Time and Work, Pipes and Cisterns, Time, Speed and Distance & Trains/Boats

Distance calculation:

• Speed of the object $= 10\text{ feet/second}$.
• Time $= 1\text{ hour} = 60 \times 60 = 3600\text{ seconds}$.
• Distance covered:
  $\text{Distance} = \text{Speed} \times \text{Time} = 10\text{ feet/s} \times 3600\text{ s} = 36,000\text{ feet}$.

Relative speed calculation:

• Since the goods train passes a single point observer (the man) in 9 seconds, the distance covered equals the length of the goods train ($270\text{ m}$).
• Relative speed in m/s:
  $\text{Relative Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{270\text{ m}}{9\text{ s}} = 30\text{ m/s}$.
• Convert relative speed to km/h:
  $\text{Relative Speed} = 30 \times \frac{18}{5} = 108\text{ km/h}$.
• Because the two trains move in opposite directions, relative speed is the sum of their speeds:
  $\text{Speed}_{\text{goods}} + \text{Speed}_{\text{man's train}} = 108\text{ km/h}$.
  $\text{Speed}_{\text{goods}} + 40 = 108 \implies \text{Speed}_{\text{goods}} = 108 - 40 = 68\text{ km/h}$.

Let the number of days required by A and B working together be $x$.
A takes $(x + 18)$ days, and B takes $(x + 8)$ days.
Using the standard algebraic time-and-work theorem:
$\frac{1}{x+18} + \frac{1}{x+8} = \frac{1}{x}$
This simplifies directly to:
$x = \sqrt{a \times b} = \sqrt{18 \times 8} = \sqrt{144} = 12\text{ days}$.

Option analysis:

• 18: The excess days taken by A alone.
• 12: Correct time required when working together.
• 16: Numerical error.
• 36: Double the excess of A.