A can do a piece of work in 10 days and B can do it in 15 days. In how many days will they finish the work together?
Answer: B
A's one day's work = 1/10. B's one day's work = 1/15. Together, their one day's work = \(1/10 + 1/15 = (3+2)/30 = 5/30 = 1/6\). So, they will finish the work together in 6 days.
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If A, B, and C can complete a piece of work in 12, 15, and 20 days respectively, in how many days can they together complete the work?
Answer: B
Total work can be assumed as the LCM of 12, 15, and 20, which is 60 units. A's efficiency = 60/12 = 5 units/day. B's efficiency = 60/15 = 4 units/day. C's efficiency = 60/20 = 3 units/day. Together, their efficiency = 5+4+3 = 12 units/day. Time taken to complete the work together = Total work / Combined efficiency = 60 / 12 = 5 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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If 1 man or 2 women or 3 boys can do a work in 44 days, then in how many days will 1 man, 1 woman and 1 boy do the work?
Answer: B
From the problem, 1M = 2W = 3B. Let's express everything in terms of boys. 1M = 3B, 1W = 3/2 B. The combined work rate of 1 man, 1 woman, and 1 boy is \(3B + 3/2B + 1B = (6+3+2)/2 B = 11/2 B\). We know 3 boys take 44 days. Using M1D1=M2D2, \(3 \times 44 = (11/2) \times D2 \Rightarrow 132 = 5.5 \times D2 \Rightarrow D2 = 132 / 5.5 = 1320/55 = 24\) days.
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A is twice as efficient as B and B is thrice as efficient as C. If they together can complete the work in 10 days, in how many days can A alone do it?
Answer: B
Let C's efficiency be x. Then B's is 3x and A's is 2*(3x)=6x. Their combined efficiency is x+3x+6x = 10x. Total work = \(10x \times 10 = 100x\). Time taken by A alone = Total work / A's efficiency = \(100x / 6x = 100/6 = 50/3 = 16.67\) days.
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A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long would it take for B alone to complete the work?
Answer: B
A completes 80% (4/5) of the work in 20 days. So, A can complete the whole work in \(20 \times (5/4) = 25\) days. A's 1-day work is 1/25. Remaining work is 20% (1/5). A and B together finish 1/5 of the work in 3 days. So, their combined 1-day work is \((1/5)/3 = 1/15\). B's 1-day work = (A+B)'s work - A's work = \(1/15 - 1/25 = (5-3)/75 = 2/75\). So, B alone can do the work in 75/2 = 37.5 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, B and C are employed to do a piece of work for Rs. 529. A and B together are supposed to do 19/23 of the work and B and C together 8/23 of the work. What amount should A be paid?
Answer: B
Wages are paid in proportion to the work done. Work done by (A+B) = 19/23. Work done by C = \(1 - 19/23 = 4/23\). Work done by (B+C) = 8/23. Work done by B = (B+C) - C = 8/23 - 4/23 = 4/23. Work done by A = (A+B) - B = 19/23 - 4/23 = 15/23. A's share = \((15/23) \times 529 = 15 \times 23 = 345\). A should be paid Rs. 345.
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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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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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