find both direction & distance Detailed Explanation And More Example

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The following question based on direction & distance sense test topic of verbal reasoning

Questions : Rasik walks 20 m North. Then he turns right and walks 30 m. Then he turns right and walks 35 m. Then he turns left and walks 15 m. Then he again turns left and walks 15 m. In which direction and how many metres away is he from his original position ?

(a) 30 metres East

(b) 15 metres West

(c) 45 metres East

(d) 30 metres West

The correct answers to the above question in:

Answer: (c)

direction-and-distance-verbal-reasoning

The movement of Rasik from A to F is as shown in Figure

Since CD = AB + EF, so F lies in the line with A.

∴ Rasik's distance from original position A

= AF = (AG + GF)

= (BC + DE) = (30 + 15)m = 45m

Also, F lies to the East of A

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Read more finding both direction distance test Based Verbal Reasoning Questions and Answers

Question : 1

Rohit walked 25 metres towards South. Then he turned to his left and walked 20 metres. He then turned to his left and walked 25 metres. He again turned to his right and walked 15 metres. At what distance is he from the starting point rind in which direction ?

a) 35 metres North

b) 35 metres East

c) 60 metres East

d) 40 metres East

e) None of these

Answer: (b)

direction-and-distance-verbal-reasoning

The movement of Rohit is as shown in Figure

∴ Rohit's distance from starting point A

= AE = (AD + DE)

= (BC + DE) = (20 + 15) = 35m

Also, E is the East of A.

Question : 2

A man leaves for his office from his house. He walks towards East. After moving a distance of 20 m, he turns towards South and walks 10 m. Then he walks 35 m towards the West and further 5 m towards the North. He then turns towards East and walks 15 m. What is the straight distance in metres between his initial and final positions ?

a) 5

b) 0

c) Cannot be determined

d) 10

e) None of these

Answer: (a)

direction-and-distance-verbal-reasoning

 The movement of the man from A to F are as shown in Figure

Clearly, DC = AB + EF.

∴ F is in line with A.

Also, AF = (BC - DE) = 5 m.

So, the man is 5 metres away from his initial position.

Question : 3

Starting from a point P, Sachin walked 20 metres towards South. He turned left and walked 30 metres. He then turned left and walked 20 metres. He again turned left and walked 40 metres and reached a point Q. How far and in which direction is the point Q from the point P ?

a) 10 metres East

b) 20 metres West

c) 10 metres North

d) 10 metres West

e) None of these

Answer: (d)

direction-and-distance-verbal-reasoning

The movement of Sachin is as shown in Figure

(P to B, B to C, C to D and D to Q)

Clearly, the distance of Q from P

= PQ = (DQ - PQ) = (DQ - BC)

= (40 - 30)m = 10m.

Also, Q is to the west of P

∴ Q is 10m West of P.

Question : 4

Kishenkant walks \0 kilometres towards North. From there, he walks 6 kilometres towards South. Then, he walks 3 kilometres towards East. How far and in which direction is he with reference to his starting point ?

a) 5 kilometres North-east

b) 5 kilometres West

c) 7 kilometres West

d) 7 kilometres East

Answer: (a)

direction-and-distance-verbal-reasoning

The movement of Kishenkant is as shown in the figure.

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

AC = (AB - BC)

= (10 - 6)km = 4km

Clearly, D is to the North-east of A.

∴ Kishenkant's distance from starting point A

= AD = $√{AC^2 + CD^2}$

= $√{4^2 + 3^2}$ = $√{25}$

= 5km.

So, Kishenkant is 5km to the North-east of his starting point.

DIRECTIONS:

The following questions are based on the diagram given below showing four persons stationed at the four corners of a square piece of plot as shown.

direction-and-distance-verbal-reasoning

Question : 5

From the original position given in the above figure, A and B move one arm length clockwise and then cross over to the corner diagonally opposite; C and D move one arm length anti-clockwise and cross over the comer diagonally opposite. The original configuration ADBC has now changed to

a) BDAC

b) CBDA

c) ACBD

d) DACB

e) BCAD

Answer: (b)

Clearly, (i), (ii), (iii) and (iv) show the movements of A, B, C and D respectively while the new arrangement so obtained is shown in (v).

So, the configuration changes to CBDA.

direction-and-distance-verbal-reasoning

Question : 6

A square field ABCD of side 90 m is so located that its diagonal AC is from north to south and the corner B is to the west of D. Rohan and Rahul start walking along the sides from B and C respectively in the clockwise and anticlockwise directions with speeds of 8 km/hr and 10 km/hr. Where shall they cross each other the second time ?

a) On BC at a distance of 10 m from B

b) On AD at a distance of 30 m from A

c) On BC at a distance of 10 m from C

d) On AD at a distance of 30 m from D

Answer: (c)

Clearly, the arrangement is as shown in the adjioning figure.

direction-and-distance-verbal-reasoning

Rohan's speed = 8 km/hr.

= $8000/{60 x 60}$ m/sec = $20/9$ m/sec.

Rahul's speed = 10 km/hr.

= $10000/{60 x 60}$ m/sec = $25/9$ m/sec.

Since Rohan and Rahul are moving in opposite directions, so they together cover a distance of $({20}/9 + {25}/9)$

i.e., $45/9$ or 5 meters in one second.

To meet at a point, they together have to cover the distance (CD + DA + AB) i.e., 270m.

Now, 5 meters is covered in 1 second.

So, 270 m will be covered in $(1/{5} x 270)$ = 54 seconds.

Now, distance covered by Rohan in 54 seconds = $({20}/9 x 54)$ m = 120 m.

distance covered by Rahul in 54 seconds = $({25}/9 x 54)$ m = 150 m.

Thus, Rohan and Rahul meet for the first time on AD at a point 30 m from A and 60 m from D.

Now, to meet again, Rohan and Rahul will have to complete one full round i.e. together move a distance of 360 m.

5 metres is covered by both together in 1 second.

Thus, 360 m will be covered by both in $({1}/5 x 360)$ = 72 seconds.

Now, distance covered by Rohan in 72 seconds = $({20}/9 x 72)$ = 160 m.

Distance covered by Rahul in 72 seconds = $({25}/9 x 72)$ = 200 m.

Thus, Rohan and Rahul meet on BC at a point 10 m from C and 80 m from B.

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