find distance only section 5 Practice Questions Answers Test with Solutions & More Shortcuts

Question : 11

A person starts from point T in east direction. Walks 6 m and turns right. Next walks 4 m and turns left. Next walks 3m and turns right. Now cycles for 8 km and stops. Find his distance from T.

a) $2√31 m$

b) 15 m

c) 17 m

d) $7√21 m$

e) 12 m

Answer: (b)

Distance = $√[(6+3)^2 + (4+8)^2 ] = 15 m$

Question : 12 [SSC Graduate Level Tier-I 2013]

A man travels 4 km due north, then travels 6 km due east and further travels 4 km due north. How far he is from the starting point ?

a) 10 km

b) 14 km

c) 8 km

d) 6 km

Answer: (a)

direction-and-distance-verbal-reasoning

Required distance

AD = $√{DE^2 + AE^2}$

= $√{8^2 + 6^2}$

= $√{64+ 36}$

= $√{100}$ = 10 km

Question : 13

Kashish goes 30 metres North, then turns right and walks 40 metres, then again turns right and walks 20 metres, then again turns right and walks 40 metres. How many metres is he from his original position ?

a) 20

b) 10

c) 0

d) 40

Answer: (b)

direction-and-distance-verbal-reasoning

The movements of Kashish are as shown in Figure.

(A to b, B to C, C to D, D to E)

∴ Kashish's distance from his original position

A = AE = (AB - BE) = (AB - CD)

= (30 - 20) = 10m.

Question : 14 [SSC CHSL LDC, DEO & PA/SA 2015]

Karthik travelled 3 km east, then took a right turn and travelled 4 kms. How far is he from starting point ?

a) 12 kms

b) 7 kms

c) 5 kms

d) 3 kms

Answer: (c)

direction-and-distance-verbal-reasoning

Required distance

AC = $√{AB^2 + BC^2}$

= $√{3^2 + 4^2}$

= $√{9 + 16}$

= $√{25}$ 

= 5 km

Question : 15 [SSC Delhi Police SI 2014]

A postman starts delivering letters 3 km southwards and then turns right. He covers 4 km on this road and again turns right. He delivers letters for 3 km and completes his daily beat. Then, he turns left for lunch at home, which is 5 km away. How far has he to travel to reach the post-office after lunch ?

a) 10 km

b) 8 km

c) 15 km

d) 9 km

Answer: (c)

direction-and-distance-verbal-reasoning

Required distance

= AE = AD + DE

= 4 km + 5 km = 9 km

IMPORTANT verbal reasoning EXERCISES

find distance only section 5 Online Quiz

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921 direction & distance sense test based verbal reasoning section 5 question answer with explanation pdf

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