|
|
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.
17 Mar 2020, 08:22
Kudos
Bookmarks
C
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:
65% (02:30) correct
35%
(02:30)
wrong
based on 298
sessions
History
Date
Time
Result
Not Attempted Yet
If x > 0 and y > 0, what is the value of x − 2y?
(1) \(16y^2 = \frac{x^4 - 240}{y^2}\)
(2) \(x^2 + 4y^2 = \frac{120}{x + 2y}\)
(1) \(16y^2 = \frac{x^4 - 240}{y^2}\)
(2) \(x^2 + 4y^2 = \frac{120}{x + 2y}\)
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
Bunuel
Asked: If x > 0 and y > 0, what is the value of x − 2y?
(1) \(16y^2 = \frac{x^4 - 240}{y^2}\)
16y^4 = x^4 - 240
x^4 - 16y^4 = 240 = 256 - 16 = 4^4 - -4*1^4
x = +- 4; y = +-1
x - 2y can not be determined
NOT SUFFICIENT
(2) \(x^2 + 4y^2 = \frac{120}{x + 2y}\)
(x^2 + 4y^2)(x+2y) = 120
(x^2 + 4y^2)(x+2y)(x-2y) = 120(x-2y)
x^4 - 16y^4 = 120(x-2y)
NOT SUFFICIENT
(1) + (2)
(1) \(16y^2 = \frac{x^4 - 240}{y^2}\)
16y^4 = x^4 - 240
x^4 - 16y^4 = 240 = 256 - 16 = 4^4 - -4*1^4
x = +- 4; y = +-1
(2) \(x^2 + 4y^2 = \frac{120}{x + 2y}\)
(x^2 + 4y^2)(x+2y) = 120
(x^2 + 4y^2)(x+2y)(x-2y) = 120(x-2y)
x^4 - 16y^4 = 120(x-2y) = 240
x - 2y = 2
SUFFICIENT
IMO C
17 Mar 2020, 23:03
Kudos
Bookmarks
Bunuel
Solution
Step 1: Analyse Question Stem
- • \( x > 0 \)
• \( y > 0\)
• We need to find the value of \(x- 2y.\)
- o So, we can find \(x – 2y \) if
- We find the exact values of x and y or
We directly find the value of \(x -2y\) using relation between x and y.
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: \(16y^2=\frac{x^4−240}{y^2}\)
- • According to this statement: \(16y^4 = x^4 – 240\)
- o Or, \(x^4 – (2y)^4 = 240\)
\(⟹(x^2 + (2y)^2)(x^2 – (2y)^2) = 240\)
\(⟹ (x^2 + 4y^2)(x+2y)(x – 2y) = 240\)
Note: we know is that x > 0 and y >0, which means they are positive, however, we don't know if they are integers or not, so we shouldn't assume only integral values here to solve the statements.
Statement 2: \(x^2+4y^2=\frac{120}{x+2y}\)
- • According to this statement:\( (x^2 + 4y^2)(x+2y) = 120\)
• From this statement we cannot find the values of x and y or the value of x- 2y.
Step 3: Analyse Statements by combining.
- • From statement 1: \((x^2 + 4y^2) (x+2y) (x – 2y) = 240\)
• From statement 2: \((x^2 + 4y^2) (x+2y) = 120\)
• Combining both the statements, we get, \(120*(x – 2y) = 240 ⟹ x – 2y = 2\)
