R takes twice as long as A and S together take to do a certain piece of work. A takes thrice as long as S and R together. The three together can do the work in 10 days. How many days would S alone take to do the work?

Let the daily work rates of R, A, and S be r, a, and s respectively. Since R takes twice as long as A and S together, we have r = (a + s)/2, which means the total rate r + a + s = 3r = 1/10, so r = 1/30. Similarly, since A takes thrice as long as S and R together, we have a = (s + r)/3, which means r + a + s = 4a = 1/10, so a = 1/40. Solving for s gives s = 1/10 - 1/30 - 1/40 = 1/24, meaning S alone takes 24 days.