Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 page. How much time will they take, working together on two different computers to type an assignment of 110 pages ?

Solution
Number of pages typed by Ronald in 1 hour = \(\frac{32}{6} = \frac{16}{6}\)
Number of pages typed by Elan in 1 hour = \(\frac{40}{5} = 8\).
Number of pages typed by both in 1 hour = \(\left ( \frac{16}{3} + 8 \right ) = \frac{40}{3}\)
∴ Time taken by both to type 110 pages = \(\left ( 110 \times \frac{3}{40} \right )\) hrs = \(8\frac{4}{1}\) hrs = 8 hrs 15 min.
A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in:

Solution
(A +B + C)’s 1 day’s work = \(\frac{1}{6}\);
(A + B)’s 1 day’s work = \(\frac{1}{8}\);
(B + C)’s 1 day’s work = \(\frac{1}{12}\).
∴ (A + C)'s 1 day’s work = \(\left ( 2 \times \frac{1}{6} \right )  \left ( \frac{1}{8} + \frac{1}{12} \right )\)
= \(\left ( \frac{1}{3}  \frac{5}{24} \right )\)
= \(\frac{3}{24}\)
= \(\frac{1}{8}\)
So, A and C together will do the work in 8 days.
A and B can do a piece of work in 72 days, B and C can do it in 120 days, A and C can do it 90 days. In what time can A alone do it ?

Solution
(A + B)’s 1 day’s work = \(\frac{1}{72}\);
(B + C)'s 1 day’s work = \(\frac{1}{120}\);
(A+C)’s 1 day’s work = \(\frac{1}{90}\)
Adding, we get; 2 (A+B+C)’s 1 day’s work = \(\left ( \frac{1}{72} + \frac{1}{120} + \frac{1}{90} \right ) \)
= \(\frac{12}{360}\)
= \(\frac{1}{30}\)
⇒ (A+B +C)’s 1 day’s work = \(\frac{1}{60}\)
So, A’s 1 day’s work = \(\left ( \frac{1}{60}  \frac{1}{120} \right ) = \frac{1}{120}\)
∴ A alone can do the work in 120 days.
A and B can do a piece of work in 5 days, B and C can do it in 7 days, A and C can do it in 4 days. Who among these will take the least time if put to do it alone ?

Solution
(A + B )’s 1 day’s work = \(\frac{1}{5}\);
(B + C)’s 1 day’s work = \(\frac{1}{7}\);
(A + C )’s 1 day’s work =\(\frac{1}{4}\)
Adding,we get : 2(A+B+C)’s 1 day’s work = \(\left ( \frac{1}{5} + \frac{1}{7} + \frac{1}{4} \right ) = \frac{83}{140}\)
(A+B+C)’s 1 day’s work = \(\frac{83}{280}\)
A’s 1 day’s work = \(\left ( \frac{83}{280}  \frac{1}{7} \right ) = \frac{43}{280}\); B’s 1 day’s work = \(\left ( \frac{83}{280}  \frac{1}{4} \right ) = \frac{13}{280}\)
C’s 1 day’s work = \(\left ( \frac{83}{280}  \frac{1}{4} \right ) = \frac{27}{280}\)
Thus time taken by A,B,C is \(\frac{280}{43}\) days, \(\frac{280}{13}\) days, \(\frac{280}{27}\) days respectively.
Clearly,the time taken by A is least.
A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it 2 hours. How long will B alone take to do it ?

Solution
A’s 1 hour’s work = \(\frac{1}{4}\);
(B + C)'s 1 hour’s work =\(\frac{1}{3}\);
(A + C)'s 1 hour’s work = \(\frac{1}{2}\)
(A + B + C)’s 1 hour’s work = \(\left ( \frac{1}{4} + \frac{1}{3} \right ) = \frac{7}{12}\).
B’s 1 hour’s work = \(\left ( \frac{7}{12}  \frac{1}{2} \right ) = \frac{1}{12}\).
∴ B alone will take 12 hours to do the work.