24 - x Step 4: Time taken with defect = 36 hours, so (1/24 - x) * 36 = 1 Step 5: Solve for x: 36/24 - 36x = 1 => 1.5 - 36x = 1 => 36x = 0.5 => x = 1/72 Step 6: Defect empties the container at 1/72 container/hour Step 7: Time to empty = 1 / (1/72) = 72 hours
15
Worker A can complete a task in 20 days, while Worker B can complete the same task in 30 days. They start working together, but Worker B leaves 10 days before the task is finished. How many days does it take to complete the task?
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Solution: Step 1: Calculate individual work rates: A's rate = 1/20 tasks/day, B's rate = 1/30 tasks/day Step 2: Combined rate = 1/20 + 1/30 = 1/12 tasks/day Step 3: Let total days to complete = x days Step 4: Work done by both = (x - 10) * (1/12) + 10 * (1/20) Step 5: Set up equation: (x - 10)/12 + 10/20 = 1 Step 6: Simplify: (x - 10)/12 + 1/2 = 1 Step 7: Convert to common denominator: (x - 10 + 6)/12 = 1 Step 8: Solve for x: x - 4 = 12 Step 9: x = 16 days
16
A can complete a task in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?
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Solution: Step 1: Determine the hourly work rates from the given information. A's 1 hour's work = `1/4`. (B + C)'s 1 hour's work = `1/3`. (A + C)'s 1 hour's work = `1/2`. Step 2: Calculate the combined work rate of A, B, and C. We know (A + B + C)'s 1 hour's work = (A's 1 hour's work) + (B + C)'s 1 hour's work. `= 1/4 + 1/3`. Find the LCM of 4 and 3, which is 12. `= (3/12) + (4/12) = 7/12`. Step 3: Calculate B's 1 hour's work. B's 1 hour's work = (A + B + C)'s 1 hour's work - (A + C)'s 1 hour's work. `= 7/12 - 1/2`. Find the LCM of 12 and 2, which is 12. `= (7/12) - (6/12) = 1/12`. Step 4: Determine the time B alone would take to complete the work. If B does 1/12 of the work in 1 hour, then B alone will take `1 / (1/12) = 12` hours to complete the work. Step 5: Conclusion: B alone will take 12 hours to do the work.
17
2 individuals working 2 hours a day assemble 2 machines in 2 days. What is the number of machines assembled by 6 individuals working 6 hours a day in 6 days?
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Solution: Step 1: Identify the variables: Persons (M), Hours per day (H), Days (D), and Machines (W). Step 2: Use the formula M1D1H1/W1 = M2D2H2/W2. Step 3: Given values: M1 = 2, H1 = 2, D1 = 2, W1 = 2. Step 4: New values: M2 = 6, H2 = 6, D2 = 6, W2 = x (unknown machines). Step 5: Substitute the values into the formula: (2 × 2 × 2) / 2 = (6 × 6 × 6) / x. Step 6: Simplify the equation: 8 / 2 = 216 / x. Step 7: Further simplify: 4 = 216 / x. Step 8: Solve for x: 4x = 216. Step 9: Calculate: x = 216 / 4 = 54. Step 10: Therefore, 54 machines can be assembled by 6 persons working 6 hours a day in 6 days.
18
A monkey attempts to climb a 60-meter high pole. In the first minute, it climbs 6 meters, but in the subsequent minute, it slips down 3 meters. How much total time will the monkey need to reach the top of the pole?
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Solution: Step 1: Analyze the monkey's progress over a 2-minute cycle: * Minute 1: Climbs 6 m. * Minute 2: Slips down 3 m. * Net climb in 2 minutes = 6 - 3 = 3 meters. Step 2: The total height of the pole is 60 m. The final climb, where the monkey does not slip, is equal to its maximum single climb, which is 6 meters. Step 3: Calculate the distance to be covered at the net climbing rate before the final 6-meter ascent: 60 meters (total height) - 6 meters (final climb) = 54 meters. Step 4: Determine how many 2-minute cycles are needed to climb these 54 meters at a net rate of 3 meters per 2 minutes: * Number of cycles = 54 meters / 3 meters/cycle = 18 cycles. Step 5: Time taken for 18 cycles = 18 cycles * 2 minutes/cycle = 36 minutes. Step 6: After 36 minutes, the monkey has climbed 54 meters. Step 7: In the next minute (the 37th minute), the monkey makes its final climb of 6 meters and reaches the top (54 + 6 = 60 meters) without slipping further. Step 8: Total time taken = 36 minutes + 1 minute = 37 minutes.
19
Person A completes a task in 17 days, Person B in 5 days. Working together, how many days to finish?
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Solution: Step 1: Calculate individual work rates A's rate = 1/17 work/day B's rate = 1/5 work/day Step 2: Calculate combined rate Combined rate = 1/17 + 1/5 = (5+17)/85 = 22/85 work/day Step 3: Calculate time for 1 complete work Time = 1 / (22/85) = 85/22 ≈ 3.86 days (rounded to 3.8 days)
20
Person A is twice as efficient as Person B, and together they complete a job in 28 days. How many days will Person A take to complete the job alone?
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Solution: Step 1: Let's assume the efficiency of Person B is x units/day. Step 2: Efficiency of Person A is 2x units/day. Step 3: Combined efficiency of A and B = 3x units/day. Step 4: Work done by A and B together in 28 days = 3x * 28 units. Step 5: Work done = 84x units. Step 6: Time taken by A alone to complete the work = 84x / 2x = 42 days.
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