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Re: Is x < y ? (1) z^2 < y (2) x < z [#permalink]
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Bunuel wrote:
Is x < y ?

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

We need to determine whether x < y.

Statement One Alone:

z^2 < y

Since we do not have any information regarding x, statement one is not sufficient to answer the question.

Statement Two Alone:

x < z

Since we do not have any information regarding y, statement two is not sufficient to answer the question.

Statements One and Two Together:

Using the information in statements one and two, we still cannot answer the question. For instance, if y = 10, z = 2, and x = 1, then x is less than y. However, if z = 1/2, x = 1/3, and y = 1/3, then x IS NOT less than y.

Answer: E
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Is x < y ? (1) z^2 < y (2) x < z [#permalink]
Bunuel wrote:
Is x < y ?

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

This is how i solved...

Statement 1 : We don't know the value of x therefore not sufficient.
Statement 2 : We don't know the value of y therefore not sufficient.

Together Statement 1 and 2, ( Statement 1 says z^2 < y ... let z be 0.5 then z^2 = 0.25 ... and y is greater than z^2 let y be 0.26 onwards.... ( SO Z= 0.5 AND Y = 0.26 )
Statement 2 says z is greater than x which means x is less than 0.5 which means X could be 0.4

In this case X > y ( 0.4 > 0.26 ) therfore is Is x < y ? ans is not true.

nOW LETS TEST THIIS with another value ..
But now if we assume the value of z = 2 then z^2 = 2^2 = 4 and y should be greater than 4 lets say 5 ...
so Z= 2 and y = 5

Statement 2 says x < Z so lets say x = 1 in that case x< y or 1 < 5 true ...

Since we cannot answer the question with certainty ans is E
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Is x < y ? (1) z^2 < y (2) x < z [#permalink]
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Bunuel wrote:
Is x < y ?

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

1) and 2) are insufficient alone. We can combine the statements by squaring 2) to get z^2.

However, depending on the sign of x and z we can have \(x^2 < z^2\) or \(x^2 > z^2\) so we must entertain both possibilities (at this point a good guess is E because we already have two cases).

In the first case, we have \(x^2 < z^2 < y\), thus \(x^2 < y\). This cannot simplify to \(x < y\) because we can have x = 0.8 and y = 0.7. Thus combining is insufficient.
Knowing that \(x^2 < y\) cannot simplify to \(x < y\) can be valuable knowledge for future problems. Since we cannot prove \(x < y\) with the given information it is insufficient.

The second case occurs happens when x is negative and has a bigger magnitude than z. Hence x must be negative in this case. From (1) we know y is positive, so \(x < y\).

However, since the first case was already insufficient, the statements combined are insufficient, without needing to work out case 2.

Ans: E
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Re: Is x < y ? (1) z^2 < y (2) x < z [#permalink]
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Re: Is x < y ? (1) z^2 < y (2) x < z [#permalink]
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