A, B, and C alone can do a piece of work in 9, 12 and 18 days, respectively. They all started the work togethe

Question

A, B, and C alone can do a piece of work in 9, 12 and 18 days, respectively. They all started the work together, but A left after 3 days. In how many days, was the total work completed?

Examoogle · Accepted Answer

The correct answer is Option B: $\frac{24}{5}$. Calculate LCM of (9, 12, 18) = 36 units. Total work done by A, B, C in 3 days is 3 	imes (4+3+2) = 27 units. Remaining work 9 units is done by (B+C) at rate (3+2)=5 units/day, taking 9/5 days. Total = 3 + 9/5 = 24/5 days.

Step-by-Step Solution & Concepts:
Shortcut Trick: Calculate LCM of (9, 12, 18) = 36 units. Total work done by A, B, C in 3 days is 3 	imes (4+3+2) = 27 units. Remaining work 9 units is done by (B+C) at rate (3+2)=5 units/day, taking 9/5 days. Total = 3 + 9/5 = 24/5 days.

Formula Used: $$Work = Time 	imes Efficiency$$

Given: A, B, and C can complete a work in 9, 12, and 18 days respectively. They all start together, but A leaves after 3 days.

Steps: Calculate Efficiency: Find the LCM of 9, 12, and 18, which is 36. Efficiencies are: A = 36/9 = 4, B = 36/12 = 3, C = 36/18 = 2.. Work done in 3 days: A, B, and C worked together for 3 days. Total work done = $3 	imes (4+3+2) = 27$ units.. Calculate Remaining Work: Remaining work is the total minus work already completed.. Final Time Calculation: B and C complete the remaining 9 units at their combined efficiency of $3+2=5$. Total days = $3 + \frac{9}{5} = \frac{15+9}{5} = \frac{24}{5}$.

Answer: B is correct because the total time taken to complete the work is 24/5 days.

Exam Tip: This approach is standard for all work-sharing problems; always define 'Total Work' as the LCM of individual times to simplify calculations.

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