Question:

Two cars, P and Q, start from a point X in India at 10 AM. Car P travels North with a speed of 25 km/h and car Q travels East with a speed of 30 km/h. Car P travels continuously but car Q stops for some time after travelling for one hour. If both the cars are at the same distance from X at 11:30 AM, for how long (in minutes) did car Q stop?

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In such problems, always calculate distances covered step by step and convert minutes into hours consistently before applying speed–time–distance relations.
Updated On: Aug 24, 2025
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The Correct Option is C

Solution and Explanation


Step 1: Distance travelled by Car P.
Car P travels continuously at 25 km/h. From 10:00 AM to 11:30 AM is 1.5 hours. \[ \text{Distance travelled by P} = 25 \times 1.5 = 37.5 \ \text{km}. \] Step 2: Motion of Car Q.
Car Q travels for 1 hour (10:00–11:00) at 30 km/h. \[ \text{Distance covered in 1 hour} = 30 \times 1 = 30 \ \text{km}. \] After 11:00 AM, Car Q stops for \(t\) minutes. Then it resumes travel until 11:30 AM. So, actual running time between 11:00 and 11:30 = \(30 - t\) minutes. Convert into hours: \[ \frac{30 - t}{60} \ \text{hours}. \] Step 3: Extra distance covered by Q after resuming.
\[ \text{Extra distance} = 30 \times \frac{30 - t}{60} = \frac{30(30 - t)}{60} = \frac{30 - t}{2}. \] Step 4: Total distance of Q.
\[ \text{Total distance} = 30 + \frac{30 - t}{2}. \] Step 5: Equating distances.
Since both are at the same distance from X at 11:30 AM: \[ 30 + \frac{30 - t}{2} = 37.5 \] \[ \frac{30 - t}{2} = 7.5 \] \[ 30 - t = 15 \] \[ t = 15 \ \text{minutes}. \] Final Answer: \[ \boxed{15} \]
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