Work & Wages Questions and Answers

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.

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.

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.

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.

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.

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.

Answer: B

Correct answer: 60 days

Take the whole job as 1 unit. A and B together do \(\tfrac{1}{30}\) of it per day, so in 20 days they complete

\[20 \times \frac{1}{30} = \frac{2}{3}\]

That leaves \(1 - \tfrac{2}{3} = \tfrac{1}{3}\) of the work, which A finishes alone in 20 days. So A’s daily rate is

\[\frac{1/3}{20} = \frac{1}{60}\]

Working at one-sixtieth per day, A alone would need 60 days for the whole job.

Answer: C

Correct answer: 18 days

Let B do 1 unit of work a day, so A does 2 and together they do 3.

\[\text{Total work} = 3 \times 12 = 36 \text{ units}\]

\[\text{A alone} = \frac{36}{2} = 18 \text{ days}\]

B alone would take 36 days, twice as long, as the ratio of ability requires.

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.

Answer: A

Total work = \(12 \times 9 = 108\) man-days. Work done in 6 days by 12 men = \(12 \times 6 = 72\) man-days. Remaining work = \(108 - 72 = 36\) man-days. Now, the number of men is \(12+6=18\). Time taken to complete the remaining work = Remaining work / Number of men = 36 / 18 = 2 days.