Practice Type 2 section 6 mcqs - verbal reasoning Online Quiz (set-1) For All Competitive Exams
DIRECTIONS:
- Point A is 8 m west of point B.
- Point E is 2 m east of point F.
- Point G is 3 m east of point H.
- Point E is 3 m north of point of point D.
- Point C is 9 m west of point D.
- Point G is 9 m north of point F.
- Point C is 6 m south of point B.
Q-1) A person starts from point G in east direction. Walks for 6 m, takes a right turn, now walks for 5 m. Now he takes a left turn, walks for 3 m, then after two consecutive right turns he reaches point E. Find the distance travelled by him to reach point E.
(a)
(b)
(c)
(d)
(e)
6 + 5+ 3 + (9-5) + (3+4) = 25 m
Q-2) If Sita walks 10 km towards West, then turned towards South and walked 10 km, then turned East walked 10 km, and turned North walked 10 km. How far she is from starting point?
(a)
(b)
(c)
(d)
Q-3) Mr. Das started his journey from his house straight to his friend’s house at a distance of 12 km. On returning he walked 8 km in the same route and turned right and walked 4 km, then he turned to his left and walked 4 km. Finally he turned to his left and walked 2 km. How far was he from his house ?
(a)
(b)
(c)
(d)
Required distance = AF
= 2 km
Q-4) Jai Prakash started walking towards south from the point A, walked 20 m and reached a point B, Again, he turned left and walked 20 m and reached a point C. Now he turned 45° anticlockwise, walked a distance of 20 √2 m and reached a point D. What approximately is the shortest distance between the point A and D?
(a)
(b)
(c)
(d)
Here, AC = $√{20^2 + 20^2} = 20$√{2}$ m
Since AB = BC
∴ ∠ACB = ∠BAC = 45°
∴ ∠ACD = 90°
Now,
AD2 = AC2 + CD2
= 800 + 800 = 1600
∴ AD = $√{1600}$ = 40 m
Q-5) A man walks 1 km towards East and then he turns to South and walks 5 km. Again he turns to East and walks 2 km, after this he turns to North and walks 9 km. Now, how far is he from his starting point ?
(a)
(b)
(c)
(d)
The movements of the man are as shown in the figure,
(A to B, B to C, C to D, D to E)
Clearly, DF = BC = 5km.
EF = (DE - DF) = (9 - 5)km = 4km.
BF = CD = 2km
AF =AB + BF = AB + CD = (1 + 2)km = 3km.
∴ Man's distance from initial position A
= AE = $√{AF^2 + FE^2}$
= $√{3^2 + 4^2}$
= $√25$ = 5m.
Q-6) Pran and Khan start from their office and walk in opposite direction, each travelling 10 km. Pran then turns left and walks 10 km. While Khan turns right and walks 10 km. How far they are now from each other?
(a)
(b)
(c)
(d)
It is clear from the diagram that they are at a distance of 20 km.
Q-7) Naseebah runs for 10 km in the eastern direction. She then turns left and starts walking for 6 km. Again, she turns left and starts running for 6 km. Then she turns left and walks again for 6 km. How far is she from the starting point ?
(a)
(b)
(c)
(d)
Required distance = 4 km
Q-8) Sohan started from point X and travelled forward 8 km up to point Y, then turned towards right and travelled 5 km up to point Z then turned right and travelled 7 km up to point A and then turned towards right and travelled 5 km up to B. What is the distance between point B and X ?
(a)
(b)
(c)
(d)
Required distance = xy– yB
= (8–7) km = 1 km
Q-9) Ram walked 4 km West of his house and then turned to South covering 8 km. Finally he moves 6 km towards East and then again 2 km West. How far is he from his initial position ?
(a)
(b)
(c)
(d)
AE = 8 km
DIRECTIONS:
- Point B is 9 km east of point A.
- Point T is 6 km south of point B.
- Point S is 4 km west of point T.
- Point T is 8 km west of point G.
Q-10) A person starts from point S, walks 3 km towards south and reaches point D. Next he takes two right turns and reaches point A. How much distance has he travelled?
(a)
(b)
(c)
(d)
(e)
Let after first right turn he is at point K. Since after turn from K he reaches at point A so point K is in south of point A.
So he has travelled SD + DK + KA = 3 + (9-4) + (3+6) = 17 km