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.
If x and y are non-negative integers such that 2x+3y=8 and z=x2+y2, wh
[#permalink]
21 Sep 2021, 01:44
21 Sep 2021, 01:44
1
Expert Reply
00:00
Question Stats:
60% (01:07) correct
40% (02:28) wrong based on 15 sessions
Hide Show timer Statistics
If x and y are non-negative integers such that \(2x+3y=8\) and \(z=x^2+y^2\), what is the maximum value of z?
A. 0
B. 5
C. 13
D. 16
E. 25
Source: manhattanreview
A. 0
B. 5
C. 13
D. 16
E. 25
Source: manhattanreview
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Re: If x and y are non-negative integers such that 2x+3y=8 and z=x2+y2, wh
[#permalink]
21 Sep 2021, 02:39
21 Sep 2021, 02:39
1
x and y can be either positive or 0
given \(2x+3y=2^3\), x can be 4 and y is 0
hence, \(z=x^2+y^2=4^2+0=16\)
answer is D
given \(2x+3y=2^3\), x can be 4 and y is 0
hence, \(z=x^2+y^2=4^2+0=16\)
answer is D
Carcass wrote:
If x and y are non-negative integers such that \(2x+3y=8\) and \(z=x^2+y^2\), what is the maximum value of z?
A. 0
B. 5
C. 13
D. 16
E. 25
Source: manhattanreview
A. 0
B. 5
C. 13
D. 16
E. 25
Source: manhattanreview
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If x and y are non-negative integers such that 2x+3y=8 and z=x2+y2, wh
[#permalink]
11 Nov 2021, 12:07
11 Nov 2021, 12:07
Carcass wrote:
If x and y are non-negative integers such that \(2x+3y=8\) and \(z=x^2+y^2\), what is the maximum value of z?
A. 0
B. 5
C. 13
D. 16
E. 25
Since x and y are non-negative integers, there aren't many solutions to the given equation, \(2x+3y=8\)
In fact, after testing in a few values, we can see that the only possible solutions are as follows:
case i) x = 1 and y = 2
case ii) x = 4 and y = 0
For each possible case, let's determine the value of z
case i) x = 1 and y = 2
In this case, \(z=x^2+y^2=1^2+2^2=1+4=5\)
case ii) x = 4 and y = 0
In this case, \(z=x^2+y^2=4^2+0^2=16+0=16\)
Answer: D
Source: manhattanreview
A. 0
B. 5
C. 13
D. 16
E. 25
Since x and y are non-negative integers, there aren't many solutions to the given equation, \(2x+3y=8\)
In fact, after testing in a few values, we can see that the only possible solutions are as follows:
case i) x = 1 and y = 2
case ii) x = 4 and y = 0
For each possible case, let's determine the value of z
case i) x = 1 and y = 2
In this case, \(z=x^2+y^2=1^2+2^2=1+4=5\)
case ii) x = 4 and y = 0
In this case, \(z=x^2+y^2=4^2+0^2=16+0=16\)
Answer: D
Source: manhattanreview
