Work & Wages Questions and Answers

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.

Answer: D

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

\[ 4\left(\frac{1}{15} + \frac{1}{20}\right) = 4 \times \frac{4 + 3}{60} = \frac{28}{60} = \frac{7}{15} \]

\[ \text{left} = 1 - \frac{7}{15} = \frac{8}{15} \]

Answer: B

LCM of 20, 30, 60 is 60. Total work = 60 units. A's rate=3, B's rate=2, C's rate=1 unit/day. A works alone for the first two days, doing \(3+3=6\) units. On the third day, A, B, and C work together, doing \(3+2+1=6\) units. So, in a 3-day cycle, \(6+6=12\) units of work are done. To complete 60 units, we need \(60/12 = 5\) such cycles. The total time taken is \(5 \times 3 = 15\) days.

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.

Answer: B

A is 7/4 times as efficient as B. The ratio of their efficiencies A:B is 7:4. Combined efficiency = 11 units (say). Total work = \(11 \times 7 = 77\) units. Time for A alone = Total work / A's efficiency = 77 / 7 = 11 days.

Answer: C

Let W be woman's work, C be child's work. From the problem, \(10W \times 7 = 10C \times 14 \Rightarrow 70W = 140C \Rightarrow W=2C\). Total work = \(10W \times 7 = 10(2C) \times 7 = 140C\). We need time for 5 women and 10 children. Their combined rate = \(5W+10C = 5(2C)+10C = 10C+10C=20C\). Time = Total work / Rate = 140C / 20C = 7 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.

\[ \frac{1}{15} - \frac{1}{20} = \frac{4 - 3}{60} = \frac{1}{60} \]

\[ t = 60 \text{ minutes} \]