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Difficulty: Sub 505 Level,   Inequalities,                        
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Re: Is x < y? (1) z < y (2) z < x [#permalink]
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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]
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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]
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]
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adkikani
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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Re: Is x < y? (1) z < y (2) z < x [#permalink]
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