Olympiad Maths Prep

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Problem 117

AMC 10/12, early questions
Algebra Difficulty 3.3 Find the answer

Let x,yRx, y \in \mathbb{R}, then "x>y>0x > y > 0" is the "xy>1\frac{x}{y} > 1" of ( )
A: Necessary and sufficient condition
B: Necessary but not sufficient condition
C: Sufficient but not necessary condition
D: Neither necessary nor sufficient condition

Official solution

1. First, let's translate the given solution from Chinese to English.

"x>y>0x > y > 0" implies "xy>1\frac{x}{y} > 1", but the converse is not true. For example, if we take x=2x = -2 and y=1y = -1, then "xy>1\frac{x}{y} > 1" holds, but "x>y>0x > y > 0" does not.

Thus, "x>y>0x > y > 0" is a sufficient but not necessary condition for "xy>1\frac{x}{y} > 1".

Therefore, the answer is (C): Sufficient but not necessary condition.

2. Now, let's format the translated solution using Markdown and LaTeX.

"x>y>0x > y > 0" implies "xy>1\frac{x}{y} > 1", but the converse is not true. For example, if we take x=2x = -2 and y=1y = -1, then "xy>1\frac{x}{y} > 1" holds, but "x>y>0x > y > 0" does not.

Thus, "x>y>0x > y > 0" is a sufficient but not necessary condition for "xy>1\frac{x}{y} > 1".

Therefore, the answer is (C): Sufficient but not necessary condition.

3. Next, let's enhance the solution by breaking it down into steps and providing additional explanations.

Step 1: Consider the given statement "x>y>0x > y > 0". This implies that both xx and yy are positive, and xx is greater than yy.

Step 2: Dividing both sides of the inequality by yy (which is positive), we get xy>1\frac{x}{y} > 1.

Step 3: Now, let's consider the converse, "xy>1\frac{x}{y} > 1". If we take x=2x = -2 and y=1y = -1, then 21=2>1\frac{-2}{-1} = 2 > 1, but 2>1>0-2 > -1 > 0 is not true.

Step 4: Therefore, "x>y>0x > y > 0" is a sufficient but not necessary condition for "xy>1\frac{x}{y} > 1".

Step 5: The answer is (C): Sufficient but not necessary condition.

4. Finally, let's highlight the final answer using \boxed.

Therefore, the answer is \boxed{(C): Sufficient but not necessary condition}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.