Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
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.
Nov 2112:00 PM EST
-11:59 PM EST
Don’t miss Target Test Prep’s biggest sale of the year! Grab 25% off any Target Test Prep GMAT plan during our Black Friday sale. Just enter the coupon code BLACKFRIDAY25 at checkout to save up to $400.
Nov 2411:00 AM IST
-01:00 PM IST
Attend this session to evaluate your current skill level, learn process skills, and solve through tough Arithmetic questions
Nov 3001:00 PM EST
-02:00 PM EST
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of how students should study. So, it's no wonder that more and more Target Test Prep students are scoring super high on the GMAT.
06 Feb 2017, 14:42
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (02:05) correct
32%
(02:07)
wrong
based on 948
sessions
History
Date
Time
Result
Not Attempted Yet
If \(Y = |X + 1| - |X-2|\), then
(a) \(-3 \leq Y \leq 0\)
(b) \(-3 \leq Y \leq 3\)
(c) \(Y \leq -3\)
(d) \(Y \geq -3\)
(e) No Solution
(a) \(-3 \leq Y \leq 0\)
(b) \(-3 \leq Y \leq 3\)
(c) \(Y \leq -3\)
(d) \(Y \geq -3\)
(e) No Solution
13 Feb 2017, 17:05
Kudos
Bookmarks
quantumliner
If Y = |X + 1| - |X-2|, then
(a) -3 <= Y <=0
(b) -3 <= Y <=3
(c) Y <=-3
(d) Y >=-3
(e) No Solution
(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
09 Feb 2017, 07:08
Kudos
Bookmarks
maxschmid
rohit8865
3 ranges to define
1) x< -1 then y= -3
2) -1<=x<2 then y = 2x-1
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
maxschmid
for such questions always do the below suggested
Ix+1I will be 0 when x= -1
similarly Ix-2I will be 0 when x=2
thus 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 = 2x-1
3) x>= 2 then y=3
hope now u can understand
