|
|
GMAT 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.
28 Mar 2023, 04:45
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
50% (02:24) correct
50%
(02:38)
wrong
based on 22
sessions
History
Date
Time
Result
Not Attempted Yet
If \(x<y\), is \(x(1+x)<y(1+y)\)?
(1) \(x>\frac{1}{2}\)
(2) \(x+y>1\)
(1) \(x>\frac{1}{2}\)
(2) \(x+y>1\)
28 Mar 2023, 05:07
Kudos
Bookmarks
Bunuel
Answer: Option
Step-by-Step Video solution by GMATinsight
28 Mar 2023, 06:08
Kudos
Bookmarks
Bunuel
Solution:
Pre Analysis:
- We are given \(x<y\) or \(x-y<0\)
- We are asked if \(x(1+x)<y(1+y)\) or not
\(⇒x+x^2<y+y^2\)
\(⇒x-y+x^2-y^2<0\)
\(⇒(x-y)+(x+y)(x-y)<0\)
\(⇒(x-y)(1+x+y)<0\) - We already know that \(x-y<0\).
- So, for \(x(1+x)<y(1+y)\) to be true \(1+x+y>0\) or \(x+y>-1\)
- We need to find if x + y > -1 or not
Statement 1: \(x>\frac{1}{2}\)
- We already know that \(y>x\)
- So, if \(x>\frac{1}{2}\), then obviously \(y>\frac{1}{2}\)
- This means sum \(x+y\) will be greater than 1/2 + 1/2 i.e., \(x+y>1\) which is greater than -1
- Thus, statement 1 alone is sufficient and we can eliminate options B, C and E
Statement 2: \(x+y>1\)
- This is exactly what we inferred from statement 1 also
- Thus, statement 2 alone is also sufficient
Hence the right answer is Option D
