A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 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.
Enter details here
If 1 man or 2 women or 3 boys can do a work in 44 days, then in how many days will 1 man, 1 woman and 1 boy do the work?
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.
Enter details here
A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long would it take for B alone to complete the work?
Answer: B
A completes 80% (4/5) of the work in 20 days. So, A can complete the whole work in \(20 \times (5/4) = 25\) days. A's 1-day work is 1/25. Remaining work is 20% (1/5). A and B together finish 1/5 of the work in 3 days. So, their combined 1-day work is \((1/5)/3 = 1/15\). B's 1-day work = (A+B)'s work - A's work = \(1/15 - 1/25 = (5-3)/75 = 2/75\). So, B alone can do the work in 75/2 = 37.5 days.
Enter details here
A and B can do a work in 12 days. B and C in 15 days. C and A in 20 days. How much time will A alone take to do it?
Answer: C
One day's work: A+B=1/12, B+C=1/15, C+A=1/20. Adding these three gives \(2(A+B+C) = 1/12 + 1/15 + 1/20 = (5+4+3)/60 = 12/60 = 1/5\). So, \(A+B+C = 1/10\). To find A's work, we subtract (B+C)'s work: \(A = (A+B+C) - (B+C) = 1/10 - 1/15 = (3-2)/30 = 1/30\). A alone will take 30 days.
Enter details here
A tap can fill a cistern in 8 hours and another can empty it in 16 hours. If both are opened together, the time (in hours) to fill the tank is:
Answer: C
The filling tap's rate is 1/8 tank/hour. The emptying tap's rate is -1/16 tank/hour. The net rate is \(1/8 - 1/16 = (2-1)/16 = 1/16\) tank/hour. So, the tank will be filled in 16 hours.
Enter details here
A is thrice as good a workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:
Answer: B
Let the time taken by A be T days. Then B takes 3T days. The difference in time is \(3T - T = 2T = 60\) days. So, \(T=30\). A takes 30 days and B takes 90 days. Their combined one day's work is \(1/30 + 1/90 = (3+1)/90 = 4/90 = 2/45\). So, together they will take 45/2 = 22.5 days.
Enter details here
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
Answer: A
Let 1 man's work = m, 1 boy's work = b. From the problem, \(10(6m+8b) = 2(26m+48b) \Rightarrow 60m+80b = 52m+96b \Rightarrow 8m=16b \Rightarrow m=2b\). Total work can be calculated from the first condition: \(10(6(2b)+8b) = 10(12b+8b) = 10(20b) = 200b\). We need time for 15 men and 20 boys. Their combined work rate is \(15m+20b = 15(2b)+20b = 30b+20b = 50b\). Time taken = Total work / Rate = 200b / 50b = 4 days.
Enter details here
A contractor undertakes to do a piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the additional men, how many days behind schedule would it be finished?
Answer: A
Let the work done by 1 man in 1 day be 1 unit. Total work = Work done in first 35 days + Work done in last 5 days = \(100 \times 35 + (100+100) \times 5 = 3500 + 200 \times 5 = 3500 + 1000 = 4500\) units. If the additional men were not engaged, the 100 men would have to do the entire work. Time taken = Total work / Number of men = 4500 / 100 = 45 days. The work would be finished 5 days behind the scheduled 40 days.
Enter details here
Three men, four women and six children can complete a work in 7 days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days?
Answer: A
Let a man's one day work be x. Then a woman's work is 2x and a child's work is x/2. Total work done in 1 day = \(3x + 4(2x) + 6(x/2) = 3x+8x+3x = 14x\). Total work = \(14x \times 7 = 98x\). We need to find how many women can complete this in 7 days. Let the number of women be W. Their one day's work is \(W \times 2x\). Total work done by them in 7 days = \(W \times 2x \times 7 = 14Wx\). Equating the total work: \(14Wx = 98x \Rightarrow 14W=98 \Rightarrow W=7\). So, 7 women are needed.
Enter details here
Two pipes can fill a cistern in 14 and 16 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom it took 32 minutes more to fill the cistern. When the cistern is full, in what time will the leak empty it?
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.
Enter details here