Consider the diagram.
The arrows could be vertical, horizontal or diagonal.


The vertical arrows are shown by the blue arrows. 5 of them will start from x = 0, 5 from x = 1 and so on till x = 9. So you have 50 of these blue arrows. You have another 50 vertical arrows which are the same arrows but with the arrow head on the opposite end (shown by the red arrow). So you have a total of 100 vertical arrows.
Similarly, you have 100 horizontal arrows.

Now check out the diagonal arrows. One co-ordinate should be of length 3 and another of 4 (so that the arrow length is 5 and all points are integers). Look at the purple arrows. The x co-ordinate is 3 and the y co-ordinate is 4. You can make 7*6 = 42 such arrows. Similarly, you can make 42 arrows with x cor-ordinate as 4 and y co-ordinate as 3. So you have 84 arrows. But you get another set of 84 arrows by keeping the arrows the same but putting the arrow head on the opposite end so you get a total of 2*84 = 168 arrows.

Similarly, you can make arrows in the opposite direction shown by the green arrows. So you have another 168 arrows.

Total = 100 + 100 + 168 + 168 = 536
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ANSWER: E
If A and B have the same x-coordinate then we have 10 pairs of y-coordinate of A and B per x-coordinate. (eg: 1-5, 2-6...) => 10*10 = 100 arrows.
Similarly, if A and B have the same y-coordinate then we have another 100 arrows.
If A(a,b) and B(c,d) don't have the same x-coordinate or y-coordinate then either |a-b|=3,|c-d|=4 or |a-b|=4,|c-d|=3
In the first case, there are 14 pairs of x-coordinate, and 12 pairs of y-coordinate. => 14*12 = 168 arrows.
Similarly in the second case, there are 168 arrows.
Therefore, We have 100 + 100 + 168 + 168 = 536 arrows.

I didn't get the explaination....
what i did was i formed the grid on xy plane with info provided. total grid points i got were 100 and we need to select 2 points to form an arrow...so 100C2 : 4950...which is nowhere near the answer....whr exactly i m making the mistake?

Posted from my mobile device

What you forgot to consider is that the length of the arrow must be 5 units.
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Thanks for the wonderful explanation Karishma. Kudos for that.
Now what my concern is, should we expect to get this type of problems in the real exam?...I mean in this problem we need to draw the figure and need to manually count the possibilities that is time consuming.

Thanks.
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This question is based on an OG question. The OG question uses this concept though it doesn't require you to manually count the different cases. I have discussed that question in this post:
http://www.veritasprep.com/blog/2011/09 ... o-succeed/

Given unlimited time, you should be able to do this question i.e. conceptually you should be clear with this. It is a time consuming laborious question so I wouldn't expect GMAT to give this. It is missing the excitement - you can do most GMAT in under a minute or perhaps even 30 secs. The fun is to be able to figure out the logical trick that makes it tick.
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How did you find the coordinate pairs?

Thanks.

From 100 hardest questions
Bumping for review and further discussion.
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Responding to a pm:

This is what this statement means: Draw an arrow starting from (0, 0) to (0, 5). Head of the arrow is at (0, 5). Then draw another one starting from (0, 1) to (0, 6). Another from (0, 2) to (0, 7). Another from (0, 3 to 0, 8). Another from (0, 4) to (0, 9). You are able to draw these 5 arrows such that x co-ordinate is 0 in each case. You cannot go higher up because y co-ordinate cannot be more than 9.

Similarly, draw an arrow starting from (1, 0) to (1, 5). Another from (1, 1) to (1, 6) and so on... You will again be able to draw 5 such arrows.

Keep increasing x co-ordinate by 1 and you will get 5 arrows each time till you reach x = 9. So you will get 10 groups of 5 vertical arrows each i.e. 50 such arrows.
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Hi, cant we make slanting arrows also in the opposite direction like vertical and horizontal arrows i.e. just reversing the coordinates of A and B.

thanks
Abhishek

Look at the highlighted part above. We have already taken care of it.
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Hi, cant we make slanting arrows also in the opposite direction like vertical and horizontal arrows i.e. just reversing the coordinates of A and B.

thanks
Abhishek

Look at the highlighted part above. We have already taken care of it.[/quote]


Thanks a lot for the reply.

There can be 4 different types of lines with 4 different slopes that may have 5 units as length (for eg. co-ordinates - (4,0)&(0,3); (3,0)&(0,4); (0,0)&(3,4); (0,0)&(4,3). Thus total number of lines 42*4=168. And if arrows are reversed number will be 168*2 = 336.

Request you to please let me know if I am going wrong somewhere.

thanks.

Note that we have already taken all 4 of them them into account in the highlighted part above.
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I guess it is more than 400. ans is E

Now check out the diagonal arrows. One co-ordinate should be of length 3 and another of 4 (so that the arrow length is 5 and all points are integers),..
Am not clear with this. Please help.
Thanks in advance1

The way I did the diagonal arrows is... I drew a 3x4 (right 3, up 4) box from the origin. For each of those, you can have 4 sets of coordinates for A and B (e.g. A(0,0)+B(3,4); A(3,4)+B(0,0); A(3,0)+B(0,4); A(0,4)+B(3,0).

Moving up, you can make 5 more boxes, so that's 6 for the first column. Using the same 3x4 boxes, you can move right 6 times, so there are 7 of those. That's 6x7=42 times the box is made, with each box having 4 distinct arrows with their respective vectors. So, that's 42x4 = 168.

You can also make 4x3 squares from the origin. Using the same idea, you get another 168 distinct arrows.

So far in total that's 336 arrows.

Combine that with the 100 vertical and 100 horizontal arrows, then the total is 536 arrows.

There is a much simpler way to do this, without having to visualise it.

The formula for distance between two points is d=Sqrt[(x2-x1)^2+(y2-y1)^2].

Now we know the distance is 5. so we get:

5=Sqrt[(x2-x1)^2+(y2-y1)^2].

25=[(x2-x1)^2+(y2-y1)^2]

Lets keep this aside, and look at our inequalities: 0 ≤ x ≤ 9 and 0 ≤ y ≤ 9.

This means the difference x2-x1 and y2-y1 can lie between 0 and 9.

Since our sum of the squares of the difference needs to be 25. The difference can be either 5 and 0, 0 and 5, 4 and 3 or 3 and 4.

The number of different coordinates for x1 and x2 which will give us a difference of:

5 is 10
0 is 10
4 is 12
3 is 14

Same for y coordinates.

So number of lines possible where the difference between x coordinates is 5 and y coordinates is 0 is 100. Interchange the differences and you get 100 more lines.

Number of lines possible where the difference between x coordinates is 4 and y coordinates is 3 is 168. Interchange the differences and you get 168 more lines.

Therefore total number of lines possible is - 100+100+168+168 = 536.

Hi Karishma,
Can't the arrows start from point (1,2), (1,3), (2,3) etc.

Yes, they can. In the solution above:

"...5 of them will start from x = 0, 5 from x = 1 and so on till x = 9. So you have 50 of these blue arrows. "

5 will start from x = 0. For these, y will be 0, 1, 2, 3 and 4.
So the arrows will start from (0, 0), (0, 1), (0, 2) etc

5 will start from x = 1. For these too, y will be 0, 1, 2, 3 and 4.
So the arrows will start from (1, 0), (1, 1), (1, 2) etc

Similarly, arrows will start from (2, 0), ... (2, 4), ... (3, 0) ... etc

Does this help?
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