Reasoning Aptitude
DIRECTION SENSE TEST MCQs
Direction & Distance Sense Test
Total Questions : 1290
| Page 6 of 129 pages
Answer: Option B. -> 50 m
Question 52. Amit starts from point A and walks 5 m towards North-East direction and reaches point B. From here, he travels 8 m in the East direction and reaches point C. From C, he travels towards South-West direction and reaches point D after travelling a distance equal to AB. At last, he turns towards the West direction and reaches point A. How much distance has been covered by Amit and which geometrical figure has been formed by the path travelled by him?
Answer: Option B. -> 26 m, parellelogram
Answer: (b)
According to the question, the direction diagram is as follows.
Therefore, Required distance = AB + BC + CD + DA
= 5 + 8 + 5 + 8 = 26 m.
and the geometrical figure formed is Parallelogram.
Answer: (b)
According to the question, the direction diagram is as follows.
Therefore, Required distance = AB + BC + CD + DA
= 5 + 8 + 5 + 8 = 26 m.
and the geometrical figure formed is Parallelogram.
Answer: Option E. -> None of these
Answer: (e)
According to the question, the direction diagram will be as follows.
A = Starting point, E = Finishing point DE = 15 m,
BC = FD = 20 m FE = FD + DE = 20 + 15 = 35 m
AF = BF - AB = 25 - 10 = 15 m
Therefore, Required distance, AE = $√{(AF)^2 + (FE)^2}$
= $√((15)^2 + (35)^2)$
= $√{(225 + 1225)}$
= $√1450$
and the direction is South-East
Answer: (e)
According to the question, the direction diagram will be as follows.
A = Starting point, E = Finishing point DE = 15 m,
BC = FD = 20 m FE = FD + DE = 20 + 15 = 35 m
AF = BF - AB = 25 - 10 = 15 m
Therefore, Required distance, AE = $√{(AF)^2 + (FE)^2}$
= $√((15)^2 + (35)^2)$
= $√{(225 + 1225)}$
= $√1450$
and the direction is South-East
Answer: Option D. -> 15 ft towards East
Answer: (d)
According to the questions, the direction diagram is as follows.
AB = CD = 12 ft
BC = AD = 10 ft
DE = 5 ft
Therefore, Required distance, AE = AD + DE = 10 + 5 = 15 ft
Hence at finishing point E, Raju is 15 ft East from starting point A.
Answer: (d)
According to the questions, the direction diagram is as follows.
AB = CD = 12 ft
BC = AD = 10 ft
DE = 5 ft
Therefore, Required distance, AE = AD + DE = 10 + 5 = 15 ft
Hence at finishing point E, Raju is 15 ft East from starting point A.
Answer: Option A. -> $10√2$ km, north - East
Answer: (a)
According to the question, the direction diagram will be as follows.
As in a square, diagonals are equal. Therefore, SH = OM = 10√2 km
Therefore, Required distance, OM = Distance between office and multiplex = 10√2 km
Also, it is clear from diagram that multiplex (M) is to the North - East of O (office)
So, the multiplex is $10√2$ km North - East from office.
Answer: (a)
According to the question, the direction diagram will be as follows.
As in a square, diagonals are equal. Therefore, SH = OM = 10√2 km
Therefore, Required distance, OM = Distance between office and multiplex = 10√2 km
Also, it is clear from diagram that multiplex (M) is to the North - East of O (office)
So, the multiplex is $10√2$ km North - East from office.
Question 56. A school bus driver starts from the school, drives 2 km towards North, takes a left turn and drives for 5 km. He, then takes a left turn and drives for 8 km before taking a left turn again and driving for further 5 km. The driver finally takes a left turn and drives 1 km before stopping. How far and towards which direction should the driver drive to reach the school again?
Answer: Option E. -> 5 km towards North
Answer: (e)
According to the question, the direction diagram will be as follows.
Now, BC = DE = 5 km, CD = BE = 8 km AB = 2 km, EF = 1 km
Therefore, Required distance, AF = BE - (AB + EF) = 8 - (2 + 1) = 8 - 3 = 5 km.
From the direction diagram, it is clear that the bus is present finally in the South direction of school.
So, the driver needs to drive 5 km towards North direction to reach the school.
Answer: (e)
According to the question, the direction diagram will be as follows.
Now, BC = DE = 5 km, CD = BE = 8 km AB = 2 km, EF = 1 km
Therefore, Required distance, AF = BE - (AB + EF) = 8 - (2 + 1) = 8 - 3 = 5 km.
From the direction diagram, it is clear that the bus is present finally in the South direction of school.
So, the driver needs to drive 5 km towards North direction to reach the school.
Question 57. In an exercise, Krishnan walked 25 m towards South and then he turned to his left and moved for 20 m. He again turned to his left and walked 25 m. Thereafter, he turned to his right and walked 15 m. What is the distance and direction of his present location with reference to the starting point?
Answer: Option C. -> 35 m, East
Answer: (c)
According to the question, the direction diagram will be as follows.
Now, AB = CD = 25 km, BC = AD = 20 m, DE = 15 m
Therefore, Required distance, AE = AD + DE AE = (20 + 15) = 35 m
Also, E is to the East of A.
Answer: (c)
According to the question, the direction diagram will be as follows.
Now, AB = CD = 25 km, BC = AD = 20 m, DE = 15 m
Therefore, Required distance, AE = AD + DE AE = (20 + 15) = 35 m
Also, E is to the East of A.
Answer: Option B. -> 15 m, East
Answer: (b)
Let Sharada starts from point A and passing through B and C, she reaches D.
Clearly, she is in East and at a distance of 15 m from her original position.
Answer: (b)
Let Sharada starts from point A and passing through B and C, she reaches D.
Clearly, she is in East and at a distance of 15 m from her original position.
Question 59.
[I.B.P.S. (PO) 2012]
If a persons walks 4 m towards the South from point L, takes a right turn and walks for another 3 m, then which of the following points would be reach?
Directions for the following questions: Read the following information carefully and answer the questions that follow.
Point H is 6 m towards the East of points G.Point R is 8 m North of point G.Point Q is exactly midway between point R and point G.Point K is 10 m to the South of point Q.Point L is 3 m towards the East of point Q.Point S is exactly midway between point G and point H.[I.B.P.S. (PO) 2012]
If a persons walks 4 m towards the South from point L, takes a right turn and walks for another 3 m, then which of the following points would be reach?
Answer: Option B. -> G
Answer: (b)
Clearly, the person starts from L goes 4 m southward and reaches point S from where he takes right turn and walks 3 m to reach G.
Answer: (b)
Clearly, the person starts from L goes 4 m southward and reaches point S from where he takes right turn and walks 3 m to reach G.
Question 60. Rajnikanth left his home foe office in car. He drove 15 km straight towards North and then turned Eastwards and covered 8 km. He, then turned to left and covered 1 km. He again turned left and drove for 20 km and reached office. How far and in what direction in his office from the home?
Answer: Option C. -> 20 km, North - West
Answer: (c)
According to the question, the direction diagram will be as follows.
Now, AB = 15 km, CD = BF = 1 km
AF = AB + BF = 15 + 1 = 16 km
DE = 20 km DF = BC = 8 km
EF = DE - DF = 20 - 8 = 12 km
Therefore, Required distance, AE = $√{(AF)^2 + (EF)^2}$
= $√{(16)^2 + (12)^2}$
= $√{(256 + 144)}$ = √400 = 20 km
So, his office is 20 km North - West.
Answer: (c)
According to the question, the direction diagram will be as follows.
Now, AB = 15 km, CD = BF = 1 km
AF = AB + BF = 15 + 1 = 16 km
DE = 20 km DF = BC = 8 km
EF = DE - DF = 20 - 8 = 12 km
Therefore, Required distance, AE = $√{(AF)^2 + (EF)^2}$
= $√{(16)^2 + (12)^2}$
= $√{(256 + 144)}$ = √400 = 20 km
So, his office is 20 km North - West.
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email