GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
x/y>1
[#permalink]
02 Sep 2024, 08:05
02 Sep 2024, 08:05
Expert Reply
1
Bookmarks
00:00
Question Stats:
76% (01:42) correct
23% (00:46) wrong based on 13 sessions
Hide Show timer Statistics
\(\frac{x}{y}>1\)
Quantity A |
Quantity B |
\( \frac{x}{y} + \frac{y}{x}\) |
\( 2\) |
- Part of the project: GRE Quant & Verbal ADVANCED Daily Challenge 2024 Edition - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
- Also replying to the unanswered questions
Re: x/y>1
[#permalink]
02 Sep 2024, 10:05
02 Sep 2024, 10:05
1
Since \(\frac{x}{y}>1\), let us write: \(\frac{x}{y}=1+k\), where \(k\) is a positive quantity.
\(\frac{x}{y}+\frac{y}{x} = 1+k + \frac{1}{1+k}\)
\(= \frac{(1+k)^2 +1}{1+k}\)
\(= \frac{(2+2k)+k^2 }{1+k}\)
\(=2 + \frac{k^2 }{1+k} > 2 \) because k is positive.
Hence, A is correct.
\(\frac{x}{y}+\frac{y}{x} = 1+k + \frac{1}{1+k}\)
\(= \frac{(1+k)^2 +1}{1+k}\)
\(= \frac{(2+2k)+k^2 }{1+k}\)
\(=2 + \frac{k^2 }{1+k} > 2 \) because k is positive.
Hence, A is correct.
gmatclubot
Re: x/y>1 [#permalink]
02 Sep 2024, 10:05
Moderators:
