find distance only Detailed Explanation And More Example

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The following question based on direction & distance sense test topic of verbal reasoning

Questions : A child is looking for his mother. He went 80 meters in the East before turning to his right. He went 20 meters before turning to his right again to trace his mother at his married sister's house, 20 meters from this point. His mother was not there. From there he went 100 meters to his north where he met his mother who was shopping there in the market. How far did the son meet from the starting point?

(a) 60 m

(b) 140 m

(c) 80 m

(d) 100 m

The correct answers to the above question in:

Answer: (d)

AE = AB – EB

= 80 – 20 = 60m

EF = DF – DE

= 100 - 20 = 80m

direction-and-distance

Now in Δ AEF,

AF = $√{(AE)^2 + (EF)^2}$

= $√{60^2 + 80^2}$

= $√{3600 + 6400}$ = $√{10000}$ = 100m

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Read more finding distance sense test Based Verbal Reasoning Questions and Answers

Question : 1

A bus starts from point A and runs 20 kms towards South, turns to its right and runs 25 km. It then turns right again runs 20 km. Afterwards it runs 5 km in the East direction and reaches point B. How far is the bus from the starting point ?

a) 20 km

b) 30 km

c) 35 km

d) 25 km

Answer: (a)

direction-and-distance-verbal-reasoning

Required distance

AB = 20 km

Question : 2

Ravi starts from his house and moves towards south. He walks 100 m, then turns left and walks 200 m, turns right and walks 500. How far is he from his house?

a) 800 m

b) 200$√{2}$ m

c) 400$√{5}$ m

d) 200$√{10}$ m

e) None of these

Answer: (d)

direction-and-distance

Ravi starts from O and finished at C.

OC = $√{OD^2 + CD^2}$ = $√{OD^2 + (DA + AC)^2}$

= $√{(200)^2 + (100 + 500)^2}$ m

= $√{200 × 200 + 600 × 600}$ m

= 100 $√{4 + 36}$ m = 100 $√{40}$ m = 200 $√{10}$ m

[OD = AB = 200 m ; DA = OB = 100 m and AC = 500 m]

Question : 3

Rahul walks 30 metres towards south. Then turns to his right and starts walking straight till he completes another 30 metres. Then again turning to his left he walks for 20 metres. He then turns to his left and walks for 30 metres. How far is he from his initial position?

a) 30 metres

b) 60 metres

c) 50 metres

d) 10 metres

Answer: (c)

direction-and-distance-verbal-reasoning

AE = (30 + 20) metres

= 50 metres

Question : 4

My friend and I started simultaneously towards each other from two places 100 m apart. After walking 30 m, my friend turns left and goes 10m, then he turns right and goes 20 m and then turns right again and comes back to the road on which he had started walking. If we walk with the same speed, what is the distance between us at this point of time ?

a) 20 m

b) 40 m

c) 50 m

d) 30 m

Answer: (a)

direction-and-distance-verbal-reasoning

Friend covers a total distance of

= (30 + 10 + 20 + 10) = 70 m

Distance between them

= 70 – 50 = 20 m

Directions :
Study the information given below carefully and answer the questions that follow: On a playing ground, Dinesh, Kunal, Nitin, Atul and Prashant are standing as described below facing the North.

    (i) Kunal is 40 metres to the right of Atul.
    (ii) Dinesh is 60 metres to the south of Kunal.
    (iii) Nitin is 25 metres to the west of Atul.
    (iv) Prashant is 90 metres to the north of Dinesh.
[NMAT, 2005]

Question : 5

If a boy walks from Nitin, meets Atul followed by Kunal, Dinesh and then Prashant, how many metres has he walked if he has travelled the straight distance all through?

a) 155 metres

b) 185 metres

c) 215 metres

d) 245 metres

e) None of these

Answer: (c)

Required distance = NA + AK + KD + DP

= (25 + 40 + 60 + 90) m = 215 m.

Directions :
Study the information given below carefully and answer the questions given below: A, B, C, D, E, F, G, H and I are nine houses. C is 2 km east of B. A is 1 km north of B and H is 2 km South of A. G is 1 km west of H while D is 3 km east of G and F is 2 km north of G. I is situated just in middle of B and C while E is just in middle of H and D.

[M.A.T. 2003]

Question : 6

Distance between A and F is

a) 1 km

b) 1.41 km

c) 2 km

d) 3 km

Answer: (a)

Since GF = AH = 2 km, so F and A lie in the same line.

Therefore, Required distance = AF = GH = 1 km.

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