Practice Finding distance sense test - verbal reasoning Online Quiz (set-1) For All Competitive Exams
Q-1) Ram goes 15 m North, then turns right and walks 20 m, then again turns right and walks 10 m, then again turns right and walks 20 m. How far is he from his original position?
(a)
(b)
(c)
(d)
(e)
DC = EB = 5 m
Therefore, Required distance = AE = AB + BE
= 10 + 5 m = 15 m.
Q-2) A man leaves for his office from his house. He walks towards East. After moving a distance of 20 m, he turns 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)
(b)
(c)
(d)
(e)
The movements of the man from A to F are as shown in Fig. 13
Clearly, DC = AB + EF. Therefore, F in line with A.
Also, AF = (BC - DE) = 5 m.
So, the man is 5 metres away from his initial position.
Q-3) Mohan walked 30 m towards South, took a left turn and walked 15 m. He, then took a right turn and walked 20 m. He again took a right turn and walked 15 m. How far is the from the starting point?
(a)
(b)
(c)
(d)
(e)
According to the question, the direction diagram will be as follows.
Therefore, Required distance, PT = PQ + QT = 30 + 20 = 50 m.
Q-4) Ankit started walking towards North. After walking 30 m, he turned towards left and walked 40 m. He, then turned left and walked 30 m. He again turned left and walked 50 m. How far is the from his original position?
(a)
(b)
(c)
(d)
(e)
According to the question, the direction diagram will be as follows. Therefore, Required distance, PT = ST - SP = 50 - 40 = 10 m.
Q-5) Amol travels a distance of 10 ft from A to B. Then, he turns to left and travels 5 ft and again turns to left and travels 2 ft. Finally, he turns to left and moves 5 ft. How far is he now from his original position?
(a)
(b)
(c)
(d)
(e)
According to the question, the direction diagram is as follows. A = Original position,
E = Final position AB = 10 ft, BC = DE = 5 ft CD = EB = 2 ft
Therefore, Required distance, AE = AB - EB = 10 -2 = 8 ft.
Q-6) A man goes 5 km East from his office. He then takes left turn and walks 3 km. He again takes left turn and walks 5 km. How far is he starting point?
(a)
(b)
(c)
(d)
(e)
According to the question, the direction diagram is as follows.
OB = CD = 5 km BC = OD = 3 km Therefore, Required distance, OD = 3 km.
Q-7) A boy rode his bicycle northwards, then turned left and rode one km and again turned left and rode 2 km. He found himself exactly one km west of his starting point. How far did he ride northwards initially?
(a)
(b)
(c)
(d)
(e)
Clearly, the boy rode from A to B, then to C and finally up to D.
Since D lies to the west of A, so required distance = AB = CD = 2 km.
Q-8) 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)
(b)
(c)
(d)
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.
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.Q-9) Distance between E and G is
(a)
(b)
(c)
(d)
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.
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.Q-10) Distance between E and I is
(a)
(b)
(c)
(d)
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.