Direction Sense Test Questions in English

(B) Let Ravi's house be at point $A$.
$1$. Ravi moves $2 \, km$ west to point $B$.
$2$. From $B$,he moves $4 \, km$ south to point $C$.
$3$. From $C$,he moves $3 \, km$ east to point $D$.
$4$. From $D$,he moves $1 \, km$ west to point $E$.
Now,we need to find the distance $AE$.
Let the horizontal displacement be $x$ and vertical displacement be $y$.
Horizontal displacement: $2 \, km$ (West) + $3 \, km$ (East) + $1 \, km$ (West) = $-2 + 3 - 1 = 0 \, km$.
Vertical displacement: $4 \, km$ (South) = $-4 \, km$.
Since the horizontal displacement is $0$,Ravi is directly south of his house.
The distance $AE$ is equal to the vertical displacement,which is $4 \, km$.

(C) Let the starting point be $A$. The man walks $6 \, km$ east to point $B$,then $5 \, km$ south to point $C$,then $6 \, km$ east to point $D$,and finally $10 \, km$ north to point $E$.
To find the distance $AE$,we construct a right-angled triangle. Extend $AB$ to a point $O$ such that $EO$ is perpendicular to $AO$.
Here,$AO = AB + BO = 6 \, km + 6 \, km = 12 \, km$ (since $BO = CD = 6 \, km$).
The vertical distance $OE = ED - OD = 10 \, km - 5 \, km = 5 \, km$.
Using the Pythagorean theorem in $\triangle AOE$:
$AE^2 = AO^2 + OE^2$
$AE^2 = 12^2 + 5^2$
$AE^2 = 144 + 25 = 169$
$AE = \sqrt{169} = 13 \, km$.
Thus,the man is $13 \, km$ away from his starting point.

(D) Arun started from point $A$ and moved $75 \, m$ towards the north to reach point $B$.
From $B$,he turned left and walked $25 \, m$ to reach point $C$.
From $C$,he turned left again and moved $80 \, m$ towards the south to reach point $D$.
At point $D$,he was facing south. He then turned to the right at an angle of $45^{\circ}$.
Turning $45^{\circ}$ to the right from the south direction leads to the south-west direction.
Therefore,he was finally moving in the south-west direction.

(A) Let the starting point be $A$. Satheesh drives $15 \, km$ north to point $B$,then $10 \, km$ west to point $C$.
From $C$,he turns south and travels $5 \, km$ to point $D$.
From $D$,he turns east and travels $8 \, km$ to point $E$.
From $E$,he turns right (which is south) and travels $10 \, km$ to point $F$.
Now,let's analyze the vertical and horizontal displacements:
Vertical displacement: $15 \, km$ (North) - $5 \, km$ (South) - $10 \, km$ (South) = $15 - 15 = 0 \, km$.
This means he is at the same horizontal level as his starting point $A$.
Horizontal displacement: $10 \, km$ (West) - $8 \, km$ (East) = $2 \, km$ (West).
Therefore,he is $2 \, km$ to the west of his starting point $A$.

(C) Let the starting point be $A$.
Kishore walks $10 \,km$ north to point $B$.
Then he walks $6 \,km$ south to point $C$.
So,the distance $AC = AB - BC = 10 \,km - 6 \,km = 4 \,km$ (towards north).
Then he walks $3 \,km$ towards east to point $D$.
Now,we have a right-angled triangle $ACD$ where $AC = 4 \,km$ and $CD = 3 \,km$.
The distance from the starting point $A$ is the hypotenuse $AD$.
$AD = \sqrt{AC^2 + CD^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,km$.
Since point $D$ is to the north and east of point $A$,the direction is north-east.
Thus,Kishore is $5 \,km$ north-east from his starting point.

(B) Let the starting point be $O$.
$1$. Man walks $20 \, m$ East to point $A$.
$2$. Turns South and walks $10 \, m$ to point $B$.
$3$. Turns West and walks $35 \, m$ to point $C$.
$4$. Turns North and walks $5 \, m$ to point $D$.
$5$. Turns East and walks $15 \, m$ to point $E$.
Let's use a coordinate system where $O = (0, 0)$.
- After $20 \, m$ East: $(20, 0)$.
- After $10 \, m$ South: $(20, -10)$.
- After $35 \, m$ West: $(20 - 35, -10) = (-15, -10)$.
- After $5 \, m$ North: $(-15, -10 + 5) = (-15, -5)$.
- After $15 \, m$ East: $(-15 + 15, -5) = (0, -5)$.
The final position is $(0, -5)$.
The initial position is $(0, 0)$.
The straight distance is $\sqrt{(0-0)^2 + (-5-0)^2} = \sqrt{25} = 5 \, m$.

(A) Let the starting point be $A$. James moves $30 \,m$ west to reach point $B$.
From $B$,he turns right (north) and walks $20 \,m$ to reach point $C$.
From $C$,he turns left (west) and walks $10 \,m$ to reach point $D$.
From $D$,he turns left (south) and walks $40 \,m$ to reach point $E$.
From $E$,he turns left (east) and walks $5 \,m$ to reach point $F$.
Finally,from $F$,he turns left. Since he was moving east,a left turn means he is now moving towards the north (direction $FG$).
Therefore,he is walking in the north direction.

(B) Let the starting point be $A$. Sudheesh walks $10 \, m$ south to point $B$.
He turns left (east) and walks $20 \, m$ to point $C$.
He turns right (south) and walks $20 \, m$ to point $D$.
He turns right (west) and walks $20 \, m$ to point $E$.
Finally,he turns right (north) and walks $10 \, m$ to point $F$.
Total vertical distance from $A$ is $10 \, m$ (south) $+ 20 \, m$ (south) $- 10 \, m$ (north) $= 20 \, m$ south.
Total horizontal distance is $20 \, m$ (east) $- 20 \, m$ (west) $= 0 \, m$.
Thus,he is $20 \, m$ south of the starting point.

(D) Let the initial position of the man be $A$.
$1$. He walks $30 \, m$ towards the south to reach point $B$.
$2$. He turns right (west) and walks $30 \, m$ to reach point $C$.
$3$. He turns left (south) and walks $20 \, m$ to reach point $D$.
$4$. He turns left (east) and walks $30 \, m$ to reach point $E$.
The total distance from the initial position $A$ to the final position $E$ is the sum of the vertical displacements:
Distance $= AB + BE$
Since $BE = CD = 20 \, m$,
Distance $= 30 \, m + 20 \, m = 50 \, m$.
Since $50 \, m$ is not among the given options,the correct answer is 'none of these'.

(B) Let the starting point be $A$. Gopal moves $1 \,km$ east to point $B$,then $5 \,km$ south to point $C$,then $2 \,km$ east to point $D$,and finally $9 \,km$ north to point $E$.
To find the distance $AE$,we construct a right-angled triangle by extending $AB$ to a point $O$ such that $EO$ is perpendicular to $AO$.
Here,$AO = AB + BO = 1 \,km + 2 \,km = 3 \,km$.
The vertical distance $OE = ED - OD = 9 \,km - 5 \,km = 4 \,km$.
Using the Pythagorean theorem in $\triangle AOE$:
$AE^2 = AO^2 + OE^2$
$AE^2 = 3^2 + 4^2 = 9 + 16 = 25$
$AE = \sqrt{25} = 5 \,km$.
Thus,Gopal is $5 \,km$ away from his starting point.

(C) Let Sumesh's house be at point $A$.
$1$. He moves $15 \, km$ north to point $B$.
$2$. He turns west and moves $10 \, km$ to point $C$.
$3$. He turns south and moves $5 \, km$ to point $D$.
$4$. Finally,he turns east and moves $10 \, km$ to point $E$.
Since the distance moved west $(10 \, km)$ is equal to the distance moved east $(10 \, km)$,his final position $E$ lies on the same vertical line as $A$ and $B$.
Since he moved $15 \, km$ north and then $5 \, km$ south,his net displacement from $A$ is $15 - 5 = 10 \, km$ north.
Therefore,he is in the north direction from his house.

(A) Let the house be at point $A$. Lakshmi walks $50 \, m$ south to point $B$.
Then she turns left (east) and walks $20 \, m$ to point $C$.
Then she turns north and walks $30 \, m$ to point $D$.
Now,she is at point $D$ and needs to go back to her house at point $A$.
Looking at the coordinate system,point $D$ is to the south-east of point $A$. Therefore,to go from $D$ to $A$,she must move in the north-west direction.

$ (A) $ Let the starting point be $A$.
$1$. Dinesh walks $20 \,m$ north to reach $B$.
$2$. Turns right and walks $30 \,m$ to reach $C$.
$3$. Turns right and walks $35 \,m$ to reach $G$.
$4$. Turns left and walks $15 \,m$ to reach $D$.
$5$. Turns left and walks $15 \,m$ to reach $E$.
$6$. Turns left and walks $15 \,m$ to reach $F$.
Now, let's analyze the horizontal and vertical displacements from $A(0,0)$:
- North/South ($y$-axis): $+20 (up) - 35 (down) + 15 (up) = 0 \,m$.
- East/West ($x$-axis): $+30 (right) - 15 (left) = +15 \,m$ (east).
Thus, he is $15 \,m$ east from his original position.

(D) Anil's house faces east,so the back-side of the house faces west.
$1$. Anil starts from point $A$ and walks $50 \,m$ towards the west to reach point $B$.
$2$. From $B$,he turns right (towards the north) and walks $50 \,m$ to reach point $C$.
$3$. From $C$,he turns left (towards the west) and walks $25 \,m$ to reach the final point $D$.
$4$. Looking at the starting point $A$,the final position $D$ is towards the north-west direction.

(D) Let the initial direction of the bus be $x$.
$1$. The bus starts from the house and moves in direction $x$.
$2$. It turns right (first right turn).
$3$. It turns right again (second right turn).
$4$. It turns left.
$5$. Finally,the bus faces North.
Working backwards from the final position:
- If the final direction is North,and the last turn was a left turn,the direction before the left turn was East (since a left turn from East leads to North).
- Before the second right turn,the direction was South (since a right turn from South leads to East).
- Before the first right turn,the direction was West (since a right turn from West leads to South).
Therefore,the bus was initially facing West when it left the house.

(A) In a card game with four players sitting around a table,they face the center of the table.
Given that $D$ faces north,$D$ is sitting on the south side of the table.
Since $A$ faces west,$A$ is sitting on the east side of the table.
$A$ and $B$ are partners,so they sit opposite each other. Thus,$B$ sits on the west side and faces east.
$C$ must sit opposite to $D$. Since $D$ is on the south side facing north,$C$ is on the north side facing south.
Therefore,$C$ faces south.

(C) $1$. According to the given information,$Q$ is to the south-west of $P$.
$2$. $R$ is to the east of $Q$ and south-east of $P$.
$3$. $T$ is to the north of $R$ and is in line with $QP$. This means if we extend the line $QP$ further,it passes through $T$.
$4$. By visualizing the positions on a compass,$P$ is the reference point. Since $T$ is positioned above and to the right of $P$ along the extended line of $QP$,$T$ is located in the north-east direction with respect to $P$.

(B) Let the initial distance between $A$ and $B$ be $200 \,m$.
$B$ walks $60 \,m$ along the road,then turns left and walks $20 \,m$,then turns right and walks $40 \,m$,and finally turns right to return to the original road.
Since $B$ turns right after walking $40 \,m$ and returns to the road,the distance covered by $B$ along the original road is $60 \,m + 40 \,m = 100 \,m$.
Since $A$ and $B$ walk at the same speed,$A$ also covers a total distance of $140 \,m$ (the total path length of $B$ is $60 + 20 + 40 + 20 = 140 \,m$).
$A$ moves $140 \,m$ towards $B$ along the original road.
Initially,$A$ and $B$ were $200 \,m$ apart. After $A$ moves $140 \,m$ and $B$ effectively moves $100 \,m$ along the road,the remaining distance between them is $200 - (140 + 100) = -40 \,m$. This implies they have crossed each other.
The distance between them is $|200 - (140 + 100)| = 40 \,m$.

(D) Let the four positions be North,East,South,and West.
Given that no lady is facing East,it implies that a man must be facing East.
Since persons sitting opposite to each other are not of the same sex,the person sitting opposite to the man facing East (i.e.,at the West position) must be a lady. Thus,one lady is facing West.
It is also given that one man is facing South. The person sitting opposite to him (i.e.,at the North position) must be a lady. Thus,the other lady is facing North.
Therefore,the ladies are facing North and West.

(D) In the morning,the sun rises in the $East$. Therefore,shadows fall towards the $West$.
Since Vishakh's shadow falls to his left,he must be facing $North$ (because if one faces $North$,the $West$ direction is to their left).
Since Vishakh and Satheesh are standing with their backs towards each other,Satheesh must be facing the opposite direction of Vishakh.
Therefore,Satheesh is facing $South$.

(C) Let the distance of the house from the school be $x$.
According to the problem,the distance of the market from the post office is also $x$.
Let the school be at the origin $(0,0)$ on a Cartesian plane.
The post office is to the east of the school,so its position is $(d, 0)$ for some distance $d$.
The house is to the south of the school,so its position is $(0, -x)$.
The market is to the north of the post office,so its position is $(d, x)$.
To find the direction of the market with respect to the school,we look at the coordinates $(d, x)$.
Since both $d$ and $x$ are positive,the market lies in the first quadrant relative to the school,which corresponds to the North-east direction.

(C) $1$. Initial positions: $A$ is at south-west,$B$ is at north-east,$C$ is at north-west,and $D$ is at south-east.
$2$. $A$ moves diagonally to the opposite corner (north-east) and then one side clockwise,which brings it to the south-east corner.
$3$. $C$ moves diagonally to the opposite corner (south-east) and then one side anticlockwise,which brings it to the north-east corner.
$4$. $D$ moves two sides clockwise,which brings it to the north-east corner.
$5$. $B$ moves two sides anticlockwise,which brings it to the south-west corner.
$6$. Therefore,$A$ is now at the south-east corner.

(C) $1$. Initial positions: $A$ is at bottom-left,$D$ is at bottom-left (Wait,looking at the figure,$A$ is bottom-right,$D$ is bottom-left,$B$ is top-left,$C$ is top-right). Let's re-evaluate based on the standard clockwise order $ADBC$ given in the prompt.
$2$. $A$ and $B$ move one arm length clockwise: $A$ moves from bottom-right to top-right,$B$ moves from top-left to bottom-left.
$3$. Then $A$ and $B$ cross over to the diagonally opposite corner: $A$ moves from top-right to bottom-left,$B$ moves from bottom-left to top-right.
$4$. $C$ and $D$ move one arm length anticlockwise: $C$ moves from top-right to top-left,$D$ moves from bottom-left to bottom-right.
$5$. Then $C$ and $D$ cross over to the diagonally opposite corner: $C$ moves from top-left to bottom-right,$D$ moves from bottom-right to top-left.
$6$. Final positions: $A$ is at bottom-left,$D$ is at top-left,$B$ is at top-right,$C$ is at bottom-right.
$7$. The new configuration is $DACB$.

(C) $1$. Let the starting point be $P$. Rajesh faces East,turns left (North),and walks $10 \, m$.
$2$. He turns left again (West) and walks $10 \, m$. At this point,he is $10 \, m$ North and $10 \, m$ West of his starting point.
$3$. He then turns $45^{\circ}$ to his right. Since he was facing West,a $45^{\circ}$ right turn means he is now facing North-West.
$4$. He walks $25 \, m$ in this North-West direction.
$5$. By analyzing the final position relative to the starting point $P$,he is in the North-West direction from $P$.

(D) Initially,the man is facing West.
$1$. He turns $45^{\circ}$ clockwise: West to North-West.
$2$. He turns another $180^{\circ}$ clockwise: North-West to South-East.
$3$. He turns $270^{\circ}$ anticlockwise: From South-East,turning $270^{\circ}$ anticlockwise is equivalent to turning $90^{\circ}$ clockwise.
$4$. Turning $90^{\circ}$ clockwise from South-East leads to South-West.
Therefore,the man is now facing South-West. The correct option is $(d)$.

(B) Let Ravi's home be at point $A$.
$1$. He cycles $10 \,km$ south to point $B$.
$2$. He turns right (west) and cycles $5 \,km$ to point $C$.
$3$. He turns right (north) and cycles $10 \,km$ to point $D$.
$4$. He turns left (west) and cycles $10 \,km$ to point $E$.
To reach home $(A)$ from point $E$ in a straight line,he needs to cover the horizontal distance and the vertical distance.
The total horizontal distance from $E$ to $A$ is the sum of the segments $5 \,km$ (from $C$ to $B$ equivalent) and $10 \,km$ (from $E$ to $D$ equivalent),which is $5 \,km + 10 \,km = 15 \,km$.

(B) Let the starting point be $A$. The child moves $90 \, m$ east to point $B$. Then he turns right and moves $20 \, m$ to point $C$. Then he turns right again and moves $30 \, m$ to point $D$ (uncle's place). From $D$,he turns right and moves $100 \, m$ north to point $E$ where he meets his father.
Let's represent this on a coordinate plane:
Starting point $A = (0, 0)$.
Point $B = (90, 0)$.
Point $C = (90, -20)$.
Point $D = (90 - 30, -20) = (60, -20)$.
Point $E = (60, -20 + 100) = (60, 80)$.
The distance from the starting point $A(0, 0)$ to the final point $E(60, 80)$ is given by the distance formula:
$AE = \sqrt{(60 - 0)^2 + (80 - 0)^2}$
$AE = \sqrt{60^2 + 80^2}$
$AE = \sqrt{3600 + 6400}$
$AE = \sqrt{10000}$
$AE = 100 \, m$.

(D) Let the starting point be $A$.
$1$. Kailash faces North and walks $25 \,m$ to his right,reaching point $B$ (East direction).
$2$. He turns left and walks $30 \,m$ to point $C$ (North direction).
$3$. He turns right and walks $25 \,m$ to point $D$ (East direction).
$4$. He turns right and walks $55 \,m$ to point $E$ (South direction).
$5$. Finally,he turns right and walks $40 \,m$ to point $F$ (West direction).
Comparing the final position $F$ with the starting point $A$:
- Total horizontal displacement: $25 \,m$ (East) + $25 \,m$ (East) - $40 \,m$ (West) = $10 \,m$ (East).
- Total vertical displacement: $30 \,m$ (North) - $55 \,m$ (South) = $-25 \,m$ (South).
Since the final position is to the East and South of the starting point,he is in the South-east direction.

(D) $1$. Deepa starts from point $A$ and moves $75 \, \text{m}$ towards the North to reach point $B$.
$2$. From $B$,she turns left (West) and walks $25 \, \text{m}$ to reach point $C$.
$3$. From $C$,she turns left again (South) and walks $80 \, \text{m}$ to reach point $D$.
$4$. At point $D$,she is moving in the South direction. She then turns to the right at an angle of $45^{\circ}$.
$5$. $A$ right turn of $45^{\circ}$ from the South direction points towards the South-west direction. Therefore,she is finally moving in the South-west direction.

(D) Let the starting point be $A$. Kunal walks $10 \,km$ North to reach point $B$.
From $B$,he walks $6 \,km$ South to reach point $C$.
Thus,the distance $AC = AB - BC = 10 \,km - 6 \,km = 4 \,km$.
From $C$,he turns East and walks $3 \,km$ to reach point $D$.
Now,we have a right-angled triangle $ACD$ where $AC = 4 \,km$ and $CD = 3 \,km$.
The distance from the starting point $A$ is the hypotenuse $AD$.
Using the Pythagorean theorem: $AD = \sqrt{AC^2 + CD^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,km$.
Since point $D$ is to the East and North of $A$,he is $5 \,km$ North-East of his starting point.

(A) Let the starting point be $A$.
$1$. Johnson drives $15 \, km$ North to point $B$.
$2$. He turns West and drives $10 \, km$ to point $C$.
$3$. He turns South and drives $5 \, km$ to point $D$.
$4$. He turns East and drives $8 \, km$ to point $E$.
$5$. Finally,he turns right (which is South) and drives $10 \, km$ to point $F$.
Now,calculate the net displacement:
- North-South direction: $15 \, km$ (North) $- 5 \, km$ (South) $- 10 \, km$ (South) $= 15 - 15 = 0 \, km$.
- East-West direction: $10 \, km$ (West) $- 8 \, km$ (East) $= 2 \, km$ (West).
Since the net North-South displacement is $0$,he is on the same horizontal line as the starting point $A$. The net displacement is $2 \, km$ towards the West.

(D) Initially,the man is facing South.
He turns $135^{\circ}$ in the anticlockwise direction. From South,a $135^{\circ}$ anticlockwise turn brings him to the North-east direction.
Then,he turns $180^{\circ}$ in the clockwise direction. $A$ $180^{\circ}$ turn from North-east brings him to the opposite direction,which is South-west.
Therefore,the man is now facing South-west.

(B) Initially,the man is facing north-west.
He turns $90^{\circ}$ clockwise,which brings him to face north-east.
Then,he turns $135^{\circ}$ anticlockwise from the north-east direction.
Since the angle between north-east and north-west is $90^{\circ}$ and the angle between north-west and west is $45^{\circ}$,a total turn of $135^{\circ}$ anticlockwise from north-east will point him exactly towards the west.
Therefore,the man is now facing west.

(D) The man is initially facing North-West.
$1$. He turns $90^{\circ}$ clockwise: North-West + $90^{\circ}$ clockwise = North-East.
$2$. He turns $180^{\circ}$ anticlockwise: North-East + $180^{\circ}$ anticlockwise = South-West.
$3$. He turns another $90^{\circ}$ in the same (anticlockwise) direction: South-West + $90^{\circ}$ anticlockwise = South-East.
Therefore,the man is now facing South-East.

(B) Initially,the person is facing East.
Turning $100^{\circ}$ clockwise from East brings the person to a position $100^{\circ}$ clockwise from the East axis.
Then,turning $145^{\circ}$ anticlockwise means moving back $145^{\circ}$ from the current position.
The net change in direction is $145^{\circ} - 100^{\circ} = 45^{\circ}$ in the anticlockwise direction from the East.
Since $45^{\circ}$ anticlockwise from East is North-east,the person is now facing North-east.

(D) Let the starting point be $A$. Deepak walks $75 \, m$ towards the east to reach point $B$.
From $B$,he turns left and walks $25 \, m$ to reach point $C$.
From $C$,he turns left and walks $40 \, m$ to reach point $D$.
From $D$,he turns left and walks $25 \, m$ to reach point $E$.
Since the distance $BC = 25 \, m$ and $DE = 25 \, m$,point $E$ lies on the line segment $AB$.
The distance $AE = AB - EB$.
Since $EB = DC = 40 \, m$,we have $AE = 75 \, m - 40 \, m = 35 \, m$.
Thus,Deepak is $35 \, m$ away from the starting point.
Since $35 \, m$ is not among the given options,the correct answer is 'None of these'.

(A) $1$. Kishenkant starts from point $A$ and walks $10 \, km$ North to reach point $B$.
$2$. From $B$,he walks $6 \, km$ South to reach point $C$. The distance from the starting point $A$ to $C$ is $10 \, km - 6 \, km = 4 \, km$ (North).
$3$. From $C$,he walks $3 \, km$ East to reach point $D$.
$4$. Now,we need to find the distance $AD$ and the direction of $D$ with respect to $A$.
$5$. In the right-angled triangle $ACD$,by Pythagoras theorem:
$AD^2 = AC^2 + CD^2$
$AD^2 = 4^2 + 3^2$
$AD^2 = 16 + 9 = 25$
$AD = \sqrt{25} = 5 \, km$.
$6$. Since $D$ is to the North and East of $A$,the direction is North-East.

(B) Let the starting point be $O$.
$1$. The man moves $20 \,m$ East to reach point $A$.
$2$. He turns South and moves $10 \,m$ to reach point $B$.
$3$. He turns West and moves $35 \,m$ to reach point $C$.
$4$. He turns North and moves $5 \,m$ to reach point $D$.
$5$. He turns East and moves $15 \,m$ to reach point $E$.
Let's use a coordinate system where the starting point $O$ is $(0, 0)$.
- After $20 \,m$ East: $(20, 0)$.
- After $10 \,m$ South: $(20, -10)$.
- After $35 \,m$ West: $(20 - 35, -10) = (-15, -10)$.
- After $5 \,m$ North: $(-15, -10 + 5) = (-15, -5)$.
- After $15 \,m$ East: $(-15 + 15, -5) = (0, -5)$.
The final position is $(0, -5)$.
The initial position is $(0, 0)$.
The straight distance is the magnitude of the displacement vector: $\sqrt{(0-0)^2 + (-5-0)^2} = \sqrt{0 + 25} = 5 \,m$.

(D) Let the starting point be $P$.
$1$. Gaurav walks $20 \, m$ North to reach point $Q$.
$2$. He turns left and walks $40 \, m$ to reach point $R$.
$3$. He turns left again and walks $20 \, m$ to reach a point $S$ (which is directly West of $P$).
$4$. He turns right and walks $20 \, m$ further away from the original line.
However,based on the standard interpretation of such problems,the total horizontal distance from the starting point $P$ is the sum of the horizontal segments.
Total distance from $P = 40 \, m + 20 \, m = 60 \, m$.

(C) Let the starting point be $A$. Radha moves $7 \, km$ to $B$ (South-East),then $14 \, km$ to $C$ (West),then $7 \, km$ to $D$ (North-West),and finally $4 \, km$ to $E$ (East).
Since the movement $AB$ ($7 \, km$ South-East) and $CD$ ($7 \, km$ North-West) are parallel and equal in magnitude but opposite in direction,they cancel each other out in terms of vertical and horizontal displacement.
The net displacement is the horizontal distance covered along the West-East axis.
She moved $14 \, km$ West and then $4 \, km$ East.
Required Distance $= 14 \, km - 4 \, km = 10 \, km$.

(A) Let Gopal's house be at point $A$.
$1$. He walks $30 \, m$ towards the West to reach point $B$.
$2$. He turns right (North) and walks $20 \, m$ to reach point $C$.
$3$. He turns left (West) and walks $10 \, m$ to reach point $D$.
$4$. He turns left (South) and walks $40 \, m$ to reach point $E$.
$5$. He turns left (East) and walks $5 \, m$ to reach point $F$.
$6$. Finally,he turns left (North) and starts walking towards point $G$.
Thus,Gopal is finally walking in the North direction.

(C) Let us trace the path of the rat step by step:
$1$. The rat starts at $A$ and runs $20$ units towards East to reach point $B$.
$2$. At $B$,it turns right (facing South) and runs $10$ units to reach point $C$.
$3$. At $C$,it turns right (facing West) and runs $9$ units to reach point $D$.
$4$. At $D$,it turns left (facing South) and runs $5$ units to reach point $E$.
$5$. At $E$,it turns left (facing East) and runs $12$ units to reach point $F$.
$6$. At $F$,it turns left (facing North) and runs $6$ units to reach point $G$.
As shown in the figure,the final movement is along the line $FG$,which is in the North direction. Therefore,the rat is facing North.

(A) Let the girl's home be at point $A$.
$1$. She walks $30 \, m$ North-west to reach point $B$.
$2$. From $B$,she walks $30 \, m$ South-west to reach point $C$.
$3$. From $C$,she walks $30 \, m$ South-east to reach point $D$.
$4$. Finally,she turns towards her home at $A$.
By observing the geometric path,the movement from $D$ to $A$ is in the North-east direction.

(B) Let the starting point be $A.$
$1$. Sanjeev walks $10 \, m$ South to point $B.$
$2$. Turning left,he walks $20 \, m$ East to point $C.$
$3$. Turning right,he walks $20 \, m$ South to point $D.$
$4$. Turning right,he walks $20 \, m$ West to point $E.$
$5$. Finally,turning right,he walks $10 \, m$ North to point $F.$
Now,let's calculate the net displacement:
Vertical movement: $10 \, m$ (South) $+ 20 \, m$ (South) $- 10 \, m$ (North) $= 20 \, m$ South.
Horizontal movement: $20 \, m$ (East) $- 20 \, m$ (West) $= 0 \, m.$
Thus,Sanjeev is $20 \, m$ to the South of his starting point $A.$

(B) Let the starting point be $A$.
$1$. Kashish walks $30 \, m$ North to point $B$.
$2$. Turns right and walks $40 \, m$ to point $C$.
$3$. Turns right and walks $20 \, m$ to point $D$.
$4$. Turns right and walks $40 \, m$ to point $E$.
Since $BC = 40 \, m$ and $DE = 40 \, m$,point $E$ lies on the line segment $AB$.
The distance from the original position $A$ is $AE = AB - BE$.
Since $BE = CD = 20 \, m$,we have $AE = 30 \, m - 20 \, m = 10 \, m$.

(D) Let the starting point be $A.$
$1$. Facing South,$I$ turn right (West) and walk $20 \, m$ to reach point $B.$
$2$. Turning right (North),$I$ walk $10 \, m$ to reach point $C.$
$3$. Turning left (West),$I$ walk $10 \, m$ to reach point $D.$
$4$. Turning right (North),$I$ walk $20 \, m$ to reach point $E.$
$5$. Turning right (East),$I$ walk $60 \, m$ to reach point $F.$
Comparing the final position $F$ with the starting point $A,$ the total displacement in the North direction is $10 + 20 = 30 \, m$ and in the East direction is $20 + 10 + 60 = 90 \, m$ (relative to the starting vertical axis). Since the final position is both North and East of the starting point,the direction is North-east.

(B) Let the starting point be $O$.
$1$. The man walks $30 \,m$ towards South to reach point $A$.
$2$. He turns right (West) and walks $30 \,m$ to reach point $B$.
$3$. He turns left (South) and walks $20 \,m$ to reach point $C$.
$4$. He turns left (East) and walks $30 \,m$ to reach point $D$.
Since he walked $30 \,m$ West and then $30 \,m$ East,his horizontal displacement is zero.
His total vertical displacement is $30 \,m$ (South) $+ 20 \,m$ (South) $= 50 \,m$.
Therefore,his distance from the initial position is $50 \,m$.

(A) Let the starting point be $A$.
$1$. Rohit walks $25 \, m$ South to reach point $B$.
$2$. He turns left and walks $20 \, m$ East to reach point $C$.
$3$. He turns left and walks $25 \, m$ North to reach point $D$.
$4$. He turns right and walks $15 \, m$ East to reach point $E$.
Total horizontal distance from $A$ to $E$ is the sum of the horizontal segments $BC$ and $DE$.
Distance $= BC + DE = 20 \, m + 15 \, m = 35 \, m$.
Since point $E$ is to the right of the vertical line starting from $A$,he is in the East direction from the starting point.
Therefore,he is $35 \, m$ away in the East direction.

(A) Let the starting point be $P.$
$1$. Sachin walks $20 \,m$ South to reach point $A.$
$2$. He turns left (East) and walks $30 \,m$ to reach point $B.$
$3$. He turns left (North) and walks $20 \,m$ to reach point $C.$
$4$. He turns left (West) and walks $40 \,m$ to reach point $Q.$
Since he walked $30 \,m$ East and then $40 \,m$ West,his net displacement in the East-West direction is $40 \,m - 30 \,m = 10 \,m$ towards the West.
Since he walked $20 \,m$ South and then $20 \,m$ North,his net displacement in the North-South direction is $0.$
Therefore,point $Q$ is $10 \,m$ to the West of point $P.$

(D) Let the starting point be $A$. Ramakant walks northwards to point $B$.
From $B$,he turns to his right and walks to point $C$.
From $C$,he turns to his left and walks $1 \ km$ to point $D$.
From $D$,he turns to his left again and walks towards point $E$.
Since he was moving northwards initially,his right turn leads him East,his left turn leads him North,and his final left turn leads him West.
Therefore,he is now moving in the West direction.