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ChandlerBong
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Is p<3? 11 Aug 2022, 07:27
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55% (hard)

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55% (02:03) correct 45% (02:19) wrong based on 51 sessions

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Is p<3?

(1) In the coordinate plane, the point (-1,p) lies inside the square S. The sides of the square are 6 each and its diagonals intersect at the origin.

(2) In the coordinate plane, the point (p,-1) lies inside the circle with equation \(x^2 + y^2 = 25\).
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Is p<3? 11 Aug 2022, 07:38
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chetan2u Bunuel nick1816 @veritaskarishma @TheDudeKnows

Can you please share your approach to this question?

Thanks in advance.
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Is p<3? 11 Aug 2022, 07:41
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KarishmaB ThatDudeKnows

Can you please share your approach to this question?

Thanks in advance.
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Is p<3? Updated on: 11 Aug 2022, 08:49
Originally posted by ThatDudeKnows on 11 Aug 2022, 08:10.
Last edited by ThatDudeKnows on 11 Aug 2022, 08:49, edited 1 time in total.
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WolfOfStark7
Is p<3?

(1) In the coordinate plane, the point (-1,p) lies inside the square S. The sides of the square are 6 each and its diagonals intersect at the origin.

(2) In the coordinate plane, the point (p,-1) lies inside the circle with equation \(x^2 + y^2 = 25\).

Show more

WolfOfStark7

Edited to add: As pointed out by gmatophobia in the post below, I made an oversight about possible orientations for the square in Statement (1). I'll leave this post as-is, so be sure to check out why it's wrong!

Looking at statement (1) alone:
Sketch a quick coordinate plane. The diagonals of a square intersect at its center. The origin is the center and the sides are 6. Sketch that onto the coordinate plane. Now sketch in a vertical line at x=-1 that's inside the square. What are the possible values for p? Looks like it can go from -3 to 3. -3<p<3 From statement (1) alone, do we have enough information to determine whether p<3? Yep. AD.

Looking at statement (2) alone:
Same idea. This time, the line is horizontal. Looks like it'll extend far enough to the right that it'll include some possible values for p that are greater than 3. From statement (2) alone, do we have enough information to determine whether p<3? Nope. A.

Answer choice A.
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Is p<3? 11 Aug 2022, 08:33
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Is p<3?

(1) In the coordinate plane, the point (-1,p) lies inside the square S. The sides of the square are 6 each and its diagonals intersect at the origin.

(2) In the coordinate plane, the point (p,-1) lies inside the circle with equation \(x^2 + y^2 = 25\).


WolfOfStark7

Looking at statement (1) alone:
Sketch a quick coordinate plane. The diagonals of a square intersect at its center. The origin is the center and the sides are 6. Sketch that onto the coordinate plane. Now sketch in a vertical line at x=-1 that's inside the square. What are the possible values for p? Looks like it can go from -3 to 3. -3<p<3 From statement (1) alone, do we have enough information to determine whether p<3? Yep. AD.

Looking at statement (2) alone:
Same idea. This time, the line is horizontal. Looks like it'll extend far enough to the right that it'll include some possible values for p that are greater than 3. From statement (2) alone, do we have enough information to determine whether p<3? Nope. A.

Answer choice A.
Show more

ThatDudeKnows - Thanks for your quick response. I assume that the square that you mentioned in Statement 1 has its sides parallel to the axis (i.e. the square is at position ABCD in the attached figure).

I am wondering if we can have the square titled (as depicted by the red square) as well. If this were the case, point A (side of the square) can lie on any point on the circle represented in green.

Therefore, point A will lie on the circle represented by \(x^2 + y^2 = (3\sqrt{2})^2\)

At x = -1

\(y^2 = 18 - 1 = 17\)

y = +- 4.XX

Is this an incorrect understanding ?
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Is p<3? 11 Aug 2022, 08:46
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Is p<3?

(1) In the coordinate plane, the point (-1,p) lies inside the square S. The sides of the square are 6 each and its diagonals intersect at the origin.

(2) In the coordinate plane, the point (p,-1) lies inside the circle with equation \(x^2 + y^2 = 25\).


WolfOfStark7

Looking at statement (1) alone:
Sketch a quick coordinate plane. The diagonals of a square intersect at its center. The origin is the center and the sides are 6. Sketch that onto the coordinate plane. Now sketch in a vertical line at x=-1 that's inside the square. What are the possible values for p? Looks like it can go from -3 to 3. -3<p<3 From statement (1) alone, do we have enough information to determine whether p<3? Yep. AD.

Looking at statement (2) alone:
Same idea. This time, the line is horizontal. Looks like it'll extend far enough to the right that it'll include some possible values for p that are greater than 3. From statement (2) alone, do we have enough information to determine whether p<3? Nope. A.

Answer choice A.

ThatDudeKnows - Thanks for your quick response. I assume that the square that you mentioned in Statement 1 has its sides parallel to the axis (i.e. the square is at position ABCD in the attached figure).

I am wondering if we can have the square titled (as depicted by the red square) as well. If this were the case, point A (side of the square) can lie on any point on the circle represented in green.

Therefore, point A will lie on the circle represented by \(x^2 + y^2 = (3\sqrt{2})^2\)

At x = -1

\(y^2 = 18 - 1 = 17\)

y = +- 4.XX

Is this an incorrect understanding ?
Show more

Great catch that the square doesn't need to be "flat," gmatophobia!! And doubly impressive given that the posted OA appears to have taken the same (incorrect!) approach that I did, so double kudos to you!! :clap: :clap:

One quick nit about the final bit of your explanation: we are told that the point lies inside the circle, not on it, so there are infinite possible values for p, some of which are greater than 3 and some of which are less than 3.
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Is p<3? 12 Aug 2022, 03:29
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WolfOfStark7

Is p<3?

(1) In the coordinate plane, the point (-1,p) lies inside the square S. The sides of the square are 6 each and its diagonals intersect at the origin.
So Center of square is the origin.
If sides of square are 6, then the diagonal will be \(6\sqrt{2}\). Thus, the Center will be at middle of diagonal that is at \(3\sqrt{2}\).
Now the farthest point of square can be anywhere \(3\sqrt{2}\) from origin, that is (-1,p) lies within a circle of radius \(3\sqrt{2}\)Since we do not know p, (-1,p) can be any point on line inside this circle.
But p could be anywhere 1 distance away from y-axis. Thus it is a line parallel to Y axis and on the left side of y-axis.
p could be anywhere on this line within the circle. It could be slightly above -3\(\sqrt{2}\) and slightly below 3\(\sqrt{2}\).
Insufficient

(2) In the coordinate plane, the point (p,-1) lies inside the circle with equation \(x^2 + y^2 = 25\).
Here the circle radius is 5, and with the same explanation as above p will be on the line parallel to x axis and 1 distance below x axis.
Again, p could be anywhere on this line within the circle. It could be slightly on right of -5 and slightly left of 5.
Insufficient


Combined
Both statements combined common range of slightly above -3\(\sqrt{2}\) and slightly below 3\(\sqrt{2}\).
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Is p<3? 12 Aug 2022, 04:50
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WolfOfStark7
Is p<3?

(1) In the coordinate plane, the point (-1,p) lies inside the square S. The sides of the square are 6 each and its diagonals intersect at the origin.

(2) In the coordinate plane, the point (p,-1) lies inside the circle with equation \(x^2 + y^2 = 25\).
Show more

(1) In the coordinate plane, the point (-1,p) lies inside the square S. The sides of the square are 6 each and its diagonals intersect at the origin.


When the square has a kite orientation, the point (-1, p) lies on the green line inside the square.
Since slope of the blue sides is 1, when x co-ordinate moves from -4.2 (which is approx 3*sqrt(2) so that side is 6 each) to -1, the y co-ordinate moves from 0 to 3.2.
So p can be anything from -3.2 to 3.2. So p may be less than 3 or more than 3. So not sufficient alone.
Attachment:
Screenshot 2022-08-12 at 5.04.02 PM.png
Screenshot 2022-08-12 at 5.04.02 PM.png [ 25.09 KiB | Viewed 1794 times ]


(2) In the coordinate plane, the point (p,-1) lies inside the circle with equation \(x^2 + y^2 = 25\)

This is the equation of a circle centred at (0, 0) and radius 5. The point (p , -1) will lie on the red line.

\(p^2 + (-1)^2 = 25\)
\(p = \sqrt{24} = 4.9\) approx because sqrt(25) = 5

So p could take values somewhere between -4.9 to 4.9. So not sufficient alone.
Attachment:
Screenshot 2022-08-12 at 5.08.14 PM.png
Screenshot 2022-08-12 at 5.08.14 PM.png [ 19.3 KiB | Viewed 1754 times ]

Using both statements, we know that p would lie between -3.2 and 3.2 so it could still be less than or more than 3. Not sufficient. If p = 0, it satisfies both statements and if p = 3.1, it satisfies both statements.

Answer (E)
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