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Question

A knight is placed somewhere on the chess board. If you have 2 consecutive chances then taking the initial position of knight as the origin which of these position can be the maximum displaced position of the knight?

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Solution

The correct option is **D**

(-4, 2)

Well we know how a knight moves 2 steps forward then one step to the left or to the right. Also from his start position the knight can go in any of the North, South, west or east direction.

Very well, let's start from what we know, I have 2 consecutive chances and my knight in shining armor is at origin. He can go in any of the 4 cardinal directions we can choose any one as its symmetric and will result in the same displacement. Let's choose north. Okay so from origin i.e., (0, 0) my knight moved 2 steps forward in the north direction then he can go either left or right, both will give me same displacement so let's pick one, let's say right, so now his position will be - - - - - -. That's right 2 steps on y-axis and 1 step on x so (1, 2)

Now I have one more chance. Again I can go north, south, west or east but now if I go south or west I will be coming closer to origin hence decreasing my displacement.

Case I: Let's go North 2 steps then 1 step right as coming left will decrease my displacement.

So, I end up at (2, 4)

Now to find the displacement let's use distance formula

√(2−0)2+(4−0)2=√20

Let's see if we can beat that

Case II: instead of going north let's start going east 2 steps and 1 step left as if I go right I will be going towards origin and hence decreasing my displacement.

So I end up at (3, 3)

Let's find my displacement

√(3−0)2+(3−0)2=√18

This is less than the previous case so (2, 4) is the correct option

Due to symmetry I can either end up on (-2, 4) (-2, -4), (2, -4), (4, 2), (-4, -2); (-4, 2), (4, -2).

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