GATE CS 2013 Q64 — General Aptitude

We can solve this problem using two simple methods.

Method 1: Assuming a Convenient Distance

Let the total distance of the journey be $120\text{ km}$ (chosen because it is easily divisible by the fractions and speeds involved).

• Train: Covers half of the journey.
  $$\text{Distance } (d_1) = 60\text{ km}, \quad \text{Speed } (v_1) = 60\text{ km/h}$$
  $$\text{Time taken } (t_1) = \frac{60}{60} = 1\text{ hour}$$
• Bus: Covers half of the remainder (remainder is $120 - 60 = 60\text{ km}$).
  $$\text{Distance } (d_2) = 30\text{ km}, \quad \text{Speed } (v_2) = 30\text{ km/h}$$
  $$\text{Time taken } (t_2) = \frac{30}{30} = 1\text{ hour}$$
• Cycle: Covers the rest of the journey.
  $$\text{Distance } (d_3) = 30\text{ km}, \quad \text{Speed } (v_3) = 10\text{ km/h}$$
  $$\text{Time taken } (t_3) = \frac{30}{10} = 3\text{ hours}$$

Now, calculate the total time and average speed:

Method 2: Algebraic Approach

Let the total distance be $D$.

• Time taken by train: $t_1 = \frac{D/2}{60} = \frac{D}{120}$
• Time taken by bus: $t_2 = \frac{D/4}{30} = \frac{D}{120}$
• Time taken by cycle: $t_3 = \frac{D/4}{10} = \frac{D}{40}$

Total time $T$ is:

Thus, the correct option is C.