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Is |x - y| \gt |x + y| ?

1) x^2 - y^2 = 9 2) x - y = 2

From S1, can we assume that x has to equal 5 and y has to equal 4 so when you square x and y, their difference would be 9. Thus S1 would be sufficient.

S2 is insufficient.

Thus the answer is A, but the Official Answer is C. Can someone please explain. Thanks.

From S1, can we assume that x has to equal 5 and y has to equal 4 so when you square x and y, their difference would be 9. Thus S1 would be sufficient.

S2 is insufficient.

Thus the answer is A, but the Official Answer is C. Can someone please explain. Thanks.

Why would you assume that? Why not x=-5 and y=-4? Or x=3 and y=0? x^2 - y^2 = 9 has infinitely many solutions for x and y, and only one of them is x=5 and y=4.

Is |x - y| \gt |x + y| ?

(1) x^2 - y^2 = 9 --> (x-y)(x+y)=9. Not sufficient. (2) x - y = 2 --> don't know the value of x+y. Not sufficient.

(1)+(2) Since from (2) x - y = 2 then from (1) (x-y)(x+y)=2(x+y)=9 --> x+y=4.5. Sufficient.

Thanks Bunuel for the reply. That make sense now. I did a similar problem on MGMAT CAT. The difference is that on the MGMAT problem there is a constrain of x < y < 0

If x and y are integers such that x < y < 0, what is x – y?

(1) (x + y)(x – y) = 5

(2) xy = 6

Official Explanation: The problem stem tells you that x and y are both negative integers, and x is less than y. You are asked for the value of x – y. Note that in order to answer this question, you might not need the separate values of x and y; however, if you can find those separate values, then you can solve for x – y or any other “combo.”

(1) SUFFICIENT: This statement is tricky. One approach is to distribute the left side to get the difference of squares:

x2 – y2 = 5

Now, since both x and y are integers, x2 and y2 are integers as well. Let’s look at the sequence of squares, and consider their differences (we can leave out zero, since neither x nor y can be zero):

1 4 9 16 25 36… Diff = 3 5 7 9 11…

As you can see, the difference between squares grows as the squares themselves get larger. The only difference between two squares that equals 5 is the difference between 4 and 9. Since x and y are both negative, this tells us that x = –3 and y = –2; therefore, x – y = –1.

Alternatively, we could note that both x + y and x – y are themselves integers. Looking at the statement, we have (x + y)(x – y) = 5 int × int = 5

The only possible integer pairs that give 5 as a product are (5, 1) and (-5, -1), since 5 is prime. Now, because both x and y are negative, the (5, 1) pair won’t work either way (either with x + y = 5 or with x + y = 1). So let’s try (-5, -1):

x + y = –5 x – y = –1

Adding these two equations, we get 2x = -6, or x = -3. Substituting back into x + y = –5, we get y = –2. (If we had assigned x + y = –1 and x – y = –5, we would have gotten y = 2, which doesn’t fit the problem constraints.)

(2) INSUFFICIENT: There are two pairs of integers that fit the constraint x < y < 0 and the statement xy = 6: (-3)(-2) = 6 AND (-6)(-1) = 6.

Thanks Bunuel for the reply. That make sense now. I did a similar problem on MGMAT CAT. The difference is that on the MGMAT problem there is a constrain of x < y < 0

If x and y are integers such that x < y < 0, what is x – y?

(1) (x + y)(x – y) = 5

(2) xy = 6

Actually you have more than one constraint: (i) x and y are integers; (ii) x<y<0

If x and y are integers such that x<y<0 what is x-y?

(1) (x+y)(x-y)=5. x and y are integers means that both x+y and x-y are integers. So, we have that the product of two integer factors equal to 5. There are only two combination of such factors possible: (1, 5) and (-1, -5). Since given that x and y are both negative then the first case is out, so x-y is either -1 or -5, but it can not be -5, because in this case x+y must be -1 and no sum of two negative integers yields -1. Hence x-y=-1. Sufficient.

(2) xy= 6. If x=-3 and y=-2 then x-y=-1 but if x=-6 and y=-1 then x-y=-5. Not sufficient.

Just for information, if a questions as such if we square on both sides, then it does not change anything..am i correct in stating this? for examle |x-y| > |x+y|...then (x-y)2 = (x=y)2?

Just for information, if a questions as such if we square on both sides, then it does not change anything..am i correct in stating this? for examle |x-y| > |x+y|...then (x-y)2 = (x=y)2?

Yes, since both sides of the inequality are non-negative then we can square it and write: (x-y)^2>(x+y)^2.

Generally: A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: 2<4 --> we can square both sides and write: 2^2<4^2; 0\leq{x}<{y} --> we can square both sides and write: x^2<y^2;

But if either of side is negative then raising to even power doesn't always work. For example: 1>-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: -2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3; x<y --> we can raise both sides to third power and write: x^3<y^3.