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Difficulty:
95%
(hard)
Question Stats:
34% (01:43) correct
66%
(01:41)
wrong
based on 282
sessions
History
Date
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A Square is to be drawn on the xy plane. If one vertex of the square needs to be at
the origin, sides of the square need to be 5 units each, and all coordinates of all
vertices need to be integers, how many different squares are possible?
A. 4
B. 6
C. 8
D. 12
E. 16
Source: Experts Global
the origin, sides of the square need to be 5 units each, and all coordinates of all
vertices need to be integers, how many different squares are possible?
A. 4
B. 6
C. 8
D. 12
E. 16
Source: Experts Global
Posts: 398
Hi,
This is a very interesting question.
Easily someone could fall for the trap answer as 4 here.
If one visualize the squares only on the x and y axis, then they end up falling for the trap.
So the vertices need not be only on the x and y axis, they have mentioned all the vertices should be integers,
So, Let’s try to visualize it,
We can see that,
Co-ordinates (-4,3) forms a side length of the square as 5 from the origin (0,0), Because 3-4-5 are pythogorean triplets.
As shown in the figure(In the attachment), we can form four squares starting with point (-4,3).
So similarly starting with the point (-3,4), we can form 4 different squares.
Already on the x and y axis there are 4 different squares.
So totally, we can form 12 different squares with side length as 5 and one vertex at the origin. So the answer is D.
Most important factor of solving a co-ordinate geometry question is visualizing.
Hope this helps.
This is a very interesting question.
Easily someone could fall for the trap answer as 4 here.
If one visualize the squares only on the x and y axis, then they end up falling for the trap.
So the vertices need not be only on the x and y axis, they have mentioned all the vertices should be integers,
So, Let’s try to visualize it,
We can see that,
Co-ordinates (-4,3) forms a side length of the square as 5 from the origin (0,0), Because 3-4-5 are pythogorean triplets.
As shown in the figure(In the attachment), we can form four squares starting with point (-4,3).
So similarly starting with the point (-3,4), we can form 4 different squares.
Already on the x and y axis there are 4 different squares.
So totally, we can form 12 different squares with side length as 5 and one vertex at the origin. So the answer is D.
Most important factor of solving a co-ordinate geometry question is visualizing.
Hope this helps.
Attachments
PS - GMAT thread.png [ 24.51 KiB | Viewed 15274 times ]
General Discussion
pushpitkc
Similar question to practice: https://gmatclub.com/forum/a-certain-sq ... 27018.html
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rahul16singh28
Joined: 31 Jul 2017
Last visit: 09 Jun 2020
Posts: 428
Location: Malaysia
Schools: INSEAD Jan '19
GPA: 3.95
WE:Consulting (Energy)
Schools: INSEAD Jan '19
Posts: 428
pushpitkc
Consider the First Quadrant -
The square can be formed with points when Y = 5,3,4 and X correspondingly will be X = 0,4,3. So, for each Quadrant there will be 3 Different Squares, hence for all the 4 Quadrants, it will be 12.
soleimanian
Joined: 11 Jun 2015
Last visit: 17 Dec 2019
Posts: 15
Location: Iran (Islamic Republic of)
Concentration: Accounting, Finance
WE:Education (Education)
Posts: 15
The square could rotate around the origin, making a cycle.
(x^2)+(y^2)=Diagonal of the square
(x^2)+(y^2)=50
Solving above Equation with Integer solutions=12
(x^2)+(y^2)=Diagonal of the square
(x^2)+(y^2)=50
Solving above Equation with Integer solutions=12
KarishmaB chetan2u KrishnakumarKA1
Will you be kind enough to show me how , starting with the point (-3,4), we can form 4 different squares ? Finding it a bit confusing to visualize..Any simpler approach..
Will you be kind enough to show me how , starting with the point (-3,4), we can form 4 different squares ? Finding it a bit confusing to visualize..Any simpler approach..
KrishnakumarKA1
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