A man is facing north. He turns 45 degree in the clockwise direction and then another 180 degree in the same direction and then 45 degree in the anticlockwise direction. Find which direction he is facing now ?

Answer: Option A

Explanation:

Please refer to the explanation image.
The man firstly faces the direction OA. On moving 45 degree clockwise[Please check carefully always if clockwise or anticlockwise], he faces the direction OB.
Now again he moved 180 degree clockwise, now he will be facing OC. From here he moved 270 degree anticlockwise, Finally he is facing OD, which is South west.

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 B

Explanation:

Now dog is facing North.

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 A

Explanation:

Please refer to figure for the movements.
Final direction will be north-east with reference to the starting position.