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 and y = (2x-y)^2
[#permalink]
05 Aug 2020, 12:11
05 Aug 2020, 12:11
Expert Reply
1
Bookmarks
00:00
Question Stats:
86% (02:14) correct
13% (03:17) wrong based on 37 sessions
Hide Show timer Statistics
\(\clubsuit x \clubsuit y \clubsuit = (2x-y)^2
\)
\(\spadesuit a \spadesuit b \spadesuit = (a+2b)^2\)
If \(\spadesuit 5 \spadesuit (n-1) \spadesuit = 5n^2+4n\) and \(n< 6\), then \(\clubsuit 2n \clubsuit 2 \clubsuit =\)
A. 4
B. 9
C. 16
D. 36
E. 49
\)
\(\spadesuit a \spadesuit b \spadesuit = (a+2b)^2\)
If \(\spadesuit 5 \spadesuit (n-1) \spadesuit = 5n^2+4n\) and \(n< 6\), then \(\clubsuit 2n \clubsuit 2 \clubsuit =\)
A. 4
B. 9
C. 16
D. 36
E. 49
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: x and y = (2x-y)^2
[#permalink]
05 Aug 2020, 15:50
05 Aug 2020, 15:50
1
Carcass wrote:
\(\clubsuit x \clubsuit y \clubsuit = (2x-y)^2
\)
\(\spadesuit a \spadesuit b \spadesuit = (a+2b)^2\)
If \(\spadesuit 5 \spadesuit (n-1) \spadesuit = 5n^2+4n\) and \(n< 6\), then \(\clubsuit 2n \clubsuit 2 \clubsuit =\)
A. 4
B. 9
C. 16
D. 36
E. 49
\)
\(\spadesuit a \spadesuit b \spadesuit = (a+2b)^2\)
If \(\spadesuit 5 \spadesuit (n-1) \spadesuit = 5n^2+4n\) and \(n< 6\), then \(\clubsuit 2n \clubsuit 2 \clubsuit =\)
A. 4
B. 9
C. 16
D. 36
E. 49
Tricky!!!!!
Given: \(\spadesuit a \spadesuit b \spadesuit = (a+2b)^2\)
So, \(\spadesuit 5 \spadesuit (n-1) \spadesuit = (5 + 2(n-1))^2\)
\( = (5 + 2n-2)^2\)
\( = (3 + 2n)^2\)
\( = 9 + 12n + 4n^2\)
Also given: \(\spadesuit 5 \spadesuit (n-1) \spadesuit = 5n^2+4n\)
So, we can write: \(9 + 12n + 4n^2 = 5n^2+4n\)
Subtract \(4n^2\) from both sides of the equation to get: \(9 + 12n = n^2+4n\)
Subtract \(12n\) from both sides of the equation to get: \(9 = n^2-8n\)
Subtract \(9\) from both sides of the equation to get: \(0=n^2-8n-9\)
Factor to get: \(0 = (n-9)(n+1)\)
So, EITHER \(n=9\) OR \(n =-1\)
Since we're told that \(n< 6\), we can eliminate the solution \(n=9\), which means \(n=-1\)
Given: \(\clubsuit x \clubsuit y \clubsuit = (2x-y)^2\)
We want the value of \(\clubsuit 2n \clubsuit 2 \clubsuit\)
Replace \(n\) with \(-1\) to get: \(\clubsuit 2(-1) \clubsuit 2 \clubsuit\)
\(= \clubsuit -2 \clubsuit 2 \clubsuit\)
\(= [2(-2)-2]^2\)
\(= [(-4)-2]^2\)
\(= [-6]^2\)
\(= 36\)
Answer: D
Cheers,
Brent
gmatclubot
Re: x and y = (2x-y)^2 [#permalink]
05 Aug 2020, 15:50
Moderators:
