Work & Wages Questions and Answers

Answer: C

One day's work: (A+B)=1/8, (B+C)=1/12, (A+B+C)=1/6. A's work = (A+B+C) - (B+C) = 1/6 - 1/12 = 1/12. C's work = (A+B+C) - (A+B) = 1/6 - 1/8 = (4-3)/24 = 1/24. (A+C)'s one day work = 1/12 + 1/24 = (2+1)/24 = 3/24 = 1/8. So, A and C can do the work in 8 days.

Answer: B

The total work is fixed in man-days.

\[ 10 \times 12 = 120 \text{ man-days} \]

\[ \frac{120}{8} = 15 \text{ men} \]

Men and days are inversely proportional for a fixed job.

Answer: C

LCM of 24, 9, 12 is 72. Total work = 72 units. A's rate=3, B's rate=8, C's rate=6 units/day. B and C work for 3 days, so work done = \((8+6) \times 3 = 14 \times 3 = 42\) units. Remaining work = \(72 - 42 = 30\) units. Time taken by A to finish the remaining work = Remaining work / A's rate = 30 / 3 = 10 days.

Answer: A

Wages are distributed in the ratio of the work done. A's 1 day work = 1/6. B's 1 day work = 1/8. (A+B+C)'s 1 day work = 1/3. C's 1 day work = \(1/3 - (1/6+1/8) = 1/3 - 7/24 = (8-7)/24 = 1/24\). Ratio of work done by A:B:C = 1/6 : 1/8 : 1/24. Multiplying by 24 gives the ratio 4:3:1. C's share = \(\frac{1}{4+3+1} \times 600 = \frac{1}{8} \times 600 = 75\). So, C's share is Rs. 75.

Answer: C

Correct answer: 30 hours

Work in tanks per hour, treating the emptying pipe as negative.

\[\frac{1}{15} + \frac{1}{20} - \frac{1}{12}\]

Over a common denominator of 60:

\[\frac{4}{60} + \frac{3}{60} - \frac{5}{60} = \frac{2}{60} = \frac{1}{30}\]

The tank fills at one-thirtieth per hour, so it takes 30 hours. The net rate is positive, which is why it fills at all - had C been faster than A and B combined, the tank would never fill.

Answer: A

Rate of Pipe A = 1/5 tank/hour. Rate of Pipe B = 1/10 tank/hour. Rate of Pipe C (emptying) = -1/7.5 = -2/15 tank/hour. When all are open, the combined rate = \(1/5 + 1/10 - 2/15\). The LCM of 5, 10, 15 is 30. Combined rate = \(\frac{6+3-4}{30} = \frac{5}{30} = \frac{1}{6}\) tank/hour. So, the tank will be filled in 6 hours.

Answer: C

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

\[ A = \frac{1}{4}, \qquad B + C = \frac{1}{3}, \qquad A + C = \frac{1}{2} \]

\[ C = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \]

\[ B = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \]

\[ t_{B} = 12 \text{ hours} \]

Answer: C

Let 1 man's 1 day's work be 'm' and 1 boy's 1 day's work be 'b'. From the problem, \(10(2m+3b)=1\) and \(8(3m+2b)=1\). So, \(20m+30b=24m+16b \Rightarrow 4m=14b \Rightarrow m=3.5b\). Substitute this in the first equation: \(2(3.5b)+3b = 1/10 \Rightarrow 7b+3b=1/10 \Rightarrow 10b=1/10 \Rightarrow b=1/100\). Then \(m=3.5/100 = 7/200\). We need to find the time for 2 men and 1 boy. Their combined work per day is \(2m+1b = 2(7/200) + 1/100 = 14/200 + 2/200 = 16/200 = 2/25\). So, they will take 25/2 = 12.5 days.

Answer: B

Let B's efficiency be 10 units/day. Then A's efficiency is 13 units/day. A alone takes 23 days, so total work = \(13 \times 23 = 299\) units. Working together, their efficiency is 10+13 = 23 units/day. Time taken together = Total work / Combined efficiency = 299 / 23 = 13 days.

Answer: C

From the problem, 5 men = 7 women. So, 1 man = 7/5 women. We need to find the earnings of 7 men and 13 women. Let's convert everything to women's earnings. \(7 \text{ men} = 7 \times (7/5) \text{ women} = 49/5 \text{ women}\). Total women = \(13 + 49/5 = (65+49)/5 = 114/5\) women. 7 women earn Rs. 5250 per day. So 1 woman earns \(5250/7 = 750\) per day. Earnings of \(114/5\) women = \((114/5) \times 750 = 114 \times 150 = 17100\). So they earn Rs. 17,100 per day.