question_answer

A child is looking for his father. He went 90 m in the East before turning to his right. He went 20 m before turning to his right again to look for his father at his uncle's place 30 m from this point. His father was not there. From here he went 100 m to the North before meeting his father in a street. How for did the son meet his father from the starting point?

A)  50 m               

B)  100 m

C)  40 m                           

D)  260 m

Correct Answer: B

Solution :

The movements of the child from A to E are as shown in figure. Clearly, the child meets his father at E.             Now, \[AF=(AB-FB)\]             \[=(AB-DC)=(90-30)m=60m\] \[EF=(DE-DF)=(DE-BC)\]             \[=(100-20)m=80\,m\] \[\therefore \] Required distance             \[=AE=\sqrt{A{{F}^{2}}+E{{F}^{2}}}=\sqrt{{{(60)}^{2}}+{{(80)}^{2}}}\]             \[=\sqrt{3600+6400}=\sqrt{10000}=100\,m\]