find distance only Practice Questions Answers Test with Solutions & More Shortcuts

direction & distance sense test PRACTICE TEST [15 - EXERCISES]

find directions only

find distance only

find both direction & distance

find directions only section 2

find directions only section 3

find directions only section 4

find directions only section 5

find directions only section 6

find distance only section 2

find distance only section 3

find distance only section 4

find distance only section 5

find distance only section 6

find both direction & distance section 2

find both direction & distance section 3

Question : 6 [R.R.B. (TC/CC) 2010]

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

Shortcut »

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 : 7 [N.I.F.T. (UG) 2013]

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

Shortcut »

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 : 8 [S.S.C. (CPO) 2013]

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

Shortcut »

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.

Question : 9 [S.S.C. (10+2) 2013]

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

Shortcut »

Answer: (a)

DC = EB = 5 m

Therefore, Required distance = AE = AB + BE

= 10 + 5 m = 15 m.

Question : 10 [S.S.C. (CGL) 2013]

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

Shortcut »

Answer: (d)

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

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