Distance and Speed Analysis

To determine who reaches first, we need to compare the time taken by each person. The time taken is calculated using the formula: $t = \frac{d}{v}$ where $t$ is time, $d$ is distance, and $v$ is speed.

Relationship Definitions

Let the distances traveled be $d_A, d_C, d_B$ and speeds be $v_A, v_C, v_B$ for Abdul, Catherine, and Binoy, respectively.

• Distance relationships:
  • $d_A = 3d_C$
  • $d_C = 2d_B$
• Speed relationships:
  • $v_C = \frac{1}{3}v_A$
  • $v_A = \frac{1}{2}v_B$

Expressing Variables

Let's express all distances and speeds in terms of a common reference, say Binoy's distance ($d_B$) and Abdul's speed ($v_A$):

• Distances:
  • Let $d_B = d$.
  • Then $d_C = 2d_B = 2d$.
  • And $d_A = 3d_C = 3(2d) = 6d$.
• Speeds:
  • Let $v_A = v$.
  • Then $v_C = \frac{1}{3}v_A = \frac{1}{3}v$.
  • And $v_B = 2v_A = 2v$.

Calculating Travel Times

Now, calculate the time taken for each person using $t = d/v$:

• Abdul's time: $t_A = \frac{d_A}{v_A} = \frac{6d}{v}$
• Catherine's time: $t_C = \frac{d_C}{v_C} = \frac{2d}{(1/3)v} = \frac{6d}{v}$
• Binoy's time: $t_B = \frac{d_B}{v_B} = \frac{d}{2v}$

Comparing Times and Conclusion

Comparing the calculated times:

• $t_A = \frac{6d}{v}$
• $t_C = \frac{6d}{v}$
• $t_B = \frac{1}{2} \times \frac{d}{v}$

Since $\frac{1}{2} < 6$, the time $t_B$ is the smallest value among the three. The person who takes the least time reaches first.

Therefore, Binoy reaches first.