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# x+y=4

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Joined: 18 Apr 2015
Posts: 6596
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Kudos [?]: 1262 [0], given: 5975

x+y=4 [#permalink]  08 Sep 2018, 22:35
Expert's post
00:00

Question Stats:

94% (00:22) correct 5% (01:06) wrong based on 17 sessions
$$x+y=4$$

$$xy=0$$

 Quantity A Quantity B $$x$$ $$y$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E
[Reveal] Spoiler: OA

_________________
GRE Instructor
Joined: 10 Apr 2015
Posts: 1751
Followers: 58

Kudos [?]: 1668 [2] , given: 8

Re: x+y=4 [#permalink]  09 Sep 2018, 07:01
2
KUDOS
Expert's post
Carcass wrote:
$$x+y=4$$

$$xy=0$$

 Quantity A Quantity B $$x$$ $$y$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

GIVEN: x + y = 4 and xy = 0

If xy = 0, then EITHER x = 0 OR y = 0
Let's examine each possible case:

Case i) if x = 0, then y = 4 (since we're also told that x + y = 4)
Case ii) if y = 0, then x = 4 (since we're also told that x + y = 4)

In case i, Quantity A is greater than Quantity B
In case ii, Quantity B is greater than Quantity A

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

Intern
Joined: 22 Mar 2018
Posts: 4
Followers: 0

Kudos [?]: 3 [1] , given: 0

Re: x+y=4 [#permalink]  09 Sep 2018, 09:11
1
KUDOS
if X+Y = 4 & XY = 0

We can write as

= (X-Y)^2

= X^2+Y^2-2XY

= (X+Y)^2-4XY ( Bifurcating the equation)

= 4^2-4*0

= 16

THEREFORE

X-Y = 16^0.5

(Taking square root of 16)

X-Y = (+/-) 4

Now we have

X+Y = 4 .....1

X-Y = 4 .....2

On solving the equation 1 & 2 we get as follows:

X = 4

Y = 0

AND

X-Y = 4 .....2

X-Y = -4 ...4

On solving the equation 3 & 4 we get as follows:

X = 0

Y = 4

Since both X & Y can take both values 0 and 4 which is contradictory statements regarding the large value among X & Y.

HENCE PROVED
Re: x+y=4   [#permalink] 09 Sep 2018, 09:11
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