find distance only section 6 Detailed Explanation And More Example

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

Questions : Mr. Das started his journey from his house straight to his friend’s house at a distance of 12 km. On returning he walked 8 km in the same route and turned right and walked 4 km, then he turned to his left and walked 4 km. Finally he turned to his left and walked 2 km. How far was he from his house ?

(a) 4 km

(b) 2 km

(c) 8 km

(d) 6 km

The correct answers to the above question in:

Answer: (b)

direction-and-distance-verbal-reasoning

Required distance = AF

= 2 km

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Read more type 2 section 6 mcqs Based Verbal Reasoning Questions and Answers

Question : 1

Rahul travelled from a point and straightly goes to point ‘Y’ at a distance of 90 metres. He turned right and walked 40 metres, then again turned right and walked 70 metres. Finally, he turned right and walked 40 metres. How far he is from the starting point ?

a) 10 metres

b) 30 metres

c) 70 metres

d) 20 metres

Answer: (d)

direction-and-distance-verbal-reasoning

Required distance = AE

= 20 metre

Question : 2

Jai Prakash started walking towards south from the point A, walked 20 m and reached a point B, Again, he turned left and walked 20 m and reached a point C. Now he turned 45° anticlockwise, walked a distance of 20 √2 m and reached a point D. What approximately is the shortest distance between the point A and D?

a) 30 m

b) 40$√{2}$ m

c) Can't say

d) 40 m

Answer: (d)

direction-and-distance

Here, AC = $√{20^2 + 20^2} = 20$√{2}$ m

Since AB = BC

∴ ∠ACB = ∠BAC = 45°

∴ ∠ACD = 90°

Now,

AD2 = AC2 + CD2

= 800 + 800 = 1600

∴ AD = $√{1600}$ = 40 m

Question : 3

A man walks 1 km towards East and then he turns to South and walks 5 km. Again he turns to East and walks 2 km, after this he turns to North and walks 9 km. Now, how far is he from his starting point ?

a) 5 km

b) 4 km

c) 3 km

d) 7 km

Answer: (a)

direction-and-distance-verbal-reasoning

The movements of the man are as shown in the figure,

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

Clearly, DF = BC = 5km.

EF = (DE - DF) = (9 - 5)km = 4km.

BF = CD = 2km

AF =AB + BF = AB + CD = (1 + 2)km = 3km.

∴ Man's distance from initial position A

= AE = $√{AF^2 + FE^2}$

= $√{3^2 + 4^2}$

= $√25$ = 5m.

Question : 4

If Sita walks 10 km towards West, then turned towards South and walked 10 km, then turned East walked 10 km, and turned North walked 10 km. How far she is from starting point?

a) 0 km

b) 20 km

c) 40 km

d) 10 km

Answer: (a)

direction-and-distance-verbal-reasoning

Question : 5

An insect is walking in a straight line. It covers a distance of 15 cm per minute. It comes back 2.5 cm after every 15 cm. How long will it take to cover a distance of 1 metre ?

a) 8 min

b) 12 min

c) 6.5 min

d) 10 min

Answer: (a)

The insect covers 15 – 2.5 = 12.5 cm in one minute.

So, it will cover 1 metre (= 100 cm) in $100/12.5$ = 8 minutes

Question : 6

Raju drives 25 km North and turns left and travels 5 km and reaches point ‘O’. He then turns right and covers another 5 kms. Afterwards turns to east and drives 5 km. How much distance he has to travel to go back to the starting point?

a) 25 kms

b) 20 kms

c) 35 kms

d) 30 kms

Answer: (d)

direction-and-distance-verbal-reasoning

Required distance AD

= (25 + 5) km = 30 km

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