find distance only section 3 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 : 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 »Answer: (a)
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 »Answer: (b)
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 »Answer: (c)
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 »Answer: (b)
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 »Answer: (c)
Required distance
= AE = AB + BE
= (10 + 5) metres
= 15 metres
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