6. Speed, Distance and Time

Hard Acceptance: 37.5%

A thief is spotted by a policeman from a distance of 200m. When the policeman starts the chase, the thief also starts running. If the speed of the thief is 10 km/hr and that of the policeman is 11 km/hr, how far will the thief have run before he is overtaken?

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Speed, Distance and Time Quantitative Aptitude

Hard

Interactive Visual Explanation

Step 1 of 3

Step 1

Find Relative Speed

Since both are running in the same direction, the relative speed is the difference between the policeman's and thief's speeds:

\text{Relative Speed} = 11 \text{ km/hr} - 10 \text{ km/hr} = 1 \text{ km/hr}

Step 2

Convert to m/s and Find Catching Time

The policeman must cover the initial 200m gap. Convert relative speed to meters per second ($1 \text{ km/hr} = 5/18 \text{ m/s}$):

\text{Time to Overtake} = \frac{\text{Gap}}{\text{Relative Speed}} = \frac{200}{\frac{5}{18}} = 720 \text{ seconds}

Step 3

Calculate Thief's Run Distance

Calculate the total distance run by the thief in 720 seconds at their speed of 10 km/hr ($10 \times 5/18 = 50/18 \text{ m/s}$):

Chasing Displacement Model पुलिस vs चोर

Relative Speed: 1 km/h (Police runs 1.1x faster)

Overtake distance: Thief: 2.0 km | Police: 2.2 km

\text{Thief Distance} = \frac{50}{18} \text{ m/s} \times 720 \text{ s} = 2000 \text{ meters} = 2.0 \text{ km}

⚡

CGL Exam Speed Trick

⚡ Ratio Shortcut: When time is constant, Distance is directly proportional to Speed. Speed Ratio (Police : Thief) = 11 : 10. Distance Ratio = 11 : 10. The difference is 1 part, which corresponds to the initial gap of 200m. Therefore, the Thief runs 10 parts = 10 × 200m = 2000m = 2.0 km. Zero unit conversions needed! Takes 6 seconds.

⚡ Speed-Run CGL Cheatsheet

Mastering these core principles will let you solve similar questions under 30 seconds.

Core Identity:

Distance Ratio = Speed Ratio (when Time is constant)

Check the interactive visualizer in the first tab to see these formulas modeled geometrically!

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