question_answer

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

A) 80 m                  

B)          100 m

C) 140 m                 

D)          280 m

E) None of these

Correct Answer: D

Solution :

Explanation Option [d] is correct. The movement of the child from A to E are as shown in figure. Finally the child meets his father at E. Now, \[\operatorname{AF}=\left( 90-30 \right)m=60m.\] \[\operatorname{EF}=\left( DE-DF \right)=\left( DE-BC \right)=\left( 100-20 \right)=80m.\] \[\therefore \] Required distance \[=AE=\sqrt{{{(AE)}^{2}}+{{\left( EF \right)}^{2}}}\]\[=\sqrt{{{(60)}^{2}}+{{\left( 80 \right)}^{2}}}\]\[=\sqrt{3600+6400}\]\[=\sqrt{10,000}=100m.\]