If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
Answer: A
Let 1 man's work = m, 1 boy's work = b. From the problem, \(10(6m+8b) = 2(26m+48b) \Rightarrow 60m+80b = 52m+96b \Rightarrow 8m=16b \Rightarrow m=2b\). Total work can be calculated from the first condition: \(10(6(2b)+8b) = 10(12b+8b) = 10(20b) = 200b\). We need time for 15 men and 20 boys. Their combined work rate is \(15m+20b = 15(2b)+20b = 30b+20b = 50b\). Time taken = Total work / Rate = 200b / 50b = 4 days.
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A contractor undertakes to do a piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the additional men, how many days behind schedule would it be finished?
Answer: A
Let the work done by 1 man in 1 day be 1 unit. Total work = Work done in first 35 days + Work done in last 5 days = \(100 \times 35 + (100+100) \times 5 = 3500 + 200 \times 5 = 3500 + 1000 = 4500\) units. If the additional men were not engaged, the 100 men would have to do the entire work. Time taken = Total work / Number of men = 4500 / 100 = 45 days. The work would be finished 5 days behind the scheduled 40 days.
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Three men, four women and six children can complete a work in 7 days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days?
Answer: A
Let a man's one day work be x. Then a woman's work is 2x and a child's work is x/2. Total work done in 1 day = \(3x + 4(2x) + 6(x/2) = 3x+8x+3x = 14x\). Total work = \(14x \times 7 = 98x\). We need to find how many women can complete this in 7 days. Let the number of women be W. Their one day's work is \(W \times 2x\). Total work done by them in 7 days = \(W \times 2x \times 7 = 14Wx\). Equating the total work: \(14Wx = 98x \Rightarrow 14W=98 \Rightarrow W=7\). So, 7 women are needed.
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A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:
Answer: C
Let their 1 day's work be A, B, C. Given A = B+C. Also, A+B=1/10 and C=1/50. Substitute C in the first equation: A = B + 1/50. Substitute this A into the second equation: \((B+1/50) + B = 1/10 \Rightarrow 2B = 1/10 - 1/50 = (5-1)/50 = 4/50 = 2/25\). So, \(B = 1/25\). B alone can do the work in 25 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 alone would take 27 hours more to complete a work than A and B together. B takes 3 hours more to complete a work alone than A and B together. In how many days can A alone do it?
Answer: B
Let the time taken by A and B together be x hours. Then A takes x+27 hours and B takes x+3 hours. Their combined one hour's work is \(\frac{1}{x+27} + \frac{1}{x+3} = \frac{1}{x}\). This gives \(x(x+3+x+27) = (x+27)(x+3) \Rightarrow x(2x+30) = x^2+30x+81 \Rightarrow 2x^2+30x = x^2+30x+81 \Rightarrow x^2=81 \Rightarrow x=9\). Time taken by A alone = x+27 = 9+27 = 36 hours.
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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 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: C
The work that (A+B) could have done in the remaining 10 days (30-20) was done by A in 20 days. So, \(10(A+B) = 20A \Rightarrow 10A+10B = 20A \Rightarrow 10B = 10A \Rightarrow A=B\). This means they have equal efficiency. If they together take 30 days, one person would take double the time, i.e., 60 days. So A alone can finish the work in 60 days.
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P can complete a work in 12 days working 8 hours a day. Q can complete the same work in 8 days working 10 hours a day. If both P and Q work together, working 8 hours a day, in how many days can they complete the work?
Answer: A
Total hours taken by P = \(12 \times 8 = 96\) hours. Total hours taken by Q = \(8 \times 10 = 80\) hours. P's 1 hour work = 1/96. Q's 1 hour work = 1/80. Together, their 1 hour work = \(1/96 + 1/80 = (5+6)/480 = 11/480\). They will take 480/11 hours to complete the work. Since they work 8 hours a day, the number of days = \((480/11) / 8 = 60/11 = 5 \frac{5}{11}\) days.
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