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 : Hari travelled 17 km to the East, he turned left and went 15 km, he again turned left and went 17 km. How far is he from the starting point?

(a) 32 km

(b) 17 km

(c) 2 km

(d) 15 km

The correct answers to the above question in:

Answer: (d)

According to the questions, the direction diagram will be as follows.

Now, AB = CD = 17 km and BC = AD = 15 km

Therefore, Required distance, AD = 15 km.

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

Question : 1

Ram goes 15 m North, then turns right and walks 20 m, then again turns right and walks 10 m, then again turns right and walks 20 m. How far is he from his original position?

a) 5

b) 10

c) 15

d) 20

e) 25

Answer: (a)

DC = EB = 5 m

Therefore, Required distance = AE = AB + BE

= 10 + 5 m = 15 m.

Question : 2

A and B both are walking away from point 'X'. A walked 3 m and B walked 4 m from it, then A walked 4 m North of X and B walked 5 m south of A. What is the distance between them now?

a) 9.5 m

b) 9 m

c) 16 m

d) 11.40 m

e) 10.40 m

Answer: (d)

Required distance, AB = $√{(72 + 92)}$ = $√{130}$ = 11.40 m

Question : 3

A boy rode his bicycle northwards, then turned left and rode one km and again turned left and rode 2 km. He found himself exactly one km west of his starting point. How far did he ride northwards initially?

a) 1 km

b) 2 km

c) 5 km

d) 8 km

e) 10 km

Answer: (b)

Clearly, the boy rode from A to B, then to C and finally up to D.

Since D lies to the west of A, so required distance = AB = CD = 2 km.

Question : 4

John travels 14 km westwards and then turns left and travels 6 km and further turns left and travels 26 km. How far is John now from the starting point?

a) $√180$ km

b) $√80$ km

c) $√100$ km

d) None of these

Answer: (a)

According to the question, the direction diagram will be as follows.

AB = CE = 14 km

BC = AE = 6 km

CD = 26 km

ED = CD - CE = 26 - 14 = 12 km

Therefore, Required distance, AD = $√{(AE)^2 + (ED)^2}$ (by pythagoras theorem)

= $√{(62 + 122)}$ = $√180$ km

Question : 5

A man goes 5 km East from his office. He then takes left turn and walks 3 km. He again takes left turn and walks 5 km. How far is he starting point?

a) 3 km

b) 4 km

c) 6 km

d) 7 km

e) 8 km

Answer: (a)

According to the question, the direction diagram is as follows.

OB = CD = 5 km BC = OD = 3 km Therefore, Required distance, OD = 3 km.

Question : 6

Shyam walks 6 m towards East, then turns right and walks 9 m. Again, he turns to his left and walks 6 m. At what distance is he now from his original point?

a) 15 m

b) 21 m

c) 18 m

d) Cannot be determined

e) None of these

Answer: (a)

According to the question, the direction diagram is as follows. O is the mid - point of BC and

AD OB = OC = $9/2$ = 4.5 m

OA = $√{(AB)^2 + (OB)^2}= √{(6)^2+ (4.5)^2}$

= $√{(36 + 20.25)} = √{(56.25)}$ = 7.5 m

OD = OA = 7.5

Therefore, Required distance, AD = AO + OD = 7.5 + 7.5 = 15 m.

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