Work & Wages Questions and Answers

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.

Answer: A

Combined one day's work = 1/3. Man's one day work = 1/6. Boy's one day work = 1/18. Woman's one day work = \(1/3 - (1/6+1/18) = 1/3 - (3+1)/18 = 1/3 - 4/18 = 1/3 - 2/9 = (3-2)/9 = 1/9\). A woman alone will take 9 days.

Answer: B

50% more efficient means X does the work in \( \tfrac{2}{3} \) of Y's time.

\[ t_{X} = \frac{2}{3} \times 30 = 20 \text{ days} \]

\[ \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{1}{12} \]

\[ t = 12 \text{ days} \]

Answer: B

Correct answer: Rs. 300

Wages divide in the ratio of work actually done, which for a common period is the ratio of the daily rates.

All three together complete the job in 5 days, so their combined rate is \(\tfrac{1}{5}\). Subtracting A and B gives C’s rate:

\[\frac{1}{5} - \frac{1}{15} - \frac{1}{12} = \frac{12}{60} - \frac{4}{60} - \frac{5}{60} = \frac{3}{60} = \frac{1}{20}\]

So the rates are \(\tfrac{1}{15} : \tfrac{1}{12} : \tfrac{1}{20}\), which on multiplying by 60 becomes \(4 : 5 : 3\).

\[\text{C’s share} = \frac{3}{4+5+3} \times 1200 = \frac{3}{12} \times 1200 = 300\]

Answer: C

Correct answer: 7.2 days

Add the daily rates rather than the times.

\[\frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36}\]

Working at five thirty-sixths of the job per day, they need

\[\frac{36}{5} = 7.2 \text{ days}\]

Answer: C

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

\[ \frac{1}{16} + \frac{1}{12} + \frac{1}{C} = \frac{1}{4} \]

\[ \frac{1}{C} = \frac{12}{48} - \frac{3 + 4}{48} = \frac{5}{48} \]

\[ C = \frac{48}{5} = 9\tfrac{3}{5} \text{ days} \]

Answer: A

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

Once \( \tfrac{1}{8} \) is done, \( \tfrac{7}{8} \) remains for A and B:

\[ \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{1}{4} \]

\[ t = \frac{7}{8} \div \frac{1}{4} = \frac{7}{8} \times 4 = 3.5 = 3\tfrac{1}{2} \text{ days} \]

C's rate never enters the calculation, since C leaves before the stretch being timed.

Answer: B

Let the original number of workers be x. Using the man-days formula M1*D1 = M2*D2. \(x \times 100 = (x-10) \times 110 \Rightarrow 100x = 110x - 1100 \Rightarrow 10x = 1100 \Rightarrow x = 110\). The original number of workers was 110.

Answer: D

Work done by A and B in 2 days = 2/10 = 1/5. Remaining work = 4/5. B completes 4/5 of the work in 12 days. So B can complete the full work in \(12 \times 5/4 = 15\) days. B's one day work is 1/15. (A+B)'s one day work is 1/10. A's one day work = \(1/10 - 1/15 = (3-2)/30 = 1/30\). A alone takes 30 days.

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.