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Bunuel
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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Bunuel
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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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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