CAT 2025 Slot 3 QA Question 9

Rahul starts at 5 pm and is supposed to reach his destination at 11 pm. That means his total journey time is 6 hours. But he stops for 20 minutes on his way. After stopping, he increases his speed by 3 km/h to reach on time. If he had stopped for 10 minutes more, he would have needed to increase his speed by 5 km/h instead. We find his initial speed.

Hmm, let's break this down. First, let's denote his initial speed as 'v' km/h. The total distance he needs to cover is the same in both scenarios, so maybe I can set up equations based on that.

In the first case, he stops for 20 minutes, which is 1/3 of an hour. So, the time he spends driving is 6 hours minus 1/3 hour, which is 5 and 2/3 hours or 17/3 hours. But wait, no, that's not right. Because he starts at 5 pm and stops for 20 minutes, the total time from start to finish is still 6 hours, but the driving time is less by the stopping time. So, driving time is 6 hours minus 20 minutes, which is 5 hours 40 minutes, or 5 + 40/60 = 5 + 2/3 = 17/3 hours. But then he increases his speed by 3 km/h for the remaining distance. he stops once for 20 minutes and then increases his speed for the rest of the journey.

Wait, maybe I should think of it as two parts: before stopping and after stopping. Let the distance he covered before stopping as D1 and after stopping as D2. The total distance is D1 + D2.

But actually, maybe it's better to think in terms of the total distance. Let's denote the total distance as D. His initial speed is v, and he stops for 20 minutes. So, the time he would have taken without stopping is D/v. But since he stops, the actual driving time is D/v - 20 minutes. Wait, no, that's not correct. Because stopping doesn't affect the driving time; it just adds to the total time.

Wait, let's clarify. The total time from 5 pm to 11 pm is 6 hours. If he stops for 20 minutes, his driving time is 6 hours minus 20 minutes, which is 5 hours 40 minutes, or 5 + 40/60 = 17/3 hours. But he increases his speed after stopping, so maybe the distance is covered in two parts: before stopping and after stopping.

Wait, no, he stops once for 20 minutes, then increases his speed. So, the total driving time is 17/3 hours, but he spends some time driving at speed v and the rest at speed v + 3.

Wait, maybe I should set up the equation based on the total distance. Let's say he drives for t hours at speed v, then stops for 20 minutes, and then drives the remaining distance at speed v + 3. The total time from start to finish is 6 hours, so t + 20 minutes + t2 = 6 hours. But t2 is the time after stopping, which is (D - v*t)/(v + 3). Hmm, this is getting a bit complicated.

Alternatively, maybe I can think of the total distance as D = v(t) + (v + 3)(t2), where t + t2 + 20 minutes = 6 hours. But I'm not sure if this is the right approach.

Wait, let's try another way. Without any stops, the time taken would be D/v = 6 hours. But he stops for 20 minutes, so the driving time is 6 - 20/60 = 6 - 1/3 = 17/3 hours. But he increases his speed after stopping, so the distance covered at speed v is v(t1), and the distance covered at speed v + 3 is (v + 3)(t2), where t1 + t2 = 17/3. But also, the total distance D = v*6, since without stopping he would take 6 hours. So, D = 6v.

So, 6v = vt1 + (v + 3)t2, and t1 + t2 = 17/3.

Similarly, in the second scenario, he stops for 30 minutes, so driving time is 6 - 30/60 = 5.5 hours, or 11/2 hours. He increases his speed by 5 km/h, so D = vt3 + (v + 5)t4, where t3 + t4 = 11/2, and D = 6v.

So, now I have two equations:

But I have four variables here: t1, t2, t3, t4. That's too many variables. Maybe I can express t1 and t2 in terms of each other.

From the first equation: t1 + t2 = 17/3, so t2 = 17/3 - t1.

Substitute into the first equation:

6v = vt1 + (v + 3)(17/3 - t1)

Let's expand this:

6v = vt1 + (v + 3)(17/3) - (v + 3)*t1

Combine like terms:

6v = [vt1 - (v + 3)t1] + (v + 3)*(17/3)

Factor t1:

6v = t1(v - v - 3) + (v + 3)(17/3)

Simplify:

6v = t1(-3) + (v + 3)(17/3)

So,

6v + 3t1 = (v + 3)(17/3)

Multiply both sides by 3 to eliminate the denominator:

18v + 9t1 = 17(v + 3)

Expand the right side:

18v + 9*t1 = 17v + 51

Subtract 17v from both sides:

v + 9*t1 = 51

So,

v = 51 - 9*t1

Similarly, let's do the same for the second scenario.

From the second equation