A takes twice as much time as B or thrice as much time to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone is:

Solution
Suppose A, B and C take x, \(\frac{x}{2}\) and \(\frac{x}{3}\) hours respectively to finish the work.
Then, \(\left ( \frac{1}{x} + \frac{2}{x} + \frac{3}{x} \right ) = \frac{1}{2}\) ⇒ x = 12.
So, B takes 6 hours to finish the work.
P, Q and R are three typists who working simultaneously can type 216 pages in 4 hours. In one hour, R can type as many pages more than Q as Q can type more than P. During a period of five hours, R can type as many pages as P can during seven hours. How many pages does each of them type per hour ?

Solution
Let the number of pages typed in one hour by P, Q and R be x, y and z respectively.
Then,
x + y + z = \(\frac{216}{4}\) ⇒ x + y + z = 54 (i)
z − y = y − x ⇒ 2y = x+ z (ii)
5z = 7x ⇒ \(x = \frac{5}{7}z\) (iii)
Solving (i) (ii) and (iii),we get x=15, y=18, z=21.
A dose a work in 10 days and B does the same work in 15 days. In how many days they together will do the same work ?

Solution
A's 1 day's work = \(\frac{1}{10}\) and B's 1 day's work = \(\frac{1}{15}\)
∴ (A + B)'s 1 day's work = \(\left ( \frac{1}{10} + \frac{1}{15} \right ) = \frac{1}{6}\)
So, both together can finish the work in 6 day's.
A takes twice as much time as B or thrice as much time to finish a piece of work working together, they can finish the work in 2 days. B can do the work alone in :

Solution
Suppose A, B and C take x, \(\frac{x}{2}\) and \(\frac{x}{3}\) hours respectively to finish the work.
Then, \(\frac{1}{x} + \frac{2}{x} + \frac{3}{x} = \frac{1}{2}\) ⇒ \(\frac{6}{x} = \frac{1}{2}\) ⇒ x = 12
So, B takes 6 hour to finish the work.
A and B can do a work in 12 days, B and C in 15 days, C and A in 20 days. If A, B and C work together, they will complete the work in:

Solution
(A + B)'s 1 day's work = \(\frac{1}{12}\); (B + C)'s 1 day's work = \(\frac{1}{15}\); (A + C)'s 1 day's work = \(\frac{1}{20}\).
Adding, we get : 2(A + B + C)'s a day's work = \(\frac{1}{12} + \frac{1}{15} + \frac{1}{20} = \frac{12}{60} = \frac{1}{5}\)
∴ (A + B + C)'s 1 day's work = \(\frac{1}{10}\)
So, A , B and C together can complete the work in 10 days.
A is thrice efficient as B and C is twice as efficient as B. what is the ratio of number of days taken by A,B and C, when they work individually?

Solution
A : B : C Ratio of efficiency 3 : 1 : 2 Ratio of No.of days \(\frac{1}{3}\) : \(\frac{1}{1}\) : \(\frac{1}{2}\) or 2 : 6 : 3 Hence A is correct.
∵ \(Time \prec \frac{1}{Efficiency}\)
A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in 55% of the time. How many workers were there in the group?

Solution
It can be solved easily through option.
\(\left ( 10 + 9 + 8 + .... + 1 \right ) = 10 \left ( 10 \times \frac{55}{100} \right )\)
55 = 55 Hence correct
Alternatively:
\(\frac{n\left ( n+1 \right )}{2} = n \times \frac{55n}{100}\)
⇒ n = 10
In Both cases total work is 55 mandays.
A Contractor employed a certain number of workers to finish constructing a road in a certain scheduled time. Sometime later, when a part of work had been completed, he realized that the work would get delayed by threefourth of the scheduled time, so he at once doubled the no of workers and thus he managed to finish the road on the scheduled time. How much work he had been completed, before increasing the number of workers?

Solution
Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100D) days.
D × x + (100  D) × 2x = 175x
⇒ D = 25 days
Now, the work done in 25 days = 25x
Total work = 175x
therefore, work done before increasing the no of workers = \(\frac{25x}{175x} \times 100 = 14\frac{2}{7}\) %
A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 20 days and C alone in 60 days, then B alone could do it in:

Solution
(A+B)'s 1 day's work = \(\frac{1}{20}\)
C's 1 day work = \(\frac{1}{60}\)
(A+B+C)'s 1 day's work = \(\left ( \frac{1}{20} + \frac{1}{30} \right ) = \frac{4}{60} = \frac{1}{15}\)
Also A's 1 day's work = (B+C)'s 1 day's work
we get: 2 × (A's 1 day 's work) = \(\frac{1}{15}\)
⇒ A's 1 day's work = \(\frac{1}{30}\)
∴ B's 1 day's work = \(\left ( \frac{1}{20}  \frac{1}{30} \right ) = \frac{1}{60}\)
So, B alone could do the work in 60 days.
A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. If they get Rs. 3000 as wages for the whole work, what are the daily wages of A, B and C respectively (in Rs):

Solution
A's 5 days work = 50%
B's 5 days work = 33.33%
C's 2 days work = 16.66% [100  (50+33.33)]
Ratio of contribution of work of A, B and C = \(50 : 33\frac{1}{3} : 16\frac{2}{3}\)
= 3 : 2 : 1A's total share = Rs. 1500
B's total share = Rs. 1000
C's total share = Rs. 500
A's one day's earning = Rs.300
B's one day's earning = Rs.200
C's one day's earning = Rs.250