find distance only section 3 Practice Questions Answers Test with Solutions & More Shortcuts

Question : 11 [SSC GL Tier-I 2014]

Deepak walks 20 metres towards North. He then turns left and walks 40 metres. He again turns left and walks 20 metres. Further he moves 20 metres after turning to the right. How far is he from his original position ?

a) 60 mts.

b) 30 mts

c) 20 mts.

d) 50 mts.

Answer: (a)

direction-and-distance-verbal-reasoning

Required distance = AE = AD + DE

= (40 + 20) metres

= 60 metres

Question : 12 [SSC CGL Tier-I 2015]

A man starts from a point and moves 3 km North, then turns to West and goes 2 km. He turns North and walks 1 km and then moves 5 km towards East. How far is he from the starting point?

a) 10 km.

b) 5 km

c) 11 km.

d) 8 km.

Answer: (b)

direction-and-distance-verbal-reasoning

Required distance

AE = $√{AF^2 + EF^2}$

= $√{4^2 + 3^2}$

= $√{16 + 9}$

= $√{25}$

= 5 km

Question : 13 [SSC GL Tier-I 2013]

X goes 15 metres North, then turns right and walks 20 metres, then again turns right and walks 10 metres then again turns right and walks 20 metres. How many metres is he from his original position ?

a) 10 m

b) 20 m

c) 5 m

d) 15 m

Answer: (c)

direction-and-distance-verbal-reasoning

Required distance

= 15 – 10 = 5m

Question : 14

Two men start walking from one point towards opposite direction. After walking 3 km straight the both turn right wards and walk straight for the distance of 4km. How far are they both from each-other?

a) 7 km

b) 10 km

c) 8 km

d) 9 km

Answer: (b)

direction-and-distance

In Δ ABC, AC = $√{(AB)^2 + (BC)^2}$

= $√{(3)^2 + (4)^2}$ = $√{9 + 16}$ = $√{25}$ = 5 km

In Δ ADE, AE = $√{(AD)^2 + (DE)^2}$

= $√{(3)^2 + (4)^2}$ = $√{9 + 16}$ = $√{25}$ = 5 km

Hence, CE = 5 + 5 = 10 km

Question : 15 (FCI Asst Grade-II 2012]

A house faces North. A man coming out of his house walked straight for 10 metres, turned left and walked 25 metres. He then turned right and walked 5 metres and again turned right and walked 25 metres. How far is he from his house?

a) 55 metres

b) 65 metres

c) 15 metres

d) 60 metres

Answer: (c)

direction-and-distance-verbal-reasoning

Required distance

= AE = AB + BE

= (10 + 5) metres

= 15 metres

IMPORTANT verbal reasoning EXERCISES

find distance only section 3 Online Quiz

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919 direction & distance sense test based verbal reasoning section 3 question answer with explanation pdf

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