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
(a) $√180$ km
(b) $√80$ km
(c) $√100$ km
(d) None of these
The correct answers to the above question in:
Answer: (a)
According to the question, the direction diagram will be as follows.
AB = CE = 14 km
BC = AE = 6 km
CD = 26 km
ED = CD - CE = 26 - 14 = 12 km
Therefore, Required distance, AD = $√{(AE)^2 + (ED)^2}$ (by pythagoras theorem)
= $√{(62 + 122)}$ = $√180$ km
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Read more finding distance sense test Based Verbal Reasoning Questions and Answers
Question : 1
Hari travelled 17 km to the East, he turned left and went 15 km, he again turned left and went 17 km. How far is he from the starting point?
a) 32 km
b) 17 km
c) 2 km
d) 15 km
Answer »Answer: (d)
According to the questions, the direction diagram will be as follows.
Now, AB = CD = 17 km and BC = AD = 15 km
Therefore, Required distance, AD = 15 km.
Question : 2
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) 5
b) 10
c) 15
d) 20
e) 25
Answer »Answer: (a)
DC = EB = 5 m
Therefore, Required distance = AE = AB + BE
= 10 + 5 m = 15 m.
Question : 3
A and B both are walking away from point 'X'. A walked 3 m and B walked 4 m from it, then A walked 4 m North of X and B walked 5 m south of A. What is the distance between them now?
a) 9.5 m
b) 9 m
c) 16 m
d) 11.40 m
e) 10.40 m
Answer »Answer: (d)
Required distance, AB = $√{(72 + 92)}$ = $√{130}$ = 11.40 m
Question : 4
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) 3 km
b) 4 km
c) 6 km
d) 7 km
e) 8 km
Answer »Answer: (a)
According to the question, the direction diagram is as follows.
OB = CD = 5 km BC = OD = 3 km Therefore, Required distance, OD = 3 km.
Question : 5
Shyam walks 6 m towards East, then turns right and walks 9 m. Again, he turns to his left and walks 6 m. At what distance is he now from his original point?
a) 15 m
b) 21 m
c) 18 m
d) Cannot be determined
e) None of these
Answer »Answer: (a)
According to the question, the direction diagram is as follows. O is the mid - point of BC and
AD OB = OC = $9/2$ = 4.5 m
OA = $√{(AB)^2 + (OB)^2}= √{(6)^2+ (4.5)^2}$
= $√{(36 + 20.25)} = √{(56.25)}$ = 7.5 m
OD = OA = 7.5
Therefore, Required distance, AD = AO + OD = 7.5 + 7.5 = 15 m.
Question : 6
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) 5 ft
b) 6 ft
c) 8 ft
d) 12 ft
e) 10 ft
Answer »Answer: (b)
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.
GET direction & distance sense test PRACTICE TEST EXERCISES
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