SSC JE Mechanical Engineering 22 Mar 2021 Shift 2 Papers

Question : 38 of 199

Marks: +1, -0

A student walked 110 m to the East to look for his class teacher, and then he turned right and walked 40 m. Then he turned right again, and after walking 50 m he reached his class teacher's staffroom. His class teacher was not there. From there he walked 120 m to his north and met his class teacher. How far from the starting point did he meet his class teacher?

80 m
100 m
260 m
120 m

Validate

Solution:

The diagram below shows the path followed by Student:

Here A is the Starting point and E is Ending point
The shortest distance to meet the teacher by student is

AE^2 = AF^2 + EF^2

AE^2 = 60^2 + 80^2

(110m - 50m = 60m = AF)

AE^2= 10000

AE = √10000 = 100m

So the shortest distance is 100m
Hence, the correct answer is "100m'.

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