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Manager  B
Joined: 18 Jul 2018
Posts: 51
Location: United Arab Emirates

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Bunuel wrote:
Official Solution:

$$x^2 + y^2 = 100$$. All of the following could be true EXCEPT

A. $$|x| + |y| = 10$$
B. $$|x| \gt |y|$$
C. $$|x| \gt |y| + 10$$
D. $$|x| = |y|$$
E. $$|x| - |y| = 5$$

A. $$|x| + |y| = 10$$ is possible if one is 0 and the other is 10.

B. $$|x| \gt |y|$$ is possible if $$|x| \gt |5\sqrt{2}|$$ and $$|y| \lt |5\sqrt{2}|$$

C. $$|x| \gt |y| + 10$$ is never possible because if $$|x| \gt 10$$, $$x^2+y^2$$ becomes greater than 100, which is wrong.

D. $$|x| = |y|$$ is possible if each is equal to $$|5\sqrt{2}|$$.

E. $$|x| - |y| = 5$$ is possible if $$|x| = |9.11|$$ and $$|y| = |4.11|$$.

Therefore all but C are possible. $$|x| \gt |y| + 10$$ means $$x$$ is greater than 10, which is not possible.

Hi Bunuel

Can we solve this in the following manner???

x^2+y^2 = 100.
If we take another look at the given statement, it can be consider an equation of right angle triangle.
We know that sum of any two sides of triangle must be always greater than the third side.

Hence, Option C can never be true.

Manager  B
Joined: 18 Jul 2018
Posts: 51
Location: United Arab Emirates

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Bunuel wrote:
Official Solution:

$$x^2 + y^2 = 100$$. All of the following could be true EXCEPT

A. $$|x| + |y| = 10$$
B. $$|x| \gt |y|$$
C. $$|x| \gt |y| + 10$$
D. $$|x| = |y|$$
E. $$|x| - |y| = 5$$

A. $$|x| + |y| = 10$$ is possible if one is 0 and the other is 10.

B. $$|x| \gt |y|$$ is possible if $$|x| \gt |5\sqrt{2}|$$ and $$|y| \lt |5\sqrt{2}|$$

C. $$|x| \gt |y| + 10$$ is never possible because if $$|x| \gt 10$$, $$x^2+y^2$$ becomes greater than 100, which is wrong.

D. $$|x| = |y|$$ is possible if each is equal to $$|5\sqrt{2}|$$.

E. $$|x| - |y| = 5$$ is possible if $$|x| = |9.11|$$ and $$|y| = |4.11|$$.

Therefore all but C are possible. $$|x| \gt |y| + 10$$ means $$x$$ is greater than 10, which is not possible.

Xylan can you help?? What if i don't want to do number plugging??
Manager  P
Status: The darker the night, the nearer the dawn!
Joined: 16 Jun 2018
Posts: 182
GMAT 1: 640 Q50 V25 ### Show Tags

2
JIAA wrote:
Bunuel wrote:
Official Solution:

$$x^2 + y^2 = 100$$. All of the following could be true EXCEPT

A. $$|x| + |y| = 10$$
B. $$|x| \gt |y|$$
C. $$|x| \gt |y| + 10$$
D. $$|x| = |y|$$
E. $$|x| - |y| = 5$$

A. $$|x| + |y| = 10$$ is possible if one is 0 and the other is 10.

B. $$|x| \gt |y|$$ is possible if $$|x| \gt |5\sqrt{2}|$$ and $$|y| \lt |5\sqrt{2}|$$

C. $$|x| \gt |y| + 10$$ is never possible because if $$|x| \gt 10$$, $$x^2+y^2$$ becomes greater than 100, which is wrong.

D. $$|x| = |y|$$ is possible if each is equal to $$|5\sqrt{2}|$$.

E. $$|x| - |y| = 5$$ is possible if $$|x| = |9.11|$$ and $$|y| = |4.11|$$.

Therefore all but C are possible. $$|x| \gt |y| + 10$$ means $$x$$ is greater than 10, which is not possible.

Xylan can you help?? What if i don't want to do number plugging??

JIAA It's completely okay if you do NOT want to do number plugging.
However, aspire to reach the CORRECT solution in the least possible time so that one can spend judicious time on 700+ Qs.

Quote:
Remember, Our arsenal should be equipped with all sorts of ammunition to tame the beast such as Reverse-solving, plugging different numbers, edge-case scenarios, etc.

The equation $$x^2 + y^2 = 100$$ is actually the locus of a circle with the origin as the center and radius of 10 units.
If possible Pre-Think the problem such as the allowable value of X and Y -
According to the question, the maximum value of either $$x^2$$ or $$y^2$$ can be 100, which implies that the greatest absolute value of either X or Y can be 10.
Thus, $$|x|$$ must be $$<= 10$$. Refer the attached picture.

Hence, $$|x|$$ CANNOT be $$> 10$$. Let alone $$|x|$$ being greater than $$|y| + 10$$.

If we take another look at the given statement $$x^2 + y^2 = 100$$, it can be considered an equation of right angle triangle with hypotenuse = 10 and perpendicular-sides as X and Y.
$$x^2 + y^2 = 100$$
And we know that the sum of two sides is always greater than the third side.
Thus: |y| + 10 > |x| : The third-side is smaller than the sum of other two-sides.
Therefore, OptionC is incorrect as it says $$|x| > |y| + 10$$, which can NEVER be true.

>> !!!

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Manager  B
Joined: 18 Jul 2018
Posts: 51
Location: United Arab Emirates

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Xylan wrote:
JIAA wrote:
Bunuel wrote:
Official Solution:

$$x^2 + y^2 = 100$$. All of the following could be true EXCEPT

A. $$|x| + |y| = 10$$
B. $$|x| \gt |y|$$
C. $$|x| \gt |y| + 10$$
D. $$|x| = |y|$$
E. $$|x| - |y| = 5$$

A. $$|x| + |y| = 10$$ is possible if one is 0 and the other is 10.

B. $$|x| \gt |y|$$ is possible if $$|x| \gt |5\sqrt{2}|$$ and $$|y| \lt |5\sqrt{2}|$$

C. $$|x| \gt |y| + 10$$ is never possible because if $$|x| \gt 10$$, $$x^2+y^2$$ becomes greater than 100, which is wrong.

D. $$|x| = |y|$$ is possible if each is equal to $$|5\sqrt{2}|$$.

E. $$|x| - |y| = 5$$ is possible if $$|x| = |9.11|$$ and $$|y| = |4.11|$$.

Therefore all but C are possible. $$|x| \gt |y| + 10$$ means $$x$$ is greater than 10, which is not possible.

Xylan can you help?? What if i don't want to do number plugging??

JIAA It's completely okay if you do NOT want to do number plugging.
However, aspire to reach the CORRECT solution in the least possible time so that one can spend judicious time on 700+ Qs.

Quote:
Remember, Our arsenal should be equipped with all sorts of ammunition to tame the beast such as Reverse-solving, plugging different numbers, edge-case scenarios, etc.

The equation $$x^2 + y^2 = 100$$ is actually the locus of a circle with the origin as the center and radius of 10 units.
If possible Pre-Think the problem such as the allowable value of X and Y -
According to the question, the maximum value of either $$x^2$$ or $$y^2$$ can be 100, which implies that the greatest absolute value of either X or Y can be 10.
Thus, $$|x|$$ must be $$<= 10$$. Refer the attached picture.

Hence, $$|x|$$ CANNOT be $$> 10$$. Let alone $$|x|$$ being greater than $$|y| + 10$$.

If we take another look at the given statement $$x^2 + y^2 = 100$$, it can be considered an equation of right angle triangle with hypotenuse = 10 and perpendicular-sides as X and Y.
$$x^2 + y^2 = 100$$
And we know that the sum of two sides is always greater than the third side.
Thus: |y| + 10 > |x| : The third-side is smaller than the sum of other two-sides.
Therefore, OptionC is incorrect as it says $$|x| > |y| + 10$$, which can NEVER be true.

Xylan this makes perfect sense! THANKS for the detailed explanation.
Much appreciated!
Intern  B
Joined: 19 Feb 2019
Posts: 12
GMAT 1: 730 Q49 V41
GPA: 3.8

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I solved it using the equation for a circle; x^2 + y^2 = r^2. In this case r = 10, and the centre of the circle is the origin (0,0).

With this in mind we can conclude that point |x|>|y|+10 will lie outside the circle. Hence not possible
Manager  S
Joined: 23 Apr 2018
Posts: 164

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Hi, what would be an efficient way to solve this in an exam, with a timer running obviously.
I was able to eliminate 2/3 choices, but I think such a question can take up a lot of time if you are -
a) zoned out and this comes around the last part of the test
b) just guessed 1 question and are low on morale

Please guide a quicker way to solve such questions, as the numbers described in the options won't come to mind swiftly.
Intern  B
Joined: 08 Dec 2019
Posts: 1

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This question can be visually comprehended in another way where xSquare +ySquare = 10Square.

The equation is in the form of Pythagoras theorem. Where x, y and 10 are sides of the triangle. We know that in no case one side can be greater than the sum of the two sides.

Option C implies just that where x > y+10, which is not possible.

This is another approach which seems easier to visualize.
Intern  B
Joined: 15 Jan 2019
Posts: 12

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Bunuel wrote:
Official Solution:

$$x^2 + y^2 = 100$$. All of the following could be true EXCEPT

A. $$|x| + |y| = 10$$
B. $$|x| \gt |y|$$
C. $$|x| \gt |y| + 10$$
D. $$|x| = |y|$$
E. $$|x| - |y| = 5$$

A. $$|x| + |y| = 10$$ is possible if one is 0 and the other is 10.

B. $$|x| \gt |y|$$ is possible if $$|x| \gt |5\sqrt{2}|$$ and $$|y| \lt |5\sqrt{2}|$$

C. $$|x| \gt |y| + 10$$ is never possible because if $$|x| \gt 10$$, $$x^2+y^2$$ becomes greater than 100, which is wrong.

D. $$|x| = |y|$$ is possible if each is equal to $$|5\sqrt{2}|$$.

E. $$|x| - |y| = 5$$ is possible if $$|x| = |9.11|$$ and $$|y| = |4.11|$$.

Therefore all but C are possible. $$|x| \gt |y| + 10$$ means $$x$$ is greater than 10, which is not possible.

Hey Guys,
I wanted to check if my process of solving this problem was right.
I Multiplied the equation with square root, so I got x +y=10
basically with this,
I eliminated A with x=5, y=5
eliminated B with x=7, y=3
eliminated D with x=5, y=5
eliminated E with x=7.5, y=2.5
One cannot eliminate C because of the lowest value of Y=0, so X cannot be greater than 10, as x+y=10
Math Expert V
Joined: 02 Sep 2009
Posts: 60644

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sankarsh wrote:
Bunuel wrote:
Official Solution:

$$x^2 + y^2 = 100$$. All of the following could be true EXCEPT

A. $$|x| + |y| = 10$$
B. $$|x| \gt |y|$$
C. $$|x| \gt |y| + 10$$
D. $$|x| = |y|$$
E. $$|x| - |y| = 5$$

A. $$|x| + |y| = 10$$ is possible if one is 0 and the other is 10.

B. $$|x| \gt |y|$$ is possible if $$|x| \gt |5\sqrt{2}|$$ and $$|y| \lt |5\sqrt{2}|$$

C. $$|x| \gt |y| + 10$$ is never possible because if $$|x| \gt 10$$, $$x^2+y^2$$ becomes greater than 100, which is wrong.

D. $$|x| = |y|$$ is possible if each is equal to $$|5\sqrt{2}|$$.

E. $$|x| - |y| = 5$$ is possible if $$|x| = |9.11|$$ and $$|y| = |4.11|$$.

Therefore all but C are possible. $$|x| \gt |y| + 10$$ means $$x$$ is greater than 10, which is not possible.

Hey Guys,
I wanted to check if my process of solving this problem was right.
I Multiplied the equation with square root, so I got x +y=10
basically with this,
I eliminated A with x=5, y=5
eliminated B with x=7, y=3
eliminated D with x=5, y=5
eliminated E with x=7.5, y=2.5
One cannot eliminate C because of the lowest value of Y=0, so X cannot be greater than 10, as x+y=10

No, that's not correct. The point is that $$\sqrt{x^2 + y^2}$$ does not equal to x + y. Ask yourself does $$\sqrt{1^2 + 2^2}$$ equal to 1 + 2? No.
_________________ Re: D01-44   [#permalink] 16 Dec 2019, 10:48

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# D01-44

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