(x−2)(2+y)=? (1) x·y=6 (2) x+y=5 : GMAT Data Sufficiency (DS)
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(x−2)(2+y)=? (1) x·y=6 (2) x+y=5

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(x−2)(2+y)=? (1) x·y=6 (2) x+y=5 [#permalink]

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New post 20 Jun 2011, 23:58
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(x−2)(2+y)=?

(1) x·y=6

(2) x+y=5

Source Master Gmat sample Test Q 20
[Reveal] Spoiler: OA

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Re: x,y and the rest of the band [#permalink]

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New post 21 Jun 2011, 00:15
Statement 1: if you substitute for 6/x for y you get 2 values for x which means there is no certain answer
Statement 2: ditto. if you substitute 5-y for x you get (3-y)(2+y), which gives you, y = 3 of y = -2.

I had some trouble working it out and the whole post got messed up (I think I did something wrong using some math signs or something:S), I'll try to do it later.
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Re: x,y and the rest of the band [#permalink]

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New post 21 Jun 2011, 00:15
The answer is E. (1) and (2) don't give single specific values for x and y. Combination of (1) and (2) gives a quadratic equation, which can give 2 values of x (or y), which in turn can result in two values of the expression that we're supposed to evaluate. Hence there is no single specific answer for the expression based on data given.
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Re: x,y and the rest of the band [#permalink]

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New post 21 Jun 2011, 06:59
shouldnt it b C?

From 1 and 2 we can get values of X and y ( 2 equations and 2 variables )

from 1 & 2 -> x = 7/2 , y=3/2

from this we can calculate the value of x−2)(2+y)=?
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Re: x,y and the rest of the band [#permalink]

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New post 21 Jun 2011, 07:12
msbinu wrote:
shouldnt it b C?

From 1 and 2 we can get values of X and y ( 2 equations and 2 variables )

from 1 & 2 -> x = 7/2 , y=3/2

from this we can calculate the value of x−2)(2+y)=?


from 1 & 2 -> x = 7/2 , y=3/2..............??????

this is wrong.. x*y = 6
your values give 21/4
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New post 21 Jun 2011, 07:16
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sorry .. my bad . .I read x*y as x-y :( ..
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Re: x,y and the rest of the band [#permalink]

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New post 21 Jun 2011, 07:17
msbinu wrote:
shouldnt it b C?

From 1 and 2 we can get values of X and y ( 2 equations and 2 variables )

from 1 & 2 -> x = 7/2 , y=3/2

from this we can calculate the value of x−2)(2+y)=?

From 1) and 2) we have answer pairs:
\((2;3) and (3;2)\)
substitute it and get:
\(1*4=4\)
or
\(0*5=0\)
two different answers,
(E)
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New post 25 Jun 2011, 18:28
1. Not sufficient
different combinations of x , y yields different values.

2 3
6 1
2. Not sufficient
different combinations of x , y yields different values.
1 4
2 3

Together .. Not sufficient

we know that x can be 2 or 3
then y can be 3 or 2.
these two combinations yields different values..

Answer is E.
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New post 20 Jul 2011, 02:36
Normally, the 2 equation system has 2 set of solution
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New post 05 Sep 2011, 08:22
Quote:
(x−2)(2+y)=?

(1) x·y=6

(2) x+y=5

Source Master Gmat sample Test Q 20



(x-2)(2+y) = xy - 2y + 2x -4 ...............[eq 1]

From Statement 1
Insert x.y=6 into [eq 1], 6 - 12/x + 2x - 4 = 2x-10
insufficient

From Statement 2
x+y=5
y=5-x

Insert into [eq 1], x(5-x)-2(5-x)+2x-4 = 5x-x^2-10+2x+2x-4 = -x^2+9x-14
insufficient

Statement 1 + 2
x(5-x)=6
5x-x^2-6=0
(-x-2)(x-3)=0
x=3 or x =2
Insufficient

Answer: E

Quote:
The answer is E. (1) and (2) don't give single specific values for x and y. Combination of (1) and (2) gives a quadratic equation, which can give 2 values of x (or y), which in turn can result in two values of the expression that we're supposed to evaluate. Hence there is no single specific answer for the expression based on data given.

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Re: (x−2)(2+y)=? (1) x·y=6 (2) x+y=5 [#permalink]

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(x−2)(2+y)=? (1) x·y=6 (2) x+y=5 [#permalink]

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New post 19 Sep 2016, 05:50
E is correct. Here's why:

(1) xy = 6

Try plugging into original equation...

= xy + 2x - 2y -4
= 6 + 2x - 2y - 4 --> 2 + 2x - 2y

NOT SUFFICIENT

(2) x+y = 5 --> rewrite as x = 5-y & plug into original equation

= ((5-y) - 2)(y+2)
= (3-y)(y+2)

NOT SUFFICIENT - we don't know the value of y

A, B, D are eliminated

Together - (1) + (2) = Plug both into original eq

=(x-2)(y+2)
=xy + 2x-2y-4
=6+2(5-y)-2y-4
=2+10-2y-2y
=12-4y

NOT SUFFICIENT - we still don't have a value for y
(x−2)(2+y)=? (1) x·y=6 (2) x+y=5   [#permalink] 19 Sep 2016, 05:50
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