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If |x| = |y| and xy = 0, which of the following must be true?

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HarveyKlaus
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If |x| = |y| and xy = 0, which of the following must be true? [#permalink] If |x| = |y| and xy = 0, which of the following must be true?   Updated on: 27 Nov 2017, 21:36
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If |x| = |y| and xy = 0, which of the following must be true?


A \(xy^2>0\)

B. \(x^2y>0\)

C. \(x+y=0\)

D. \(\frac{x}{(y+1)}=2\)

E. \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\)
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C

Originally posted by HarveyKlaus on 22 Jan 2016, 12:51.
Last edited by Bunuel on 27 Nov 2017, 21:36, edited 5 times in total.
Reformatted the question
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Re: If |x| = |y| and xy = 0, which of the following must be true? [#permalink] If |x| = |y| and xy = 0, which of the following must be true?   22 Jan 2016, 13:04
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HarveyKlaus
If lXl = lYl and XY=0, which of the following must be true?

A \(xy^2>0\)
B. \(x^2y>0\)
C. \(x+y=0\)
D. \(x/(y+1)=2\)
E. \(1/x+1/y=1/2\)

Follow posting guidelines, including proper formatting of the question.

For the question, confirm that you have transcribed option E correctly.

You are given that |x| = |y| and xy = 0. For a MUST BE TRUE question, make sure to use POE for the options as the only option remaining will be true for ALL possible cases.

xy=0 ---> either x=0 and y \(\neq\) 0 or y=0 and x \(\neq\) 0 or both x =y=0. for the sake of simplicity, I will choose the case when x=y=0 and this also satisfies |x|=|y|.

Substitute x=y=0 in the options A-D and see which one remains true.

A \(xy^2>0\) . Not true. Eliminate
B. \(x^2y>0\). Not true. Eliminate
C. \(x+y=0\). True . Keep.
D. \(x/(y+1)=2\). x/(y+1) = 0 \(\neq\) 2. Eliminate.
E. \(1/x+1/y=1/2\). Not true. Eliminate. The only way you are going to get 1/x + 1/y = 1/2 is when you have x=y=1 . Although you will satisfy |x|=|y| , xy \(\neq\) 0. Thus eliminate this option.

Hence C is the correct answer.

Hope this helps.
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Re: If |x| = |y| and xy = 0, which of the following must be true? [#permalink] If |x| = |y| and xy = 0, which of the following must be true?   13 Oct 2017, 08:22
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HarveyKlaus
If |x| = |y| and xy = 0, which of the following must be true?

A \(xy^2>0\)
B. \(x^2y>0\)
C. \(x+y=0\)
D. \(x/(y+1)=2\)
E. \(1/x+1/y=1/2\)


Any number multiplied by zero, gives the product zero. Since the absolute value of 0 is 0, and |x| = |y|, x,y = 0

Therefore, x + y = 0 is the only answer that must be true.
(C)
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Re: If |x| = |y| and xy = 0, which of the following must be true? [#permalink] If |x| = |y| and xy = 0, which of the following must be true?   27 Nov 2017, 19:08
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HarveyKlaus
If |x| = |y| and xy = 0, which of the following must be true?

A \(xy^2>0\)
B. \(x^2y>0\)
C. \(x+y=0\)
D. \(x/(y+1)=2\)
E. \(1/x+1/y=1/2\)

Since |x| = |y| and xy = 0, both x and y must be zero.

Thus, we see that A and B are not true, because both answers are equal to zero.

C, however, must be true because 0 + 0 = 0.

Answer: C
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If |x| = |y| and xy = 0, which of the following must be true? [#permalink] If |x| = |y| and xy = 0, which of the following must be true?   27 Mar 2018, 06:00
Bunuel niks18 chetan2u pushpitkc

Quote:
If |x| = |y| and xy = 0, which of the following must be true?


A \(xy^2>0\)

B. \(x^2y>0\)

C. \(x+y=0\)

D. \(\frac{x}{(y+1)}=2\)

E. \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\)

Can you explain solving modulus on both sides of equality if it was a PS problem.

|x| = |y|

Do I definitely need to know if x or y is positive or negative?

I approached in below manner:

If product of two numbers x and y is zero, then EITHER of x or y is zero, or
BOTH x and y are zero.

With this logic, A,B and E are out. (Since this is PS problem ALL of above
conditions must be satisfied.)

For D, if x = 0 then RHS = 2 is Not satisfied, since LHS = 0
If y = 0 , then |0| = 0 LHS = 0/1 or 0, RHS = 2. Not satisfied.

Let me know if my steps are correct.
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Re: If |x| = |y| and xy = 0, which of the following must be true? [#permalink] If |x| = |y| and xy = 0, which of the following must be true?   27 Mar 2018, 06:29
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Bunuel niks18 chetan2u pushpitkc

Quote:
If |x| = |y| and xy = 0, which of the following must be true?


A \(xy^2>0\)

B. \(x^2y>0\)

C. \(x+y=0\)

D. \(\frac{x}{(y+1)}=2\)

E. \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\)

Can you explain solving modulus on both sides of equality if it was a PS problem.

|x| = |y|

Do I definitely need to know if x or y is positive or negative?

I approached in below manner:

If product of two numbers x and y is zero, then EITHER of x or y is zero, or
BOTH x and y are zero.

With this logic, A,B and E are out. (Since this is PS problem ALL of above
conditions must be satisfied.)

For D, if x = 0 then RHS = 2 is Not satisfied, since LHS = 0
If y = 0 , then |0| = 0 LHS = 0/1 or 0, RHS = 2. Not satisfied.

Let me know if my steps are correct.

Hi adkikani

|x|=|y| implies magnitude of x and y are equal irrespective of their sign. For e.g. if x=2 and y=-2 then also |x|=|y|.

As you rightly mentioned xy =0 so either of the two has to be 0 but here as their magnitude are same so both x & y has to be 0

Only option C is valid in this case

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Re: If |x| = |y| and xy = 0, which of the following must be true? [#permalink] If |x| = |y| and xy = 0, which of the following must be true?   10 Nov 2023, 10:23
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Re: If |x| = |y| and xy = 0, which of the following must be true? [#permalink]
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