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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
D01-43  [#permalink]

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4
24 00:00

Difficulty:   95% (hard)

Question Stats: 39% (02:33) correct 61% (02:10) wrong based on 282 sessions

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If $$5x = y + 7$$, is $$(x - y) \gt 0$$?

(1) $$xy = 6$$

(2) $$x$$ and $$y$$ are consecutive integers with the same sign

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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re D01-43  [#permalink]

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3
8
Official Solution:

Statement 1: $$xy = 6$$ is not sufficient to answer the question because solving the equation gives $$y = 3$$ when $$x$$ is 2 and $$y$$ is -10 when $$x$$ is $$-\frac{3}{5}$$.

$$5x = y + 7$$

$$x = \frac{y + 7}{5}$$ ................i

$$xy = 6$$ ........................ ii

Replacing the value of $$x$$ on eq ii: $$\frac{y(y + 7)}{5} = 6$$

$$y^2 + 7y = 30$$

$$y^2 + 10y - 3y - 30= 0$$

$$y (y + 10) - 3 (y + 10) = 0$$

$$(y + 10) (y - 3) = 0$$

$$y = - 10$$ or $$3$$

If $$y = - 10$$, $$x = -\frac{3}{5}$$. In this case, $$x - y = -\frac{3}{5} - (-10) = \frac{47}{5}$$. Yes.

If $$y = 3$$, $$x = 2$$. In this case, $$x - y = 2 - 3 = -1$$. No.

In each case $$(x-y) \gt 0$$ and $$\lt 0$$. Hence statement 1 is not sufficient.

Statement 2: If $$x$$ and $$y$$ are consecutive integers, $$x = 2$$ and $$y = 3$$.

When $$x$$ and $$y$$ are consecutive integers, either $$x = y+1$$ or $$y = x+1$$ is possible.

(i) If $$x = y+1$$

$$5x = y + 7$$

$$5(y+1) = y + 7$$

$$4y = 2$$

$$y = \frac{1}{2}$$. Then $$x = \frac{3}{2}$$. However this is not possible because $$x$$ and $$y$$ are even not integers. So this option is not valid.

(ii) If $$y = x+1$$

$$5x = y + 7$$

$$5x = x + 1 + 7$$

$$4x = 8$$

$$x = 2$$

Then, $$y = 3$$. This is valid because $$x$$ and $$y$$ are consecutive integers.

So $$(x-y) = 2-3 = -1$$. Hence statement 2 is sufficient.

Answer: B
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Manager  Joined: 08 Feb 2014
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Location: United States
Concentration: Finance
GMAT 1: 650 Q39 V41 WE: Analyst (Commercial Banking)
Re: D01-43  [#permalink]

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How did we get to y(y+10)-3(y+10) from the step just before this.

Thanks!
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: D01-43  [#permalink]

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1
JackSparr0w wrote:
How did we get to y(y+10)-3(y+10) from the step just before this.

Thanks!

y^2 + 10y - 3y - 30= 0

Factor out y from y^2 + 10y and -3 from - 3y - 30: y(y + 10) - 3(y + 10) = 0.

Hope it's clear.
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Manager  Joined: 08 Feb 2014
Posts: 204
Location: United States
Concentration: Finance
GMAT 1: 650 Q39 V41 WE: Analyst (Commercial Banking)
Re: D01-43  [#permalink]

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Got it! thanks!
Intern  Joined: 31 May 2013
Posts: 11
Re: D01-43  [#permalink]

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1
if x and y are consecutive then x=y+1 and y=x+1 both can be true either ways both the numbers are consecutive
my solution was like this
given
5x=7+y
x-y=7-4x
therefore, x-y>0 when 7-4x>0
so when x is negative x-y is postive
Intern  Joined: 24 Nov 2014
Posts: 18
D01-43  [#permalink]

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7
1
How can you actually do that in <2min? Alone the writing of the official solution without thinking would take more time.

Therefore I thought about it that way: 5x = y + 7 /*5 /-y
So x-y = (-4y+7)/5

Question: (-4y+7)/5 > 0 ?
Rephrased: y < 7/4 ?

1) Not sufficient: many ways y < or > 7/4

2) In the original equation the only way the integers are consecutive is x=2/y=3, everything else would increase the distance between the two sides of the equation. -> Sufficient

Is that right?

I always think that those 750+ test taker find a very efficient solution and I believe they give us on purpose this crappy marker and a slippery foil, which I find kind of sadistic, because not much should be written down.. That is why I sometimes wonder about those solutions which are incredible long.
Manager  Joined: 05 Jul 2015
Posts: 98
Concentration: Real Estate, International Business
GMAT 1: 600 Q33 V40 GPA: 3.3
Re: D01-43  [#permalink]

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2
The way I thought of it:

I rearranged the question to say 0= y+7-5x

Looking at the statements I skipped straight to B. Since they have to be consecutive integers Y=3 and X = 2 works, . Sufficient.
With statement A) x could be 3 and y could be 2 or vice versa so not sufficient.
Manager  Joined: 24 May 2013
Posts: 79
Re: D01-43  [#permalink]

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If 5x=y+7, is (x−y)>0?

(1) xy=6
x(5x-7)=6 => 5x^2-7x-6=0 => x=(7+/- Sqrt (49-4*5*(-6)))/10 => (7+/-13)/10 => 2 or -3/5
x=2,y=3, x-y=-1<0
x=-3/5, y=-10, x-y= 47/5=9.4 >0
Insufficient

(2) x and y are consecutive integers with the same sign

Let x=y+1
5x=y+7 => 4y=2 => y=0.5 not integer

Let y=x+1
5x=y+7 => 4x=8 =>x=2 , y=3
x-y= -1<0
SUFFICIENT

Hence B

Thanks
Manager  B
Joined: 23 Apr 2014
Posts: 57
Location: United States
GMAT 1: 680 Q50 V31 GPA: 2.75
Re D01-43  [#permalink]

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I think this is a high-quality question and I agree with explanation.
Intern  Joined: 29 Jul 2015
Posts: 27
Location: India
Concentration: Other
GMAT 1: 720 Q50 V38 GPA: 2.9
D01-43  [#permalink]

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rephrase as
x- y =7- 4x ( eq.1)
thus .. we can say...
is 7 - 4x > 0..
i.e. x < 7/4..
x < 1.75?

1. x = 2 NO... X = 1.5 YES
2 * 3 = 6
1.4 * 4 =6
INSUFFICIENT

2. x = y + 1
substitu in eq. 1
x - x + 1 = 7 - 4x => x= 3/2= 1.5 so y = 0.5 ... NOT AN INTEGER... reject this case

y = x+1
substiute in eq.1
x - x -1 = 7 - 4x => x = 2 ... so y = 3.. BOTH ARE CONSECUTIVE INTGERS....
Definite NO .. for x < 1.75?

(2) is sufficient
so B Current Student G
Joined: 28 Nov 2014
Posts: 839
Concentration: Strategy
Schools: Fisher '19 (M\$)
GPA: 3.71
Re: D01-43  [#permalink]

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I like this question. This is, for sure, a great question. Thanks team at GC!
Intern  Joined: 02 Aug 2016
Posts: 7
Concentration: International Business, Sustainability
Re: D01-43  [#permalink]

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from statement 1 we have xy=6
then x=1,y=6 or x=6,y=1 or x=2,y=3 or x=3,y=2
.if we sub them in given eq 5x=y+7,we get x=2 and y=3.so why statement 1 is not sufficent?
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: D01-43  [#permalink]

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phanikrishna wrote:
from statement 1 we have xy=6
then x=1,y=6 or x=6,y=1 or x=2,y=3 or x=3,y=2
.if we sub them in given eq 5x=y+7,we get x=2 and y=3.so why statement 1 is not sufficent?

Please read the solution carefully. There are TWO sets of (x,y) given which satisfy the equations: x=2 and y=3 AND x=-5/3 and y=-10.

Also, the red part above is wrong. xy=6 has infinitely many solutions for x and y. You considered only positive integers there but x and y can be negative and fractions too.
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Joined: 17 Aug 2016
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D01-43  [#permalink]

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madmax1000 wrote:
How can you actually do that in <2min? Alone the writing of the official solution without thinking would take more time.

Therefore I thought about it that way: 5x = y + 7 /*5 /-y
So x-y = (-4y+7)/5

Question: (-4y+7)/5 > 0 ?
Rephrased: y < 7/4 ?

1) Not sufficient: many ways y < or > 7/4

2) In the original equation the only way the integers are consecutive is x=2/y=3, everything else would increase the distance between the two sides of the equation. -> Sufficient

Is that right?

I always think that those 750+ test taker find a very efficient solution and I believe they give us on purpose this crappy marker and a slippery foil, which I find kind of sadistic, because not much should be written down.. That is why I sometimes wonder about those solutions which are incredible long.

Bunuel, could you please suggest me if the highlight is correct? That "many ways" is without any proof, and as a matter of fact there are only two solutions to the system of equations XY=6 and 5X=Y-7, which in the specific case are one with Y>7/4 and one with Y<7/4. But with the above approach how could I say that the first stm is not suff?

Thank you in advance
Intern  B
Joined: 26 Dec 2016
Posts: 19
Re: D01-43  [#permalink]

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Is there a shortcut for this question ?
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: D01-43  [#permalink]

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1
BoomHH wrote:
Is there a shortcut for this question ?

You can check alternative solutions here: if-5x-y-7-is-x-y-164140.html

Hope it helps.
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Intern  Joined: 22 Sep 2015
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Re: D01-43  [#permalink]

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Statement 2 gives , both integers negative, both are positive, 0/+ and -/0. Then we have so many options to pick, how will it be answer B then?
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: D01-43  [#permalink]

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chidananda3891 wrote:
Statement 2 gives , both integers negative, both are positive, 0/+ and -/0. Then we have so many options to pick, how will it be answer B then?

Which values did you get for (2) and how? There is only set possible as shown in the solution.
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Intern  B
Joined: 27 Jul 2017
Posts: 3
Re: D01-43  [#permalink]

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why does it half to be an integer? isnt this an assumption Re: D01-43   [#permalink] 28 Sep 2018, 21:34

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