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07 Aug 2019, 03:27
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
49% (01:26) correct
51%
(01:51)
wrong
based on 71
sessions
History
Date
Time
Result
Not Attempted Yet
If x and y are positive integers, is x > y?
(1) x^2 < y
(2) The square root of x is less than y
(1) x^2 < y
(2) The square root of x is less than y
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
07 Aug 2019, 04:20
Kudos
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If x and y are positive integers, is x > y?
(1) \(x^2\) < y
(2) The square root of x is less than y
Since x > 0 and y > 0.
Statement 1) \(x^2 < y\)
x*x < y so x would definitely be less than y always.
ALTERNATIVELY:
Verifying using numbers and fractions.
Case 1: x = \(\frac{1}{2}\) and y > 1(2 OR 3 OR any value).
Case 2: x = \(\frac{1}{2}\) and 0 < y < 1 such that \(x^2\) < y always (y = \(\frac{2}{3}\) OR \(\frac{4}{9}\))
Case 3: x > 1 and y > 1 such that \(x^2\) < y always. Take x = 2 so that \(x^2\) = 4 and y ≥ 5
Only Case 3 is required here as per question
Sufficient.
Statement 2) \(\sqrt{x}\) < y
As both x and y are positive we can square the inequality
x < \(y^2\)
Take x = 1 and y = 2, So x > y ? NO
OR
x = 2 and y = 2, So x > y ? NO
OR
x = 3 and y = 2, So x > y ? YES
Insufficient.
Answer (A)
(1) \(x^2\) < y
(2) The square root of x is less than y
Since x > 0 and y > 0.
Statement 1) \(x^2 < y\)
x*x < y so x would definitely be less than y always.
ALTERNATIVELY:
Verifying using numbers and fractions.
Case 1: x = \(\frac{1}{2}\) and y > 1(2 OR 3 OR any value).
Case 2: x = \(\frac{1}{2}\) and 0 < y < 1 such that \(x^2\) < y always (y = \(\frac{2}{3}\) OR \(\frac{4}{9}\))
Case 3: x > 1 and y > 1 such that \(x^2\) < y always. Take x = 2 so that \(x^2\) = 4 and y ≥ 5
Only Case 3 is required here as per question
Sufficient.
Statement 2) \(\sqrt{x}\) < y
As both x and y are positive we can square the inequality
x < \(y^2\)
Take x = 1 and y = 2, So x > y ? NO
OR
x = 2 and y = 2, So x > y ? NO
OR
x = 3 and y = 2, So x > y ? YES
Insufficient.
Answer (A)
13 Aug 2019, 23:09
Kudos
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Bunuel
x and y are positive integers
Possible values of x = {1, 2, 3, 4, . . . .. }
Possible values of y = {1, 2, 3, 4, . . . .. }
(1) x^2 < y
If x = 1, y > 1 --> x < y (Answer - Definite NO)
If x = 2, y > 4 --> x < y (Answer - Definite NO)
If x = 3, y > 9 --> x < y (Answer - Definite NO)
--> Sufficient
(2) The square root of x is less than y
√x < y
If x = 1, y > 1 --> x < y (Answer - Definite NO)
If x = 4, y > 2 --> Possible values (x, y) = (4, 3), (4, 5) --> Can't say
--> Insufficient
IMO Option A
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