Direction Sense Test Questions in English

(D) Initial positions: $A$ (North-West),$B$ (North-East),$C$ (South-East),$D$ (South-West).
$1$. $C$ moves diagonally to North-West and then one side clockwise to North-East.
$2$. $A$ moves diagonally to South-East and then one side anticlockwise to North-East.
$3$. $B$ moves two sides clockwise from North-East to South-West.
$4$. $D$ moves two sides anticlockwise from South-West to North-West.
Therefore,$D$ is at the North-West corner.

(C) The perimeter of the square field is $4 \times 90 = 360 \,m$.
Initially,Rohan is at $B$ and Rahul is at $C$. The distance between them along the path (clockwise from $B$ to $C$) is $90 \,m$.
Since they move in opposite directions (Rohan clockwise,Rahul anticlockwise),their relative speed is $8 + 10 = 18 \,km/hr$.
For the $1st$ meeting,they must cover the initial distance of $90 \,m$. Time taken $t_1 = 90 \,m / 18 \,km/hr = 0.09 \,km / 18 \,km/hr = 0.005 \,hr = 18 \,seconds$.
Distance covered by Rohan = $8 \,km/hr \times 0.005 \,hr = 0.04 \,km = 40 \,m$.
For the $2nd$ meeting,they must cover the total perimeter of $360 \,m$ additionally.
Total distance covered by both together = $90 + 360 = 450 \,m$.
Time taken for $2nd$ meeting $t_2 = 450 \,m / 18 \,km/hr = 0.45 \,km / 18 \,km/hr = 0.025 \,hr = 90 \,seconds$.
Total distance covered by Rohan = $8 \,km/hr \times 0.025 \,hr = 0.2 \,km = 200 \,m$.
Rohan starts at $B$ and moves clockwise $(B \rightarrow C \rightarrow D \rightarrow A \rightarrow B)$.
After $200 \,m$: $B$ to $C$ $(90 \,m)$,$C$ to $D$ $(90 \,m)$,$D$ to $A$ $(20 \,m)$.
Total distance covered by Rahul = $10 \,km/hr \times 0.025 \,hr = 0.25 \,km = 250 \,m$.
Rahul starts at $C$ and moves anticlockwise $(C \rightarrow B \rightarrow A \rightarrow D \rightarrow C)$.
After $250 \,m$: $C$ to $B$ $(90 \,m)$,$B$ to $A$ $(90 \,m)$,$A$ to $D$ $(70 \,m)$.
Both are at the same point on $AD$ at a distance of $20 \,m$ from $D$ or $70 \,m$ from $A$.
Re-evaluating the meeting point: The relative distance covered for the second meeting is $360 \,m$ after the first meeting. Total distance $450 \,m$. Rohan covers $200 \,m$ from $B$ clockwise,reaching $20 \,m$ past $D$ on $AD$. Rahul covers $250 \,m$ from $C$ anticlockwise,reaching $70 \,m$ past $A$ on $AD$. Both meet at $20 \,m$ from $D$ on $AD$.

(B) Balu's birthday is on $27^{th}$ January,which is a Wednesday.
To find the first Friday after $27^{th}$ January: Wednesday $(27^{th})$,Thursday $(28^{th})$,Friday $(29^{th})$.
So,the first Friday is on $29^{th}$ January.
The fifth Friday will be $4$ weeks after the first Friday.
Number of days = $29^{th} \text{ (first Friday)} + (4 \times 7 \text{ days}) = 29 + 28 = 57^{th}$ day of the sequence.
Since January has $31$ days,the date of the fifth Friday is $57 - 31 = 26^{th}$ February.
Difference in days = $(31 - 27) \text{ (remaining days in Jan)} + 26 \text{ (days in Feb)} = 4 + 26 = 30$ days.
Therefore,Mohan is $30$ days younger than Balu.

(B) Shailesh saw the movie on Monday.
Nitin saw the movie $3$ days after Shailesh,so Nitin saw it on Monday + $3$ days = Thursday.
Nitin saw the movie $2$ days prior to Vikas,which means Vikas saw the movie $2$ days after Nitin.
Therefore,Vikas saw the movie on Thursday + $2$ days = Saturday.

(D) $1$. Pravin starts from a point and walks $30 \, m$ towards the East.
$2$. He takes a right turn and walks $20 \, m$ towards the South.
$3$. He takes another right turn and walks $30 \, m$ towards the West.
$4$. Now,he is exactly $20 \, m$ South of his starting point.
$5$. Therefore,the distance from the starting point is $20 \, m$.

(D) According to the given information:
$1$. $R$ is to the West of $P$ (i.e.,$R - P$ horizontally).
$2$. $P$ is to the North of $S$ (i.e.,$P$ is above $S$).
$3$. $T$ is to the East of $S$ (i.e.,$S - T$ horizontally).
Combining these,we get a layout where $P$ is North-East of $S$,and $R$ is West of $P$. Since $S$ is South of $P$ and $T$ is East of $S$,$T$ is South-East of $P$.
Relative to $R$,$P$ is to the East. Since $T$ is South-East of $P$,$T$ lies in the South-East direction with respect to $R$.

(B) Given that the $3^{rd}$ day of the month is Tuesday.
To find the day on the $27^{th}$,we calculate the difference in days: $27 - 3 = 24$ days.
Since there are $7$ days in a week,we divide $24$ by $7$: $24 = 7 \times 3 + 3$.
This means the $27^{th}$ day will be $3$ days after Tuesday,which is Wednesday,Thursday,Friday. So,the $27^{th}$ is Friday.
The $4^{th}$ day before the $27^{th}$ day is $27 - 4 = 23^{rd}$ day.
Since the $27^{th}$ is Friday,the $26^{th}$ is Thursday,the $25^{th}$ is Wednesday,the $24^{th}$ is Tuesday,and the $23^{rd}$ is Monday.
Therefore,the correct answer is Monday.

(D) $1$. Let us represent the positions on a coordinate plane:
$2$. Sushil is North of Rajesh.
$3$. Rajesh is West of Kamlesh.
$4$. This implies Sushil is North-West of Kamlesh.
$5$. Arun lives to the South of Sushil.
$6$. Since the exact distance between Sushil and Rajesh is not given,Arun's position relative to Kamlesh could be North-West,West,or South-West depending on how far South of Sushil he lives.
$7$. Therefore,the exact direction cannot be determined.

(B) Sachin is free from $10^{th}$ April to $11^{th}$ April (since he leaves on $12^{th}$ April).
Vinod is free from $11^{th}$ April onwards (since his exams end on $10^{th}$ April).
Therefore,the common dates when both are free are $11^{th}$ April.
However,looking at the options provided,if we consider the overlap of their availability: Sachin is free on $10^{th}$ and $11^{th}$. Vinod is free on $11^{th}$ and onwards. Thus,they can definitely meet on $11^{th}$ April. Given the options,$11^{th}$ is the only day they are both free. Since the question asks for the dates they can meet,and $11^{th}$ is the only common date,the provided options might be slightly ambiguous. Re-evaluating: Sachin is free $10, 11$. Vinod is free $11, 12, ...$. The only common date is $11^{th}$ April.

(D) The watch digits are $12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1$.
Following the pattern of replacing with alternate letters starting from $A$ $(12 \rightarrow A, 11 \rightarrow C, 10 \rightarrow E, 9 \rightarrow G, 8 \rightarrow I, 7 \rightarrow K, 6 \rightarrow M, 5 \rightarrow O)$:
At $5:00$,the small hand (hour hand) points to $5$.
According to the replacement pattern,$5$ is replaced by $O$.
Therefore,the small hand will be on the alphabet $O$.

(A) $1$. Ram is facing South.
$2$. Ramesh is walking towards Ram,which means Ramesh is walking in the North direction.
$3$. Ramesh stops and turns to his right. Since he was moving North,turning to his right means he is now facing East.
$4$. Ramesh sees Umesh standing before him,facing him. This means Umesh is standing in the East direction and facing West (towards Ramesh).
$5$. Therefore,Umesh is facing West.

(A) Satish read the book on Sunday.
Sudha read the book $4$ days after Satish. Since Sunday is the $1^{st}$ day,$4$ days after Sunday is Thursday ($1+4=5$,where $1$ is Sunday,$2$ is Monday,$3$ is Tuesday,$4$ is Wednesday,$5$ is Thursday).
Sudha read the book one day prior to Anil. This means Anil read the book one day after Sudha.
Since Sudha read it on Thursday,Anil read it on Friday.

(A) $1$. Suresh is at the $10^{th}$ position from the left end.
$2$. Manish is $4^{th}$ to the left of Suresh,so Manish's position from the left is $10 - 4 = 6^{th}$.
$3$. Nisha is $2^{nd}$ to the right of Suresh,so Nisha's position from the left is $10 + 2 = 12^{th}$.
$4$. Nisha is also $8^{th}$ from the right end of the row.
$5$. The total number of children in the row can be calculated using the formula: $\text{Total} = (\text{Position from left} + \text{Position from right}) - 1$.
$6$. Total = $(12 + 8) - 1 = 20 - 1 = 19$.

(D) Saroj is available only after her lunch break, but the duration of her lunch break is not specified.
Similarly, Rupa is free after her meeting, but the duration of her meeting is not specified.
Since the end times for these activities are unknown, it is impossible to determine the exact time they are both available to attend the seminar together.
Therefore, the answer cannot be determined.

(C) $1$. Initially,Saroj is facing West.
$2$. She takes a left turn. $A$ left turn from West leads to the South direction.
$3$. She walks some distance and takes another left turn. $A$ left turn from South leads to the East direction.
$4$. Therefore,Saroj is now facing East.

(A) $1$. Raman starts from $P$ and walks $5 \, km$ South to reach $Q.$
$2$. At $Q,$ he takes a right turn (facing West) and walks $5 \, km$ to reach a point.
$3$. Then he takes a left turn (facing South) and walks $5 \, km$ to reach point $R.$
$4$. Finally,he takes a left turn (facing East) and walks $5 \, km$ to reach point $S.$
$5$. Point $S$ is $5 \, km$ South of point $Q.$ Therefore,to reach point $Q$ from point $S,$ Raman must walk in the North direction.

(B) Let us represent the heights in descending order based on the given conditions:
$1$. $P$ is taller than only $Q$,which means $Q$ is the shortest and $P$ is the second shortest: $\_ > \_ > \_ > \_ > P > Q$.
$2$. $S$ is shorter only than $L$,which means $L$ is the tallest and $S$ is the second tallest: $L > S > \_ > \_ > P > Q$.
$3$. $N$ is taller than $Q$ and $P$ but shorter than $M$: $M > N > P$.
$4$. Combining these,the order from tallest to shortest is: $L > S > M > N > P > Q$.
Therefore,$L$ is the tallest and $Q$ is the shortest.

(D) According to Nitin,the birthday is after Tuesday and before Friday,which means it could be on Wednesday or Thursday.
According to Derek,the birthday is after Wednesday and before Saturday,which means it could be on Thursday or Friday.
To find the day that satisfies both conditions,we look for the common day in both sets: {Wednesday,Thursday} $\cap$ {Thursday,Friday} = {Thursday}.
Therefore,Nidhi's birthday definitely falls on Thursday.

(C) Based on the provided diagram,the path from point $A$ to point $E$ follows the shortest route indicated by the arrows.
Starting from point $A$,the path goes directly towards point $F$.
Therefore,the first point the man will pass through is point $F$.

(A) Let us represent the positions on a coordinate plane,starting with point $B$ at $(0, 0).$
Point $A$ is $12 \,meters$ north of $B$,so $A = (0, 12).$
Point $C$ is $24 \,meters$ east of $B$,so $C = (24, 0).$
Point $D$ is $8 \,meters$ south of $C$,so $D = (24, -8).$
Point $D$ is $12 \,meters$ east of $E$,which means $E$ is $12 \,meters$ west of $D$. So,$E = (24 - 12, -8) = (12, -8).$
Point $F$ is $8 \,meters$ north of $E$,so $F = (12, -8 + 8) = (12, 0).$
Now,we need to find the position of $F$ relative to $C$.
$C = (24, 0)$ and $F = (12, 0).$
The distance between $C$ and $F$ is $24 - 12 = 12 \,meters.$
Since the $x$-coordinate of $F$ is less than the $x$-coordinate of $C$,$F$ is to the west of $C$.
Therefore,point $F$ is $12 \,meters$ west of point $C$.

(D) Let us represent the points on a Cartesian coordinate system,taking point $A$ as the origin $(0, 0)$.
$A = (0, 0)$
$D$ is $14 \,m$ West of $A$,so $D = (-14, 0)$.
$B$ is $4 \,m$ South of $D$,so $B = (-14, -4)$.
$F$ is $9 \,m$ South of $D$,so $F = (-14, -9)$.
$E$ is $7 \,m$ East of $B$,so $E = (-14 + 7, -4) = (-7, -4)$.
$C$ is $4 \,m$ North of $E$,so $C = (-7, -4 + 4) = (-7, 0)$.
$G$ is $4 \,m$ South of $A$,so $G = (0, -4)$.
Now,check the collinearity of the points in the options:
For option $D$: $E(-7, -4), G(0, -4), B(-14, -4)$. Since all these points have the same $y$-coordinate $(-4)$,they lie on the horizontal line $y = -4$. Thus,$E, G, B$ are in a straight line.

(A) $1$. Point $D$ is $14 \,m$ West of $A.$
$2$. Point $B$ is $4 \,m$ South of $D.$
$3$. Point $E$ is $7 \,m$ East of $B.$
$4$. Point $C$ is $4 \,m$ North of $E.$
$5$. Since $D$ is $14 \,m$ West of $A$ and $E$ is $7 \,m$ East of $B$ (where $B$ is $4 \,m$ South of $D$),point $C$ lies exactly midway between $D$ and $A$ horizontally.
$6$. Therefore,$C$ is $7 \,m$ West of $A.$
$7$. Consequently,$A$ is towards the East of $C.$

(C) Let us represent the positions on a coordinate plane,taking point $A$ as $(0, 0).$
Point $A = (0, 0).$
Point $D$ is $14 \,m$ West of $A$,so $D = (-14, 0).$
Point $B$ is $4 \,m$ South of $D$,so $B = (-14, -4).$
Point $F$ is $9 \,m$ South of $D$,so $F = (-14, -9).$
Point $E$ is $7 \,m$ East of $B$,so $E = (-14 + 7, -4) = (-7, -4).$
Point $C$ is $4 \,m$ North of $E$,so $C = (-7, -4 + 4) = (-7, 0).$
Point $G$ is $4 \,m$ South of $A$,so $G = (0, -4).$
Now,the person starts at $F(-14, -9)$ and walks $5 \,m$ North. The new position is $(-14, -9 + 5) = (-14, -4)$,which is point $B.$
From point $B$,the person takes a right turn. Since the person was facing North,a right turn means they are now facing East.
Moving East from $B(-14, -4)$ towards $E(-7, -4)$,the person will reach point $E$ first.

(A) Initially,the directions are North,South,East,and West.
When the directional post is turned such that the arrow originally pointing East now points South,the entire orientation shifts by $90^{\circ}$ clockwise.
Consequently:
- East becomes South
- South becomes West
- West becomes North
- North becomes East
Since the passerby thinks they are travelling West,they are actually moving in the direction that has replaced West,which is North.
Wait,let us re-evaluate: If the arrow showing East now shows South,the rotation is $90^{\circ}$ clockwise.
Original West direction is now pointing North.
Therefore,if the person thinks they are going West,they are actually travelling North.

(A) Let the starting point be $O$.
$Q$ walks $20 \, m$ West to point $A$.
Then,$Q$ takes a left turn and walks $20 \, m$ South to point $B$.
Then,$Q$ takes a right turn and walks $20 \, m$ West to point $C$.
Then,$Q$ takes a right turn and walks $20 \, m$ North to point $D$.
The total horizontal distance from the starting point is $20 \, m + 20 \, m = 40 \, m$ (West).
The total vertical displacement is $20 \, m$ (South) - $20 \, m$ (North) = $0 \, m$.
Therefore,the final distance from the starting point is $40 \, m$.

(B) According to Mohan,the birthday is between $16^{th}$ and $20^{th}$ January,which means the possible dates are $17, 18,$ and $19$ January.
According to Mohan's sister,the birthday is between $18^{th}$ and $23^{rd}$ January,which means the possible dates are $19, 20, 21,$ and $22$ January.
The common date in both sets is $19$ January.
Therefore,their father's birthday is on $19^{th}$ January.

(C) $1$. Initially,the postman is facing North towards the Post Office,which is $100 \, m$ away.
$2$. He turns left (West) and walks $50 \, m$ to reach Shanti Villa.
$3$. He continues in the same direction (West) for another $40 \, m$.
$4$. He then turns right (North) and walks $100 \, m$.
$5$. The total horizontal distance covered from the original vertical path is $50 \, m + 40 \, m = 90 \, m$.
$6$. Since he moved $100 \, m$ North,he is now at the same latitude as the Post Office,but $90 \, m$ to the West of it.
$7$. Therefore,his distance from the Post Office is $90 \, m$.

(D) The time elapsed from midnight to $11:00 \, \text{AM}$ is $11 \, \text{hours}$.
In $11 \, \text{hours}$, the first clock gains $11 \times 2 = 22 \, \text{minutes}$.
In $11 \, \text{hours}$, the second clock loses $11 \times 1 = 11 \, \text{minutes}$.
The difference between the two clocks is the sum of the gain and the loss: $22 \, \text{minutes} + 11 \, \text{minutes} = 33 \, \text{minutes}$.
Therefore, the second clock will be $33 \, \text{minutes}$ behind the first clock.

(D) The boys are standing in a row facing East. Since they face East,their left is towards North and their right is towards South. However,in a linear arrangement facing a direction,'left' and 'right' are relative to the row.
Given:
$1$. Deepak is to the left of Sameer,Tushar,and Shailendra.
$2$. Sameer,Tushar,and Shailendra are to the left of Sushil.
$3$. Shailendra is between Sameer and Tushar.
Combining these,the order from left to right is: Deepak,Sameer,Shailendra,Tushar,Sushil.
If Tushar is fourth from the left,the arrangement is: $(1)$ Deepak,$(2)$ Sameer,$(3)$ Shailendra,$(4)$ Tushar,$(5)$ Sushil.
Counting from the right: $(1)$ Sushil,$(2)$ Tushar,$(3)$ Shailendra,$(4)$ Sameer.
Therefore,Sameer is fourth from the right.

(B) Let the starting point (my house) be $O$.
$1$. Laxman moves $15 \, km$ West to point $A$.
$2$. He turns left (South) and walks $20 \, km$ to point $B$.
$3$. He turns East and walks $25 \, km$ to point $C$.
$4$. He turns left (North) and walks $20 \, km$ to point $D$.
Since the southward distance $(20 \, km)$ and the final northward distance $(20 \, km)$ are equal,he is now at the same latitude as his starting point.
The horizontal distance from the house is the difference between the eastward movement and the westward movement: $25 \, km - 15 \, km = 10 \, km$.

(B) Anuj reached the meeting place $15 \text{ minutes}$ before $8:30 \text{ AM}$,so his arrival time is $8:30 - 0:15 = 8:15 \text{ AM}$.
Anuj arrived half an hour $(30 \text{ minutes})$ earlier than the other man. Therefore,the other man arrived at $8:15 + 0:30 = 8:45 \text{ AM}$.
Since the other man was $40 \text{ minutes}$ late for the meeting,the scheduled time of the meeting is $8:45 - 0:40 = 8:05 \text{ AM}$.

(D) Let the starting point be $O$.
$1$. Kailash faces North and turns right, walking $25\; m$ East.
$2$. He turns left and walks $30\; m$ North.
$3$. He turns right and walks $25\; m$ East.
$4$. He turns right and walks $55\; m$ South.
$5$. Finally, he turns right and walks $40\; m$ West.
Comparing the final position with the starting point:
Total North-South displacement: $30\; m (N) - 55\; m (S) = 25\; m$ South.
Total East-West displacement: $25\; m (E) + 25\; m (E) - 40\; m (W) = 10\; m$ East.
Since he is $25\; m$ South and $10\; m$ East of the starting point, he is in the South-East direction.

(D) Let the starting point be $O$.
$1$. Rajneesh walks $35 \, m$ towards the East to reach point $A$.
$2$. He turns right (South) and walks $20 \, m$ to reach point $B$.
$3$. He turns right (West) and walks $35 \, m$ to reach point $C$. At this point,he is directly South of his starting point $O$ by $20 \, m$.
$4$. Finally,he turns left (South) and walks $20 \, m$ to reach his destination $D$.
$5$. The total distance from the starting point $O$ to the destination $D$ is the sum of the vertical distances covered in the South direction: $20 \, m + 20 \, m = 40 \, m$.

(D) The $1^{st}$ day of the month was a Tuesday.
Since the $1^{st}$ is a Tuesday,the $2^{nd}$ is Wednesday,the $3^{rd}$ is Thursday,the $4^{th}$ is Friday,and the $5^{th}$ is Saturday.
Saturdays in the month occur at intervals of $7$ days: $5, 12, 19, 26$.
Rama met her brother on a Saturday after the $20^{th}$ day.
The only Saturday after the $20^{th}$ is the $26^{th}$.
Therefore,Rama met her brother on the $26^{th}$.

(B) $1$. Ram starts from his house and walks $10\,m$ towards the South.
$2$. He turns left (facing East) and walks $23\,m$.
$3$. He turns left again (facing North) and walks $40\,m$.
$4$. Finally,he turns right (facing East) and walks $5\,m$ to reach his school.
$5$. Let the starting point (house) be at the origin $(0,0)$.
$6$. After $10\,m$ South,position is $(0, -10)$.
$7$. After $23\,m$ East,position is $(23, -10)$.
$8$. After $40\,m$ North,position is $(23, -10 + 40) = (23, 30)$.
$9$. After $5\,m$ East,position is $(23 + 5, 30) = (28, 30)$.
$10$. Since both coordinates are positive relative to the house,the school is in the North-East direction.