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 in 2 hours. how long will b alone take to do it?

Let's denote:

• The work rate of A as 14\frac{1}{4}41​ (since A alone takes 4 hours).
• The combined work rate of B and C as 13\frac{1}{3}31​ (since B and C together take 3 hours).
• The combined work rate of A and C as 12\frac{1}{2}21​ (since A and C together take 2 hours).

From these:

• A=14A=\frac{1}{4}A=41​
• B+C=13B+C=\frac{1}{3}B+C=31​
• A+C=12A+C=\frac{1}{2}A+C=21​

From A+C=12A+C=\frac{1}{2}A+C=21​ and A=14A=\frac{1}{4}A=41​, we get:

• C=12−14=14C=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}C=21​−41​=41​

From B+C=13B+C=\frac{1}{3}B+C=31​ and C=14C=\frac{1}{4}C=41​, we get:

• B=13−14=412−312=112B=\frac{1}{3}-\frac{1}{4}=\frac{4}{12}-\frac{3}{12}=\frac{1}{12}B=31​−41​=124​−123​=121​

Therefore, the time taken by B alone to complete the work is the reciprocal of BBB's work rate:

• Time taken by B = 1B=12\frac{1}{B}=12B1​=12 hours.

So, B alone will take 12 hours to do the work. This matches established solutions for this problem.