Pipe A can fill a tank in 5 hours, Pipe B in 10 hours and Pipe C can empty it in 7.5 hours. If all three are opened together, the tank will be filled in:
Answer: A
Rate of Pipe A = 1/5 tank/hour. Rate of Pipe B = 1/10 tank/hour. Rate of Pipe C (emptying) = -1/7.5 = -2/15 tank/hour. When all are open, the combined rate = \(1/5 + 1/10 - 2/15\). The LCM of 5, 10, 15 is 30. Combined rate = \(\frac{6+3-4}{30} = \frac{5}{30} = \frac{1}{6}\) tank/hour. So, the tank will be filled in 6 hours.
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Pipe A can fill a tank in 5 hours, Pipe B in 10 hours and Pipe C can empty it in 7.5 hours. If all three are opened together, the tank will be filled in:
Answer: C
Net work done in 1 hour when all are open = \(1/5 + 1/10 - 1/7.5 = 1/5 + 1/10 - 2/15\). LCM of 5, 10, 15 is 30. So, \((6+3-4)/30 = 5/30 = 1/6\). But this means it fills in 6 hours. Let me re-read the question. 7.5 hours is 15/2. So C empties at a rate of 2/15. My calculation is correct. Why is the answer 30? Let me check LCM again. 5, 10, 15 -> 30. Correct. (6+3-4)/30 = 5/30 = 1/6. This is 6 hours. Let me check the options. Maybe the numbers were different. What if C empties in 20 hours? Then \(1/5+1/10-1/20 = (4+2-1)/20 = 5/20 = 1/4\). 4 hours. What if A fills in 10, B in 15, C empties in 6? \(1/10+1/15-1/6 = (3+2-5)/30 = 0\). Tank never fills. Let me re-calculate my initial expression: \(1/5+1/10-2/15 = (6+3-4)/30=5/30=1/6\). 6 hours. The options are wrong. Let's make one of the options 6 hours.
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A and B undertake to do a piece of work for Rs. 600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. Find the share of C.
Answer: A
Wages are distributed in the ratio of the work done. A's 1 day work = 1/6. B's 1 day work = 1/8. (A+B+C)'s 1 day work = 1/3. C's 1 day work = \(1/3 - (1/6+1/8) = 1/3 - 7/24 = (8-7)/24 = 1/24\). Ratio of work done by A:B:C = 1/6 : 1/8 : 1/24. Multiplying by 24 gives the ratio 4:3:1. C's share = \(\frac{1}{4+3+1} \times 600 = \frac{1}{8} \times 600 = 75\). So, C's share is Rs. 75.
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A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:
Answer: C
LCM of 24, 9, 12 is 72. Total work = 72 units. A's rate=3, B's rate=8, C's rate=6 units/day. B and C work for 3 days, so work done = \((8+6) \times 3 = 14 \times 3 = 42\) units. Remaining work = \(72 - 42 = 30\) units. Time taken by A to finish the remaining work = Remaining work / A's rate = 30 / 3 = 10 days.
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10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?
Answer: C
Let W be woman's work, C be child's work. From the problem, \(10W \times 7 = 10C \times 14 \Rightarrow 70W = 140C \Rightarrow W=2C\). Total work = \(10W \times 7 = 10(2C) \times 7 = 140C\). We need time for 5 women and 10 children. Their combined rate = \(5W+10C = 5(2C)+10C = 10C+10C=20C\). Time = Total work / Rate = 140C / 20C = 7 days.
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A and B can do a work in 45 and 40 days respectively. They began the work together, but A left after some days and B finished the remaining work in 23 days. After how many days did A leave?
Answer: C
Let the total work be LCM(45,40)=360 units. A's rate=8, B's rate=9 units/day. B worked alone for the last 23 days, so work done by B alone = \(23 \times 9 = 207\) units. Work done by A and B together = \(360 - 207 = 153\) units. Combined rate of A and B = 8+9=17 units/day. Time they worked together = 153/17 = 9 days. So, A left after 9 days.
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A and B can do a work in 10 days. They start the work together but A leaves after 2 days and B completes the rest of the work in 12 days. In how many days can A do the work alone?
Answer: D
Work done by A and B in 2 days = 2/10 = 1/5. Remaining work = 4/5. B completes 4/5 of the work in 12 days. So B can complete the full work in \(12 \times 5/4 = 15\) days. B's one day work is 1/15. (A+B)'s one day work is 1/10. A's one day work = \(1/10 - 1/15 = (3-2)/30 = 1/30\). A alone takes 30 days.
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Two pipes can fill a cistern in 14 and 16 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom it took 32 minutes more to fill the cistern. When the cistern is full, in what time will the leak empty it?
Answer: B
Time taken by two pipes to fill the tank = \(1/14 + 1/16 = (8+7)/112 = 15/112\). So they take 112/15 hours. Due to leakage, time taken = \(112/15 + 32/60 = 112/15 + 8/15 = 120/15 = 8\) hours. Let the leak empty the tank in x hours. So, \(1/14 + 1/16 - 1/x = 1/8 \Rightarrow 15/112 - 1/x = 1/8 \Rightarrow 1/x = 15/112 - 1/8 = (15-14)/112 = 1/112\). So, x = 112 hours.
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A and B are working on a job. A is twice as efficient as B. If A can finish the job in 30 days, in how many days can B finish the job alone?
Answer: C
Efficiency is inversely proportional to the time taken. If A is twice as efficient as B, A will take half the time B takes. So, Time taken by B = 2 * Time taken by A = 2 * 30 = 60 days.
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A can do a job in 16 days, and B can do the same job in 12 days. With the help of C, they did the job in 4 days only. In how many days can C do the job alone?
Answer: A
Work done by (A+B+C) in 1 day = 1/4. Work done by A in 1 day = 1/16. Work done by B in 1 day = 1/12. Work done by C in 1 day = \(1/4 - (1/16 + 1/12) = 1/4 - (3+4)/48 = 1/4 - 7/48 = (12-7)/48 = 5/48\). So, C alone can do the work in 48/5 = 9.6 days.
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