Kunal walks 10 km towards North. From there he walks 6 Km towards South. Then, he walks 3 Km towards east. How far and in which direction is he with reference to his starting point ?

Answer: Option C

Explanation:

Clearly, the child moves from A to B 90 metres eastwards upto B, then turns right and moves 20 metre upto C, then turns right and moves upto 30 metre upto D. Finally he turns right and moves upto 100 metre upto E.

So AB = 90 metre, BF = CD = 30 metre,
So, AF = AB - BF = 60 metre

Also DE = 100 metre, DF = BC = 20 metre
So, EF = DE - DF = 80 metre

as we can see in image that triangle AFE is a right angled triangle and we are having two sides, need to calculate third one, so we can apply Pythagoras theorem here

\begin{aligned}
A = AE = \sqrt{AF^2 + EF^2} \\
= \sqrt{(60)^2+(80)^2} \\
= \sqrt{3600+6400} \\
= \sqrt{10000} = 100
\end{aligned}

So from starting point his father was 100 metre away.

Answer: Option C

Explanation:

The movements of Rohit are shown in figure.
Rohit's distance from the starting point A will be
AE = AD+DE = 20 + 15 = 35 metre

And direction with reference to the starting point is east.

Answer: Option D

Explanation:

Please refer to the explanation image.
The man firstly faces the direction OA. On moving 45 degree clockwise, he faces the direction OB.
Now again he moved 180 degree clockwise, now he will be facing OC. From here he moved 45 degree anticlockwise, Finally he is facing OD, which is South direction.

Answer: Option D

Explanation:

Clearly, Kunal moves from A 10 Km northwards upto B, then moves 6 Km southwards upto C, turns towards east and moves 3 km upto D.
Then AC = (AB-BC) = 4 Km

So Kunal distance from starting point A
\begin{aligned}
AD = \sqrt{AC^2 + CD^2} \\
= \sqrt{4^2+3^2} \\
= \sqrt{25} = 5
\end{aligned}

So AD is 5 Km also with reference to starting point Kunal's direction is North-East.

Answer: Option D