If 5x = y +7, is (x y) > 0?

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.

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

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.

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

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.

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.

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?

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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I think most of us are making the same mistake in eliminating statement 1. I did the same mistake too.
You cannot test values for y=1 and x=6 and just dismiss statement 1.
You have to check with the main equation. \(5x=y+7\)
So if u take y=1 and x=6 then main equation is \(5(6) = 1 + 7\).
The above is NOT possible. You cannot just take any value. It has to satisfy the main equation.


Solution :

Statement 1 : xy = 6
\(y = 6/x\)

Substitute above in main equation -
\(5x=y+7\)
\(\frac{30}{y} = y + 7\)
\(30 = y^2 + 7y\)
\(y^2 + 7y - 30 = 0\)
Solving we get y = 3 and y = -10
You can stop here because we are getting two different values of y so obviously we have different values of x.
Hence statement 1 is insufficient.

Can anyone please explain how the highlighted part derived?
Thanks

Statement 1 is INSUFFICIENT - AGREED

Statement 2 is also INSUFFICIENT.
Give x and y are two consecutive integer.
In that case x can be 2 and y can be 3. whereas x can be -2 and y can be -3.

In case 1 x- y is less than 0(2-3=-1) . whereas in 2nd case x-y is greater than 0 ( -2 +3=1)

Any views ?

if x is -2 and y is -3 then x-y is 1
but if x is -3 and y is -2 then x-y is -1

how can B be right?

I have the same doubt as you.