Two workers A and B working together completed a job in 5 days. If A worked twice as efficiently as he actually did and B worked $\frac{1}{3}$ as efficiently as he actually did, the work would have been completed in 3 days. A alone could complete the work in :

### Concept & Simultaneous Equations When work completion changes based on modified individual efficiencies, we can model the scenario using a system of linear equations representing the 1-day work of each individual. ### Step-by-Step Solution * **Given:** A and B take 5 days normally. With A at $2 \times$ efficiency and B at $1/3 \times$ efficiency, they take 3 days. * **Calculation:** Let A's 1-day work be $x$ and B's 1-day work be $y$. * From the first condition: $x + y = 1/5$ --- (Equation 1) * From the second condition: $2x + (1/3)y = 1/3$ --- (Equation 2) * We need to find the time A alone takes, which requires finding $x$. Let's eliminate $y$. * Multiply Equation 1 by $1/3$: $(1/3)x + (1/3)y = 1/15$ --- (Equation 3) * Subtract Equation 3 from Equation 2: * $(2x - (1/3)x) + ((1/3)y - (1/3)y) = 1/3 - 1/15$ * $(5/3)x = (5 - 1)/15$ * $(5/3)x = 4/15$ * $x = (4/15) \times (3/5) = 12/75 = 4/25$ * Since A's 1-day work is $4/25$, A alone will take $25/4$ days. * Convert to mixed fraction: $25/4 = 6\frac{1}{4}$ days. ### Exam Strategy & Shortcut Use algebraic manipulation effectively. Once the equations $x + y = 1/5$ and $6x + y = 1$ (multiplying the second equation by 3 to clear fractions) are set up, subtracting the first from the modified second instantly yields $5x = 4/5 \implies x = 4/25$. Thus, time is $25/4$. ### Common Pitfall A common mistake is directly substituting efficiency ratios as time. Students might write $2x + y/3 = 3$ instead of using the reciprocal ($1/3$) for the one-day work equivalent. ### Final Answer Therefore, the correct answer is **$6\frac{1}{4}$ days**.