GATE CS 2015 Set 2 Q.7 (GA Q.7): Spatial Aptitude

Four branches of a company are located at $M, N, O,$ and $P$ . $M$ is north of $N$ at a distance of 4 km; $P$ is south of $O$ at a distance of 2 km; $N$ is southeast of $O$ by 1 km. What is the distance between $M$ and $P$ in km?

5.34

6.74

28.5

45.49

Official answer and solution

Official answer: A
From: Official GATE 2015 answer key

Solution

Set up a 2D Cartesian coordinate system with point $O$ at the origin and compute the coordinates of $M$ and $P$ .

Let $O = (0, 0)$ .

$P$ is south of $O$ by 2 km:

P = (0, -2).

$N$ is southeast of $O$ by 1 km.
Southeast corresponds to an angle of $45^\circ$ south of east (fourth quadrant):

x_N = 1 \cdot \cos(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.7071\text{ km}

y_N = -1 \cdot \sin(45^\circ) = -\frac{1}{\sqrt{2}} \approx -0.7071\text{ km}

So $N = (0.7071, -0.7071)$ .

$M$ is north of $N$ by 4 km:

x_M = x_N = 0.7071\text{ km}

y_M = y_N + 4 = 4 - 0.7071 = 3.2929\text{ km}

So $M = (0.7071, 3.2929)$ .

Now compute the Euclidean distance between $M$ and $P$ :

\Delta x = x_M - x_P = 0.7071 - 0 = 0.7071

\Delta y = y_M - y_P = 3.2929 - (-2) = 5.2929

\text{Distance} = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{0.7071^2 + 5.2929^2} = \sqrt{0.5000 + 28.0148} = \sqrt{28.5148} \approx 5.34\text{ km}.

Final answer

Common trap

Using $28.5$ directly without taking the square root, leading to option C.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 2
Question: 7 of 65, printed as GA Q.7
Section: General Aptitude

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