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In an examination, Mary was asked to calculate the maximum area of a s 12 Jul 2019, 22:49
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40% (02:28) correct 60% (02:25) wrong based on 60 sessions

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In an examination, Mary was asked to calculate the maximum area of a square whose co-ordinates (on the XY plane) satisfied the relation \(x^2*y^2 = 36\). If she misread 36 as 26, and solved the question, what is the difference, in sq. units, between the actual area and Mary's answer?

A) 10 sq. units
B) 40 sq. units
C) 52 sq. units
D) 92 sq. units
E) 104 sq. units

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In an examination, Mary was asked to calculate the maximum area of a s Updated on: 13 Jul 2019, 03:00
Originally posted by directnick on 13 Jul 2019, 01:40.
Last edited by directnick on 13 Jul 2019, 03:00, edited 1 time in total.
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lakshyaag
In an examination, Mary was asked to calculate the maximum area of a square whose co-ordinates (on the XY plane) satisfied the relation \(x^2*y^2 = 36\). If she misread 36 as 26, and solved the question, what is the difference, in sq. units, between the actual area and Mary's answer?

A) 10 sq. units
B) 40 sq. units
C) 52 sq. units
D) 92 sq. units
E) 104 sq. units
Show more

I suppose there's a mistake in the Question. The formula should be \(x^2+y^2 = 36\) and not \(x^2*y^2 = 36\)

Because if it's \(x^2*y^2 = 36\) then there's no answer to the task...

\(x^2+y^2 = 36\) that's a formula of a circle with (0,0) as center and a radius of 6 (square root of 36).
If we are looking for the biggest square that intersects the circle -> then we should look at a circle square, where circle intersects the square in square's centers of the sides.



This way the area of the bigggest square is 6*2*6*2 => 144

If Mary made a mistake then the side of the second square is \(\sqrt[]{26}\)*2*\(\sqrt[]{26}\)*2=104

So the answer is 144-104 = 40, hence Option B


Please give KUDDOOOSSSSS if the explanation was clear for you!
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In an examination, Mary was asked to calculate the maximum area of a s 13 Jul 2019, 02:46
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directnick
lakshyaag
In an examination, Mary was asked to calculate the maximum area of a square whose co-ordinates (on the XY plane) satisfied the relation \(x^2*y^2 = 36\). If she misread 36 as 26, and solved the question, what is the difference, in sq. units, between the actual area and Mary's answer?

A) 10 sq. units
B) 40 sq. units
C) 52 sq. units
D) 92 sq. units
E) 104 sq. units

\(x^2*y^2 = 36\) that's a formula of a circle with (0,0) as center and a radius of 6 (square root of 36).
If we are looking for the biggest square that intersects the circle -> then we should look at a circle square, where circle intersects the square in square's centers of the sides.



This way the area of the bigggest square is 6*2*6*2 => 144

If Mary made a mistake then the side of the second square is \(\sqrt[]{26}\)*2*\(\sqrt[]{26}\)*2=104

So the answer is 144-104 = 40, hence Option B


Please give KUDDOOOSSSSS if the explanation was clear for you!
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The equation of a circle with center (0,0) and radius 6 would be x^2+ y^2=36 and not x^2*y^2=36

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In an examination, Mary was asked to calculate the maximum area of a s 03 Jun 2021, 10:15
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directnick

Quote:
This way the area of the bigggest square is 6*2*6*2 => 144

If Mary made a mistake then the side of the second square is 26−−√26*2*26−−√26*2=104
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isn't the area of square = \(\frac{ d1*d1}{2}\) ? where d1 and d2 are diagonals
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In an examination, Mary was asked to calculate the maximum area of a s 04 Jun 2021, 09:16
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lakshyaag
In an examination, Mary was asked to calculate the maximum area of a square whose co-ordinates (on the XY plane) satisfied the relation \(x^2*y^2 = 36\). If she misread 36 as 26, and solved the question, what is the difference, in sq. units, between the actual area and Mary's answer?

A) 10 sq. units
B) 40 sq. units
C) 52 sq. units
D) 92 sq. units
E) 104 sq. units

I suppose there's a mistake in the Question. The formula should be \(x^2+y^2 = 36\) and not \(x^2*y^2 = 36\)

Because if it's \(x^2*y^2 = 36\) then there's no answer to the task...

\(x^2+y^2 = 36\) that's a formula of a circle with (0,0) as center and a radius of 6 (square root of 36).
If we are looking for the biggest square that intersects the circle -> then we should look at a circle square, where circle intersects the square in square's centers of the sides.



This way the area of the bigggest square is 6*2*6*2 => 144

If Mary made a mistake then the side of the second square is \(\sqrt[]{26}\)*2*\(\sqrt[]{26}\)*2=104

So the answer is 144-104 = 40, hence Option B


Please give KUDDOOOSSSSS if the explanation was clear for you!
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Hi directnick

How does your assumption (that plus sign should be there instead of multiplication) meet the criteria that coordinates of the square satisfied the relation x^2+y^2=36?

lakshyaag: Can you please post the answer/ solution?

Warm Regards,
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In an examination, Mary was asked to calculate the maximum area of a s 05 Jun 2021, 07:15
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HI When he mentions coordinates of the square on the equation (assuming it is circle equation), then does that mean vertices of the Square should be on a circle instead of midpoints of the square on the square as mentioned in the above solutions?
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In an examination, Mary was asked to calculate the maximum area of a s 05 Jun 2021, 10:46
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Seems incorrect to me.

The question says that the coordinates of the square satisfy the equation.

Implies that the points where the circle cuts the axes (plural) are the coordinates of the square.

for the equation x^2 + y^2 = 36

(6,0) (0,6) (-6,0) (0,-6)

using distance formula:

side = ((6-0)^2+(0-6)^2))^(1/2) = (72)^(1/2)
Area 1= [(72)^(1/2)]*2 =72

Similarly,

for the equation x^2 + y^2 = 26

((26)^(1/2),0) (0, (26)^(1/2)) (-(26)^(1/2),0) (0, -(26)^(1/2))
side = (52)^(1/2) using distance formula as above
Area 2 = 52

Difference: 72-52 = 20
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In an examination, Mary was asked to calculate the maximum area of a s 06 Jun 2021, 00:47
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This seems to me like what the question is asking as well....

The thing I’m seeing is that if you follow the top answer and make the square that circumscribes the circle that is given by (X)^2 + (Y)^2 = 36

Then the vertices of that square will not satisfy the equation. The vertices are outside the circumference of the circle.


Only simpler step I would add is that when you look at the square connected together by:

(0 , 6)
(0 , -6)
(6 , 0)
(-6 , 0)

Since each side is equal, and the diagonals of a square bisect each other, etc.

You can say that a square is made up of 4 smaller 45-45-90 Right Isosceles Triangles, in which each leg is (1/2) the diagonal along the Y and X Axis ——-> 6 units

So side = 6 * sqrt(2)

Area = [ 6 * sqrt(2) ]^2 = 72

(Or you can use the Formula: Area of Square = (1/2) * (Diagonal 1) * (Diagonal 2) )

Using the same logic for

(X)^2 + (Y)^2 = 26

Each side will be = sqrt(26) * sqrt(2) = sqrt(52)

Squaring this side to get the Area: Area = 52

Answer:
72 - 52 = 20 square units




maheswariviresh
Seems incorrect to me.

The question says that the coordinates of the square satisfy the equation.

Implies that the points where the circle cuts the axes (plural) are the coordinates of the square.

for the equation x^2 + y^2 = 36

(6,0) (0,6) (-6,0) (0,-6)

using distance formula:

side = ((6-0)^2+(0-6)^2))^(1/2) = (72)^(1/2)
Area 1= [(72)^(1/2)]*2 =72

Similarly,

for the equation x^2 + y^2 = 26

((26)^(1/2),0) (0, (26)^(1/2)) (-(26)^(1/2),0) (0, -(26)^(1/2))
side = (52)^(1/2) using distance formula as above
Area 2 = 52

Difference: 72-52 = 20
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In an examination, Mary was asked to calculate the maximum area of a s 06 Jun 2021, 09:02
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As written, this question makes no sense. The coordinates of a square do not satisfy the equation x^2 + y^2 = 36 (that's the equation of a circle), nor the equation x^2 * y^2 = 36 (that's the equation of a right hyperbola, along with its reflections through the axes -- though GMAT test takers need to know literally nothing about hyperbolas). Perhaps the question means instead to say merely that the four vertices of the square satisfy the equation (rather than the whole square). If that's the case, then if the equation is meant to be x^2 + y^2 = 36, then the solutions posted by maheswariviresh and Fdambro294 above are correct, and the answer is 20, not 40.

If instead the question means to say x^2 * y^2 = 36, as is originally written in the OP, then the question does have an answer (someone said otherwise above). The points (√6, √6), (√6, -√6), (-√6, -√6) and (-√6, √6) all satisfy that equation, and are the only quartet of such points that can be the corners of a square. When you change '36' to '26', you get a similar set of points, though it looks messy -- the first quadrant point is (√√26, √√26), for example. The difference in areas is then straightforward to compute and turns out to be roughly 3.6.

My best guess is that the question meant to say that the four vertices of the square satisfy the equation x^2 * y^2 = 36^2. Then the points (6, 6), (-6, 6), (-6, -6) and (6, -6) become the four vertices of the square, and when we change the 36^2 to 26^2, the points (√26, √26), (-√26, √26), etc, become the new vertices of the square. The difference in areas is then 40, which is the OA.

But really, if the question wanted equations satisfied by the "coordinates of a square", which means that all of the points along the boundary of the square satisfy the equation (not only the four corners), the equations should have looked like this:

|y| = -|x| + 6

and, when Mary misreads it,

|y| = -|x| + √26

Then both equations represent squares, and the question makes mathematical sense (though with these two specific equations, the difference in area is 20, not 40).

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