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

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Bunuel
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Is x < y? (1) z < y (2) z < x [#permalink] New post  15 Jan 2014, 02:54
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E

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93% (00:29) correct 7% (00:35) wrong based on 898 sessions

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Is x < y?

(1) z < y
(2) z < x
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Bunuel
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Re: Is x < y? (1) z < y (2) z < x [#permalink] New post  15 Jan 2014, 02:54
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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.

Answer: E.
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Re: Is x < y? (1) z < y (2) z < x [#permalink] New post  15 Jan 2014, 09:46
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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? (1) z < y (2) z < x [#permalink] New post  15 Jan 2014, 10:26
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i) No information about 'x'
ii) No information about 'y'
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? (1) z < y (2) z < x [#permalink] New post  16 Jan 2014, 03:15
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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.

Answer: (E)
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Re: Is x < y? (1) z < y (2) z < x [#permalink] New post  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

Adding x to both sides:
x<y.
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Re: Is x < y? (1) z < y (2) z < x [#permalink] New post  04 Apr 2018, 21:59
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adkikani wrote:
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

Adding x to both sides:
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? (1) z < y (2) z < x [#permalink]
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