GATE CS 2015 Set 2 QGA-7 — General Aptitude

We can solve this problem by setting up a 2D Cartesian coordinate system. Let the branch $O$ be located at the origin $(0, 0)$.

Step 1: Find the coordinates of $N$
$N$ is southeast of $O$ at a distance of $1\text{ km}$. Since southeast is at an angle of $45^\circ$ south of east: $$x_N = 1 \cdot \cos(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.707\text{ km}$$ $$y_N = -1 \cdot \sin(45^\circ) = -\frac{1}{\sqrt{2}} \approx -0.707\text{ km}$$ Thus, the coordinates of $N$ are $(0.707, -0.707)$.

Step 2: Find the coordinates of $M$
$M$ is north of $N$ by $4\text{ km}$. Therefore, we add $4$ to the $y$-coordinate of $N$: $$x_M = 0.707$$ $$y_M = -0.707 + 4 = 3.293$$ Thus, the coordinates of $M$ are $(0.707, 3.293)$.

Step 3: Find the coordinates of $P$
$P$ is south of $O$ by $2\text{ km}$. Since $O$ is at $(0, 0)$: $$x_P = 0$$ $$y_P = -2$$ Thus, the coordinates of $P$ are $(0, -2)$.

Step 4: Calculate the distance between $M$ and $P$
Using the distance formula: $$\text{Distance } (M, P) = \sqrt{(x_M - x_P)^2 + (y_M - y_P)^2}$$ $$\text{Distance } (M, P) = \sqrt{(0.707 - 0)^2 + (3.293 - (-2))^2}$$ $$\text{Distance } (M, P) = \sqrt{(0.707)^2 + (5.293)^2}$$ $$\text{Distance } (M, P) \approx \sqrt{0.5 + 28.016} = \sqrt{28.516} \approx 5.34\text{ km}$$

Therefore, the distance between $M$ and $P$ is approximately $5.34\text{ km}$.