If xy = 12, what is the value of x – y?

xy = 12
#1
x > y:
1: x = 4, y = 3, xy = 12, x - y = 1
2: x = 6, y = 2, xy = 12, x - y = 4
insufficient

#2 - x+y = 7:
x = 4, y = 3, xy = 12, x - y = 1
x = 3, y = 4, xy = 12, x - y = -1

insufficient

#1+#2
the system of equations xy = 12 and #2 yields 2 pairs of (x;y), namely (3;4) and (4;3). If we add the condition #1, we'll have only 1 pair remaining which is (4;3) which explicitly lets us answer our question.

C is the answer

X will have only one value. But stat 2 gives us two values, stat 1 help us to discard one and keep one.
Ans:C

My first though was Option C is correct

But then it struck me that there is nowhere mentioned that x& y are integer
so how can we choose c
IMO E is correct

Experts please correct my reasoningBunuel

No, you should atleast look at the discriminant of a quadratic equation.

D (discriminant) = \({b^2-4ac}\), where a,b,c are coefficients of the quadratic equation, \(ax^2+bx+c=0\)

D can be <0, >0 or =0

Now, for 2 UNIQUE (or unequal real roots), D should be >0

For 2 EQUAL roots, D=0

For complex roots (not applicable for GMAT) , D<0

Thus, just knowing that an equation is quadratic IS NOT sufficient to say that you will get 2 different roots.

Example, \(x^2+5x+6=0\) has 2 unique roots, x=2, x=3, D = \({6^2-4*1*6}\) = 12 >0

But, \(x^2+6x+9=0\), has 2 EQUAL roots and thus only 1 value, x=-3, D = \({6^2-4*1*9}\) = 0

Another thing to look at in questions like these is whether the question stems prohibit 1 of 2 values that you obtain from a quadratic equation. In this case, although you did get 2 values, only 1 is allowed and this quadratic equation will thus be sufficient.
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xy=12

x-y=?

1. x-y>0

Not Sufficient

2. x+y=7
x=12/y
12+y^2=7y
y^2-7y+12=0
(y-3)(y-4)=0
y=3 OR 4
x=4 OR 3
x-y=1 OR -1

Not Sufficient

1 and 2:
x>y
Therefore x=4; y=3 x-y=1

Sufficient

Answer: C

St 1) X>Y -> 6X2 ; 4X3 etc.. NS
St 2) X+Y = 7 -> from ques X=12/Y; 12/Y+Y=7 So 12+Y^2=7Y -> Y=4 OR Y=3 Hence X=3 OR X=4
Combining 1&2
Only one unique value of X = 4
Answer C

I've picked B, here is why,

In 2] X+Y=7

Then X=7-Y

Then XxY= 12 becomes
[7-Y]xY= 12

The only solution is Y=3

If Xx3=12 then X = 4

X-Y=1.

Correct me if i'm wrong

mike34170 When we take [7-Y]xY= 12, this can be expanded as
\(y^2\)-7y+12=0
(y-3)(y-4)=0
Therefore,
y can be 3 OR 4.

Replacing the values of y in the initial equation of xy=12 or x+y=7,
If y=3 then x=4 and x-y=1
If y=4 then x=3 and x-y= -1

Since there can be 2 possible values of x-y, namely 1 and -1, statement 2 is insufficient.

When we take both the statements together we know for sure that the instance 1 is true. That is,
If y=3 then x=4 and x-y=1

Since we need both the statements to get a unique value of x-y, answer is C.

Hello AmoyV.

Thanks for you reply.

Exact, Y has two roots 3 and 4.

MAGOOSH OFFICIAL SOLUTION

In this case, the question includes an “if” statement, xy = 12.

Once again there is no obviously useful way to rephrase the question. We could, say, use the “if” equation to solve for y in terms of x, and substitute the resulting expression for y in the question. That’s not vey promising, though, so let’s turn to (1).

Even if we restrict ourselves to positive integers, the “if” statement and (1) together allow a number of solutions, each of which yields a different value for x – y. Specifically, x and y could be 12 and 1, 6 and 2, or 4 and 3, and could therefore be 11, 4, or 1. Eliminate A and D.

Statement (2) and the “if” statement together might seem to allow just one solution, x=4, y=3, and x-y=1. In fact, though, (2) and the “if” statement also allow x=3, y=4, =-1. So (2) is also not sufficient. Eliminate B.

Together (1) and (2) (along with the “if” statement) allow just one solution, x=4, y=3, and x-y=1.

The correct answer is C.
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xy = 12 and x + y = 7 gives x = 3 and y = 4 OR x = 4 and y = 3 but since it's given that x > y, then the first case is out and x - y = 1.
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Thanks Bunuel
Got it
Maybe i over exercised my brain on this one!

Would it be enough to know that the equation out of 2 is a quadratic and will therefore almost always yield two different results? Bc. then x-y could be two different things.

or would you proceed and calculate it out? thx

HI, couldn't I just isolate "y" in the second statement, so that y=7-x, and then calculate the so asked "x-y" as: x-(7-x)=-7? so that statement 2 is actually sufficient?
thank you

x-(7-x) = x-7-(-x) = x-7+x = 2x-7 .
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The answer will be C... x.y=12 ; also both x and y have to be a factor of 12...

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