5 questions from actual SSC CGL papers · all shown free · click option to reveal solution
SPEED, DISTANCE AND TIME — COMPLETE EXAM GUIDE ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
CORE CONCEPT ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Speed tells you how fast something moves. Distance is how far it travels. Time is how long it takes. These three are always connected. If you know any two, you can always find the third. Think of it this way: A car travels 120 km in 2 hours. Its speed is 60 km/h. Simple. That is the entire foundation of this topic.
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The three master formulas (all from one triangle):
Unit Conversion (VERY important for SSC):
km/h is BIGGER than m/s, so to go from bigger to smaller, multiply by the smaller fraction (5/18). ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
When a person travels the SAME distance at two different speeds (s1 and s2): Average Speed = 2 x s1 x s2 / (s1 + s2) WARNING: Do NOT simply add and divide by 2. That is wrong for equal distances
Two objects moving in the SAME direction → Relative Speed = (s1 - s2) Two objects moving in OPPOSITE directions → Relative Speed = (s1 + s2) ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
(What Gets Asked) ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ SSC CGL almost always tests: 1. Average speed problems (equal distance, two speeds) 2. Unit conversion traps (km/h vs m/s) 3. A person is late or early — find original speed 4.
Two trains crossing each other 5. Meeting point problems ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ SHORTCUTS AND TRICKS ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Shortcut 1 — Late and Early Trick: A man is late by T1 minutes if he walks at s1 and early by T2 minutes if he walks at s2. Distance = s1 x s2 x (T1 + T2) / (s2 - s1) (Always make sure units match — convert minutes to hours if speed is in km/h) Shortcut 2 — If speed is increased by a fraction: New time = Old time x (original speed / new speed) Time saved = Old time - New time Example: Speed increases by 25% → New speed = 5/4 of old. New time = 4/5 of old. Time saved = 1/5 of original time. Shortcut 3 — Two trains crossing a stationary object vs each other: Crossing a pole/person: Time = Length of train / Speed of train Crossing each other (opposite): Time = (L1 + L2) / (s1 + s2) Crossing each other (same direction): Time = (L1 + L2) / (s1 - s2) ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Both distances are equal, so use the harmonic mean formula. Average Speed = 2 x 40 x 60 / (40 + 60)
Numerator = 2 x 40 x 60 = 4800 Denominator = 40 + 60 = 100
Average Speed = 4800 / 100 = 48 km/h Answer: 48 km/h Note: Many students wrongly calculate (40 + 60) / 2 = 50 km/h. That is WRONG. Never use simple average for speed when distances are equal. ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ WORKED EXAMPLE 2 — Late and Early ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Question: A student walks to school at 4 km/h and is late by 10 minutes. If he walks at 6 km/h, he reaches 5 minutes early. Find the distance to school.
Use the Late and Early formula. Distance = s1 x s2 x (T1 + T2) / (s2 - s1) Here s1 = 4, s2 = 6, T1 = 10 min, T2 = 5 min
T1 + T2 = 15 minutes = 15/60 hours = 1/4 hour s2 - s1 = 6 - 4 = 2
Distance = 4 x 6 x (1/4) / 2 = 24 x (1/4) / 2 = 6 / 2 = 3 km Answer: Distance = 3 km ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
— THE NUMBER ONE TRAP ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Students use simple average (s1 + s2) / 2 for average speed problems where EQUAL DISTANCES are covered at two different speeds. This is always WRONG. The correct formula is 2s1s2 / (s1 + s2).
Simple average only works when EQUAL TIME is spent at both speeds. SSC examiners set questions deliberately to trap students who use the wrong formula. Always ask yourself first: is the distance equal or is the time equal?
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