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Re: GMAT Diagnostic Test Question 44 [#permalink]
14 Jun 2009, 08:26

Explanation Official Answer: C

A. |x| + |y| = 10 is possibl;e if one is 0 and the other is 10. B. |x| > |y| is possible if |x| > |5sqrt2| and |y| < |5sqrt2| C. |x| > |y| + 10 is never possible because if |x| > 10, (x^2+y2) becomes >100, which is wrong. D. |x| = |y| is possible if each is equal to |5sqrt2|. E. |x| - |y| = 5 is possible if if |x| = |9.11| and |y| = |4.11|.

Therefore all but C are possible. \(|x| > |y| + 10\) means \(x\) is greater than 10, which is not possible. So C is best. _________________

A. |x| + |y| = 10 is possibl;e if one is 0 and the other is 10. B. |x| > |y| is possible if |x| > |5sqrt2| and |y| < |5sqrt2| C. |x| > |y| + 10 is never possible because if |x| > 10, (x^2+y2) becomes >100, which is wrong. D. |x| = |y| is possible if each is equal to |5sqrt2|. E. |x| - |y| = 5 is possible if if |x| = |9.11| and |y| = |4.11|.

Therefore all but C are possible. \(|x| > |y| + 10\) means \(x\) is greater than 10, which is not possible. So C is best.

Hi GMAT TIGER,

I don't quite understand some of the answers. The question stem states all of the following MUST be true, except 1. And if x^2 + y^2 = 100 then x=6 and y=8 is one potential solution as is x=0 and y=10.

For statement a) |x| + |y| = 10 is only true if x=0 and y=10. But if x = 6 and y=8 then it wouldn't always be true?

Re: GMAT Diagnostic Test Question 44 [#permalink]
18 Oct 2009, 13:35

yangsta8 wrote:

GMAT TIGER wrote:

Explanation

Rating:

Official Answer: C

A. |x| + |y| = 10 is possibl;e if one is 0 and the other is 10. B. |x| > |y| is possible if |x| > |5sqrt2| and |y| < |5sqrt2| C. |x| > |y| + 10 is never possible because if |x| > 10, (x^2+y2) becomes >100, which is wrong. D. |x| = |y| is possible if each is equal to |5sqrt2|. E. |x| - |y| = 5 is possible if if |x| = |9.11| and |y| = |4.11|.

Therefore all but C are possible. \(|x| > |y| + 10\) means \(x\) is greater than 10, which is not possible. So C is best.

Hi GMAT TIGER,

I don't quite understand some of the answers. The question stem states all of the following MUST be true, except 1. And if x^2 + y^2 = 100 then x=6 and y=8 is one potential solution as is x=0 and y=10.

For statement a) |x| + |y| = 10 is only true if x=0 and y=10. But if x = 6 and y=8 then it wouldn't always be true?

Maybe I'm missing something?

I agree here too - how come 5 ^ 2 + 5 ^ 2 = 100? Maybe I missed something - Unlike to see questions on GMAT with more than one answer

Re: GMAT Diagnostic Test Question 44 [#permalink]
19 Oct 2009, 00:45

I used Pythagoras principle here:

let x and y be the sides of a triangle and 10 be the hypotenuse.

Now, A and C both are not possible. A: sum of two sides cannot be equal to 10, the sum of two sides should be > 10. C: x should be less than sum of y and 10.

Re: GMAT Diagnostic Test Question 44 [#permalink]
21 Nov 2009, 08:12

GMAT TIGER wrote:

Explanation

Rating:

Official Answer: C

A. |x| + |y| = 10 is possibl;e if one is 0 and the other is 10. B. |x| > |y| is possible if |x| > |5sqrt2| and |y| < |5sqrt2| C. |x| > |y| + 10 is never possible because if |x| > 10, (x^2+y2) becomes >100, which is wrong. D. |x| = |y| is possible if each is equal to |5sqrt2|. E. |x| - |y| = 5 is possible if if |x| = |9.11| and |y| = |4.11|.

Therefore all but C are possible. \(|x| > |y| + 10\) means \(x\) is greater than 10, which is not possible. So C is best.

I have a hard time finding a way to prove E. How did you came up with the number?

Re: GMAT Diagnostic Test Question 44 [#permalink]
22 Dec 2009, 05:02

lonewolf wrote:

GMAT TIGER wrote:

Explanation

Rating:

Official Answer: C

A. |x| + |y| = 10 is possibl;e if one is 0 and the other is 10. B. |x| > |y| is possible if |x| > |5sqrt2| and |y| < |5sqrt2| C. |x| > |y| + 10 is never possible because if |x| > 10, (x^2+y2) becomes >100, which is wrong. D. |x| = |y| is possible if each is equal to |5sqrt2|. E. |x| - |y| = 5 is possible if if |x| = |9.11| and |y| = |4.11|.

Therefore all but C are possible. \(|x| > |y| + 10\) means \(x\) is greater than 10, which is not possible. So C is best.

I have a hard time finding a way to prove E. How did you came up with the number?

Just solve the system of equations: x^2+y^2=100 x - y = 5 and you'll get the numbers to prove E.

Re: GMAT Diagnostic Test Question 44 [#permalink]
10 Feb 2010, 18:36

Igor010 wrote:

Just solve the system of equations: x^2+y^2=100 x - y = 5 and you'll get the numbers to prove E.

I don't think that'll work buddy:

x - y = 5 x = y + 5, substitute into first eq yields:

(y+5)^2 + y^2 = 100

rearranging gives: 2y^2 + 10y - 75 = 0.... there's no solution for the above: (check: b^2 - 4ac term is negative, which implies imaginary roots). _________________

Re: GMAT Diagnostic Test Question 44 [#permalink]
11 Feb 2010, 09:25

adalfu wrote:

Igor010 wrote:

Just solve the system of equations: x^2+y^2=100 x - y = 5 and you'll get the numbers to prove E.

I don't think that'll work buddy:

x - y = 5 x = y + 5, substitute into first eq yields:

(y+5)^2 + y^2 = 100

rearranging gives: 2y^2 + 10y - 75 = 0.... there's no solution for the above: (check: b^2 - 4ac term is negative, which implies imaginary roots).

I'm sorry adalfu, \(2y^2+10y-75=0\) gives us \(100-(4*2*(-75))\) which gives us +700. You can get square root and solve for Y. Even if you know approx. value of Y, you can find approx. value of X and then plug in to prove your results. Seems like you forgot to change the sign... Hope this helped.

Re: GMAT Diagnostic Test Question 44 [#permalink]
11 Feb 2010, 09:55

doh! brilliant... thanks Igor010

i feel that all my quant errors are due to some sort of simple arithmetic mistakes (i'm a math/stats major but i've forgotten my basic algebra... sometimes i forget what 7x9 or 6x9 is -- i have to seriously think about it; it just doesn't come natural anymore).

let x and y be the sides of a triangle and 10 be the hypotenuse.

Now, A and C both are not possible. A: sum of two sides cannot be equal to 10, the sum of two sides should be > 10. C: x should be less than sum of y and 10.

But the question didn't say X, Y, and 10 are the vertices of a triangle, so X or Y can be zero (and they don't need to follow the Pythagoras principle). So, just C is the right answer.

A. |x|+|y|=10 is possible if one is 0 and the other is 10. B. |x|>|y| is possible if |x|>|52√| and |y|<|52√| C. |x|>|y|+10 is never possible because if |x|>10, x2+y2 becomes greater than 100, which is wrong. D. |x|=|y| is possible if each is equal to |52√|. E. |x|−|y|=5 is possible if |x|=|9.11| and |y|=|4.11|. Therefore all but C are possible. |x|>|y|+10 means x is greater than 10, which is not possible

Is there any alternate way to tackle this problem besides just plugging in numbers? It's very challenging to figure out 9.11 and 4.11 could be two possibilities to rule out E.

Here's what I did to solve this problem. Got stuck at this step and did not have a clue to proceed further...Any algebraic help will get kudos \(x^2+y^2=100\) \((x+y)^2-2xy = 10^2\) \((x+y)^2 = \frac{10^2}{-2xy}\) \(|x +y| = \frac{|10|}{\sqrt{-2xy}}\) => Root of negative number will result in imaginary and therefore it's likely that this condition might be the answer. But I'm not even sure if this is the correct approach and could not relate |x+y| and |10|.

A. |x|+|y|=10 is possible if one is 0 and the other is 10. B. |x|>|y| is possible if |x|>|52√| and |y|<|52√| C. |x|>|y|+10 is never possible because if |x|>10, x2+y2 becomes greater than 100, which is wrong. D. |x|=|y| is possible if each is equal to |52√|. E. |x|−|y|=5 is possible if |x|=|9.11| and |y|=|4.11|. Therefore all but C are possible. |x|>|y|+10 means x is greater than 10, which is not possible

Is there any alternate way to tackle this problem besides just plugging in numbers? It's very challenging to figure out 9.11 and 4.11 could be two possibilities to rule out E.

Here's what I did to solve this problem. Got stuck at this step and did not have a clue to proceed further...Any algebraic help will get kudos \(x^2+y^2=100\) \((x+y)^2-2xy = 10^2\) \((x+y)^2 = \frac{10^2}{-2xy}\) \(|x +y| = \frac{|10|}{\sqrt{-2xy}}\) => Root of negative number will result in imaginary and therefore it's likely that this condition might be the answer. But I'm not even sure if this is the correct approach and could not relate |x+y| and |10|.

For option E: Why making so much complicated?

The max difference for |X| - |Y| will be 10 and minimum difference 0.

Whether the difference is 3, 5, 8, it doesnt matter at all. since they all will come under the above range, possibly making the given equation valid.

To solve in 15 secs: Just check for option in which X or Y is greater than 10. Only C satisfies that.

gmatclubot

Re: Quantitative :: Problem solving :: Algebra :: D01-44
[#permalink]
14 Dec 2013, 19:53