find distance only Detailed Explanation And More Example

MOST IMPORTANT verbal reasoning - 15 EXERCISES

Top 10,000+ Verbal Memory Based Exercises

Directions :
Study the information given below carefully and answer the questions given below: A, B, C, D, E, F, G, H and I are nine houses. C is 2 km east of B. A is 1 km north of B and H is 2 km South of A. G is 1 km west of H while D is 3 km east of G and F is 2 km north of G. I is situated just in middle of B and C while E is just in middle of H and D.

[M.A.T. 2003]

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

Questions : Distance between E and G is

(a) 1 km

(b) 1.5 km

(c) 2 km

(d) 5 km

The correct answers to the above question in:

Answer: (c)

Since E lies in middle of H and D, so HE = ED. But HD = 2 km.

So, HE = ED = 1 km.

Therefore, Required distance = GE = GH + HE

= (1 + 1) km = 2 km.

Practice direction & distance sense test (find distance only) Online Quiz

Discuss Form

Valid first name is required.
Please enter a valid email address.
Your genuine comment will be useful for all users! Each and every comment will be uploaded to the question after approval.

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

Question : 1

Distance between E and I is

a) 1 km

b) 2 km

c) 3 km

d) 4 km

Answer: (a)

I lies in middle of B and C.

So, BI = IC. But bC = 2 km.

Then, BI = IC = 1 km.

I lies directly above E.

Therefore, Required distance = EI = HB = 1 km.

Question : 2

Distance between A and F is

a) 1 km

b) 1.41 km

c) 2 km

d) 3 km

Answer: (a)

Since GF = AH = 2 km, so F and A lie in the same line.

Therefore, Required distance = AF = GH = 1 km.

Directions :
Study the information given below carefully and answer the questions that follow: On a playing ground, Dinesh, Kunal, Nitin, Atul and Prashant are standing as described below facing the North.

    (i) Kunal is 40 metres to the right of Atul.
    (ii) Dinesh is 60 metres to the south of Kunal.
    (iii) Nitin is 25 metres to the west of Atul.
    (iv) Prashant is 90 metres to the north of Dinesh.
[NMAT, 2005]

Question : 3

If a boy walks from Nitin, meets Atul followed by Kunal, Dinesh and then Prashant, how many metres has he walked if he has travelled the straight distance all through?

a) 155 metres

b) 185 metres

c) 215 metres

d) 245 metres

e) None of these

Answer: (c)

Required distance = NA + AK + KD + DP

= (25 + 40 + 60 + 90) m = 215 m.

Question : 4

X and Y start moving towards each other from two places 200 m apart. After walking 60 m, Y 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 X and Y walk with the same speed, what is the distance between then now?

a) 20 m

b) 30 m

c) 40 m

d) 50 m

Answer: (c)

Clearly, Y moves 60 m from Q up to A, then 20 m up to B, 40 m up to C and then up to D.

So, AD = BC = 40 m. QD = (60 + 40) m = 100 m.

Since X and Y travel in the same time i.e. (60 + 20 + 40 + 20) m = 140 m.

So, X travels 140 m up to A.

Therefore, Distance between X and Y = AD = (100 - 60) m = 40 m.

Question : 5

Two houses start from the opposite points of a main road, 150 km apart. The first bus runs for 25 km and takes a right turn and then runs for 14 km. It then turns left and runs for another 25 km 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 km along the main road. What would be the distance between the two buses at this point?

a) 65 km

b) 75 km

c) 80 km

d) 85 km

e) 90 km

Answer: (a)

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 up to E.

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

= [150 - (50 + 35)] km = 65 km.

Question : 6

Raj travelled from a point X straight to Y at a distance of 80 metres. He turned right and walked 50 metres, then again turned right and walked 70 metres. Finally, he turned right and walked 50 metres. How far is he from the starting point?

a) 10 metres

b) 20 metres

c) 50 metres

d) 70 metres

e) 90 metres

Answer: (a)

The movements of Raj are as shown in Fig . 30.

(X to Y, Y to A, A to B, B to C).

Therefore, Raj's distance from the starting point = XC = (XY - YC)

= (XY - BA)

= (80 - 70) m = 10 m.

Recently Added Subject & Categories For All Competitive Exams

100+ Quadratic Equation Questions Answers PDF for Bank

Quadratic Equation multiple choice questions with detailed answers for IBPS RRB SO. more than 250 Attitude practice test exercises for all competitive exams

03-Jul-2024 by Careericons

Continue Reading »

IBPS Aptitude Linear Equations MCQ Questions Answers PDF

Linear equations multiple choice questions with detailed answers for IBPS RRB SO. more than 250 Attitude practice test exercises for all competitive exams

03-Jul-2024 by Careericons

Continue Reading »

New 100+ Compound Interest MCQ with Answers PDF for IBPS

Compound Interest verbal ability questions and answers solutions with PDF for IBPS RRB PO. Aptitude Objective MCQ Practice Exercises all competitive exams

02-Jul-2024 by Careericons

Continue Reading »

100+ Mixture and Alligation MCQ Questions PDF for IBPS

Most importantly Mixture and Alligation multiple choice questions and answers with PDF for IBPS RRB PO. Aptitude MCQ Practice Exercises all Bank Exams

02-Jul-2024 by Careericons

Continue Reading »