SSC CGL Preparation – Day 20

Quantitative Aptitude – Speed, Time, and Distance (Advanced)

✅ 1. Key Concepts Recap

🔹 Basic Formula:

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}} \quad \text{or} \quad \text{Distance} = \text{Speed} \times \text{Time}$$

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🔹 Units Conversion:

• $$1 \text{ km/hr} = \frac{5}{18} \text{ m/s}$$
• $$1 \text{ m/s} = \frac{18}{5} \text{ km/hr}$$

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🔹 Average Speed:

If a body covers a certain distance at two different speeds: $$\text{Avg. Speed} = \frac{2xy}{x + y}$$

Where xxx and yyy are two different speeds for equal distance.

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✅ 2. Common Problem Types

🔸 A. Relative Speed:

• Same direction: Subtract speeds
• Opposite direction: Add speeds

🔸 B. Meeting Point Problems:

If two people start from opposite ends and move towards each other: $$\text{Meeting Time} = \frac{\text{Total Distance}}{\text{Relative Speed}}$$

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🔸 C. Train Problems:

• A train of length LLL crossing:
  • A pole: Time = $$\frac{L}{S}$$
  • A platform of length PPP: Time = $$\frac{L + P}{S}$$
  • A man running: Use relative speed

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🔸 D. Boat and Stream:

• Downstream speed: u+vu + vu+v
• Upstream speed: u−vu – vu−v
  Where:
  • uuu = Speed of boat in still water
  • vvv = Speed of stream

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

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✅ 3. Advanced Cases & Tricks

🔸 1. Crossing Each Other

Two trains of lengths L1​ and L2​ and speeds S1, S2​ cross each other in:

$$\text{Time} = \frac{L_1 + L_2}{S_1 + S_2}$$ (if in opposite directions) — ###

Late/Early Arrival Let: – Actual speed = ( S) – Time difference = (t) – Distance = (D)

Then: D = $$\frac{(S_1 \cdot S_2 \cdot t)}{|S_1 – S_2|}$$

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✅ 4. Examples

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Q1. A train 120 m long crosses a pole in 6 seconds. What is its speed?

Solution:
Speed = $$\frac{120}{6} = 20 \text{ m/s} = 72 \text{ km/hr}$$

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Q2. A train 150 m long crosses a platform 150 m long in 15 seconds. What is the speed?

Solution:
Distance = 150 + 150 = 300 m
Speed = $$\frac{300}{15} = 20 \text{ m/s} = 72 \text{ km/hr}$$

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Q3. A man travels 60 km at 20 km/hr and returns at 30 km/hr. Find average speed.

Solution: $$\text{Avg Speed} = \frac{2xy}{x + y} = \frac{2 \times 20 \times 30}{20 + 30} = \frac{1200}{50} = 24 \text{ km/hr}$$

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Q4. A boat covers 24 km downstream in 3 hours and the same distance upstream in 6 hours. Find speed of boat and stream.

Solution:
Downstream speed = 8 km/hr
Upstream speed = 4 km/hr
Boat speed = $$\frac{8 + 4}{2} = 6$$
Stream speed = $$\frac{8 – 4}{2} = 2$$

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Q5. Two trains of length 100 m and 120 m moving in opposite directions at 50 km/hr and 60 km/hr. Find the time to cross each other.

Solution:
Relative speed = 50 + 60 = 110 km/hr = $$\frac{110 \times 1000}{3600} = 30.56 \text{ m/s}$$
Total length = 220 m
Time = $$\frac{220}{30.56}$$ ​≈7.2 seconds

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✅ 5. SSC CGL Tips & Shortcuts

• Always convert km/hr to m/s when working with meters and seconds.
• For trains, total length is important — always add if crossing a platform or another train.
• Use relative speed logic for boats, trains, and people walking/running.