Problems Involving All Directions Turning


 
 
Concept Explanation
 

Problems Involving All Directions Turning

Direction Question: This section of verbal reasoning revolves around the various directions. It consist of questions involving all sort of direction puzzle. Candidate is required to make a successive follow-up of directions and ascertain the final direction or the distance between two points. The test is meant to judge the candidate,s ability to trace and follow correctly and sense the directions correctly.

Directions: There are four basic directions North N, South S, East E, West W and four cardinal directions North East NE, North West NW, South East SE, South West SW to help the candidates know the directions. The figure shows all the directions

Problems Involving All Directions: These type of questions are solved by taking the right and left movement of a person is always with reference to the body moving in the scenario. It is not with respect to the person who is solving the questions.A diagram is drawn showing the movement of the object. On analysing the diagram the candidate reaches the desired result

Illustration: If you are facing north-east and move 10 m forward, turn left and move 7.5 m, then you are

A. North of your initial position               B. South of your initial position           C. East of your initial position               D. 12 m from your initial position.

Answer: A

Solution: Clearly, the narrator starts from A, moves towards north-east a distance of 10 m and reaches B, turns left  (90^{circ} anti-clockwise ) and moves 7.5 m and reaches C.

Clearly, C lies to the north of A.

Also, large Delta ABC  is right-angled at B. So by Pythagoras Theorem

So, AC^{2}=AB^{2}+BC^{2}=(10)^{2}+(7.5)^{2}

                 = 100 + 56.25 = 156.25.

Rightarrow ;;;AC=(sqrt{156.25})m=12.5;m

Thus, the narrator is 12.5 m to the north of his initial position. Hence option A is correct.

Illustration: A person starts from a point A and travels 3 km eastwards to B and then turns left and travels thrice that distance to reach C. He again turns left and travels five times the distance he covered between A and B and reaches his destination D. The shortest distance between the starting point and the destination is

A. 12 km           B. 15 km           C. 16 km          D. 18 km

Answer: B

Solution: The movements of the person are as shown in fig.

Clearly, AB = 3 km,

BC = 3AB = (3 large times 3) km = 9 km,

CD = 5AB = (5 large times 3) km = 15 km.

Draw AE large perp CD.

Then, CE = AB = 3 km and AE = BC = 9 km.

DE = (CD - CE) = (15 - 3) km = 12 km.

In large DeltaAED, AD^{2} = AE^{2}  + DE^{2}

large Rightarrow  AD = (sqrt{9^{2}+(12)^{2}})  km

              =(sqrt{81+144}) Km

             = sqrt{225} km = 15 km.

large therefore Required distance = AD = 15 km

Hence the correct option is B

Sample Questions
(More Questions for each concept available in Login)
Question : 1

You go South, turn left, then right again and then go to the right. In which direction are you now ?

Right Option : B
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Explanation
Question : 2

I am standing at the centre of a circular field. I go down south to the edge of the field and then turning left I walk along the boundary of the field equal to three-eighths of its length. Then I turn west and go right across to the opposite point on the boundary. In which direction am I  from the starting point?

Right Option : C
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Explanation
Question : 3

A watch shows time 4.30 p.m. If the minute hand point towards East, in which direction will the hour hand point? 

Right Option : D
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Explanation
 
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