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Question 1 / 20
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1
A completes a task in 8 days by working 5 hours daily. B completes the same task in 10 days by working 6 hours daily. If A and B work together for 8 hours a day, in how many days will they jointly complete the work?
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Solution: Step 1: Calculate the total number of hours A takes to complete the work. - A's total hours = 8 days × 5 hours/day = 40 hours. Step 2: Calculate the total number of hours B takes to complete the work. - B's total hours = 10 days × 6 hours/day = 60 hours. Step 3: Calculate their 1-hour work rates. - A's 1-hour work rate = 1/40 of the work per hour. - B's 1-hour work rate = 1/60 of the work per hour. Step 4: Calculate their combined 1-hour work rate. - (A + B)'s 1-hour work rate = (1/40) + (1/60) - Find the LCM of 40 and 60, which is 120. - (A + B)'s 1-hour work rate = (3/120) + (2/120) = 5/120 = 1/24 of the work per hour. Step 5: This implies that A and B together can complete the entire work in 24 total hours. Step 6: If they work 8 hours a day, calculate the number of days needed to complete the work. - Number of days = Total hours needed / Hours worked per day = 24 hours / 8 hours/day = 3 days.
2
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.
3
A tank is filled in 5 hours by three pipes A, B, and C. Pipe C is twice as fast as B, and Pipe B is twice as fast as A. How much time will pipe A alone take to fill the tank?
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Solution: Step 1: Define the time taken by pipe A. Let Pipe A alone take 'X' hours to fill the tank. Step 2: Determine the time taken by pipes B and C based on their relative speeds. Pipe B is twice as fast as A, so Pipe B takes half the time of A = X/2 hours. Pipe C is twice as fast as B, so Pipe C takes half the time of B = (X/2)/2 = X/4 hours. Step 3: Determine the individual filling rates (parts of the tank filled per hour). Rate of A = 1/X tank/hour. Rate of B = 1/(X/2) = 2/X tank/hour. Rate of C = 1/(X/4) = 4/X tank/hour. Step 4: The combined filling time for all three pipes is 5 hours. So, the combined rate = 1/5 tank/hour. Step 5: Set up the equation for the combined rates. (Rate of A) + (Rate of B) + (Rate of C) = Combined Rate 1/X + 2/X + 4/X = 1/5 Step 6: Solve for X. (1 + 2 + 4) / X = 1/5 7 / X = 1/5 X = 7 * 5 X = 35. Step 7: Pipe A alone will take 35 hours to fill the tank.
4
A tap drips at a rate of one drop per second. Given that 600 drops constitute 100 ml, calculate the total volume of water, in liters, wasted over a period of 300 days.
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Solution: Step 1: Calculate the total number of seconds in 300 days. Seconds in 1 day = 24 hours * 60 minutes/hour * 60 seconds/minute = 86,400 seconds. Total seconds in 300 days = 86,400 seconds/day * 300 days = 25,920,000 seconds. Step 2: Calculate the total number of drops in 300 days (since it's 1 drop/second). Total drops = 25,920,000 drops. Step 3: Calculate the total volume in milliliters. Given: 600 drops = 100 ml. Volume per drop = 100 ml / 600 drops = 1/6 ml/drop. Total volume in ml = Total drops * Volume per drop = 25,920,000 drops * (1/6) ml/drop = 4,320,000 ml. Step 4: Convert the total volume from milliliters to liters. 1 liter = 1000 ml. Total volume in liters = 4,320,000 ml / 1000 ml/liter = 4320 liters.
5
6 men can complete a task in 12 days, 8 women can do the same task in 18 days, and 18 children can do it in 10 days. A team of 4 men, 12 women, and 20 children works for 2 days. If the remaining work is to be completed by men alone in 1 day, how many men will be needed?
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Solution: Step 1: Calculate the 1 day's work for 1 man, 1 woman, and 1 child. 1 man's 1 day's work = 1 / (6 * 12) = 1/72. 1 woman's 1 day's work = 1 / (8 * 18) = 1/144. 1 child's 1 day's work = 1 / (18 * 10) = 1/180. Step 2: Calculate the combined daily work of the initial team (4 men, 12 women, 20 children). (4 * 1/72) + (12 * 1/144) + (20 * 1/180) = 4/72 + 12/144 + 20/180 = 1/18 + 1/12 + 1/9. Step 3: Find the LCM for the denominators (18, 12, 9), which is 36. Combined daily work = (2/36) + (3/36) + (4/36) = 9/36 = 1/4. Step 4: This team works for 2 days. Work done in 2 days = 2 * (1/4) = 1/2. Step 5: Calculate the remaining work = 1 - 1/2 = 1/2. Step 6: This remaining 1/2 work must be completed by men only in 1 day. Step 7: Let 'N' be the number of men required. Their work in 1 day = N * (1 man's 1 day's work). Step 8: N * (1/72) = 1/2. Step 9: Solve for N: N = (1/2) * 72 = 36. Step 10: Therefore, 36 men will be required.
6
A is twice as efficient as B. If A and B together finish a task in 22 days, in how many days will A alone complete the same task?
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Solution: Step 1: Establish the efficiency ratio between A and B. Given A is twice as good a workman as B. So, Efficiency A : Efficiency B = 2 : 1. Step 2: Calculate their combined efficiency. Let A's efficiency be 2 units/day and B's efficiency be 1 unit/day. Combined efficiency of A + B = 2 + 1 = 3 units/day. Step 3: Calculate the total work. Total work = Combined efficiency * Days taken together. Total work = 3 units/day * 22 days = 66 units. Step 4: Calculate the time A alone takes to complete the work. Time taken by A alone = Total work / Efficiency of A = 66 units / 2 units/day = 33 days. Therefore, A alone will finish the same work in 33 days.
7
A takes 4 times the amount of time B requires, or 5 times the amount of time C requires, to complete a task. If A, B, and C work collaboratively, they can finish the work in 4 days. How many days would B alone need to complete the task?
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Solution: Step 1: Establish the relationship between their times. Time(A) = 4 * Time(B) Time(A) = 5 * Time(C) Step 2: Express their times in a common ratio. To find a common time unit, take the LCM of the coefficients of B and C for Time(A), i.e., LCM(4, 5) = 20. Let Time(A) = 20 units of time. Then Time(B) = Time(A) / 4 = 20 / 4 = 5 units of time. And Time(C) = Time(A) / 5 = 20 / 5 = 4 units of time. So, the ratio of times A:B:C = 20:5:4. Step 3: Determine their efficiency ratio (inverse of time ratio). Efficiency(A) : Efficiency(B) : Efficiency(C) = 1/20 : 1/5 : 1/4. Multiply by LCM(20,5,4)=20 to clear fractions: 1:4:5. Step 4: Calculate their combined efficiency. Let their efficiencies be 1k, 4k, and 5k units/day respectively. Combined efficiency (A+B+C) = 1k + 4k + 5k = 10k units/day. Step 5: Calculate the total work. They complete the task in 4 days together. Total work = Combined efficiency * Days = 10k * 4 = 40k units. Step 6: Calculate the time B alone takes to finish the work. Time for B = Total work / Efficiency(B) = 40k units / 4k units/day = 10 days.
8
A alone can complete a piece of work in 6 days, and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
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Solution: Step 1: Determine the individual daily work rates from the given information. A's 1 day's work = `1/6`. B's 1 day's work = `1/8`. (A + B + C)'s 1 day's work = `1/3` (since they complete the job in 3 days). Step 2: Calculate C's 1 day's work. C's 1 day's work = (A + B + C)'s 1 day's work - (A's 1 day's work + B's 1 day's work). `C's 1 day's work = 1/3 - (1/6 + 1/8)`. Step 3: Calculate the sum of A's and B's daily work. `1/6 + 1/8`. Find the LCM of 6 and 8, which is 24. `= (4/24) + (3/24) = 7/24`. Step 4: Substitute and calculate C's daily work. `C's 1 day's work = 1/3 - 7/24`. Find the LCM of 3 and 24, which is 24. `= (8/24) - (7/24) = 1/24`. Step 5: Determine the ratio of work done by A, B, and C over the 3 days they worked together. Wages are distributed in proportion to the work done. A's work in 3 days = `(1/6) * 3 = 3/6 = 1/2`. B's work in 3 days = `(1/8) * 3 = 3/8`. C's work in 3 days = `(1/24) * 3 = 3/24 = 1/8`. Ratio of work done (A:B:C) = `1/2 : 3/8 : 1/8`. To simplify, multiply by the LCM of denominators (8): `(1/2)*8 : (3/8)*8 : (1/8)*8 = 4 : 3 : 1`. Step 6: Calculate C's share of the total wages (Rs. 3200). Total ratio parts = `4 + 3 + 1 = 8`. C's share = `(C's ratio part / Total ratio parts) * Total Wages` C's share = `(1 / 8) * 3200 = Rs. 400`. Step 7: Conclusion: C should be paid Rs. 400.
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
A can write 75 pages in 25 hours. A and B combined can write 135 pages in 27 hours. How much time will B alone take to write 42 pages?
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Solution: Step 1: Calculate A's writing rate (efficiency). - A's rate = Total pages / Total hours = 75 pages / 25 hours = 3 pages/hour. Step 2: Calculate the combined writing rate of A and B. - (A + B)'s rate = Total pages / Total hours = 135 pages / 27 hours = 5 pages/hour. Step 3: Calculate B's individual writing rate. - B's rate = (A + B)'s rate - A's rate = 5 pages/hour - 3 pages/hour = 2 pages/hour. Step 4: Calculate the time B will take to write 42 pages. - Time for B = Desired pages / B's rate = 42 pages / 2 pages/hour = 21 hours.
11
A pump can fill a tank with water in 2 hours. However, due to a leak, it took 2 and 1/3 hours (7/3 hours) to fill the tank. How long would it take for the leak alone to drain all the water from the tank?
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Solution: Step 1: Calculate the pump's filling rate per hour: 1/2 of the tank per hour. Step 2: Calculate the effective filling rate per hour when the pump and leak are both active. The total time taken is 2 1/3 hours = 7/3 hours. So, the combined rate is 1 / (7/3) = 3/7 of the tank per hour. Step 3: The leak's emptying rate is the difference between the pump's filling rate and the combined filling rate (since the leak reduces the net filling speed): Leak's rate = (Pump's rate) - (Combined rate) Leak's rate = (1/2) - (3/7). Step 4: Find a common denominator (LCM of 2 and 7 is 14): Leak's rate = (7/14) - (6/14) = 1/14 of the tank per hour. Step 5: The time taken for the leak to drain the entire tank is the reciprocal of its emptying rate: Time = 1 / (1/14) = 14 hours. Step 6: The leak can drain the full tank in 14 hours.
12
A single tap can fill a tank in 6 hours. Once half of the tank is filled, three additional taps, identical to the first, are opened. What is the total duration required to fill the tank completely from an empty state?
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Solution: Step 1: The time taken by one tap to fill the entire tank is 6 hours. Step 2: The time taken by one tap to fill half of the tank is 6 hours / 2 = 3 hours. Step 3: After 3 hours, half the tank is filled, and the remaining half (1/2) needs to be filled. Step 4: Three more similar taps are opened, making a total of 1 + 3 = 4 taps working together. Step 5: The work rate of one tap is 1/6 of the tank per hour. Step 6: The combined work rate of 4 taps is 4 * (1/6) = 4/6 = 2/3 of the tank per hour. Step 7: Time taken by 4 taps to fill the remaining half of the tank = (Remaining part) / (Combined work rate). Step 8: Time = (1/2) / (2/3) = (1/2) * (3/2) = 3/4 hours. Step 9: Convert 3/4 hours to minutes: (3/4) * 60 minutes = 45 minutes. Step 10: Total time taken to fill the tank = (Time for first half) + (Time for second half) = 3 hours + 45 minutes = 3 hours 45 minutes.
13
A man's work rate is twice that of a woman, and a woman's work rate is twice that of a boy. If a man, a woman, and a boy all work together, they can complete a task in 7 days. How many days would it take for a boy to complete the work alone?
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Solution: Step 1: Establish the efficiency ratio for Man : Woman : Boy. - Let Boy's efficiency be 1 unit/day. - Woman's efficiency = 2 * Boy's efficiency = 2 units/day. - Man's efficiency = 2 * Woman's efficiency = 2 * 2 = 4 units/day. - Ratio of Efficiencies (Man : Woman : Boy) = 4 : 2 : 1. Step 2: Calculate their combined efficiency: 4 + 2 + 1 = 7 units/day. Step 3: Calculate the total work. They finish the work in 7 days. - Total work = Combined efficiency * Days = 7 units/day * 7 days = 49 units. Step 4: Calculate the time a boy alone would take to complete the work: Days for Boy = Total work / Boy's efficiency = 49 units / 1 unit/day = 49 days.
14
Person A can complete a work in 15 days, and Person B can complete it in 20 days. If they work together for 4 days, what fraction of the work remains unfinished?
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Solution: Step 1: Calculate A's 1-day work: 1/15. Step 2: Calculate B's 1-day work: 1/20. Step 3: Calculate their combined 1-day work: (1/15) + (1/20). Step 4: Find a common denominator (LCM of 15 and 20 is 60): (4/60) + (3/60) = 7/60. Step 5: Calculate the work done by A and B together in 4 days: (7/60) * 4 = 28/60 = 7/15. Step 6: Calculate the remaining work: 1 - (7/15) = (15/15) - (7/15) = 8/15.
15
A group of 4 men and 6 women can complete a work in 8 days. Another group of 3 men and 7 women can complete the same work in 10 days. How many days will 10 women alone take to complete the work?
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Solution: Step 1: Let 'm' be the 1-day work rate of 1 man and 'w' be the 1-day work rate of 1 woman. Step 2: Formulate equations based on the given information: Equation 1: 4m + 6w = 1/8 (since 4 men and 6 women complete the work in 8 days). Equation 2: 3m + 7w = 1/10 (since 3 men and 7 women complete the work in 10 days). Step 3: To eliminate 'm', multiply Equation 1 by 3 and Equation 2 by 4: (4m + 6w) * 3 = (1/8) * 3 => 12m + 18w = 3/8. (New Equation 1) (3m + 7w) * 4 = (1/10) * 4 => 12m + 28w = 4/10 = 2/5. (New Equation 2) Step 4: Subtract New Equation 1 from New Equation 2: (12m + 28w) - (12m + 18w) = (2/5) - (3/8). 10w = (16/40) - (15/40) = 1/40. Step 5: The expression '10w' represents the 1-day work of 10 women. So, 10 women's 1-day work is 1/40. Step 6: If 10 women's 1-day work is 1/40, then 10 women will complete the entire work in the reciprocal of this rate, which is 40 days.
16
One pipe can fill a container in 6 hours, while another pipe can empty the same container in 12 hours. If both pipes are opened simultaneously, what part of the container is filled by both pipes in 1 hour?
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Solution: Step 1: Calculate filling rates: Pipe 1 = 1/6 per hour, Pipe 2 = -1/12 per hour Step 2: Net rate when both pipes are open = 1/6 - 1/12 = (2/12 - 1/12) = 1/12 per hour Step 3: Part filled in 1 hour = 1/12 of the container
17
25 men and 10 boys complete a certain amount of work in 6 days. The same amount of work can be completed by 21 men and 30 boys in 5 days. How many boys are needed to assist 40 men to complete the same work in 4 days?
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Solution: Step 1: Let 1 man's 1 day's work be 'm' and 1 boy's 1 day's work be 'b'. Step 2: Total work from the first group: 6 * (25m + 10b). Step 3: Total work from the second group: 5 * (21m + 30b). Step 4: Equate the total work: 6(25m + 10b) = 5(21m + 30b). Step 5: Expand: 150m + 60b = 105m + 150b. Step 6: Rearrange to find the man-boy efficiency ratio: 45m = 90b. Step 7: Simplify the ratio: m = 2b (1 man's efficiency equals 2 boys' efficiency). Step 8: Calculate the Total Work in terms of 'b' using the first condition: Total Work = 6 * (25(2b) + 10b) = 6 * (50b + 10b) = 6 * 60b = 360b units. Step 9: We want to find the number of boys ('z') required to help 40 men complete the work in 4 days. Step 10: The new workforce's daily work = (40m + zb). Convert men to boys: (40 * 2b + zb) = (80b + zb) = (80 + z)b. Step 11: Total work done by this group in 4 days = 4 * (80 + z)b. Step 12: Equate this to the Total Work: 4 * (80 + z)b = 360b. Step 13: Divide by 'b': 4 * (80 + z) = 360. Step 14: Solve for z: 80 + z = 360 / 4 => 80 + z = 90. Step 15: z = 90 - 80 = 10. Step 16: Therefore, 10 boys must help 40 men.
18
Pipes A and B can fill a tank in 20 hours and 30 hours, respectively. Both pipes start filling the tank simultaneously. When the tank is one-third full, a leak forms that causes one-third of the water supplied by both pipes to drain out. What is the total time required to completely fill the tank?
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Solution: Step 1: Pipe A's filling rate = 1/20 tank per hour. Step 2: Pipe B's filling rate = 1/30 tank per hour. Step 3: Combined filling rate of (A + B) = (1/20) + (1/30). Step 4: Find a common denominator (LCM of 20 and 30 is 60): (3/60) + (2/60) = 5/60 = 1/12 tank per hour. Step 5: Time taken to fill the first 1/3 of the tank by (A + B) = (1/3) / (1/12) = (1/3) * 12 = 4 hours. Step 6: Remaining part of the tank to be filled = 1 - (1/3) = 2/3 of the tank. Step 7: After 4 hours, a leak develops such that one-third of the water supplied by both pipes goes out. Step 8: This means the effective filling rate is now 2/3 of the combined rate of (A + B). Step 9: New effective filling rate = (2/3) * (1/12) = 2/36 = 1/18 tank per hour. Step 10: Time taken to fill the remaining 2/3 of the tank with the new effective rate = (2/3) / (1/18) hours. Step 11: Calculate this time: (2/3) * 18 = 2 * 6 = 12 hours. Step 12: Total time taken to fill the tank = Time for first 1/3 + Time for remaining 2/3. Step 13: Total time = 4 hours + 12 hours = 16 hours.
19
Pipe A fills a tank in 38 hours, Pipe B fills it in 19 hours, and Pipe C empties the full tank in 133 hours. If all pipes are opened together, how long will it take to fill the tank?
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Solution: Step 1: Calculate individual rates: A = 1/38, B = 1/19, C = -1/133 tanks/hour Step 2: Combined rate = A + B + C = 1/38 + 1/19 - 1/133 Step 3: Find common denominator and simplify: Combined rate = (133 + 266 - 38) / (38 * 19 * 133) Step 4: Combined rate = 361 / (38 * 19 * 133) = 1/14 tanks/hour Step 5: Time to fill tank = 1 / (1/14) = 14 hours
20
Three pumps A, B, and C can fill a reservoir in 6 hours. After working together for 2 hours, C is shut off, and A and B complete the remaining work in 7 hours. How many hours would C take to fill the reservoir alone?
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Solution: Step 1: Combined rate (A+B+C) = 1/6 reservoir/hour Step 2: Work done in 2 hours = 2 * (1/6) = 1/3 Step 3: Remaining work = 1 - 1/3 = 2/3 Step 4: A+B rate = 2/3 work in 7 hours → rate = (2/3)/7 = 2/21/hour Step 5: C's rate = (1/6) - (2/21) = (7-4)/42 = 3/42 = 1/14/hour Step 6: Time for C alone = 1 / (1/14) = 14 hours
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