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

Question : 81 [SSC Com Matric 2000]

I go 5 km. cast then turn right and go 8 km. Then turn left and go 5 km. and then I turn left and I go 8 km. At what distance I am from the starting point?

a) 10 km.

b) 6 km

c) 13 km.

d) 7 km.

Answer: (a)

direction-and-distance-verbal-reasoning

Required Distance

= 5 + 5 = 10 km

Question : 82 [SSC CGL Tier-I 2016]

I go 5 km East, then turn right and go 8 km. Then I turn left and go 5 km and then I turn left and go 8 km. At what distance am I from the starting point?

a) 0

b) 5

c) 8

d) 10

Answer: (d)

direction-and-distance-verbal-reasoning

AE = (5 + 5) km = 10 km.

Question : 83

X and Y start moving towards each other from two places 200 m apart. After walking 60 m, B turns left and goes 20 m, then he turns right and goes 40 m. He then turns right again and comes back to the road on which he had started walking. If A and B walk with the same speed, what is the distance between them now ?

a) 40 m

b) 30 m

c) 20 m

d) 50 m

Answer: (a)

direction-and-distance-verbal-reasoning

Clearly Y moves 60m from Q upto A, then 20m upto b, 40m upto c and then upto D.

So, AD = BC = 40m.

QD = (60 + 40)m = 100m.

Since A and B travel with the same speed, A will travel the same speed along the horizontal as B travels in the same time.

i.e. (60 + 20 + 40+ 20) = 140m

So, X travels 140 m upto A.

∴ Distance between X and Y = AD

= (100 - 60)m = 40m.

Question : 84

Namita walks 14 metres towards west, then turns to her right and walks 14 metres and then turns to her left and walks 10 metres. Again turning to her left she walks 14 metres. What is the shortest distance (in metres) between her starting point and the present position?

a) 24

b) 38

c) 10

d) 28

Answer: (a)

direction-and-distance

The movements of Namita are shown in figure

(A to B, B to C, C to D and D to E).

Clearly, Namita’s distance from his initial position

= AE = (AB + BE) = (AB + CD) = (14 + 10) m

= 24 m.

Question : 85 [SSC CGL Tier-1 2010]

A child is looking for his father. He went 90 metres in the east before turning to his right. He went 20 metres before turning to his right again to look for his father at his uncle’s place 30 metres from this point. His father was not there. From here he went 100 metres to his north before meeting his father in a street. How far did the son meet his father from the starting point?

a) 100 m

b) 140 m

c) 80 m

d) 260 m

Answer: (a)

direction-and-distance-verbal-reasoning

Required distance = AF

= $√{80^2 + 60^2}$

 = $√{6400 + 3600}$

= $√{10000}$ = 100 km

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