CUET UG 2025 — General Test Arithmetic

The police officer is 50 metres behind the chain snatcher. The chain snatcher runs at 2 m/s and the police officer runs at 4 m/s.

The police officer closes the gap at the difference of their speeds:

Gap closing speed $= 4 - 2 = 2$=4−2=2 m/s

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The time required to close the 50 metre gap:

Time $= \dfrac{\text{Distance gap}}{\text{Gap closing speed}}$=Gap closing speedDistance gap​

Time $= \dfrac{50}{2} = 25$=250​=25 seconds

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The distance covered by the chain snatcher in 25 seconds:

Distance $= \text{Speed} \times \text{Time}$=Speed×Time

Distance $= 2 \times 25$=2×25

Distance $= 50$=50 metres

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At the moment of catching:

• Chain snatcher position: $50 + 50 = 100$50+50=100 metres from starting point
• Police officer position: $0 + (4 \times 25) = 100$0+(4×25)=100 metres from starting point

Therefore, the chain snatcher covers 50 metres before being caught.