Direction Test — BSSC General Intelligence And Reasoning

Explanation:

• Point A is the starting point.

• The man walks 15 m East (horizontal distance).

• He turns right (facing South) and walks 8 m (vertical distance).

• Applying Pythagoras theorem for the shortest distance: $AB = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17\text{ m}$.

• The final position B is East and South of the starting point A, hence South-East.

Exam Trick:

• Recognize the standard Pythagorean triplet $(8, 15, 17)$. Net displacement is East + South $\implies$ 17 m South-East. No long calculation needed.

Explanation:

• Initial direction = North-West.

• Total Clockwise (CW) rotation = $90^\circ$.

• Total Anticlockwise (ACW) rotation = $135^\circ$.

• Net rotation = $\text{ACW} - \text{CW} = 135^\circ - 90^\circ = 45^\circ$ ACW.

• From North-West, a $45^\circ$ rotation in the anticlockwise direction leads to West.

Exam Trick:

• Write CW on one side and ACW on the other. Subtract the smaller from the larger: $135^\circ \text{ (ACW)} - 90^\circ \text{ (CW)} = 45^\circ \text{ (ACW)}$. One step ($45^\circ$) back from NW is West.

Explanation:

• Initial direction = South.

• Total Left (L) turns = 2.

• Total Right (R) turns = 1.

• Net turn = $2\text{L} - 1\text{R} = 1\text{L}$ (one left turn).

• Taking 1 Left turn from South leads to East.

Exam Trick:

• Cancel pairs of Left and Right turns. $1\text{L}$ and $1\text{R}$ cancel out. You are only left with $1\text{L}$. South $+ 1\text{L} =$ East.

Explanation:

• Time is morning, so the Sun is in the East.

• All shadows fall towards the West in the morning.

• Amit's shadow falls to the West.

• The problem states the shadow falls to the right of Sumit.

• Therefore, Sumit's right side is West $\implies$ Sumit is facing South.

• Since they are facing each other, Amit is facing the opposite of South, which is North.

Exam Trick:

• Use the "Time - Shadow - Person" three-column matrix.

• Morning (Right) $\rightarrow$ Shadow Right $\rightarrow$ Person facing South. Sumit faces South. Amit faces opposite (North).

Explanation:

• Original map: North $= 0^\circ$, North-East $= 45^\circ$, South-East $= 135^\circ$, South $= 180^\circ$.

• South-East ($135^\circ$) is called North ($0^\circ$). This is a shift of $135^\circ$ Anticlockwise.

• Let's verify: North-East ($45^\circ$) shifted by $135^\circ$ ACW becomes $270^\circ$ (West). The rule holds.

• Apply the same $135^\circ$ ACW shift to South ($180^\circ$): $180^\circ - 135^\circ = 45^\circ$.

• $45^\circ$ corresponds to North-East.

Exam Trick:

• Count the 8-cardinal gaps: South-East moves 3 steps anticlockwise to become North.

• Move South 3 steps anticlockwise $\rightarrow$ South-East $\rightarrow$ East $\rightarrow$ North-East.

Explanation:

• Let starting point be $(0, 0)$.

• Walks 10 m North $\rightarrow (0, 10)$.

• Turns right (East) and walks 20 m $\rightarrow (20, 10)$.

• Turns right (South) and walks 10 m $\rightarrow (20, 0)$.

• His current position is 20 m East of his starting point.

• To return to his starting point $(0,0)$, he must walk 20 m West.

Exam Trick:

• Using the N-E-S-W line method: N: 10 | E: 20 | S: 10.

• North (10) and South (10) cancel out, leaving a net displacement of 20 m East.

• To reverse the displacement and return home, he must go 20 m West.

Explanation:

• Let Point Q be origin $(0, 0)$.

• P is 20 m East of Q $\implies P = (20, 0)$.

• R is 12 m North of Q $\implies R = (0, 12)$.

• S is 5 m East of R $\implies S = (5, 12)$.

• T is 20 m South of S $\implies T = (5, 12 - 20) = (5, -8)$.

• Coordinate of T is $(5, -8)$ and P is $(20, 0)$.

• Distance from P to T: horizontal distance is $20 - 5 = 15\text{ m}$ (West). Vertical distance is $8\text{ m}$ (South).

• Shortest distance $= \sqrt{15^2 + (-8)^2} = \sqrt{225 + 64} = \sqrt{289} = 17\text{ m}$.

• Since $X$ decreases (West) and $Y$ decreases (South) from P to T, T is South-West of P.

Exam Trick:

• Net offsets from P to T: $P$ is $+20$ East. $T$ is $+5$ East, $-8$ South. Difference from P to T is $15$ West, $8$ South. Distance $= \sqrt{15^2 + 8^2} = 17$. West and South give South-West.

Explanation:

• Decode the chain P \text{ @ } Q \text{ \% } R \text{ \ } S$ right to left to build coordinates.

• Let $S = (0, 0)$.

• R \text{ \ } S \implies R$ is 6 m West of $S \rightarrow R = (-6, 0)$.

• $Q \text{ \% } R \implies Q$ is 8 m South of $R \rightarrow Q = (-6, -8)$.

• $P \text{ @ } Q \implies P$ is 6 m East of $Q \rightarrow P = (-6 + 6, -8) = (0, -8)$.

• Comparing $S (0, 0)$ and $P (0, -8)$, we see they lie on the same vertical line.

• Distance $= |-8 - 0| = 8\text{ m}$.

• The $y$-coordinate of $P$ is $-8$, which is lower (South) than $S$'s $y$-coordinate of $0$.

• Thus, $P$ is 8 m South of $S$.

Exam Trick:

• Summing offsets from S to P:
$R$ is $-6$ (West).
$Q$ is $-6$ (West), $-8$ (South).
$P$ is $+6$ (East) from $Q \implies (-6+6) = 0$ (West/East cancels out). $P$ is just 8 m South of $S$.
Hence, $P$ is 8 m South of $S$.

Explanation:

• On a normal clock at 3:00, the minute hand points to 12 (North) and the hour hand points to 3 (East).

• The problem states the minute hand points to North-East. This implies the entire clock is rotated 45° clockwise (from North to North-East).

• Since the entire orientation shifts uniformly, we must also rotate the hour hand's original direction (East) by 45° clockwise.

• East + 45° clockwise = South-East.

Exam Trick:

• Map the shift directly: North $\rightarrow$ North-East means a $+45^\circ$ turn. Apply the same shift to East: East $+ 45^\circ =$ South-East.

Explanation:

• Initial facing direction = North ($0^\circ$).

• Clockwise turn (CW) = $135^\circ$.

• Anticlockwise turn (ACW) = $90^\circ$.

• Net turn = $\text{CW} - \text{ACW} = 135^\circ - 90^\circ = 45^\circ$ CW.

• A $45^\circ$ clockwise turn from North results in North-East.

Exam Trick:

• Do not draw multiple lines. Write CW and ACW values, subtract the smaller from the larger: $135 - 90 = 45\text{ CW}$. One step ($45^\circ$) right from North is North-East.