find distance only Detailed Explanation And More Example
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The following question based on direction & distance sense test topic of verbal reasoning
(a) 80 kms
(b) 75 kms
(c) 65 kms
(d) 85 kms
The correct answers to the above question in:
Answer: (c)
Let X and Y be two buses.
Bus X travels along the path
PA, AB, BC, CD
Now, AD = BC = 25 km.
So, PD = PA + AD = 50km.
Bus Y travels 35 km upto E.
∴ Distance between two buses = PQ - (PD - QE)
= [150 - (50 + 35)] = 65km.
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Read more finding distance sense test Based Verbal Reasoning Questions and Answers
Question : 1
Sunil walks towards the East from point A turns right at point B and walks the same distance as he walked towards the East. He now turns left, walks the same distance again and finally makes a left turn and stops at point C after walking the same distance. The distance between A and C is how many times as that of A and B?
a) Two
b) Four
c) Can't be determined
d) Three
e) None of these
Answer »Answer: (a)
Movements of Sunil are as shown in figure.
Hence, AC = AB + BC = 2AB
Question : 2
Sohan started from point X and travelled forward 8 km up to point Y, then turned towards right and travelled 5km up to point Z, then turned right and travelled 7km up to point A and then turned towards right and travelled 5km up to B. What is the distance between points B and X?
a) 2 km
b) 4 km
c) 1 km
d) 3 km
Answer »Answer: (c)
Clearly, the route of Sohan is as shown in the diagram given below :
Here, XB = XY – YB = XY – AZ (∴ YB = AZ)
= 8 km – 7 km = 1 km
Question : 3
A person starts from a point A and travels 3 km eastwards to B and then turns left and travels thrice that distance to reach C. He again turns left and travels five times the distance he covered between A and B and reaches his destination D. the shortest distance between the starting point and the destination is
a) 15 km
b) 18 km
c) 12 km
d) 16 km
Answer »Answer: (a)
The movements of the person are as shown in the figure.
Clearly, AB = 3 km; BC = 3AB = (3 × 3) km = 9 km;
CD = 5 AB = (5 × 3) km = 15 km.
Draw AE
CD. Then, CE = AB = 3 km and AE = BC = 9 km.
DE = (CD – CE) = (15 – 3) km = 12 km.
In ΔAED, AD2 = AE2 + DE2 ; AD = $√{9^2 + 12^2}$
$√{225}$ = 15 km. So, required distance
= AD = 15 km.
Question : 4
Ram and Shyam started walking in opposite directions from a point. Ram covered 7 km and Shyam covered 5km. Ram turned right and walked 3 km. Shyam turned left and walked 3km. How far are they from each other?
a) 10 km
b) 14 km
c) 8 km
d) 12 km
Answer »Answer: (d)
Clearly, they are 12 km apart.
Question : 5
Starting from a point O, Mahesh walks a distance of 5 km. South, then turns to his right and walks 3 km. From there he again turns right and walks 5 km. He then turns to his left and walks 5 km. How far is he from the starting point?
a) 8 km.
b) 15 km
c) 13 km.
d) 5 km.
Answer »Answer: (a)
Required distance
= OS = OR + RS
= (3 + 5) km = 8 km.
Question : 6
A man drives his car 50 km towards eastward direction. He turned right went for 30 km, then he turned west and drive for 10km. How far is he from the starting point ?
a) 60 km
b) 20 km
c) 50 km
d) 100 km
Answer »Answer: (c)
Required distance
AD = $√{AE^2 + DE^2}$
= $√{40^2 + 30^2}$
= $√{1600 + 900}$
= $√{2500}$ = 50 km
GET direction & distance sense test PRACTICE TEST EXERCISES
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find both direction & distance section 3
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