A can do a piece of work in 10 days and B can do it in 15 days. In how many days will they finish the work together?
Answer: B
Work in rates: someone who finishes a job in \( n \) days does \( \tfrac{1}{n} \) of it a day, and rates add.
\[ \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{1}{6} \]
\[ t = 6 \text{ days} \]
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If A, B, and C can complete a piece of work in 12, 15, and 20 days respectively, in how many days can they together complete the work?
Answer: B
Total work can be assumed as the LCM of 12, 15, and 20, which is 60 units. A's efficiency = 60/12 = 5 units/day. B's efficiency = 60/15 = 4 units/day. C's efficiency = 60/20 = 3 units/day. Together, their efficiency = 5+4+3 = 12 units/day. Time taken to complete the work together = Total work / Combined efficiency = 60 / 12 = 5 days.
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A can do a job in 16 days, and B can do the same job in 12 days. With the help of C, they did the job in 4 days only. In how many days can C do the job alone?
Answer: A
Work done by (A+B+C) in 1 day = 1/4. Work done by A in 1 day = 1/16. Work done by B in 1 day = 1/12. Work done by C in 1 day = \(1/4 - (1/16 + 1/12) = 1/4 - (3+4)/48 = 1/4 - 7/48 = (12-7)/48 = 5/48\). So, C alone can do the work in 48/5 = 9.6 days.
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A and B are working on a job. A is twice as efficient as B. If A can finish the job in 30 days, in how many days can B finish the job alone?
Answer: C
Twice the efficiency means half the time, so B takes twice as long as A.
\[ 2 \times 30 = 60 \text{ days} \]
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Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both pipes are opened simultaneously, how long will it take to fill the tank?
Answer: B
Work in rates: someone who finishes a job in \( n \) days does \( \tfrac{1}{n} \) of it a day, and rates add.
\[ \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{1}{12} \]
\[ t = 12 \text{ minutes} \]
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A works twice as fast as B. If both of them can together finish a work in 12 days, then in how many days can B alone do it?
Answer: D
Let the efficiency of B be x units/day. Then the efficiency of A is 2x units/day. Their combined efficiency is 3x units/day. Total work = Efficiency * Time = 3x * 12 = 36x units. Time taken by B alone = Total work / B's efficiency = 36x / x = 36 days.
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A and B can do a work in 8 days, B and C can do the same work in 12 days, and A, B, and C can do it in 6 days. In how many days can A and C do it?
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.
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If 10 men can do a piece of work in 12 days, how many men are needed to do the same 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.
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A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:
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.
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A and B undertake to do a piece of work for Rs. 600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. Find the share of C.
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.
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