x and y

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x and y

by gmatblood » Mon Oct 31, 2011 1:23 am
Is x @ y = y @ x, for all real values of x and y?

Stmt 1: x @ y = 1/x + 1/y
Stmt 2: x @ y = x - y
Source: — Data Sufficiency |

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by shankar.ashwin » Mon Oct 31, 2011 1:35 am
Edited: '0' is a real number as well

(1) x @ y = 1/x + 1/y
y @ x = 1/y + 1/x
Addition will give you the same value, except for x or y =0, when the equation becomes indeterminable. Sufficient

(2) x @ y = x - y
y @ x = y - x

x-y = y-x, only when x=y.

A IMO
Last edited by shankar.ashwin on Mon Oct 31, 2011 5:34 am, edited 2 times in total.

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by pemdas » Mon Oct 31, 2011 1:36 am
the click here is @, it must be @ in all cases :)

st(1) x@y=(x+y)/xy, we quickly note this statement cannot be solved if x and/or y=0 and 0 is real value Sufficient to answer No
st(2) x@y=x-y, here we have no undefined solution areas so we can conclude Sufficient to answer Yes

d
gmatblood wrote:Is x @ y = y @ x, for all real values of x and y?

Stmt 1: x @ y = 1/x + 1/y
Stmt 2: x @ y = x - y
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by vaibhavgupta » Mon Oct 31, 2011 5:32 am
gmatblood wrote:Is x @ y = y @ x, for all real values of x and y?

Stmt 1: x @ y = 1/x + 1/y
Stmt 2: x @ y = x - y
IMO D

Whts OA?

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by saketk » Tue Nov 01, 2011 2:29 am
gmatblood wrote:Is x @ y = y @ x, for all real values of x and y?

Stmt 1: x @ y = 1/x + 1/y
Stmt 2: x @ y = x - y
stmt 1: @ can be any mathematical expression here.

let's take few numbers here.
X=1 and Y=1

so, we have 1@1 = 2


from stmt 2: if I take same values then

1@1 = 0

--

If I combine the 2 statements then -- then I get

1@1 =2 and 1@1 = 0

two contradicting statements?

or am I missing something?

what's the source of this question?

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by mankey » Tue Nov 01, 2011 10:12 am
Looks A, since first one will hold for all real values, whereas second one will fail at x=y=0.

What is the OA?

Thanks.