In this section the questions are based on successive follow up of direction and/or distance is formulated and calculated. Candidate need to ascertain the final direction with respect to starting point or the distance between the starting and final point.Thus these test are consist of Distance and Direction puzzle. These questions test the ability of candidate to trace , follow and perceive the direction . For this section you are required knowledge of direction on the plane of a paper and it is necessary to sketch the direction as per the instruction given in a question in a sequence. $\\$ Directions are as drawn 
Frequent right and left turning are used to confuse the candidate. keep in mind on the surface of paper the direction of right turn is always clockwise and of left turn is anticlockwise.
To find the shortest distance PYTHAGORUS THEOREM is used.$\\$
Where H = $\sqrt{{X}^2\,+{Y}^2\,} $
Example 1 :- One evening before sunset two people Vijay and Sanjay were talking to each other face to face. If Vijay shadow was exactly to his left side , then which direction was Sanjay facing ?$\\$ (a) North $\qquad \qquad (b)$ East $\qquad \qquad (c)$South $\qquad \qquad (d)$ West. $\\$ Answer : (a) $\\$ Vijay’s shadow was on left hence Vijay was facing South Since Both were talking face to face hence Sanjay was facing North.
Example 2:- Starting from a point ‘S’ Mahesh walks walks 25 mts. towards south . He turned to his left and walked 50 mts. He then again turned to his left and walked 25 mts. He again turned to his left and walked 60 mts. and reached at point T , then distance ST IS ? $\\$ (A) 10 mts. West $\qquad (b)$ 25 mts North $\quad (c) $ 10 mts East $\quad (d) $ 25 mts. West $\\$ Answer : (a) 10 mts West. 
Example 3 :- A child is looking for his father , goes 15 Km towards east and turns left and goes 10 Km towards , Again turns right and goes 5 km , again he turns right and goes 25 Km and meet his father there . How far is he from his starting point. $\\$(a) 65 Km $\qquad \qquad \quad (b)$ 50 Km $\qquad\qquad (c) $ 25 Km $\qquad \qquad (d)$ 40 Km . $\\$ Answer : (c) $\\$ Required Distance AE $= \sqrt {(AF)^2\;+(EF)^2}$ $\\$ $\qquad \qquad\qquad=\sqrt{625}\; =\; 25 $ 