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X and Y are distinct non negative real numbers. What is the sum (X+Y)?

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X and Y are distinct non negative real numbers. What is the sum (X+Y)?  [#permalink]

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New post 17 Jan 2019, 22:10
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A
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D
E

Difficulty:

  35% (medium)

Question Stats:

60% (00:58) correct 40% (01:08) wrong based on 63 sessions

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X and Y are distinct non negative real numbers. What is the sum (X+Y)?

(1) (X^2 - Y^2)/(X-Y) = 6

(2) (X+Y)^2 = 36
NUS School Moderator
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Concentration: Finance, Marketing
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X and Y are distinct non negative real numbers. What is the sum (X+Y)?  [#permalink]

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New post 17 Jan 2019, 22:20
1
From statement 1:

\(\frac{x^2-y^2}{x-y} = 6\)

\(\frac{(x+y)(x-y)}{x-y} = 6\)

\(x+y = 6\)
Sufficient.

From statement 2:

\((x+y)^2 = 36\)
x+y = +6 or -6
Insufficient.

A is the answer.
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Re: X and Y are distinct non negative real numbers. What is the sum (X+Y)?  [#permalink]

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New post 17 Jan 2019, 23:41
1
Afc0892 wrote:
From statement 1:

\(\frac{x^2-y^2}{x-y} = 6\)

\(\frac{(x+y)(x-y)}{x-y} = 6\)

\(x+y = 6\)
Sufficient.

From statement 2:

\((x+y)^2 = 36\)
x+y = +6 or -6
Insufficient.

A is the answer.


Answer should be D if you account for the fact that x and y are non negative, therefore x+y=-6 is not a viable option.

Posted from my mobile device
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Re: X and Y are distinct non negative real numbers. What is the sum (X+Y)?  [#permalink]

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New post 17 Jan 2019, 23:43
truplayer257 wrote:
Afc0892 wrote:
From statement 1:

\(\frac{x^2-y^2}{x-y} = 6\)

\(\frac{(x+y)(x-y)}{x-y} = 6\)

\(x+y = 6\)
Sufficient.

From statement 2:

\((x+y)^2 = 36\)
x+y = +6 or -6
Insufficient.

A is the answer.


Answer should be D if you account for the fact that x and y are non negative, therefore x+y=-6 is not a viable option.

Posted from my mobile device


Damn, completely missed it. Thank you :)
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Re: X and Y are distinct non negative real numbers. What is the sum (X+Y)?   [#permalink] 17 Jan 2019, 23:43
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X and Y are distinct non negative real numbers. What is the sum (X+Y)?

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