Reasoning Aptitude
DIRECTION SENSE TEST MCQs
Direction & Distance Sense Test
Total Questions : 1290
| Page 8 of 129 pages
Answer: Option A. -> 4 ft
Answer: (a)
According to the question, the direction diagram will be as follows. Therefore, Required distance, BE = 4 ft.
Answer: (a)
According to the question, the direction diagram will be as follows. Therefore, Required distance, BE = 4 ft.
Answer: Option A. -> 3 km
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.
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.
Answer: Option D. -> 15 km
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.
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.
Answer: Option D. -> 11.40 m
Answer: (d)
Required distance, AB = $√{(72 + 92)}$ = $√{130}$ = 11.40 m
Answer: (d)
Required distance, AB = $√{(72 + 92)}$ = $√{130}$ = 11.40 m
Answer: Option A. -> $√180$ km
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
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
Answer: Option A. -> 5
Answer: (a)
DC = EB = 5 m
Therefore, Required distance = AE = AB + BE
= 10 + 5 m = 15 m.
Answer: (a)
DC = EB = 5 m
Therefore, Required distance = AE = AB + BE
= 10 + 5 m = 15 m.
Answer: Option B. -> 2 km
Answer: (b)
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.
Answer: (b)
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.
Question 78. 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?
Answer: Option B. -> 5
Answer: (b)
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.
Answer: (b)
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.
Answer: Option A. -> 10 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.
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.
Answer: Option C. -> 50 metres
Answer: (c)
The movements of the man are as shown in Fig. 25.
Therefore, Man's distance from initial position A.
= AE = (AB + BE) = (AB + CD)
= (30 + 20) m = 50 m.
Answer: (c)
The movements of the man are as shown in Fig. 25.
Therefore, Man's distance from initial position A.
= AE = (AB + BE) = (AB + CD)
= (30 + 20) m = 50 m.
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