TCS Company Numerical Ability

Related to direction & distance: a person is travelling in so and so direction, eg: 5km North then 5km East then 10km North then so and so. Find the distance between original & final positions

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Let the distance be x. Hence x^2=15^2+5^2=250 .so x=15.8km
10 years agoHelpfull: Yes(3) No(0)

ans:15.8

5km east,15km north
sqrt[(5*5)=(15*15)]=15.8
10 years agoHelpfull: Yes(1) No(1)
as the question goes,the distance would be 10 km north+ 5km north+5km east=20
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please sumbit sol??
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plzee explain..??
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sqrt((5+10)^2+5^2)=15.8113
10 years agoHelpfull: Yes(0) No(0)

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