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If x and y are positive integers, is x > y ?

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If x and y are positive integers, is x > y ?  [#permalink]

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New post 15 Feb 2018, 23:32
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Question Stats:

75% (01:45) correct 25% (01:55) wrong based on 69 sessions

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Re: If x and y are positive integers, is x > y ?  [#permalink]

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New post 15 Feb 2018, 23:52
Bunuel wrote:
If x and y are positive integers, is x > y ?


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

(2) \(\sqrt{x} < y\)



Given x and y are Z+ i.e 1,2,3,.. etc

ST 1 - x^2 < Y

X^2=1 , Y = 2 X>Y no
x^2=4 y = 5 x>y no
x^2=9 y=10 x>y no

Sufficient

ST 2- sqrt x <y
squaring on both sides
x < y^2


X=7 Y^2=9 X< Y^2 X>Y yes
x=1 Y^2=9 x < y^2 x>y no

insufficient

(A) imo
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If x and y are positive integers, is x > y ?  [#permalink]

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New post 16 Feb 2018, 15:48
Bunuel wrote:
If x and y are positive integers, is x > y ?


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

(2) \(\sqrt{x} < y\)



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

We are dealing with +ve INTEGERS....then

x < x^2 < y........the x < y............Answer is always NO

Sufficient

(2) \(\sqrt{x} < y\)

We are dealing with +ve INTEGERS....then

x < \(\sqrt{x}\)........but y > \(\sqrt{x}\)...........Here we can't establish a relationship like above. check numbers

Let x = 9, y = 4...........\(\sqrt{9}\) < 4 ......3 < 4.............Answer is Yes

Let x = 9, y = 10...........\(\sqrt{9}\) < 10 ......3 < 10.............Answer is NO

Insufficient

Answer: A
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Re: If x and y are positive integers, is x > y ?  [#permalink]

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New post 18 Feb 2018, 05:28
Bunuel wrote:
If x and y are positive integers, is x > y ?


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

(2) \(\sqrt{x} < y\)


Question: is x > y ?

Statement 1: \(x^2 < y\)

If we try to check the information with a few numbers then we realise that for every value of x for which x^2 is less than y, x must be less than y Hence

SUFFICIENT


Statement 1: \(\sqrt{x} < y\)

Case 1: x = 4, y = 3, x > y
Case 1: x = 4, y = 5, x < y

NOT SUFFICIENT

Answer: option A
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Re: If x and y are positive integers, is x > y ?  [#permalink]

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New post 19 Feb 2018, 18:31
Bunuel wrote:
If x and y are positive integers, is x > y ?


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

(2) \(\sqrt{x} < y\)


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

We can modify the question as follows.
Since \(x\) and \(y\) are positive integers, we have \(x ≥ 1\) and \(y ≥ 1\).
Since \(x ≥ 1\), we have \(1 ≤\sqrt{x} ≤ x ≤ x^2\).

Condition 1)
From \(x ≤ x^2\) and \(x^2 < y\), we have \(x ≤ x^2 < y\) or \(x < y\).
Condition 1) is sufficient.

Condition 2):
\(x = 4, y = 3\): Yes
\(x = 2, y = 3\): No
Since we don't have a unique answer from the condition 2), it is not sufficient.

Therefore, A is the answer.
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Re: If x and y are positive integers, is x > y ?  [#permalink]

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New post 19 Feb 2018, 21:17
Question: is x > y ?
Additional Info: X & Y are positive integers

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

If x = 1, y = 2:
\(x^2\) = 1 < 2

x can never be greater than y if not condition 1 will fail. Sufficient.

(2) \(\sqrt{x}\)<y

if x is 4, y is 3 -> x is greater than y
if x is 1, y is 2 -> x is smaller than y

Hence not sufficient.

IMO (A) as well. Took 3 mins to solve though :?
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Re: If x and y are positive integers, is x > y ?   [#permalink] 19 Feb 2018, 21:17
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