Please submit your concern

Question:

Three workers working together need 1 hour to construct a wall. The first worker, working alone, can construct the wall twice as fast at the third worker, and can complete the task an hour sooner than the second worker. Then, the average time in hours taken by the three workers, when working alone, to construct the wall is

Explanation:

Let the time taken by the first worker be x hours.
∴ Time taken by the second worker = x + 1 hours.
∴ Time taken by the third worker = 2x hours.

Together they take 1 hour to complete the task.

⇒ $\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}+1}+\frac{1}{2\mathrm{x}}$ = $\frac{1}{1}$

⇒ $\frac{3}{2\mathrm{x}}+\frac{1}{\mathrm{x}+1}$ = $\frac{1}{1}$

⇒ 3(x + 1) + 2x = 2x × (x + 1)

⇒ 5x + 3 = 2x² + 2x

⇒ 2x² - 3x - 3 = 0

⇒ x = $\frac{3\pm \sqrt{9-4\times 2\times -3}}{2\times 2}$ = $\frac{3\pm \sqrt{33}}{4}$

We will accept only +ve value.

∴ x = $\frac{3+\sqrt{33}}{4}$

⇒ Required average = $\frac{\mathrm{x}+\mathrm{x}+1+2\mathrm{x}}{3}$ = $\frac{4\mathrm{x}+1}{3}$ = $\frac{4+\sqrt{33}}{3}$

Hence, option (a).