If x and y are positive, is x < y?

If x and y are positive, is x < y?

(1) \(\sqrt{x} < \sqrt{y}\). Since both sides of the inequality are positive (the square root from a positive number is positive), then we can safely square: x < y. Directly answers the question. Sufficient.

(2) \((x-3)^2 < (y-3)^2\). If \(x=3\) and \(y\neq 3\), the inequality will hold true: the left hand side will be 0, while the right hand side will be more than 0. Thus, if \(x=3\), y can be less than 3, giving a NO answer to the question, as well as more than 3, giving an YES answer to the question. Not sufficient.

Answer: A.

Hope it's clear.
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MUST KNOW: \(\sqrt{x^2}=|x|\):

The point here is that since square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE QUESTION:

According to the above if you take the square root from \((x-3)^2 < (y-3)^2\) you'll get |x - 3| < |y - 3|, which means that the distance between x and 3 is less than the distance between y and 3, which is not sufficient to say whether x < y.

Hope it helps.
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Hi,
we know that x and y are positive..
is x<y?

lets see the statements

(1) \(\sqrt{x} < \sqrt{y}\)
we can say that ans is YES,
but lets solve it algebrically too..
is x<y can be written as x-y<0..
\(\sqrt{x}^2-\sqrt{y}^2\)<0..
\((\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) <0\)..
\((\sqrt{x}+\sqrt{y}) >0\), as x and y are +ive,
so we have to find
if \((\sqrt{x}-\sqrt{y})<0\) or \(\sqrt{x}<\sqrt{y}\)..
statement 1 tells us exactly this..
SUFF

(2) \((x-3)^2 < (y-3)^2\)
It will hold in many cases : two case
a) if x = 1 and y= 7.. y>x
b) if x= 4 and y=1.. y<x.
two different answers
Insuff

A
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Hi guys, thank you for your explanations Bunuel chetan2u . I have a question, I wanted to understand why we cant use an algebraic approach on the second statement? Since we have squares on either side, can't we take take the square root on either side, which would give us x - 3 < y - 3, then by adding 3 on either side we get x < y .. ? This is exactly what I did and marked D as the answer when I saw this question for the first time. From both the posts, my understanding is that we cant do what i mentioned earlier because we wouldn't know after taking the root whether x-3 and y-3 are positive or negative... Is this right? or am I mistaken? I'm asking this because I do not want to be confused in the exam under time pressure.. for example what would have happened had the question not specified that x and y are positive... and the first statement was x^2<y^2... ? would the answer in that case be E... Sorry about the long post, I just wanted clarity on this.. I do NOT want to get a 600-700 level question wrong in the exam.

Hi, could u pls explain that statement? According to me -5x-5 gives 25, so does 5x5

Posted from my mobile device

You should go through basics again before attempting the questions.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

P.S. You might find the following post useful: All You Need for Quant
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BACK TO THE QUESTION:

According to the above if you take the square root from \((x-3)^2 < (y-3)^2\) you'll get |x - 3| < |y - 3|, which means that the distance between x and 3 is less than the distance between y and 3, which is not sufficient to say whether x < y.

Hope it helps.

Dear Bunuel,

Stem tells x>0 and y>0, so why cannot we take (x-3)<(y-3) as it is since we know the signs for both x and y.

In that case, I thought x was indeed less than y. Pls let me know why this is wrong?

Thnak you

We know that |x| = x, when \(x \geq{0}\) (so |something| = something, when that something is >=0) and |x| = -x, when \(x \leq{0}\) (so |something| = -something, when that something is =<0).

Know for positive x, x-3 (expression in modulus) can be positive (when x>3) as well as negative (when x<3), thus |x-3| = x-3, when x>3 and |x-3| = -(x-3), when x<3. Thus knowing that x>0 is not enough to say that |x-3| = x-3.

Hope it's clear.
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THANK YOU BUNUEL !!! QUANT GOD!

Hi All,

This prompt is based on a couple of Number Property rules - and you can TEST VALUES to solve it.

We're told that X and Y are POSITIVE. We're asked if X is less than Y. This is a YES/NO question.

1) √X < √Y

Since we know that X and Y are both POSITIVE, squaring or square-rooting those values will NOT change the "order" of them. Even if you're dealing with positive fractions, the 'order' will not change.

For example:
√X = 1/4 and √Y = 1/2
X = 1/2 and Y = about .71

Thus, the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

2) (X-3)^2 < (Y-3)^2

While X and Y are both POSITIVE, we could end up with an (X-3) or (Y-3) that is negative though... and that will impact the answer to the question.

IF... X = 2, Y = 10.... then (-1)^2 is less than (7)^2 and the answer to the question is YES
IF... X = 2, Y = 1.... then (-1)^2 is less than (-2)^2 and the answer to the question is NO
Fact 2 is INSUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

x> 0
y> 0

We don't know if they are integers or not, but we know they are positive.

Stat (1)
We know that x and y are positive, therefore we can square the integers. This is only permitted when we know the value of the sign, otherwise if we don't we cannot do this. The reason is that if we have a negative number as one of those variables, then the sign may change.

Alternative approach: test numbers
0<x<y<1 - both variables are proper fractions

x=1/16
sq.root of x = 1/4

y= 1/4
sq.root y = 1/2
x<y

test integers that satisfy the stem
x= 4, y=16
x<y

Sufficient

(2)expand the fraction
x^2 -6x +9 < y^2 -6y +9
x^2 -6x < y^2-6y

x=3
y=4
-18 < -8 x is less than y - Yes

x=3, y=2
-9 < -8 but x is greater than y -No

Insufficient

What if we took fractions?

Root x = 1/3 and Root y = 1/4
Then x is 1/9 and y is 1/16

In this case x > y

Posted from my mobile device

But you are taking \(\sqrt{x}>\sqrt{y}\) as 1/3>1/4
Statement I says the opposite, sqrt{x}<sqrt{y}. Thus will always mean x<y.
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Hi muskaanmalhotra,

TESTing VALUES is a great way to approach this question - and you CAN use fractions for X and Y if you like. However, you have made some mistakes in your post.

First, the √(1/9) is 1/3.... since (1/3)(1/3) = 1/9 and the √(1/16) is 1/4 (you have the numbers 'backwards').

Second, you have to choose values that 'fit' what you are told in each Fact (and in the prompt). With this pair of fractions, X would be 1/16 and Y would be 1/9 (in Fact 1:√(1/16) < √(1/9) ....meaning that 1/4 < 1/3). The answer to the question (Is 1/16 less than 1/9?) is still "YES."

GMAT assassins aren't born, they're made,
Rich

Bunuel chetan2u @

I have a doubt for second statement.

\((x-3)^2 < (y-3)^2\)

subtract \((y-3)^2\) on both sides.

\((x-3)^2-(y-3)^2 <0\) --- is of the form \(a^2-b^2 = (a+b)*(a-b)\)

\((x-3+y-3) * (x-3-y+3) < 0\)

\((x+y) (x-y) < 0 \)----- > \((a+b) (a-b)\)

\(x^2-y^2 < 0\) --- add y^2 both sides ----> \( x^2 < y^2 \)----- > \(x < y.\)

What and where am I going wrong ?

The coloured portion is wrong.

It becomes (x+y-6)(x-y)<0 and not (x+y)(x-y)<0
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Hi ueh55406,

While this DS question has a certain degree of 'math concepts' built into it, the information in Fact 2 boils down to one simple concept:

If X^2 is less than Y^2, then that does NOT necessarily mean that X is less than Y.

For example...
IF... X = 2 and Y = 3, then 2^2 < 3^2 and X IS less than Y.
IF... X = 2 and Y = -3, then 2^2 < (-3)^2 and X is NOT less than Y.

This is ultimately why Fact 2 is INSUFFICIENT.

GMAT assassins aren't born, they're made,
Rich

Hello expert chetan2u

I have aquery.
for statement 2 why can not take square root of both the sides (especially they have said both x and y are positive )?
such as
(x-3)< ( y-3)
x-3+3< y-3+3
x<Y

Thanks and regards

The problem is that we do not know whether what is under square is positive.
y-3 may be negative.
For example x=4 and y=1 will give (4-3)^2<(1-3)^2….but x>y
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