📚 Puducherry PSC – Quantitative Aptitude
Topic: Time – Speed – Distance (T‑S‑D)
Objective – Master the basic concept of Time–Speed–Distance and solve typical PSC style questions in a few minutes.
1. Simple Explanation
| Symbol | Meaning | SI Unit | Example |
|---|---|---|---|
| S | Speed (rate of motion) | km / h | 60 km/h |
| T | Time taken | hours | 2 h |
| D | Distance covered | km | 120 km |
The fundamental relationship:
[ \boxed{S = \frac{D}{T}} ]
or equivalently
[ D = S \times T \quad\text{and}\quad T = \frac{D}{S} ]
Key takeaway – Speed = Distance ÷ Time.
Change one quantity → compute the other.
2. Key Points to Remember
| # | Point | Why It Matters |
|---|---|---|
| 1 | Unit Consistency | All three must be in the same units (e.g., km and h). Convert if necessary. |
| 2 | Speed vs Velocity | For PSC, speed = magnitude only. Direction is irrelevant. |
| 3 | Fractional Time | ½ h = 30 min; 1/3 h = 20 min. Convert to minutes when needed. |
| 4 | Average Speed | For varying speeds, use total distance ÷ total time. |
| 5 | Relative Speed | When two objects move towards/away from each other, add/subtract speeds. |
| 6 | Algebraic Manipulation | Keep the formula in the form you need; avoid unnecessary steps. |
| 7 | Check Reasonableness | E.g., 120 km in 0.5 h → 240 km/h → unrealistic for a car – flag the answer. |
3. Worked Example
Question (typical PSC style)
“A train leaves City A at 8:00 am heading towards City B, covering a distance of 240 km. Its average speed is 60 km/h. Another train leaves City B at 10:00 am heading towards City A along the same route. Its average speed is 80 km/h. At what time do the two trains meet?”
Step‑by‑Step Solution
-
Determine the distance each train travels until they meet.
Let the meeting time be t hours after 8:00 am.- Train A travels for t hours at 60 km/h → distance (D_A = 60t).
- Train B starts 2 h later, so it travels for ((t-2)) hours at 80 km/h → distance (D_B = 80(t-2)).
-
Set up the distance‑sum equation.
The trains together cover the entire 240 km:
[ D_A + D_B = 240 ] [ 60t + 80(t-2) = 240 ] -
Solve for t.
[ 60t + 80t - 160 = 240 \ 140t = 400 \ t = \frac{400}{140} = \frac{20}{7} \text{ h} \approx 2.857 \text{ h} ] -
Convert t into hours and minutes.
(0.857\text{ h} = 0.857 \times 60 \approx 51.4 \text{ min}).
So t ≈ 2 h 51 min. -
Add to the departure time.
8:00 am + 2 h 51 min = 10:51 am.
Answer
The two trains meet at 10:51 am.
4. Quick Revision Summary
| Concept | Formula | Typical Question | Quick Check |
|---|---|---|---|
| Speed | (S = D/T) | “If a car covers 150 km in 3 h, find speed.” | Compute (150/3 = 50) km/h. |
| Distance | (D = S \times T) | “A train travels at 90 km/h for 4 h. Find distance.” | (90 \times 4 = 360) km. |
| Time | (T = D/S) | “Distance 200 km, speed 50 km/h. Find time.” | (200/50 = 4) h. |
| Relative Speed | (S_{\text{rel}} = S_1 \pm S_2) | Two cars heading towards each other. | Add speeds. |
| Average Speed | (\text{Avg} = \frac{\text{Total Distance}}{\text{Total Time}}) | Car travels 80 km in 1 h and 120 km in 2 h. | ((80+120)/(1+2)=200/3 ≈ 66.7) km/h. |
| Unit Check | Always use km & h (or m & s). | “50 km in 30 min.” | Convert 30 min → 0.5 h → speed = 100 km/h. |
| Reasonableness | Ensure speed < 300 km/h for cars, < 2500 km/h for jets. | 240 km in 0.5 h → 480 km/h (unrealistic). | Flag and re‑check. |
Final Tips
- Write the formula first before plugging numbers.
- Keep track of time offsets (e.g., second train starts later).
- Always convert minutes to hours before applying the formula.
- Practice 5–10 mixed T‑S‑D questions daily to build speed.
Good luck with your preparation! 🚀