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.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
What is the maximum and minimum possible value of the expression
[#permalink]
11 Mar 2022, 00:11
11 Mar 2022, 00:11
1
4
Bookmarks
00:00
Question Stats:
49% (01:09) correct
49% (01:42) wrong based on 14 sessions
Hide Show timer Statistics
What is the maximum and minimum possible value of the expression \(\frac{|x+y|}{|x|+|y|} + \frac{|y+z|}{|y|+|z|} + \frac{|z+x|}{|z|+|x|}\) ?
A. 2 and 1
B. 3 and 0
C. 3 and 1
D. 4 and 0
E. 4 and 1
A. 2 and 1
B. 3 and 0
C. 3 and 1
D. 4 and 0
E. 4 and 1
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
What is the maximum and minimum possible value of the expression
[#permalink]
25 Mar 2022, 05:00
25 Mar 2022, 05:00
1
1
Bookmarks
OA Explanation:
Remember: |a+b| ≤ |a|+|b|
[https://gre.myprepclub.com/forum/gre-quant-absolute-values-modulus-theory-18389.html#p52039]
All three fraction terms separated by + sign can have maximum value as 1, which means the maximum value of the entire expression is 3
When is this possible?
|a+b| = |a|+|b|
Now, what can be the minimum value of each fraction term?
The answer is: It can be equal to 0
When is this possible?
\(x= -y\) or \(y = -z\) or \(x = -z\)
But wait - Among x, y, and z at least two will have the same sign
Here is why?
if \(x= -y\) and \(y = -z\) then \(x ≠ -z\), because \(x\) will become equal to \(z\)
Therfore, out of the three fraction terms - atmost two terms can be zero and atleast one will have to be 1
So, the minimum value of the entire expression would be 1
Hence, option C
Remember: |a+b| ≤ |a|+|b|
[https://gre.myprepclub.com/forum/gre-quant-absolute-values-modulus-theory-18389.html#p52039]
All three fraction terms separated by + sign can have maximum value as 1, which means the maximum value of the entire expression is 3
When is this possible?
|a+b| = |a|+|b|
Now, what can be the minimum value of each fraction term?
The answer is: It can be equal to 0
When is this possible?
\(x= -y\) or \(y = -z\) or \(x = -z\)
But wait - Among x, y, and z at least two will have the same sign
Here is why?
if \(x= -y\) and \(y = -z\) then \(x ≠ -z\), because \(x\) will become equal to \(z\)
Therfore, out of the three fraction terms - atmost two terms can be zero and atleast one will have to be 1
So, the minimum value of the entire expression would be 1
Hence, option C
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1115
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
What is the maximum and minimum possible value of the expression
[#permalink]
09 Dec 2023, 02:10
09 Dec 2023, 02:10
1
Theory: |A + B| ≤ |A| + |B|
=> Maximum value of \(\frac{|x+y|}{|x|+|y|} + \frac{|y+z|}{|y|+|z|} + \frac{|z+x|}{|z|+|x|}\) will be when all x , y and z have the same sign and numerator and denominator will cancel out to give us
Maximum value = 1 + 1 + 1 = 3
Ideally Minimum value should be zero if we had two different variables in each part of the absolute value. But here we have 3 variables shared across three Absolute values. So, two of them will have the same sign and value and other can have a different sign and same value.
Ex: x = 1, y = 1 and z = -1
=> \(\frac{|1+1|}{|1|+|1|} + \frac{|1-1|}{|1|+|-1|} + \frac{|-1+1|}{|-1|+|1|}\)
=> Minimum value = \(\frac{2}{2}\) + 0 + 0 = 1
So, Answer will be C
Hope it helps!
Watch the following video to learn the Basics of Absolute Values
- - As |A| + |B| will always add up, irrespective of the signs of A and B, as after coming out of Absolute value they will become non-negative
- But, |A +B| can get reduced if both have oppositive signs
=> Maximum value of \(\frac{|x+y|}{|x|+|y|} + \frac{|y+z|}{|y|+|z|} + \frac{|z+x|}{|z|+|x|}\) will be when all x , y and z have the same sign and numerator and denominator will cancel out to give us
Maximum value = 1 + 1 + 1 = 3
Ideally Minimum value should be zero if we had two different variables in each part of the absolute value. But here we have 3 variables shared across three Absolute values. So, two of them will have the same sign and value and other can have a different sign and same value.
Ex: x = 1, y = 1 and z = -1
=> \(\frac{|1+1|}{|1|+|1|} + \frac{|1-1|}{|1|+|-1|} + \frac{|-1+1|}{|-1|+|1|}\)
=> Minimum value = \(\frac{2}{2}\) + 0 + 0 = 1
So, Answer will be C
Hope it helps!
Watch the following video to learn the Basics of Absolute Values
