Is x > y ? (1) x + y > x − y (2) 3x > 2y

I messed up on how i interpreted statement two, the fact that x=y=1 would create the right condition for statement 2 to be correct, also means that when answering the prompt, x>y, we get the answer NO; trying another set of numbers- x=2 and y=1- where statement two meets the condition, 2>1/3, means that the answer to the question x>y is YES.

Bunuel explains this but I didn't quite understand the process because I wasn't registering the prompt and statement criteria clearly.

Hipingala,

here is how i would approach this question

we are asked that IS x>y
so is x-y>0
what could be the possible senarios
x>0, y>0 and |x|>|y|
x<0, y<0 and |y|>|x|
x>0, y<0
x<0, y>0 and |y|>|x|

in above all scenarios we would get the answer to our question.
Now statement A:
It says y>0. It could be x>0 or x<0, but still we don't about |x| and |y|
So insufficient.

Now Statement B:
3x>2y
or we can have
\(x>\frac{2}{3} y\)
if we read along it says x is greater than two-thirds of y.
We will have two senarios
y>0 even then we can't conclude that x> y. why say y>1 then x can be \(>\frac{2}{3}\) . So its possible that the range of of x is \(\frac{2}{3}<x<\infty\) and range of y would be \(1<y< \infty\) . Now if you imagine a number line with these two ranges then you can either conclude that x<y or x>y

Now if y<0 even then we can't conclude that x> y. Then x can be \(>-\frac{2}{3}\) . So its possible that the range of of x is \(-\frac{2}{3}<x<\infty\) and range of y would be \(-\infty<y<0\). Now if you imagine a number line with these two ranges then you can either conclude that x<y or x>y

Now even if we combine both the statements still we don't have any new information to conclusively answer if x>y

hence option E
let me know if this helps
Probus

St1 :- x + y > x - y = 2y > 0 ( here we can only conclude that Y is positive. As no information is given about X, St1 is not sufficient )

St2 :- 3x > 2y = 1.5x > y
The above statement can be hold true in following case :-
X = 2, and Y = 2,
X =3, and Y = 2, or
X = 1, and Y = 2

As there's multiple value of x, this statement is also sufficient

Combining St1, and St2
by adding both Statement, we get 3x>0 (here we can only conclude that X, and Y are positive. But can't conclude that Whether X is greater than Y)

Hence, Answer E

This can be rearranged in different ways x > y or x - y > 0 or x/y > 1 (if y is positive)

statement 1) x + y > x - y can be rearranged to 2y > 0 then y > 0, insufficient we have no information on x

Eliminate A/D

Statement 2) 3x > 2y

if x = 1 and y = 1 then 3 > 2 then back to the stem 1 > 1 is not true.

if x = 3 and y = 1, then 9 > 2 then back to the stem 3 > 1 is true

Insufficient

Now try combined

given that y > 0

then we can divide by y, \(\frac{x}{y} > \frac{2}{3}\) we don't know for sure whether it is larger than 1? it could be anything larger than 2/3

So insufficient.

Answer choice E

VeritasKarishma can you please explain this question?

Posted from my mobile device

Ques: Is x > y?

(1) x + y > x − y

This inequality is independent of x. It gives us y > -y (when we subtract x from both sides). This means y is positive.
But we don't know whether x is greater than y.

(2) 3x > 2y
We know that 3 times of x is greater than 2 times of y. This will be true when x is less than y, equal to y or greater than y.
x = 3, y = 4
x = 3, y = 3
x = 3, y = 2

Using both statements, y is positive so dividing both sides of second inequality by 3y we get
x/y > 2/3

Now, x/y could be 3/4 in which case x < y or x/y could be 1 in which case both are equal or x/y could be 4/3 in which case x > y.

Answer (E)

How could 2 variances yield the same number?

Is not this against the law of math?

If 1 = A, how could B is also equal to 1? Then what is the purpose of using variable.

If all variables a to z yield the same number

Is x > y ?


(1) x + y > x − y
2y > 0
y > 0

INSUFFICIENT.

(2) 3x > 2y
x > 2/3y

INSUFFICIENT.

(1&2) We can't conclude x > y. INSUFFICIENT.

Answer is E.

There is no rule that states variables must have different values. Often on the GMAT, 2 variables represent the same value.

Solution:

We need to determine whether x > y.

Statement One Alone:

Simplifying, we have:

y > -y

2y > 0

y > 0

We see that y is positive. However, since we don’t know anything about x, statement one alone is not sufficient.

Statement Two Alone:

Dividing both sides by 3, we have:

x > 2y/3

This does not necessarily mean x > y. If x = 4 and y = 3, then x > y. However, if x = 3 and y = 3, then x is not greater than y. Statement two alone is not sufficient.

Statements One and Two Together:

With the two statements, we still don’t have enough information to determine whether x > y. (Again, we can use x = 4 and y = 3 for the case x > y, but we can also use x = 3 and y = 3 against the case x > y.)

Answer: E

To determine whether x is greater than y, we can evaluate the statements (1) and (2) separately.

Statement (1) x + y > x - y:
This statement implies that the sum of x and y is greater than their difference. By rearranging the terms, we have y > -y. Since y is always greater than or equal to -y, we can conclude that y > -y. However, this statement does not provide any direct information about the relationship between x and y. Therefore, we cannot determine whether x is greater than y based solely on statement (1).

Statement (2) 3x > 2y:
This statement indicates that three times x is greater than two times y. By dividing both sides of the inequality by 2, we get (3/2)x > y. This implies that x is greater than y divided by (3/2). However, we don't have a specific comparison between x and y, and we cannot determine whether x is greater than y based solely on statement (2).

Combining the two statements:
Neither statement (1) nor statement (2) provides enough information to determine the relationship between x and y. Therefore, we cannot definitively say whether x is greater than y or not.

In conclusion, neither statement (1) nor statement (2) alone is sufficient to determine if x is greater than y.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.