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

Questions : Two buses start from the opposite point of a main road, 150 kms apart. The first bus runs for 25 kms and takes a right turn and then runs for 15 kms. It then turns left and runs for another 25 kms and takes the direction back to reach the main road. In the meantime, due to a minor breakdown, the other bus has run only 35 kms along the main road. What would be the distance between the two buses at this point ?

(a) 75 kms

(b) 85 kms

(c) 65 kms

(d) 80 kms

The correct answers to the above question in:

Answer: (c)

direction-and-distance

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 = 50 km

Bus Y travels 35 km upto E

∴ Distance between two buses = PQ - (PD + QE)

= {150 - (50+35)} = 65 km

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Read more finding distance sense test Based Verbal Reasoning Questions and Answers

Question : 1

Ricky after travelling for 5 km. took right turn and travelled 6 km. before taking left turn and then travelled for 3 km. find his final distance from home.

a) 9 km

b) 10 km

c) 12 km

d) 11 km

e) None of these

Answer: (b)

direction-and-distance

Question : 2

Seeta and Ram both start from a point towards North. Seeta turns to left after walking 10 km. Ram turns to right after walking the same distance. Seeta waits for sometime and then walks another 5 km, whereas Ram walks only 3 km. They both then return to their respective South and walk 15 km forward. How far is Seeta from Ram ?

a) 12 km

b) 10 km

c) 8 km

d) 15 km

Answer: (c)

direction-and-distance-verbal-reasoning

Required distance EF

= (5 + 3) km

= 8 km

Question : 3

Peter walked 8 kms. west and turned right and walked 3 kms. The again he turned right and walked 12 kms. How far is he from the starting point ?

a) 8

b) 5

c) 7

d) 4

Answer: (b)

direction-and-distance-verbal-reasoning

Required distance

AE = $√{AD^2 + DE^2}$

= $√{3^2 + 4^2}$

= $√{9+ 16}$

= $√{25}$ = 5 km

Question : 4

Kishore starts from a point and walks 6 km towards East and turning to his left he moves 3 km. After this, he again turns to his left and moves 6 km. Now, how far is he from his starting point ?

a) 2 km

b) 5 km

c) 3 km

d) 4 km

Answer: (c)

direction-and-distance-verbal-reasoning

AD = 3 km

Question : 5

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: (a)

direction-and-distance

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 : 6

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: (c)

direction-and-distance

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

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