A can do 1/3 of a work in 5 days and B can do 2/5 of the work in 10 days. In how many days both A and B together can do the work?
Answer: C
A can do the full work in \(5 \times 3 = 15\) days. B can do the full work in \(10 \times 5/2 = 25\) days. A's 1 day work = 1/15. B's 1 day work = 1/25. Together, their 1 day work = \(1/15 + 1/25 = (5+3)/75 = 8/75\). They will take 75/8 = 9 3/8 days.
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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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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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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 and B can finish a work in 30 days. They worked at it for 20 days and then B left. The remaining work was done by A alone in 20 more days. A alone can finish the work in:
Answer: B
Work done by A and B in 20 days = \(20/30 = 2/3\). Remaining work = \(1 - 2/3 = 1/3\). This remaining work (1/3) is done by A in 20 days. So, A can do the entire work in \(20 \times 3 = 60\) days. Wait, let me re-check. A does 1/3 work in 20 days, so full work in 60 days. Why is the answer 50? Let me re-read. (A+B)'s 1 day work = 1/30. They work for 20 days, doing 20/30 = 2/3 of the work. Remaining work is 1/3. A does this 1/3 work in 20 days. So A's capacity is (1/3)/20 = 1/60 of the work per day. So A alone takes 60 days. The answer should be 60. Let me adjust the options.
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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 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 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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