Mr. Stanley employed a certain number of typist for his project. 8 days later 20% of the typist left the job and it was found that it took as much time to complete the rest work from then as the entire work needed with all the employed typists. The average speed of a typist is 20 pages/hour. Minimum how many typist could be employed?

Solution
Since 20% (i.e. 1/5) typists left the job. So, there can be any value which is multiple of 5 i.e., whose 20% is always an integer. Hence, 5 is the least possible value.
A is twice efficient as B and together they do the same work in as much time as C and D together. If C and D can complete the work in 20 and 30 days respectively, working alone, then in how many days A can complete the work individually:

Solution
Ratio of efficiency
A + B = C + D     10x + 5x 9x + 6x ________ _________ 15x 15x
Therefore, ratio of efficiency of A : C = 10 : 9
Therefore, ratio of days taken by A : C = 9 : 10
Therefore, number of days taken by A = 18 days
There are three boats B1, B2 and B3 working together they carry 60 people in each trip. One day an early morning B1 carried 50 people in few trips alone. When it stopped carrying the passengers B2 and B3 started carrying the people together. It took a total of 10 trips to carry 300 people by B1, B2 and B3. It is known that each day on an average 300 people cross the river using only one of the 3 boats B1, B2 and B3. How many trips it would take to B1, to carry 150 passengers alone?

Solution
Combined efficiency of all the three boats = 60 passenger/trip
Now, consider option(a)
15 trips and 150 passengers means efficiency of B1 = 10 p/t
which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10 − 5 = 5) in which B2 and B3 together carried remaining 250 (300 − 50 = 250) Passengers.
Therefore the efficiency of B2 and B3 = \(\frac{250}{5}\) = 50 p/t
Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct
A can finish a work in 15 days and B can do the same work in 12 days. B worked for 8 days and left the job. In how many days, A alone can finish the remaining work?

Solution
B's 8 days work= \(\frac{1}{2} \times 8 = \frac{2}{3}\)
Remaining work = \(1  \frac{2}{3} = \frac{1}{3}\)
Now, \(\frac{1}{15}\) work is done by A in 1 day
∴ \(\frac{1}{3}\) work is done by A in (15 × 1/3 ) = 5 days
Pipe A can fill the tank in 4 hours,while pipe B can fill it in 6 hours working separately.pipe C can empty whole the tank in 4 hours. He opened the pipe A and B simultaneously to fill the empty tank. He wanted to adjust his alarm so that he could open the pipe C when it was halffilled, but he mistakenly adjusted his alarm at a time when his tank would be 3/4th filled. what is the time difference between both the cases, to fill the tank fully:

Solution
In ideal Case:
Time taken to fill the half tank by A and B = \(\frac{50}{41.66} = \frac{6}{5}\) hours
Time taken by A,B and C to fill rest half of the tank = \(\frac{50}{16.66}\) = 3 hours
Total time = \(\frac{6}{5} + 3\) = 4 hours 12 min
In second case:
Time taken to fill \(\frac{3}{4}\) tank by A and B = \(\frac{75}{41.66} = \frac{3}{2}\) hours
Time taken by A,B and C to fill rest \(\frac{1}{4}\) tank = \(\frac{25}{16.66} = \frac{3}{2}\) hours
Total time = \(\frac{9}{5} + \frac{3}{2}\) =3 hours 18 min
Therefore, difference in time = 54 minutes
A can do a piece of work in 18 days, B in 27 days, C in 36 days. They start worked together . But only C work till the completion of work. A leaves 4 days and B leaves 6 days before the completion of work. In how many days work be completed?

Solution
Let the work be completed in x days
(x − 4) days of A + (x − 6)days of B + x days of C = 1
⇒ \(\frac{x  4}{18} + \frac{x  6}{27} + \frac{x}{36} = 1\)
⇒ \(\frac{13x  48}{108} = 1\)
⇒ x = 12
∴ Total time = 12 days
A and B can do a piece of work in 40 and 50 days. If they work at it an alternate days with A beginning in how many days, the work will be finished ?

Solution
(A + B)'s two days work = \(\frac{1}{40} + \frac{1}{50} = \frac{9}{200}\)
Evidently, the work done by A and B during 22 pairs of days
i.e. in 44 days = \(22 \times \frac{9}{200} = \frac{198}{200}\)
Remaining work = \(1  \frac{9}{200} = \frac{1}{100}\)
Now on 45th day A will have the turn to do \(\frac{1}{200}\) of the work and this work A will do in \(40 \times \frac{1}{100}  \frac{2}{3}\)
∴ Total time taken = \(44\frac{2}{5}\) days
Two pipes A and B can fill a cistern in 4 minutes and 6 minutes respectively. If these pipes are turned on alternately for 1 minute each how long will it take to the cistern to fill?

Solution
As the pipes are operating alternatively, thus their 2 minutes job is = \(\frac{1}{4} + \frac{1}{6} = \frac{5}{12}\)
In the next 2 minutes the pipes can fill another \(\frac{5}{12}\) part of cistern.
∴ In 4 minutes the two pipes which are operating alternatively will fill \(\frac{5}{12} + \frac{5}{12} = \frac{5}{6}\)
Remaining part = \( 1  \frac{5}{6} = \frac{1}{6}\)
Pipe A can fill \(\frac{1}{4}\) of the cistern in 1 minute
Pipe A can fill \(\frac{1}{6}\) of the cistern in = \(4 \times \frac{1}{6} = \frac{2}{3}\) min
∴ Total time taken to fill the Cistern
\(4 + \frac{2}{3} = 4\frac{2}{3}\) minutes
Two pipes A and B can fill a tank in 24 hours and \(\frac{120}{7}\) hours respectively. Harihar opened the pipes A and B to fill an empty tank and some times later he closed the taps A and B , when the tank was supposed to be full. After that it was found that the tank was emptied in 2.5 hours because an outlet pipe “C” connected to the tank was open from the beginning. If Harihar closed the pipe C instead of closing pipes A and B the remaining tank would have been filled in :

Solution
Efficiency of Inlet pipe A = 4.16% \(\left ( \frac{100}{24} \right )\)
Efficiency of Inlet pipe B = 5.83% \(\left ( \frac{100 \times 7}{120} \right )\)
∴ Efficiency of A and B together = 100 %
Now, if the efficiency of outlet pipe be x % then in 10 hours the capacity of tank which will be filled
= 10 × (10 − x)Now, since this amount of water is being emptied by C at x % per hour, then
\(\frac{10 \times \left ( 10  x \right )}{x} = 2.5\) hours ⇒ x = 8 %
Therefore, in 10 hours 20% tank is filled only. Hence, the remaining 80% of the capacity will be filled by pipes A and B in \(\frac{80}{10}\) = 8 hours
Amit can do a piece of work in 45 days, but Bharath can do the same work in 5 days less, than Amit, when working alone. Amit and Bharath both started the work together but Bharath left after some days and Amit finished the remaining work in 56 days with half of his efficiency but he did the work with Bharath with his complete efficiency. For how many days they had worked together?

Solution
Amit Bharath No. of days 45 40 Efficiency 2.22 % (= 1/45) 2.5 % (= 1/40) Amit did the work in 56 days = \(56 \times \frac{1}{45 \times 2} = \frac{28}{45}\)
∴ Rest work \(\left ( \frac{17}{45} \right )\) was done by Amit and Bharath = \(\frac{17 \times 360}{45 \times 17}\) = 8 days
since Amit and Bharath do the work in one day = \(\frac{1}{45} + \frac{1}{40} = \frac{17}{360}\)