Work & Wages Questions and Answers

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.

Answer: C

One day's work: A+B=1/12, B+C=1/15, C+A=1/20. Adding these three gives \(2(A+B+C) = 1/12 + 1/15 + 1/20 = (5+4+3)/60 = 12/60 = 1/5\). So, \(A+B+C = 1/10\). To find A's work, we subtract (B+C)'s work: \(A = (A+B+C) - (B+C) = 1/10 - 1/15 = (3-2)/30 = 1/30\). A alone will take 30 days.

Answer: C

Work in rates: someone who finishes a job in \( n \) days does \( \tfrac{1}{n} \) of it a day, and rates add.

\[ \frac{1}{8} - \frac{1}{16} = \frac{2 - 1}{16} = \frac{1}{16} \]

\[ t = 16 \text{ hours} \]

Answer: B

Let the time taken by A be T days. Then B takes 3T days. The difference in time is \(3T - T = 2T = 60\) days. So, \(T=30\). A takes 30 days and B takes 90 days. Their combined one day's work is \(1/30 + 1/90 = (3+1)/90 = 4/90 = 2/45\). So, together they will take 45/2 = 22.5 days.

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.