A can do a piece of work in 4 hours, B and C together in 3 hours, and A and C together in 2 hours. How long will B alone take to do it?

### Concept & Formula / Logic The problem requires finding an individual's work rate by strategically subtracting known work rates of individuals or combinations from other given combinations. $$ \text{Rate of B} = \text{Rate of (B+C)} - \text{Rate of C} $$ $$ \text{Rate of C} = \text{Rate of (A+C)} - \text{Rate of A} $$ ### Step-by-Step Solution 1. **Identify the given rates (1-hour work):** A's 1-hour work = $\frac{1}{4}$ (B + C)'s 1-hour work = $\frac{1}{3}$ (A + C)'s 1-hour work = $\frac{1}{2}$ 2. **Find C's 1-hour work:** We know A's rate and (A + C)'s rate. Subtracting them will give C's rate. C's 1-hour work = (A + C) - A $C = \frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}$ 3. **Find B's 1-hour work:** We now know C's rate and (B + C)'s rate. Subtracting C's rate from (B + C)'s rate gives B's rate. B's 1-hour work = (B + C) - C $B = \frac{1}{3} - \frac{1}{4}$ Find the LCM of 3 and 4, which is 12. $B = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}$ Since B's 1-hour work is $\frac{1}{12}$, B alone will take 12 hours to complete the work. ### Exam Strategy & Shortcut Using the total work (LCM) method is much faster. Let Total Work = LCM of (4, 3, 2) = 12 units. Efficiency of A = $\frac{12}{4} = 3$ units/hr. Efficiency of $(B+C) = \frac{12}{3} = 4$ units/hr. Efficiency of $(A+C) = \frac{12}{2} = 6$ units/hr. Since $A=3$ and $A+C=6$, then $C = 6-3=3$ units/hr. Since $B+C=4$ and $C=3$, then $B = 4-3=1$ unit/hr. Time for B = $\frac{\text{Total Work}}{\text{B's Efficiency}} = \frac{12}{1} = 12$ hours. ### Common Pitfall A common mistake is trying to solve for B directly without systematically finding C first. You must isolate the unknown variable step-by-step. ### Final Answer Therefore, the correct answer is **12 hours**.