A dog runs 20 metre towards East and turns Right, runs 10 metre and turns to right, runs 9 metre and again turns to left, runs 5 metre and then turns to left, runs 12 metre and finally turns to left and runs 6 metre. Now which direction dog is facing ?

Answer: Option B

Explanation:

Please refer to explanation image,
Raviraj starts from home at A, moves 20 Km in south upto B. Then he turns right and moves 10 Km upto C, then he turns right and moves 20 Km upto D, then he turns lefts and moves 20 Km upto E.

So from image it is clear that, if he moves straight then he will have to move AD+DE, AD = BC = 10 Km
So, he will have to move 10 + 20 = 30 Km

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 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:

Please refer to the image for movements.

Distance from the P to Q is 10 metres and direction of Q with reference to to P is west.