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CAT 2025Slot 3QAQues & Sol

ArithmeticHard

Question

Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was

Options

Solution

1. Concept Used

  • Topic: Time, Speed and Distance — forming simultaneous equations from journey conditions
  • Formula: $$ \text{Time} = \frac{\text{Distance}}{\text{Speed}}, \quad \text{Total Travel Time} = \text{Total Journey Time} - \text{Stop Time} $$

2. Calculation

Let Rahul's initial speed be $x$ km/hr. Since he normally travels from 5 pm to 11 pm, his total journey time is 6 hours, making the total distance $= 6x$ km.

Let $k$ = the distance Rahul covers before he stops (this is the same in both scenarios).

Scenario 1: Stops for 20 minutes, then increases speed by 3 km/hr

Stop duration $= 20 \text{ min} = \frac{1}{3}$ hr, so available travel time $= 6 - \frac{1}{3} = \frac{17}{3}$ hr.

The time equation becomes: $\frac{k}{x} + \frac{6x - k}{x + 3} = \frac{17}{3}$

Multiplying out the LHS: $\frac{k(x+3) + (6x-k) \cdot x}{x(x+3)} = \frac{17}{3}$ $\frac{kx + 3k + 6x^2 - kx}{x^2 + 3x} = \frac{17}{3}$ $\frac{6x^2 + 3k}{x^2 + 3x} = \frac{17}{3}$

Cross-multiplying: $3(6x^2 + 3k) = 17(x^2 + 3x)$ $18x^2 + 9k = 17x^2 + 51x$ $x^2 + 9k = 51x$ $x^2 = 51x - 9k \quad \cdots (1)$

Scenario 2: Stops for 30 minutes, then increases speed by 5 km/hr

Stop duration $= 30 \text{ min} = \frac{1}{2}$ hr, so available travel time $= 6 - \frac{1}{2} = \frac{11}{2}$ hr.

The time equation becomes: $\frac{k}{x} + \frac{6x - k}{x + 5} = \frac{11}{2}$

Simplifying the LHS: $$\frac{k(x+5) + (6x-k) \cdot x}{x(x+5)} = \frac{11}{2}$$ $$\frac{kx + 5k + 6x^2 - kx}{x^2 + 5x} = \frac{11}{2}$$ $$\frac{6x^2 + 5k}{x^2 + 5x} = \frac{11}{2}$$

Cross-multiplying: $$2(6x^2 + 5k) = 11(x^2 + 5x)$$ $$12x^2 + 10k = 11x^2 + 55x$$ $$x^2 = 55x - 10k \quad \cdots (2)$$

Solving equations (1) and (2) simultaneously:

Since both equal $x^2$: $$51x - 9k = 55x - 10k$$ $$10k - 9k = 55x - 51x$$ $$k = 4x$$

Substituting $k = 4x$ back into equation (1): ->$x^2 = 51x - 9(4x)$ ->$x^2 = 51x - 36x$ ->$x^2 = 15x$ ->$x = 15 \quad (\text{since } x > 0)$

3. Solution

Answer = Option B ✅

The initial speed of Rahul was 15 km/hr.