Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 24 Apr 2016
Posts: 333

If Y = X + 1  X2, then [#permalink]
Show Tags
06 Feb 2017, 13:42
1
This post received KUDOS
8
This post was BOOKMARKED
Question Stats:
69% (01:31) correct 31% (01:37) wrong based on 338 sessions
HideShow timer Statistics
If \(Y = X + 1  X2\), then (a) \(3 \leq Y \leq 0\) (b) \(3 \leq Y \leq 3\) (c) \(Y \leq 3\) (d) \(Y \geq 3\) (e) No Solution
Official Answer and Stats are available only to registered users. Register/ Login.



Director
Joined: 05 Mar 2015
Posts: 962

Re: If Y = X + 1  X2, then [#permalink]
Show Tags
06 Feb 2017, 18:22
1
This post received KUDOS
2
This post was BOOKMARKED
quantumliner wrote: If Y = X + 1  X2, then
(a) 3 <= Y <=0 (b) 3 <= Y <=3 (c) Y <=3 (d) Y >=3 (e) No Solution 3 ranges to define 1) x< 1 then y= 3 2) 1<=x<2 then y = 2x1 3) x>= 2 then y=3 as second condition dependent on value of x thus from range 1 & 2 3 <= Y <=3 Ans B



Intern
Joined: 16 Jan 2017
Posts: 5

Re: If Y = X + 1  X2, then [#permalink]
Show Tags
09 Feb 2017, 05:30
rohit8865 wrote: 3 ranges to define
1) x< 1 then y= 3 2) 1<=x<2 then y = 2x1 3) x>= 2 then y=3
as second condition dependent on value of x thus from range 1 & 2 3 <= Y <=3
Ans B
How did you came up with the 3 ranges and how to define them? I dont understand what you are doing. Thanks



Director
Joined: 05 Mar 2015
Posts: 962

Re: If Y = X + 1  X2, then [#permalink]
Show Tags
09 Feb 2017, 06:08
2
This post received KUDOS
maxschmid wrote: rohit8865 wrote: 3 ranges to define
1) x< 1 then y= 3 2) 1<=x<2 then y = 2x1 3) x>= 2 then y=3
as second condition dependent on value of x thus from range 1 & 2 3 <= Y <=3
Ans B
How did you came up with the 3 ranges and how to define them? I dont understand what you are doing. Thanks maxschmidfor such questions always do the below suggested Ix+1I will be 0 when x= 1similarly Ix2I will be 0 when x= 2thus we have boundaries defined as 1 and 2 so i have checked on both sides of boundaries defined in three steps for value y 1) x< 1 then y= 3 2) 1<=x<2 then y = 2x1 3) x>= 2 then y=3 hope now u can understand



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1975

Re: If Y = X + 1  X2, then [#permalink]
Show Tags
13 Feb 2017, 16:05
2
This post received KUDOS
Expert's post
4
This post was BOOKMARKED
quantumliner wrote: If Y = X + 1  X2, then
(a) 3 <= Y <=0 (b) 3 <= Y <=3 (c) Y <=3 (d) Y >=3 (e) No Solution Since the expressions in the absolute value will equal zero when X = 1 and X = 2, we need to investigate the following three cases: i) X < 1, ii) 1 < X < 2, and iii) X > 2 Now let’s analyze each case. i) If X < 1, we see that both of the expressions inside the absolute values are negative, and thus: Y = (X  1)  [(X  2)] Y =  X 1 + X  2 Y = 3 ii) If 1 < X < 2, then the first expression is positive but the second expression is negative; thus: Y = X + 1  [(X  2)] Y = X + 1 + X  2 Y = 2X 1 Now, recall that 1 < X < 2, so 2 < 2X < 4 3 < 2X  1 < 3 Since Y = 2X  1, 3 < Y < 3. iii) If X > 2, then both expressions inside the absolute value are positive; thus: Y = X + 1  (X  2) Y = X + 1  X + 2 Y = 3 Combining the results of the three cases, we see that 3 ≤ Y ≤ 3. Answer: B
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Chat Moderator
Joined: 04 Aug 2016
Posts: 596
Location: India
Concentration: Leadership, Strategy
GPA: 4
WE: Engineering (Telecommunications)

Re: If Y = X + 1  X2, then [#permalink]
Show Tags
15 Feb 2017, 03:45
JeffTargetTestPrep wrote: quantumliner wrote: If Y = X + 1  X2, then
(a) 3 <= Y <=0 (b) 3 <= Y <=3 (c) Y <=3 (d) Y >=3 (e) No Solution Since the expressions in the absolute value will equal zero when X = 1 and X = 2, we need to investigate the following three cases: i) X < 1, ii) 1 < X < 2, and iii) X > 2 Now let’s analyze each case. i) If X < 1, we see that both of the expressions inside the absolute values are negative, and thus: Y = (X  1)  [(X  2)] Y =  X 1 + X  2 Y = 3 ii) If 1 < X < 2, then the first expression is positive but the second expression is negative; thus: Y = X + 1  [(X  2)] Y = X + 1 + X  2 Y = 2X 1 Now, recall that 1 < X < 2, so 2 < 2X < 4 3 < 2X  1 < 3 Since Y = 2X  1, 3 < Y < 3. iii) If X > 2, then both expressions inside the absolute value are positive; thus: Y = X + 1  (X  2) Y = X + 1  X + 2 Y = 3 Combining the results of the three cases, we see that 3 ≤ Y ≤ 3. Answer: B I have a query regarding the range. Shouldn't it be 1 <= X < 2 and X >= 2



Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1276
Location: Malaysia

Re: If Y = X + 1  X2, then [#permalink]
Show Tags
15 Feb 2017, 05:04
2
This post received KUDOS
warriorguy wrote: I have a query regarding the range. Shouldn't it be 1 <= X < 2 and X >= 2
Dear warriorguy, Graphical illustration will clear your doubt. When \(x = 2\) \(Y = x+1  x2 = 2+1  22 = 3 + 0 = 3\) Coordinate 1 \((2, 3)\) When \(x = 1\) \(Y = x+1  x2 = 1+1  12 = 0  3 = 3\) Coordinate 2 \((1, 3)\) Therefore, the solution is \(3 ≤ y ≤ 3\)
Attachments
Untitled.jpg [ 93.72 KiB  Viewed 2174 times ]
_________________
"Be challenged at EVERY MOMENT."
“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”
"Each stage of the journey is crucial to attaining new heights of knowledge."
Rules for posting in verbal forum  Please DO NOT post short answer in your post!



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11036
Location: United States (CA)
GRE 1: 340 Q170 V170

Re: If Y = X + 1  X2, then [#permalink]
Show Tags
16 Feb 2017, 13:13
Hi All, This question can be solved by TESTing VALUES. We're told that Y = X + 1  X2. We're asked to find the range of value for Y. Let's start with something easy.... IF.... X = 0, then Y = 1  2 = 1, so Y can equal 1. Eliminate Answers C and E. Considering the three remaining answers, we know that Y either has an "upper limit" or it does not, so let's see what happens if we make X really big or really small.... IF... X = 100, then Y = 101  98 = 3. This is interesting, since we've appeared to now randomly hit the 'upper limit' in Answer B. Eliminate Answer A. What if we try something even bigger.... X = 1000, then Y = 1001  998 = 3. This is the exact SAME value... It certainly looks like we've found the upper limit, but just to be sure, I'll do one more Test... IF.... X = 50, then Y = 49  52 = 3. We clearly have the range now. Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************



Manager
Joined: 30 May 2012
Posts: 217
Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)

Re: If Y = X + 1  X2, then [#permalink]
Show Tags
23 Sep 2017, 12:15
1
This post received KUDOS
EMPOWERgmatRichC wrote: ...
IF.... X = 0, then Y = 1  2 = 1, so Y can equal 1.
... Rich How so? How does one approach this via a number line?



Manager
Joined: 30 May 2012
Posts: 217
Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)

Re: If Y = X + 1  X2, then [#permalink]
Show Tags
23 Sep 2017, 12:23
mikemcgarry  Could you help me with plotting a solution on the number line, please?



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11036
Location: United States (CA)
GRE 1: 340 Q170 V170

Re: If Y = X + 1  X2, then [#permalink]
Show Tags
24 Sep 2017, 12:29
Hi Blackbox, Thanks for catching the error; I've updated my explanation. GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************




Re: If Y = X + 1  X2, then
[#permalink]
24 Sep 2017, 12:29






