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# If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?

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Math Expert
Joined: 02 Sep 2009
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If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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10 Aug 2017, 00:41
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If ⊙ denotes a mathematical operation, does $$x⊙y = y⊙x$$ for all x and y?

(1) For all x and y, $$x ⊙ y = 2(x^2 + y^2)$$

(2) For all y, $$0 ⊙ y = 2y^2$$

DS17602.01

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Re: If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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10 Aug 2017, 00:58
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2
Bunuel wrote:
If ⊙ denotes a mathematical operation, does $$x⊙y = y⊙x$$ for all x and y?

(1) For all x and y, $$x ⊙ y = 2(x^2 + y^2)$$

(2) For all y, $$0 ⊙ y = 2y^2$$

(1) since x and y can be interchanged
Sufficient

(2) if ⊙ denotes +
then
0 ⊙ y = 0 + 2y^2
x ⊙ y = x + 2y^2
Here x and y cannot be interchanged
but if
0 ⊙ y = 2*0^2 + 2y^2
x ⊙ y = 2*x^2 + 2y^2
Here x and y can be interchanged
Not Sufficient

A
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Joined: 11 May 2019
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Re: If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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04 Jan 2020, 22:02
(1) For all x and y, x⊙y=2(x2+y2)

Here X and Y are interchangeable and gives the same value. Hence sufficient.

(2) For all y, 0⊙y=2y2

here y⊙0=0 , the values are not equal if y has values other than 0. Hence insufficient.
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Re: If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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14 Jan 2020, 23:38
Bunuel wrote:
If ⊙ denotes a mathematical operation, does $$x⊙y = y⊙x$$ for all x and y?

(1) For all x and y, $$x ⊙ y = 2(x^2 + y^2)$$

(2) For all y, $$0 ⊙ y = 2y^2$$

DS17602.01

Bunuel can u explain this qsn ,sply 2nd stmnt?
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Re: If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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26 Jan 2020, 20:13
1
vanam52923 wrote:
Bunuel wrote:
If ⊙ denotes a mathematical operation, does $$x⊙y = y⊙x$$ for all x and y?

(1) For all x and y, $$x ⊙ y = 2(x^2 + y^2)$$

(2) For all y, $$0 ⊙ y = 2y^2$$

DS17602.01

Bunuel can u explain this qsn ,sply 2nd stmnt?

Can any one explain this question?
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GMAT 1: 670 Q46 V36
GMAT 2: 690 Q47 V38
Re: If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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13 Mar 2020, 04:26
vanam52923 wrote:
vanam52923 wrote:
Bunuel wrote:
If ⊙ denotes a mathematical operation, does $$x⊙y = y⊙x$$ for all x and y?

(1) For all x and y, $$x ⊙ y = 2(x^2 + y^2)$$

(2) For all y, $$0 ⊙ y = 2y^2$$

DS17602.01

Bunuel can u explain this qsn ,sply 2nd stmnt?

Can any one explain this question?

VeritasKarishma Bunuel Can you please explain the solution for this question? Thanks!
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Joined: 02 Apr 2020
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Re: If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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24 Apr 2020, 10:25
Could we replace "operation" by "function" here ? I cannot understand what's going on.

I understand that if we have x o y = 2(x^2+y^2) we can change the order and come up with x o y = y o x. But which of "+", "-", "/" and "*" could work with "all x and all y". I cannot understand why we use an equal sign.
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Re: If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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06 Jun 2020, 21:14
For st 2, we can see that 0 to any nonzero power = 0 and thus 0•2y^2=2y^2
Thus when we replace 0 with x, then x can technically take any power.

However, when x takes any power, then sometimes it CANNOT be interchanged with y eg. 2(X^3+y^2)≠2(y^3+x^2) for say y=2 and x=3 [ie. 2(27+4)≠2(8+9)]

But sometimes it CAN be interchanged with y^2, ie. When x has the power 2. Not Sufficient

Posted from my mobile device
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Joined: 31 May 2020
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If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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18 Jun 2020, 08:34
OFFICIAL EXPLANATION

1) For all x and y, $$x⊙y = 2(x^2 + y^2)$$ is equal to $$y⊙x = 2(y^2 + x^2)$$; SUFFICIENT.

2) If $$x⊙y = 2(x^2+y^2)$$ for all x and y, then $$0⊙y = 2y^2$$ for all y and x⊙y = y⊙x for all x and y. However, if $$x⊙ y = 2(x^3 + y^2)$$ for all x and y, then $$0⊙y = 2y^2$$ for all y, but$$1⊙2 = 2(1^3) + 2(2^2) = 10$$ and $$2⊙1 = 2(2^3) + 2(1^2) = 18;$$NOT sufficient.

statement 1 alone is sufficient.
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Re: If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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24 Jun 2020, 13:19
Bunuel wrote:
If ⊙ denotes a mathematical operation, does $$x⊙y = y⊙x$$ for all x and y?

(1) For all x and y, $$x ⊙ y = 2(x^2 + y^2)$$

(2) For all y, $$0 ⊙ y = 2y^2$$

DS17602.01

anyone got a PROPER. COMPREHENSIBLE. CLEAR solution for this. none of the posts in the thread deliver
Manager
Joined: 27 Mar 2017
Posts: 177
If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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08 Jul 2020, 23:56

For statement 2, isn't this logic sufficient that the expression only holds true when one of the variables (be it x or y) is 0 whereas the questions asks if it is true for all values of x and y ?

On a related note, is the expression in statement 2 even interchangeable ? I mean the expression hold strictly when x is 0 and y is a variable. Since one of the values is fixed, doesn't that mean that the expression can't be interchanged (i.e x for y and y for as can be done in Statement 1).

Will appreciate some comments to improve my logic here.
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Re: If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?  [#permalink]

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13 Jul 2020, 08:26

For statement 2, isn't this logic sufficient that the expression only holds true when one of the variables (be it x or y) is 0 whereas the questions asks if it is true for all values of x and y ?

On a related note, is the expression in statement 2 even interchangeable ? I mean the expression hold strictly when x is 0 and y is a variable. Since one of the values is fixed, doesn't that mean that the expression can't be interchanged (i.e x for y and y for as can be done in Statement 1).

Will appreciate some comments to improve my logic here.

(2) For all y,
0⊙y=2y^2

The user defined operator could stand for anything. It is possible that

x⊙y = 2y^2 + x (not symmetrical)
Then 0⊙y = 2y^2

It is also possible that x⊙y = 2y^2 + 2x^2 (symmetrical)
Then also 0⊙y = 2y^2

In one case, x⊙y=y⊙x, in one it is not. Hence not sufficient.
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Re: If ⊙ denotes a mathematical operation, does x⊙y = y⊙x for all x and y?   [#permalink] 13 Jul 2020, 08:26