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Arrow AB which is a line segment exactly 5 units along with an arrowhead at A is to be constructed in the xy-plane. The x and y coordinates of A and B are to be integers that satisfy the inequalities 0 ≤ x ≤ 9 and 0 ≤ y ≤ 9. How many different arrows with these properties can be constructed ?
A. 50
B. 168
C. 200
D. 368
E. 536
A. 50
B. 168
C. 200
D. 368
E. 536
harikris
Consider the diagram.
The arrows could be vertical, horizontal or diagonal.
Attachment:
Ques4.jpg [ 25.55 KiB | Viewed 70631 times ]
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.
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.
General Discussion
daviesj
Joined: 23 Aug 2012
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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 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
daviesj
What you forgot to consider is that the length of the arrow must be 5 units.
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daviesj
Joined: 23 Aug 2012
Last visit: 09 May 2025
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Status:Never ever give up on yourself.Period.
Location: India
Concentration: Finance, Human Resources
Schools: MBS '17 (A$) Smurfit FT'16 (A)
GMAT 1: 570 Q47 V21
GMAT 2: 690 Q50 V33
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WE:Information Technology (Finance: Investment Banking)
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.
_________________
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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Beat it, no one wants to be defeated My journey from 570 to 690.
daviesj
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.
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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VeritasPrepKarishma
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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VeritasPrepKarishma
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.
Attachments
Untitled.jpg [ 21.32 KiB | Viewed 63664 times ]
Quote:
Note that we have already taken all 4 of them them into account in the highlighted part above.
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Posts: 12
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.
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.
NaeemHasan
Joined: 06 Oct 2015
Last visit: 04 Feb 2019
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Location: Bangladesh
Concentration: Accounting, Leadership
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Schools: Iowa State '21 Marshall '21 Stanford '21
Posts: 61
VeritasPrepKarishma
Hi Karishma,
Can't the arrows start from point (1,2), (1,3), (2,3) etc.
NaeemHasan
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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"The x co-ordinate is 3 and the y co-ordinate is 4. You can make 7*6 = 42 such arrows. "
Hi Karishma,
Many thanks in advance...
Will you please tell me how you got this ?
" You can make 7*6 = 42 such arrows.[/color]"
How you got 7*6 ?
Hi Karishma,
Many thanks in advance...
Will you please tell me how you got this ?
" You can make 7*6 = 42 such arrows.[/color]"
How you got 7*6 ?
sayan640
x is between 0 and 9 (inclusive) and y is between 0 and 9 (inclusive).
So if we need the length of x to be 3, we can start it anywhere from x = 0 to x = 6 (7 cases).
If we start the arrow at x = 7, with a length of 3, it will mean that the end point of the arrow will have x co-ordinate of 10 (which is not allowed).
So for the x co-ordinate, you have 7 options.
Similarly, the length of y should be 4 so we can start it anywhere from y = 0 to y = 5 (6 cases)
If we start the arrow at y = 6, with a length of 4, it will mean that the end point of the arrow will have y co-ordinate of 10 (which is not allowed).
So for the y co-ordinate, you have 6 options.
Hence you get a total of 7 * 6 = 42 acceptable cases.
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ShukhratJon
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Solving without graph.
Attachments
File comment: Solving without graph.
photo_2019-04-23_18-21-36.jpg [ 115.74 KiB | Viewed 9663 times ]
Hello!
We can solve this question by using Combinatorics as well.
My approach for this question is below:
Given two points with their co-ordinates: A(x1, y1), B(x2, y2). Then the length of AB should be:
AB^2 = 5^2 =25 = (x1-x2)^2 + (y1-y2)^2 = X^2 + Y^2 in which I denote X = x1-x2 and Y = y1-y2
Our goal is to determine X and Y such that X^2 + Y^2 = 25. Let's find out all the pairs (X^2, Y^2) that meet such a requirement, and also keep in mind that the square root of Y^2 and X^2 has to be integers. One example below does not meet this requirement such as the pair (1,24) as square root of 24 is not an integer:
(X^2,Y^2) = { (0,25), (1,24), ..., (9,16),...} Feel Free to exploit the whole set but we have found only one candidate in this list that meets the requirements above: (9,16) = (3^2, 4^2)
Next, we need to find out all the pairs (x1, x2) that has a difference of 3 and all the pairs (y1,y2) that has a difference of 4.
1. All the pairs (x1, x2) that has a difference of 3: (0,3), (1,4), ..., (6,9) and you flip the coordinates = 14 pairs
AND all the pairs (y1,y2) that has a difference of 4: (0,4), (1,5),...,(5,9) and you flip the coordinates = 12 pairs
2. Question: How many ways for you to choose a pair from 7 pairs AND a pair from 6 pairs? 14 x 12 = 168
Next: Don't forget, there are 2 ways to arrange 2 chose pairs once we choose one from (y1,y2) and one from (x1,x2) so we have a total of 168 x 2 = 336
3. Do the same thing for special case: AB^2 = 5^2 =25 = (x1-x2)^2 + (y1-y2)^2 = X^2 + Y^2 = 0 + 25
Find All the pairs (x1, x2) that has a difference of 0: (0,0), (1,1), ..., (9,9) and you do not flip the coordinates in this case as you will get identical pairs = 10 pairs
AND all the pairs (y1,y2) that has a difference of 5: (0,5), (1,6),...,(4,9) and you flip the coordinates = 10 pairs
4. Question: How many ways for you to choose a pair from 10 pairs AND a pair from 10 pairs? 10 x 10 = 100
Next: Don't forget, there are 2 ways to arrange 2 chose pairs once we choose one from (y1,y2) and one from (x1,x2) so we have a total of 100 x 2 = 200
The FINAL Answer is: 336 + 200 = 536
We can solve this question by using Combinatorics as well.
My approach for this question is below:
Given two points with their co-ordinates: A(x1, y1), B(x2, y2). Then the length of AB should be:
AB^2 = 5^2 =25 = (x1-x2)^2 + (y1-y2)^2 = X^2 + Y^2 in which I denote X = x1-x2 and Y = y1-y2
Our goal is to determine X and Y such that X^2 + Y^2 = 25. Let's find out all the pairs (X^2, Y^2) that meet such a requirement, and also keep in mind that the square root of Y^2 and X^2 has to be integers. One example below does not meet this requirement such as the pair (1,24) as square root of 24 is not an integer:
(X^2,Y^2) = { (0,25), (1,24), ..., (9,16),...} Feel Free to exploit the whole set but we have found only one candidate in this list that meets the requirements above: (9,16) = (3^2, 4^2)
Next, we need to find out all the pairs (x1, x2) that has a difference of 3 and all the pairs (y1,y2) that has a difference of 4.
1. All the pairs (x1, x2) that has a difference of 3: (0,3), (1,4), ..., (6,9) and you flip the coordinates = 14 pairs
AND all the pairs (y1,y2) that has a difference of 4: (0,4), (1,5),...,(5,9) and you flip the coordinates = 12 pairs
2. Question: How many ways for you to choose a pair from 7 pairs AND a pair from 6 pairs? 14 x 12 = 168
Next: Don't forget, there are 2 ways to arrange 2 chose pairs once we choose one from (y1,y2) and one from (x1,x2) so we have a total of 168 x 2 = 336
3. Do the same thing for special case: AB^2 = 5^2 =25 = (x1-x2)^2 + (y1-y2)^2 = X^2 + Y^2 = 0 + 25
Find All the pairs (x1, x2) that has a difference of 0: (0,0), (1,1), ..., (9,9) and you do not flip the coordinates in this case as you will get identical pairs = 10 pairs
AND all the pairs (y1,y2) that has a difference of 5: (0,5), (1,6),...,(4,9) and you flip the coordinates = 10 pairs
4. Question: How many ways for you to choose a pair from 10 pairs AND a pair from 10 pairs? 10 x 10 = 100
Next: Don't forget, there are 2 ways to arrange 2 chose pairs once we choose one from (y1,y2) and one from (x1,x2) so we have a total of 100 x 2 = 200
The FINAL Answer is: 336 + 200 = 536
Engineer1
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Engineer1
We most certainly are to consider them.
We found that we get 100 vertical arrows. So then we will have 100 horizontal arrows too (see the highlighted above). This is so because 0 ≤ x ≤ 9 and 0 ≤ y ≤ 9.
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Quote:
I have question if there was no arrow head.
I think the significance of arrow head is that (0,0)<-(5,0) is different from (0,0)->(5,0)
The equation (x1-x2)^2 + (y1-y2)^2 = 5^2
can be considered
1) 0+25=25
2) 25+0=25
3) 16+9=25
4) 9+16=25
1) x1,x2 can be 0-0,1-1,...,9-9. but, x1 and x2 cant be interchanged since they will be the same. 10.
while y1,y2 can be 0-5,1-5,..4-9. y1 and y2 can be interchanged since square is used in equation. 5*2=10
Total =10*10=100
2) similar as 1. Total 100
3)x1,x2 can be 0-4,1-5,...5-9. x1 and x2 can be interchanged since square. 6*2=12
y1,y2 can be 0-3,1-4,...,6-9. y1 and y2 can be interchanged. 7*2=14
Total=14*12=168
4)Similar as 3. Total 168.
Final=200+336=536
So if the arrow head was not significant then the reversing that I did in multiple steps may not have been possible?
it would be 50+50+42+42=184 ?
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