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- (d);
- (a);
- (a);
- (b);
- (c);
- (b); Obviously, Z is facing south
- (c);
Clearly, Alok is in South – West direction from starting point. - (b); As shown in Fig. the man initially faces in the direction OP. On moving 90° clockwise the man faces in the direction OQ. On further moving 135° anticlockwise, he faces in the direction OR, which is West.
- (d); As shown in Fig. the man initially faces in the direction OP. On moving 90° clockwise, he faces in the direction OX. On further moving 180° anticlockwise, he faces in the direction OY. Finally, on moving 90° anticlockwise, he faces in the direction OZ, which is South-east.
- (b); As shown in Fig. the man initially faces towards east i.e., in the direction OA. On moving 100° clockwise he faces in the direction OB. On further moving 145°clockwise, he faces in the direction OC, Clearly, OC makes an angle of (145°-100°) i.e. 45° with OA and as such points in the direction North-east.
- (b); The movements of Kishankant are as shown in Fig. (A to B, B to C and C to D).
AC = (AB-BC) =(10-6) km = 4 km
Kishankant’s distance from starting point A
= AD
= 5km
So Kishankant is 5 km to the North-east of his starting point.
- (d); The movements of Gaurav are as shown in Fig. Clearly, Gaurav’s distance from his initial position
P = PX = (PS + SX)
= (QR + SX) = (40 + 20)m = 60
- (c); The movements of Radha are as shown in figure.
Clearly, Radha’s distance from the starting point O = OD = (OC – CD) = (AB – CD) = (14 – 4)m = 10 m
- (c); The movements of rat from A to G are as shown in Fig.
Clearly, it is finally walking in the direction FG, ie North.
- (d); The movements of the person are from A to F, as shown in figure.
Clearly, the final position if F which is to the North-east of the starting point.
- (a); The movements of Rohit are as shown in fig.
Rohit’s distance from starting point A
= AE = (AD + DE) = (BC + DE) = (20 + 15) m = 35 m.
Also, E is to the East of A
- (c); The movements of Sachin are as shown in figure
(P to B, B to C, C to D and D to Q).
Clearly, distance of Q from P
= PQ = (DQ – PD) = (DQ –BC) = (40 – 30) m = 10 m.
Also, Q is to the West of P. So, Q is 10 m West of P.
- (c); The movements of the man are as shown in Fig.
(A to B, B to C, C to D and D to E)
Clearly, DF = BC = 5 km
EF= (DE-DF) = (9-5) km = 4 km
BF = CD = 2 km
AF = AB + BF = AB + CD = (1 + 2) km = 3 km
Man’s distance from starting point A = AE
- (c); The movements of Lokesh are as shown in Fig.
(A to B, B to C, C to D and D to E)
Clearly, his final position is E which is to the North of his house at A.
- (a); Let X and Y be two buses.
Bus X travels along the path PA, AB, BC, CD.
Now, AD = BC = 25km
So, PD = PA + AD = 50km
Bus Y travels 35 km upto E
– Distance between two buses = PQ – (PD + QE)
= [150 – (50+35)] = 65 km - (a); Clearly, the route of Sohan is as shown in the diagram given below:
Here, XB = XY – YB = XY – AZ (∴ YB = AZ)
= 8km – 7km =1km
- (d); From the figure given below, Raj is finally in the North – West direction from the starting point.
- (d); From the figure, it is clear that Alok is in the South-East direction with reference to starting Point Q.
- (d);
- (a);
Finally he is facing North direction. - (a);
Clearly the distance between A and B = 1km - (b);
- (b);
Let his starting point is S.
(29 – 30)
AB = SC =15 m, BC = AS = 15km
BD – BC = 35 – 15 = 20 km
SD = √SC2+√CD2 = √152+√202
= √225+400 = √625 = 25km
- (d); Total distance walked by the boy.
= 40+50+60+80=230m - (c); CE = √BC2 + √BE2 = √8500 = 92.19m
- (a);
- (d); The movements of the person are from A to F, as shown in figure.
Clearly, the final position if F which is to the North-east of the starting point.
- (a); The movements of Aditya are shown in figure from A to G. Clearly,
Gopal is finally walking in the direction FG i.e., North.
- (c); The movements indicated are as shown in figure.
Thus, the final movements in the direction indicated by DE, which is east.
- (d); The movements is shown in figure— Clearly, EB
= DC = 40 m.
So, distance from the starting point A = (AB – EB) = (75 – 40)m = 35m
- (b); : The movements is as shown in figure.
AC = (AB – BC) = (10 – 6) km = 4 km.
Clearly, D is to the North-east of A.
So, Kunal’s distance from starting point A = AD
= √AC2 + √CD2 = “ √42 + √32 = √25 = 5 km.
So, it is 5 km to the North-east of his starting point.
- (a); The movements of Rohan are as shown in figure
(A to B, B to C, C to D and D to E). Clearly, AD =
BC = 2 km
So, required distance = AE = (DE – AD) = (3 – 2)
= 1 km.
- (d); The movements of Manick are as shown in figure (A to B, B to C and C to D). Clearly, ABCD is a rectangle and so AD = BC = 20m. Thus, D is 20m to the west of A.
- (b); The movements of Namita are shown in figure (A to B, B to C, C to D and D to E). Clearly, Namita’s distance from his initial position = AE = (AB + BE) = (AB + CD) = (14 + 10) m = 24 m.
- (b); Clearly, DC = AB + FE. So, F is in the line with A. Also, AF = (BC – DE) = 5 m. So, the man is 5 metres away from his initial position.
- (c); The movements of Radha are as shown in figure.
Clearly, Radha’s distance from the starting point O = OD = (OC – CD)
= (AB – CD) = (14 – 4)m = 10 m.
- (c); The movements of Amit are shown in figure (P to Q, Q to R and R to S). Clearly, his final position is S which is to the South-east of the starting point P.
- (b); By Pythagorean Theorem:
AD = √92 + √122
AD = √81 + √144
AD = √225
AD = 15 km
- (a); The movements of the girl are as shown in figure (A to B, B to C, C to D, D to A).
Clearly, she is finally moving in the direction DA i.e., Northeast.
- (b); The movements of Sanjeev from A to F are as shown in Figure Clearly,
Sanjeev’s distance from starting point A
= AF = (AB + BF) = AB + ( BE – EF ) = AB + (CD – EF)
= [10 + (20 – 10)] m = (10 + 10) m = 20 m.
Also, F lies to the South of A. So, Sanjeev is 20 meters to the south of his starting point.
- (b); The movements of Kashish are as shown in figure (A to B, B to C, C to D, D to E). So, Kashish’s distance from his original position A = AE = (AB – BE) = (AB – CD) = (30 – 20) m = 10 m.
- (d); The movements of the man are as shown in figure. So, Man’s distance from initial position A = AE = (AB + BE) = (AB + CD) = (30 + 20) m = 50 m.
- (a); The movements of Rohit are as shown in the figure. So, Rohit’s distance from starting point A = AE = (AD + DE) = (BC + DE) = (20 + 15) m = 35 m.
Also, E is to the East of A
- (c); The movements of Sachin are as shown in figure (P to B, B to C, C to D and D to Q). Clearly, distance of Q from P = PQ = (DQ – PD) = (DQ – BC) = (40 – 30) m = 10 m. Also, Q is to the West of P. So, Q is 10 m West of P.
- (d); The movements of Ramakant are as shown in figure. Clearly, he is finally walking in the direction DE i.e., West.
- (d); As shown in figure, the man initially faces in the direction OA. On moving 135° anticlockwise, he faces in the direction OB. On further moving 180° clockwise, he faces in the direction OC, which is South – West.
- (b); As shown in figure, the man initially faces in the direction OP on moving 90° clockwise, the man faces in the direction OQ. On further moving 135° anti clockwise, he faces in the direction OR, which is West.
- (d);
- (b); As shown in fig, the river flows eastwards from A towards B, turns left and follows a semi-circular path to reach C where it turns left and flows east wards towards D.
- (c); The movements are as indicated in figure(O to A, A to B and b to C). Clearly, C lies to the South-West of O.
- (b); As shown in the figure, the man initially faces towards east i.e., in the direction OA. On moving 100∘ clockwise. he faces in the direction OB. On further moving 145∘ anticlockwise, he is facing the direction OC. Clearly, OC makes an angle of (145∘−100∘) i.e. 45∘ with OA and so, the man faces the direction North-east.
- (c); The movements of rat from A to G are shown in figure. Clearly, it is finally walking in the direction FG i.e., North.
- (b); The movements of Maya from T to R are as shown in figure.
RT = √(RV)2+√(TV)2 = √16 + √9 = 5ft - (b);
- (c);
