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# Is x < y?

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Math Expert
Joined: 02 Sep 2009
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15 Jan 2014, 02:54
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

Is x < y?

(1) z < y
(2) z < x

Data Sufficiency
Question: 40
Category: Algebra Inequalities
Page: 156
Difficulty: 500

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Re: Is x < y?  [#permalink]

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15 Jan 2014, 02:54
SOLUTION

Is x < y?

(1) z < y
(2) z < x

Even when we combine the statements we just have that x and y are greater than some number z, we clearly cannot say whether x<y.

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Re: Is x < y?  [#permalink]

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15 Jan 2014, 09:46
1
As per my understanding best way will be to eliminate choices.
Since both (1) and (2) does not talk about the relation between x and y so we can eliminate option a,b,d.
We are left with option c and option e .

Now we take following two cases satisfying both given condition z<x and z<y -
x=2, y=3, z=1 -> x<y

x= 10, y=5, z =1 -> x>y

we can see that we can not guarantee the statement x>y by using both condition so answer is (E)-Statements (1) and (2) TOGETHER are NOT sufficient to answer the question
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Re: Is x < y?  [#permalink]

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15 Jan 2014, 10:26
1
Combining (i) and (ii), we know that z is less than, both x and why, but we do not know which one of the two (x or y) is greater than the other.
Hence (E)
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Re: Is x < y?  [#permalink]

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16 Jan 2014, 03:15
1
Is x < y?

(1) z < y
(2) z < x

Statement 1

Given that z < y, the relationship between z and x can not be determined, hence it is not possible to evaluate the inequality x < y .........Not Sufficient.

Statement 2

Given that z < x, the relationship between z and y can not be determined, hence it is not possible to evaluate the inequality x < y .........Not Sufficient.

Combining statements (1) and (2), we have z less than both y and x and this statement is not sufficient to evaluate x < y.
As an example, the sets (x=3, y=4, z=2) and (x=4, y=3, z=2) satisfy the given inequalities, whereas x and y relation is not unique.

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Re: Is x < y?  [#permalink]

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18 Jan 2014, 05:14
SOLUTION

Is x < y?

(1) z < y
(2) z < x

Even when we combine the statements we just have that x and y are greater than some number z, we clearly cannot say whether x<y.

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Re: Is x < y?  [#permalink]

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11 Sep 2015, 02:14
we do not have specific details about Z,X, Y

hence E
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Re: Is x < y?  [#permalink]

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27 Aug 2017, 05:17
For this question, I found using the number line a bit easier to think through this logically

Stem: Does X lie to the left of y on the number line?

S1: Z lies to the left of Y on the number line. Dont know the position of X. Insuff

S2: Z lies to the left of X on the number line. Dont know the position of Y. insuff

S1+S2: Z lies to the left of X and Y on the number line. We still dont know relative position of X and Y. X could be to the left of Y or Vice Versa. Insuff

Number line attached, just for better understanding
Attachments

Screen Shot 2017-08-27 at 5.41.47 PM.png [ 28.3 KiB | Viewed 1235 times ]

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04 Apr 2018, 21:55
Bunuel niks18 amanvermagmat

Quote:
Is x < y?

(1) z < y
(2) z < x

Combining both statements, can we subtract (2) from (1)?

0 < y-x

x<y.
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Re: Is x < y?  [#permalink]

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04 Apr 2018, 21:59
Bunuel niks18 amanvermagmat

Quote:
Is x < y?

(1) z < y
(2) z < x

Combining both statements, can we subtract (2) from (1)?

0 < y-x

x<y.

No.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

For more check Manipulating Inequalities.
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Re: Is x < y?  [#permalink]

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04 Apr 2018, 22:42
Bunuel niks18 amanvermagmat

Quote:
Is x < y?

(1) z < y
(2) z < x

Combining both statements, can we subtract (2) from (1)?

0 < y-x

x<y.

Hello

As Bunuel has already explained, we can add here and we will get:
z+z < y+x OR 2z < y+x

If you want to subtract the inequalities then the signs should be opposite, so lets re-write the second inequality to get the two as:
z < y
x > z

NOW we can subtract second from first, and the sign will be that of the first inequality, so we will get:
z - x < y - z OR 2z < y+x
This is the same result when we added the two. So the answer would be E.
Re: Is x < y? &nbs [#permalink] 04 Apr 2018, 22:42
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# Is x < y?

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