Work & Wages Questions and Answers

Answer: B

Let B's efficiency be 2 units/day. Then A's is 1 unit/day. A+B together do 3 units/day. C does half of this, so C's efficiency is 1.5 units/day. C takes 20 days, so total work = \(1.5 \times 20 = 30\) units. All three together have an efficiency of 1+2+1.5 = 4.5 units/day. Time taken together = 30 / 4.5 = 300 / 45 = 20/3 = 6 2/3 days.

Answer: A

From the problem, 3 men = 6 women, so 1 man = 2 women. We need to find the time for 12 men and 8 women. Convert men to women: 12 men = 24 women. Total equivalent women = 24+8=32 women. We know 6 women take 16 days. Using M1D1=M2D2, \(6 \times 16 = 32 \times D2 \Rightarrow 96 = 32 \times D2 \Rightarrow D2 = 3\) days.

Answer: B

LCM of 6 and 4 is 12. Total work = 12 units. A's rate=2, B's rate=3 units/hour. In the first hour, A fills 2 units. In the second hour, B fills 3 units. In a 2-hour cycle, 5 units are filled. After 2 cycles (4 hours), 10 units are filled. Remaining work is 2 units. It is A's turn. A will fill the remaining 2 units in 1 hour. Total time = 4 + 1 = 5 hours.

Answer: C

A's writing rate = 75 pages / 25 hours = 3 pages/hour. (A+B)'s writing rate = 135 pages / 27 hours = 5 pages/hour. B's writing rate = (A+B)'s rate - A's rate = 5 - 3 = 2 pages/hour. Time for B to write 42 pages = Total pages / B's rate = 42 / 2 = 21 hours.

Answer: A

LCM of 18 and 24 is 72. Total work = 72 units. A's rate=4, B's rate=3 units/day. Work done in 8 days together = \((4+3) \times 8 = 56\) units. Remaining work = 72 - 56 = 16 units. Time for B to finish this = 16 / 3 = 5 1/3 days.

Answer: A

Pay follows the share of work done.

\[ \text{B's share} = 1 - \frac{19}{23} = \frac{4}{23} \]

\[ 575 \times \frac{4}{23} = \text{Rs. } 100 \]

Answer: B

From the problem, 1M = 2W = 3B. Let's express everything in terms of boys. 1M = 3B, 1W = 3/2 B. The combined work rate of 1 man, 1 woman, and 1 boy is \(3B + 3/2B + 1B = (6+3+2)/2 B = 11/2 B\). We know 3 boys take 44 days. Using M1D1=M2D2, \(3 \times 44 = (11/2) \times D2 \Rightarrow 132 = 5.5 \times D2 \Rightarrow D2 = 132 / 5.5 = 1320/55 = 24\) days.

Answer: B

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

All three together finish in 3 days, so their combined rate is \( \tfrac{1}{3} \).

\[ \frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24} \]

\[ C = \frac{1}{3} - \frac{7}{24} = \frac{8 - 7}{24} = \frac{1}{24} \]

C's share of the work is \( \tfrac{1/24}{1/3} = \tfrac{1}{8} \):

\[ \frac{3200}{8} = \text{Rs. } 400 \]

Answer: B

Time taken by two pipes to fill the tank = \(1/14 + 1/16 = (8+7)/112 = 15/112\). So they take 112/15 hours. Due to leakage, time taken = \(112/15 + 32/60 = 112/15 + 8/15 = 120/15 = 8\) hours. Let the leak empty the tank in x hours. So, \(1/14 + 1/16 - 1/x = 1/8 \Rightarrow 15/112 - 1/x = 1/8 \Rightarrow 1/x = 15/112 - 1/8 = (15-14)/112 = 1/112\). So, x = 112 hours.

Answer: C

A's share is ₹ 400.

Payment is shared in proportion to the work each does. In the 5 days the work takes, A does 5/12 of it and B does 5/15 = 1/3:

\[ \begin{aligned} C &= 1 - \tfrac{5}{12} - \tfrac{4}{12} = \tfrac{3}{12} \\ A &: B : C = 5 : 4 : 3 \end{aligned} \]

A's share is \(960 \times 5/12 = 400\), with B getting 320 and C 240. Sharing by the ratio of days (12 : 15) instead of work done gives the wrong amounts.