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If 5x = y +7, is (x – y) > 0?

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If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 04 Dec 2013, 17:23
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If 5x = y +7, is (x – y) > 0?

(1) xy = 6
(2) x and y are consecutive integers with the same sign

D01-43
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 19 Aug 2014, 04:51
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smartyman wrote:
If 5x = y +7, is (x – y) > 0?

(1) xy = 6
(2) x and y are consecutive integers with the same sign

D01-43


The given question can be reduced to

5x=y+7 so \(x=\frac{y+7}{5}\) ----->\(\frac{y+7}{5}-y>0\)

\(\frac{7-4y}{5}\)>0 or \(7-4y>0\)-----> Is y<1.75 ??

St 1 says xy=6...we have multiple case x=2,y=3 (Ans to question is Y) but if y=1,x=6 then Ans to Q. is no

St 2 says x,y are consecutive integers. Let x=y+1 so we have
5y+5=y+7 or 4y=2 or y=1/2 and x=3/2------> But x and y are fractions here so this case is not possible

Consider y=x+1 so we have 5x=x+1+7 or 5x=x+8 or 4x=8 x=2, y=3...Answer to the question is Y.

Ans is B
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Re: If 5x = y +7, is (x – y) > 0? 1) xy = 6 2) x and y are conse  [#permalink]

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New post 04 Dec 2013, 17:24
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This is my version of explanation, please correct my concept if there is any. Many Thanks in advance.

If 5x = y +7, is (x – y) > 0?
1) xy = 6
2) x and y are consecutive integers with the same sign

Statement 1:
Because the given statement xy = 6, and 5x = y + 7; we know that there will be 2 answer since xy = 6 is non-linear.
Therefore, statement 1 is not sufficient.

Statement 2:
Since x and y are consecutive integers of the same sign, we can assign them x = x, y = x+1.
Therefore by plugging into the equation: we get -1<0, which means that (x-y) NOT > 0.
Thus, statement 2 is sufficient.

Answer: B
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 18 Aug 2014, 11:54
I think 5x-y-7 = 0. xy=6, so it may be any of the following: 2*3=6, -2*-3=6, 1*6=6, -1*-6=6, but Here only 2 and 3 will make it the equation 0. Statement 2 is definitely sufficient. Answer is D. Correct me if I am wrong. Thanks.
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 18 Aug 2014, 12:13
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sunaimshadmani wrote:
I think 5x-y-7 = 0. xy=6, so it may be any of the following: 2*3=6, -2*-3=6, 1*6=6, -1*-6=6, but Here only 2 and 3 will make it the equation 0. Statement 2 is definitely sufficient. Answer is D. Correct me if I am wrong. Thanks.


Yes, you are wrong.

We are not told that x and y are integers, so there exist other solutions. For example, y=-10 and x=-3/5.
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If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 19 Aug 2014, 07:09
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1
5x=y+7... given

statement 1:
xy=6
x=6/y
substitute above value in given

5*6/y=y+7
30=y^2+7y
y^2+7y-30=0
(y+10)(y-3)=0
y=-10 or y=3
Therefore
x=-6/10 or x=2

for x=-6/10 and y=-10
(x-y)>0 ?
(-6/10+10)>0.... true

for x=2= and y=3
(x-y)>0 ?
2-3>0... false

statement insufficient

statement 2:
x and y are consecutive integers with the same sign
There are two possibilities
x>y or y>x
If x>y, then x=y+1
substitute in given
5(y+1)=y+7
4y=2
y=1/2.. not an integer... possibility discarded

Therefore, y>x
y=x+1
substitute in given
5x=x+8
x=2
y=3
we can answer definitely (x-y)>0 ?

statement sufficient

Ans - B
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 24 Feb 2017, 04:47
How can it be B? Even in B, the two numbers could be x, x+1 or y+1,y

This would give us both positive and negative answers. Please help.

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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 25 Feb 2017, 06:05
OreoShake wrote:
How can it be B? Even in B, the two numbers could be x, x+1 or y+1,y

This would give us both positive and negative answers. Please help.

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Hmm i guess in B, the only way to have two integers is by taking y = x+1. Hence is B is sufficient.
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 10 May 2017, 14:36
On gmat, when we are told that x and y are consecutive integers, does it always mean that they are ordered exactly as given (x, y=x+1)? I don't know why I am confused about this :)
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 10 May 2017, 21:19
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smartyman wrote:
If 5x = y +7, is (x – y) > 0?

(1) xy = 6
(2) x and y are consecutive integers with the same sign

D01-43


Solution.

From given condition, we have \(y=5x-7\)

(1) \(xy=6 \implies x(5x-7)=6 \implies 5x^2-7x-6=0 \\
\implies 5x^2-10x+3x-6=0 \implies 5x(x-2)+3(x-2)=0 \\
\implies (x-2)(5x-3)=0 \implies x=2 \,\,\, \text{or} \,\,\, x=\frac{3}{5}\)

If \(x=2 \implies y=3 \implies x-y=-1 <0\)
If \(x=\frac{3}{5} \implies y=-4 \implies x-y=\frac{23}{5} >0\)

Hence insufficient.

(2) \(x\) and \(y\) are consecutive integers.

If \(y=x+1 \implies 5x=(x+1)+7 \implies x=2 \implies y=3\).
In this case, we have \(x-y=-1 <0\)

If \(y=x-1 \implies 5x=(x-1)+7 \implies x=\frac{3}{2}\). Eliminate this case because \(x\) must be integer.

Hence, in (2), we only have \(x-y<0\). Sufficient.

The answer is B.
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 14 Oct 2017, 15:29
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kivalo wrote:
On gmat, when we are told that x and y are consecutive integers, does it always mean that they are ordered exactly as given (x, y=x+1)? I don't know why I am confused about this :)



No one seems to have discussed this yet. I am unsure whether when the GMAT says X and Y are consecutive integers that always means X = X and Y = X + 1

BUT

In this problem,

You can plug Consecutive integers X, X+1 into the given equation.

5X = Y + 7
X = X and Y = X+1
-----------------------
5x = y +7
5X = (X+1) + 7
5X = X + 8
4X = 8
X = 2, X+1 = 3

Now if you plug X=X and Y= X-1 into the equation:

5X = (X-1) + 7
4X = 6
X = 6/4

Note this is not an integer, so this solution does not work. Thus only X, X+1 should be considered.
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 15 Oct 2017, 01:32
subsauce wrote:
kivalo wrote:
On gmat, when we are told that x and y are consecutive integers, does it always mean that they are ordered exactly as given (x, y=x+1)? I don't know why I am confused about this :)



No one seems to have discussed this yet. I am unsure whether when the GMAT says X and Y are consecutive integers that always means X = X and Y = X + 1

BUT

In this problem,

You can plug Consecutive integers X, X+1 into the given equation.

5X = Y + 7
X = X and Y = X+1
-----------------------
5x = y +7
5X = (X+1) + 7
5X = X + 8
4X = 8
X = 2, X+1 = 3

Now if you plug X=X and Y= X-1 into the equation:

5X = (X-1) + 7
4X = 6
X = 6/4

Note this is not an integer, so this solution does not work. Thus only X, X+1 should be considered.


x and y are consecutive integers does not necessarily mean that y = x + 1, it could be that x = y + 1. But yes you are right 5x = y +7 and x = y + 1 does not give integer solution for x and y, so this case is not possible while 5x = y +7 and x = y - 1 give x = 2 and y = 3. So, the answer is B.
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 15 Oct 2017, 08:07
Question stem: 5x - y = 7 - (I)
Statement 1: Insufficient since xy could be fractions

Statement 2: either x-y = 1 - (II) or y-x = 1 - (III) (note that the signs of x and y do not matter) and we are told that x and y are integers. Now subtracting II from I we get x = 6/4 which is not possible since x is an integer. Adding I with II we get x as 2 and y as 3. Hence statement 2 is sufficient.

Ans - B

Does this make sense?
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 24 Jun 2018, 12:26
Is there a reason that x = -1 y = -2 would not also be a suitable pair for the second statement.

5x = y + 7 would become 5*-1 = -2 + 7 and (x - y) > 0

Using x = 2, y = 3, we get 5*2 = 3 + 7 and (x - y) < 0.

Wouldn't these two pairs make statement 2 insufficient and require both statements for sufficiency? Thanks!
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Re: If 5x = y +7, is (x – y) > 0?  [#permalink]

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New post 24 Jun 2018, 13:46
ncs58 wrote:
Is there a reason that x = -1 y = -2 would not also be a suitable pair for the second statement.

5x = y + 7 would become 5*-1 = -2 + 7 and (x - y) > 0

Using x = 2, y = 3, we get 5*2 = 3 + 7 and (x - y) < 0.

Wouldn't these two pairs make statement 2 insufficient and require both statements for sufficiency? Thanks!


-1 and -2 doesnt satisfy the equation.
5x=y+7
5(-1)= -2+7
-5=5 which is no possible, this we cannot use -1,-2

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Re: If 5x = y +7, is (x – y) > 0? &nbs [#permalink] 24 Jun 2018, 13:46
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